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AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry InText Questions and Answers.

10th Class Maths 7th Lesson Coordinate Geometry InText Questions and Answers

Do these

Question 1.
From the figure write coordinates of the points A, B, C, D, E, F, G, H.          (Page No. 159)
Answer:
Given: Knight is at the origin (0, 0).

Therefore, A (- 1, 2), B (1, 2), C (2, 1), D (2, – 1), E (1, – 2), F (-1, -2), G (-2, -1) and H (-2, 1).

Question 2.
Find the distance covered by the Knight in each of its 8 moves i.e., find the distance of A, B, C, D, E, F, G and H from the origin.    (Page No. 159)
Answer:
Origin (0, 0).
Points A, B, C, D, E, F, G and H.
Distance of any point P(x, y) from the

Question 3.
What is the distance between two points H and C? And also find the distance between two points A and B.        (Page No. 159)
Answer:
Given: H (- 2, 1), C (2, 1), A (- 1, 2), B (1, 2).
Distance between any two points P(x₁, y₁) and Q(x₂, y₂) is
\(\overline{\mathrm{PQ}}\) = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
∴ Distance between H and C is HC
= \(\sqrt{[2-(-2)]^{2}+(1-1)^{2}}\)
= \(\sqrt{4^{2}+(0)^{2}}\)
= \(\sqrt{16}\)
= 4 units
Distance between A and B is
AB = = \(\sqrt{[1-(-1)]^{2}+(2-2)^{2}}\)
= \(\sqrt{2^{2}+0^{2}}\)
= \(\sqrt{4+0}\)
= 2 units
[!! H, C are points on a line parallel to X – axis.
∴ HC = |x₂ – x₁| = |2 – (- 2)| = 4 Similarly,
AB = |x₂ – x₁| = |-1-1| = 2 ]

Question 4.
Where do these following points lie (-4, 0), (2, 0), (6, 0), (-8, 0) on coordinate plane?    (Page No. 160)
Answer:
Given points are (- 4, 0), (2, 0), (6, 0), (- 8, 0).
All these points have their y-coordinates = 0
∴ These points lie on X-axis.

Question 5.
What is the distance between points (- 4,0) and (6, 0) on coordinate plane?    (Page No. 160)
Answer:
Given points = (- 4, 0) and (6, 0).
These two points lie on the X – axis.
∴ Distance between them = |x₂ – x₁| = |16 – (-4)| = 16 + 41 = 10 units.

Question 6.
Find the distance between the following pairs of points:    (Page No. 162)
i) (3, 8), (6, 8).
Answer:
Given points = A (3, 8) and B (6, 8)
These two points lie on the line parallel to X – axis.
Distance between A (3, 8) and B (6, 8) = |x₂ – x₁|
= |6 – 3| = 3 units.

ii) (-4, -3), (-8, -3).
Answer:
Given points = A (- 4, – 3) and B (- 8, – 3)
These two points lie on the line parallel to X – axis.
∴ Distance between A (- 4, – 3) and B (- 8, – 3) = |x₂ – x₁|
= |-8 – (-4)|
= |-8 + 4|
= |-4| = 4 units.

iii) (3, 4), (3, 8).
Answer:
Given points = A (3, 4) and B (3, 8) These two points lie on the line parallel to Y – axis.
∴ Distance between A (3, 4) and B (3, 8)= |y₂ – y₁|
= |8 – 4| = 4 units.

iv) (-5, -8), (-5, -12).
Answer:
Given points = A (-5, -8) and B (-5, -12)
These two points lie on the line parallel to Y – axis.
∴ Distance between A (-5, -8) and B (-5, -12) = |y₂ – y₁|
= |-12 – (-8)|
= |-12 + 8|
= |-4| = 4 units.

Question 7.
Find the distance between the following points.    (Page No. 162)
i) A = (2, 0) and B(0, 4)
Answer:
Given points = A (2, 0) and B (0, 4)

ii) P(0, 5) and Q(12, 0).
Answer:
Given points = P (0, 5) and Q (12, 0)

Question 8.
Find the distance between the following pair of points.    (Page No. 164)
i) (7, 8) and (- 2, 3)
Answer:
Given points = (7, 8) and (- 2, 3)
Distance formula

ii) (- 8, 6) and (2, 0)
Answer:
Given points = (- 8, 6) and (2, 0).
Distance formula

Try these

Question 1.
Where do these following points lie – (0, -3), (0, -8), (0, 6), (0, 4) on coordinate plane?    (Page No. 161)
Answer:
As the x – coordinate of all these points is zero, all points lie on Y – axis.

Question 2.
What is the distance between (0, -3), (0, -8) and justify that the distance between two points on Y – axis is |y₂ – y₁| on coordinate plane?    (Page No. 161)
Answer:
As the given two points lie on Y-axis, distance between them is
|y₂ – y₁| = |-3 + 8| = |5| = 5 units.
Let (0, y₁) and (0, y₂) be any two points on Y-axis, then distance between them

[∵ distance can’t be negative]

Question 3.
Find the distance between points ‘O’(origin) and ‘A’ (7, 4).    (Page No. 162)
Answer:
Given: Origin and a point (7, 4).
Distance of a point (x, y) from the origin is \(\sqrt{x^{2}+y^{2}}\)
= \(\sqrt{7^{2}+4^{2}}\) = \(\sqrt{49+16}\) = √65 units.

Question 4.
Find the distance between A(1, -3) and B(-4, 4) and rounded to two decimals.      (Page No. 164)
Answer:

Think & Discuss

Question 1.
How will you find the distance between two points in which x or y coordinates are same but not zero?  (Page No. 161)
Answer:
Let the points be A(2, 3), B(2, 5).
Here the x-coordinates are same, then the distance between the points A and B is |y₂ – y₁| = |5 – 3| = 2 units.
If the points are P (4, 3), Q (- 8, 3),
here the y-coordinates are same. In such a case, the distance is given by |x₂ – x₁| = |-8-4| = |-12|
= 12 units.
i.e.,

Question 2.
Ramu says the distance of a point P(x, y) from the origin O(0, 0) is \(\sqrt{x^{2}+y^{2}}\). Do you agree with Ramu or not? Why?    (Page No. 163)
Answer:
Yes. The distance between O(0, 0) and
P(x, y) is \(\sqrt{(x-0)^{2}+(y-0)^{2}}\)
= \(\sqrt{x^{2}+y^{2}}\)

Question 3.
Ramu also writes the distance formula as AB = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\). Why?    (Page No. 163)
Answer:
(x₁ – x₂)² is same as (x₂ – x₁)²
and (y₁ – y₂)² is same as (y₂ – y₁)²

Question 4.
Sridhar calculated the distance between T(5, 2) and R(-4, -1) to the nearest tenth is 9.5 units. Now you find the distance between P (4, 1) and Q (-5, -2). Do you get the same answer that Sridhar got? Why?    (Page No. 164)
Answer:
Given points (4, 1), (-5, -2)

We got the same answer.
∵ (x)² = (-x)²
(x₁ – x₂)² = (x₂ – x₁)² or
The given points (4, 1) and (-4, -1) are images to each other and (5, 2), (-5, -2) are also images to each other.

Do these

Question 1.
Find the point which divides the line segment joining the points (3, 5) and (8, 10) internally in the ratio 2 : 3.  (Page No. 171)
Answer:
Let P (x, y) be the required point then, P (x, y) =

∴ P (x, y) = (5, 7)

Question 2.
Find the midpoint of the line segment joining the points (2, 7) and (12, -7).  (Page No. 171)
Answer:
Midpoint of the line joining the points (x₁, y₁) and (x₂, y₂) is

Question 3.
Find the centroid of the triangle whose vertices are (-4, 6), (2, -2) and (2, 5) respectively.    (Page No. 173)
Answer:
Given points: (- 4, 6), (2, – 2) and (2,-5).
The coordinates of the centroid


∴ the centroid is (0, 3)

Question 4.
Find the trisectional points of line joining (2, 6) and (-4, 8).  (Page No. 175)
Answer:
A (2, 6) and B (- 4, 8) be the given points.
Let P, Q divide the line joining of \(\overline{\mathrm{AB}}\) in the ratio 1 : 2 and 2 : 1.
Section formula (x, y)

For P(x, y) the ratio is 1 : 2.
P(x,y).

For Q (x, y) the ratio is 2 : 1.
Q (x, y)

∴ The points of trisection are \(\left(0, \frac{20}{3}\right)\) and \(\left(-2, \frac{22}{3}\right)\).

Question 5.
Find the trisectional points of line joining (-3, -5) and (-6, -8).  (Page No. 175)
Answer:
Given: A (- 3, – 5) and B (- 6, 8).
Let P and Q be the points of trisection of \(\overline{\mathrm{AB}}\), then P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 2.

Q (x, y) divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1
Q (x, y)
∴ The points of trisection are P (-4, -6) and Q (-5, -7).

Try these

(Page No. 172)

Let A(4,2), B(6, 5) and C(l, 4) be the vertices of △ABC.

Question 1.
The median from A meets BC at D. Find the coordinates of the point D.
Answer:
D is the midpoint of BC

Question 2.
Find the coordinates of the point P on AD such that AP : PD = 2 : 1.      (Page No. 172)
Answer:
P is a point on AD which divides AD in the ratio 2 : 1
∴ P(x,y)

Midpoint of the line joining the points (x₁, y₁) and (x₂, y₂) is

Question 3.
Find the points which divide the line segment BE in the ratio 2 : 1 and also that divide the line segment CF in the ratio 2 : 1.    (Page No. 172)
Answer:
Given: B (6, 5), E\(\left(\frac{5}{2}, 3\right)\)
Let it be P(x, y) =

Similarly, let P divide C(1, 4) and F\(\left(5, \frac{7}{2}\right)\) in the ratio 2:1.

Question 4.
What do you observe? Justify the point that divides each median in the ratio 2 : 1 is the centriod of a triangle.    (Page No. 172)
Answer:
From the above problems, we conclude that the point ‘P’ divides each median in the ratio 2:1.
i.e., the three medians are concurrent at P, which is called the centroid.
A centriod divides each median in the ratio 2 : 1.

Question 5.
The points (2, 3) (x, y) (3, -2) are the vertices of a triangle. If the centroid of this triangle is origin then find (x, y).   (Page No. 173)
Answer:
Given vertices of triangle are (2, 3) (x, y) (3, -2)
now formula for centroid

∴ 5 + x = 0 and 1 + y = 0
⇒ x = -5 and y = -1
∴ (x, y) = (-5, -1)

Think & Discuss

(Page No. 174)

Question 1.
The line joining points A(6, 9) and B(-6, -9) are given.
a. In which ratio does origin divide \(\overline{\mathrm{AB}}\)? And what it is called for \(\overline{\mathrm{AB}}\)?
Answer:
Given : A (6, 9), B (-6, -9)
Let origin O(0, 0) divides \(\overline{\mathrm{AB}}\) in the ratio k : 1 internally.
[By section formula]

The ratio is 1 : 1.
Here the origin bisects \(\overline{\mathrm{AB}}\).
∴ Origin is called the midpoint of \(\overline{\mathrm{AB}}\).

b. In which ratio does the point P(2, 3) divide \(\overline{\mathrm{AB}}\)?
Answer:
Given: A (6, 9), B (-6, -9) and P (2, 3) divide \(\overline{\mathrm{AB}}\) internally in the ratio say k : 1. [By section formula]
Then P (2, 3) =

⇒ 2k + 2 = -6k + 6 and 3k + 3 = -9k + 9
⇒ 8k = 6 – 2 and 3k + 9k = 9 – 3
⇒ k = \(\frac{4}{8}\) and 12k = 6
⇒ k = \(\frac{1}{2}\) and k = \(\frac{6}{12}\)
⇒ k = \(\frac{1}{2}\)
∴ The ratio (k : 1) = \(\left(\frac{1}{2}: 1\right)\) = 2 : 1

c. In which ratio does the point Q(-2, -3) divide \(\overline{\mathrm{AB}}\)?
Answer:
Let Q divide \(\overline{\mathrm{AB}}\) in the ratio say k : 1 internally, then

[By section formula]

⇒ 6k + 6 = -2k-2 and -9k + 9 = -3k-3
⇒ -6k + 2k = -2-6 and -9k + 3k = -3-9
⇒ -4k = -8 and -6k = -12
⇒ k = \(\frac{-8}{-4}\) = 2 and k = \(\frac{-12}{-6}\) = 2
∴ The ratio is k : 1 = 2 : 1

d. Into how many equal parts is \(\overline{\mathrm{AB}}\) divided by P and Q?
Answer:
Since P, Q divide \(\overline{\mathrm{AB}}\) in the ratio 1 : 2 and 2 : 1, \(\overline{\mathrm{AB}}\) is divided into three equal parts by P and Q.

e. What do we call P and Q for \(\overline{\mathrm{AB}}\)?
Answer:
P and Q are the points of trisection of \(\overline{\mathrm{AB}}\).

Do these

Question 1.
Find the area of the triangle whose vertices are
(5, 2) (3,-5) and (-5,-1).  (Page No. 180)
Answer:
Given: The vertices of the triangle are (5, 2), (3, -5) and (-5, -1).
Area of the triangle

Question 2.
(6, -6), (3, -7) and (3, 3).    (Page No. 180)
Answer:
Given: The vertices of a*triangle are (6, -6), (3, -7) and (3, 3).
Area of a triangle

Question 3.
Verify whether the following points are collinear or not.  (Page No. 182)
i) (1, -1), (4, 1), (-2, -3).
Answer:
Given: Three points (1, -1), (4, 1), (-2, -3).
Area of the triangle

As the area of the triangle is ‘O’, the three points are collinear.

ii) (1, -1), (2, 3), (2, 0).
Answer:
Given points are (1, -1), (2, 3), (2, 0).

Area of the triangle formed by the given three points is

△ ≠ 0.
Hence the points are not collinear.

iii) (1, -6), (3, -4), (4, -3).
Answer:
The given points are (1, -6), (3, -4), (4, -3).
Area of a triangle

Area of the triangle formed by the given three points is

As △ = 0, the points are collinear.

Question 4.
Find the area of the triangle whose lengths of sides are 15 m, 17 m, 21 m. (use Heron’s Formula)  (Page No. 183)
Answer:
Given: The sides of a triangle
a = 15 m; b = 17 m and c = 21 m.

Note: As the ‘height’ is not given we can’t verify the above answer by A = \(\frac{1}{2}\)bh

Question 5.
Find the area of the triangle formed by the points (0, 0), (4, 0), (4, 3) by using Heron’s formula.  (Page No. 183)
Answer:
The given points are O (0, 0), A (4, 0) and B (4, 3).
Then the sides

Heron’s formula

Verification:
If the sides are 3, 4 and 5 units, clearly the triangle is a right triangle.

Try these

Question 1.
Take a point A on X-axis and B on Y-axis and find area of the triangle AOB. Discuss with your friends what did they do.      (Page No. 178)
Answer:

[∴ Axes are perpendicular to each other]
Consider the points A(5, 0) and B(0, 6).
△AOB = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × 6 × 5 = 15 sq. units.
Area of a triangle A (x, 0), O (0, 0) and
B (0, y) is \(\frac{1}{2}\)xy.

Question 2.
Find the area of the square formed by (0, -1), (2, 1), (0, 3) and (-2, 1) taken in order are as vertices.    (Page No. 178)
Answer:

Let A (0, -1), B (2, 1), C (0, 3) and D (- 2, 1) are the vertices of a square.
Area of the square ABCD = side²
= AB²
But, AB = \(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
= \(\sqrt{(2-0)^{2}+(1+1)^{2}}\)
= \(\sqrt{2^{2}+2^{2}}\)
= \(\sqrt{4+4}\)
= √8
Area of square = √8 × √8
= 8 sq. units.

Think & Discuss

Question 1.
Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃). Then find the area of the following triangles in a plane. And discuss with your friends in groups about the area of that triangle.    (Page No. 178)

Answer:
i) Given △AOB where A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃)

But we know that the origin O is (0, 0) which is given as C.
\(\overline{\mathrm{CB}}\) = x₂ – x₃ = x₂ – 0 = x₂
\(\overline{\mathrm{AB}}\) = y₁ – y₂ = y₁ – 0 = y₁
∴ Area of △ABC = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\)x₂y₁

ii)

A(x₁, y₁) = (x₁, y₁)
B(x₂, y₂) = (0, y₂)
C(x₃, y₃) = (0, 0)
∴ Area of △ABC = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × y₂ × x₁
= \(\frac{1}{2}|\)x₁y₂| sq. units.

iii)

∴ AB = x₁ – x₂
BC = y₂ – y₃
A(x₁, y₁) = (-x₁, y₁)
B(x₂, y₂) = (x₂, y₂)
C(x₃, y₃) = (x₃, -y₃)
∴ Area of △ABC

iv)

AB = (y₂ – y₁)
BC = (x₁ – x₃)
A(x₁, y₁) = (x₁, y₁)
B(x₂, y₂) = (x₁, y₂)
C(x₃, y₃) = (x₃, y₂)
∴ Area of △ABC = \(\frac{1}{2}\) × base × height

Question 2.
Find the area of the triangle formed by the following points    (Page No. 181)
i) (2, 0), (1, 2), (1, 6)
Answer:
Take the third point as (-1, 6).

ii) (3, 1), (5, 0), (1, 2)
Answer:

iii) (-1.5, 3), (6, 2), (-3, 4)
Answer:
Take the second point as (6, -2)

Question 3.
What do you observe?          (Page No. 181)
Answer:
We observe that the area formed by above all triangles is zero.

Question 4.
Plot these points on three different graphs. What do you observe?      (Page No. 181)
Answer:

We observe that the points are collinear.

Question 5.
Can we have a triangle having area zero square units area?    (Page No. 181)
Answer:
No.

Question 6.
What does it mean?      (Page No. 181)
Answer:
If the area of the triangle formed by any three points is zero, it means the points are collinear.

Do these

Question 1.
Plot these points on the coordinate axis and join them:
Which gives a straight line? Which is not? Why?    (Page No. 185)
i) A(1, 2), B(-3, 4), C(7, -1)
Answer:
Points ABC gives a straight line.

ii) P(3, -5), Q(5, -1), R(2, 1), S(1, 2)
Answer:
Points PQRS doesn’t give a straight line.

Question 2.
Find the slope of \(\overleftrightarrow{\mathbf{A B}}\) with the given end points,      (Page No. 188)
i) A(4, -6), B(7, 2)
Answer:

ii) A(8, -4), B(-4, 8)
Answer:

iii) A(-2, -5), B(l, -7)
Answer:

Try these

(Page No. 188)

Question 1.
Find the slope of AB with the points lying on
i) A(2, 1), B(2, 6)
Answer:

Question 2.
A(-4, 2), B(-4, -2)
Answer:

Question 3.
A(-2, 8), B(-2, -2)
Answer:

Question 4.
Justify that the line AB line segment formed by given points is parallel to Y-axis. What can you say about their slope? Why?
Answer:
In the above problem, all points are of the form (K, y) where K is a fixed number and y is a variable.
∴ All lines in the above problem are parallel to Y – axis. Slope of lines parallel to y – axis are not defined.

Think & Discuss

Question 1.
Does y = x + 7 represent a straight line? Draw the line on the coordinate plane. At which point does this line intersect Y – axis?
How much angle does it make with X – axis? Discuss with your friends.        (Page No. 185)
Answer:
Yes. y = x + 7 represents a straight line.

Angle made by y = x + 7 with X-axis is 45°. (∵ (0, 7) and (- 7, 0) are equidistant from the origin and hence the triangle formed is right isosceles triangle.)

Question 2.
Find the slope AB with the points lying on A(3, 2), B(- 8, 2). When the line AB parallel to X-axis? Why? Think and discuss with your friends in groups.      (Page No. 188)
Answer:
Given : A (3, 2), B (-8, 2)

Yes. The line is parallel to X-axis as the points are of the form (x₁, K), (x₂, K)

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions InText Questions and Answers.

10th Class Maths 6th Lesson Progressions InText Questions and Answers

Question 1.
Write three examples for finite A.P and three for infinite A.P. (Page No. 130)
Answer:
Examples for finite A.P
i) 3, 5, 7, 9, 11 where a = 3; d = 2.
ii) x, x + a, x + 2a, x + 3a, x + 4a, x + 5a where a = x; d = a.
iii) \(\frac{1}{9}\), \(\frac{2}{9}\), \(\frac{3}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{6}{9}\), \(\frac{7}{9}\), ….
where a = \(\frac{1}{9}\); d = \(\frac{1}{9}\).

Examples for infinite A.P
i) 10, 20, 30, 40, ……
where a = 10, d = 10.
ii) 5.5, 6.6, 7.7, 8.8, 9.9,……
where a = 5.5; d = 1.1.
iii) -100, -95, -90, -85,…..
where a = – 100, d = 5.

Do these

Question 1.
Write three examples for finite A.P and three for infinite A.P. (Page No. 130)
Answer:
Examples for finite A.P
i) 3, 5, 7, 9, 11 where a = 3; d = 2.
ii) x, x + a, x + 2a, x + 3a, x + 4a, x + 5a where a = x; d = a.
iii) \(\frac{1}{9}\), \(\frac{2}{9}\), \(\frac{3}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{6}{9}\), \(\frac{7}{9}\), ….
where a = \(\frac{1}{9}\); d = \(\frac{1}{9}\).
Examples for infinite A.P
i) 10, 20, 30, 40, ……
where a = 10, d = 10.
ii) 5.5, 6.6, 7.7, 8.8, 9.9,……
where a = 5.5; d = 1.1.
iii) -100, -95, -90, -85,…..
where a = – 100, d = 5.

(Page Nos. 131, 132)

Question 2.
Take any Arithmetic Progression.
Answer:
4, 7, 10, 13, 16, ……

Question 3.
Add a fixed number to each and every tetm of A.P. Write the resulting numbers as a list.
Answer:
4, 7, 10, 13, 16, …….
Adding ‘5’ to each term of the above A.P. we get
4 + 5, 7 + 5, 10 + 5, 13 + 5, 16 + 5,…
9, 12, 15, 18, 21, ……
In the list obtained the first term
a₁ = 9; a₂ = 12, a₃ = 15, a₄ = 18,
Also a₂ – a₁ = 12 – 9 = 3
a₃ – a₂ = 15 – 12 = 3
a₄ – a₃ = 18 – 15 = 3
……………………………………
i.e.,
d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = …. = 3
∴ The resulting list forms an A.P.

Question 4.
Similarly subtract’a fixed number from each and every term of A.P. Write the resulting numbers as a list.
Answer:
4, 7, 10, 13, 16, ……
Subtracting ‘2’ from the each term of A.P in given series, we get
4 – 2, 7 – 2, 10- 2, 13 – 2, 16 – 2, ……
2, 5, 8, 11, 14, ……
In the list obtained, the first term
a₁ = 2 , a₂ = 5, a₃ = 8, a₄ = 11,
Also a₂ – a₁ = 5 – 2 = 3
a₃ – a₂ = 8 – 5 = 3
a₄ – a₃ = 11 – 8 = 3
……………………………………
i.e., d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = 3
∴ The resulting list forms an A.P.

Question 5.
Multiply and divide each term of A.P by a fixed number and write the resulting numbers as a list.
Answer:
4, 7, 10, 13, 16, ……
Multiplying each term by 3, we get
4 × 3, 7 × 3, 10 × 3, 13 × 3, 16 × 3, ……
12, 21, 30, 39, 48, …….
In the list obtained the first term a₁ = 12 and a₂ = 21, a₃ = 30, ….
Also a₂ – a₁ = a₃ – a₂ = …… = 9
∴ The resulting list also forms an A.P.
Now divide every term by 7, we get
\(\frac{4}{7}\), \(\frac{7}{7}\), \(\frac{10}{7}\), \(\frac{13}{7}\), \(\frac{16}{7}\), …… is the resulting list.

Question 6.
Check whether the resulting lists are AP in each case.
Answer:
The first term
∴ d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = ….. = \(\frac{3}{7}\)
and the above list forms an A.P.

Question 7.
What is your conclusion?
Answer:
If a₁, a₂, a₃, …… are in A.P, then

i.e., “If each term of an A.P is added/ multiplied / divided by a fixed number, the resulting terms also form an A.P” and fixed term is subtracted from each term of an A.P, then the resulting terms also form an A.P.

Try these

Question 1.
i) Which of these are arithmetic progressions and why?  (Page No. 128)
a) 2, 3, 5, 7, 8, 10, 15, ……
Answer:
2, 3, 5, 7, 8, 10, 15, …… is not an A.P.
∵ a₂ – a₁ = 3 – 2 = 1
a₃ – a₂ = 5 – 3 = 2
a₄ – a₃ = 7 – 5 = 2
i.e., The difference between any two successive terms is not same throughout the series.
(or)
Every number is not formed by adding a fixed number to its preceding term.

b) 2, 5, 7, 10, 12, 15 ……
Answer: The given list does not form an A.P, since each term is not obtained by adding a fixed number to its preceding term.

c) -1,-3,-5,-7, ……
Answer: -1,-3,-5,-7,….. is an A.P.
a₂ – a₁ = – 3 – (- 1) = -3 + 1 = -2
a₃ – a₂ = – 5 – (-3) = -5 + 3 = -2
a₄ – a₃ = – 7 – (- 5) = -7 + 5 = -2
Every number is formed by adding a fixed number to its preceding term,

ii) Write 3 more Arithmetic Progressions.
Answer:
a) a = -7;d = -3 and
A.P. is-7, – 10, – 13, – 16, …….
b) a = 15; d = 4 and
A.P. is 15, 19, 23, 27, 31, …….
c) a = 100; d = 50 and
A.P. is 100, 150, 200, 250, ……..

Think & Discuss

(Page No. 129)

Question 1.
Think how each of the list given above form an A.P. Discuss with your friends.
a) Heights (in cm) of some students of a school standing in a queue in the morning assembly are 147,148, 149,…, 157.
Answer:
The given list forms an A.P, since each term starting from the second is obtained by adding a fixed number + 1 to its preceding term.

b) Minimum temperatures (in degree ‘ Celsius) recorded for a week, in the month of January in a city, arranged in ascending order are -3.1, -3.0, -2.9, -2.8, -2.7, -2.6, -2.5, …….
Answer:
The given list forms an A.P, since every term starting from the second is obtained by adding a fixed number +0.1 to its preceding term.

c) The balance money (in Rs.) after paying 5% of the total loan of Rs. 1000 every month is 950, 900, 850, 800, …, 50.
Answer:
The given list forms an A.P, since each term starting from the second is obtained by adding a fixed number (-50) to its preceding term.

d) Cash prizes (in Rs.) given by a school to the toppers of Classes I to XII are 200, 250, 300, 350, ….., 750 respectively.
Answer:
The given list forms an A.P, since each term starting from the second is obtained by adding a fixed number 50 to its preceding term.

e) Total savings (in Rs.) after every month for 10 months when Rs. 50 are saved each mouth are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
Answer:
The given list forms an A.P, since every term starting from the second term is obtained by adding a fixed number 50 to its preceding term.

Question 2.
Find the common difference of each of the above lists. Think when is it positive?
Answer:
Common difference d = a₂ – a₁
a) 148 – 147 = 1
b) -3.0 – (-3.1) = 0.1
c) 900 – 950 = -50
d) 250 – 200 = 50
e) 100 – 50 = 50
Common difference is positive when a₂ > a₁

Question 3.
Make a positive Arithmetic Progression in which the common difference is a small positive quantity.
Answer:
a = 50 ; d = 0.5 then A.P is 50, 50.5, 51, 51.5, 52, ……

Question 4.
Make an A.P in which the common difference is big (large) positive quantity.
Answer:
a = 100; d = 1000 then A.P. is 100, 1100, 2100, 3100, 4100, ……

Question 5.
Make an A.P in which the common difference is negative.
Answer:
a = 80, d = -7
then A.P. is 80, 73, 66, 59, 52, ……

Do these

(Page No. 143)

Find the sum of indicated number of terms in each of the following A.Ps.
i) 16, 11, 6, …..; 23 terms.
Answer:
Given: 16, 11, 6, …..; S₂₃
t₁ = a = 16; t₂ = 11; t₃ = 6,
d = t₂ – t₁ = 11 – 16 = -5

= -23 × 39 = -897

ii) -0.5, -1.0, -1.5,…..; 10 terms.
Answer:
Given : -0.5, -1.0, -1.5, …. S₁₀
a = – 0.5
d = t₂ – t₁ = (-1.0) – (-0.5)

iii) -1, \(\frac{1}{4}\), \(\frac{3}{2}\), …… ;10 terms.
Answer:
Given: -1, \(\frac{1}{4}\), \(\frac{3}{2}\), …… ;S₁₀.
a = – 1
d = t₂ – t₁ = \(\frac{1}{4}\) – (-1) = 1 + \(\frac{1}{4}\) = \(\frac{5}{4}\)

Do these

(Page No. 149)

Find which of the following are not G.P.

Question 1.
6, 12, 24, 48, ……
Answer:
Given: 6, 12, 24, 48, ……
a₁ = a = 6; a₂ = 12; a₃ = 24, a₄ = 48,…

The given list is of the form
a, ar, ar², ar³,
∴ The given numbers are in G.P.

Question 2.
1, 4, 9, 16, ……
Answer:
Given: 1, 4, 9, 16, …..
a₁ = a = 1
a₂ = 4; a₃ = 9, a₄ = 16

∴ The given numbers do not form a G.P.

Question 3.
1, -1, 1, -1, …..
Answer:
Given: 1, -1, 1, -1, …….
a₁ = a = 1
a₂ = -1; a₃ = 1, a₄ = -1, …..

∴ The given list forms a G.P.

Question 4.
-4, -20, -100, -500, ……
Answer:
Given: -4, -20, – 100, -500, ……
a₁ = a = -4, a₂ = -20, a₃ = -100, a₄ = -500,

∴ The given list forms a G.P.

Think & Discuss

(Page No. 149)

Question 1.
Explain why each of the lists above is a G.P.
i) 1, 4, 16, 64, 256, …….
Answer:
Here
a = 1 = a₁; a₂ = 4; a₃ = 16; a₄ = 64,….

i.e., Common ratio r = 4.

ii) 550, 605, 665.5, ……..
Answer:
The given series is in G.P. Since every term can be obtained by multiplying its preceding term by a fixed number ‘1.1’.

iii) 256, 128, 64, 32,…….
Answer:
The given series forms a G.P.
Since every term, starting from the second can be obtained by multiplying its preceding term by a fixed number \(\frac{1}{2}\).

iv) 18, 16.2, 14.58, 13.122, …….
Answer:
The given list forms a G.P.
Since each term, starting from the second can be obtained by multiplying its preceding term by a fixed number 0.9.
here

Question 2.
To know about a G.P. what is minimum information that we need?
Answer:
To know whether a number pattern forms a G.P or not, we should check that the ratio between the successive terms is equal or not.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations InText Questions and Answers.

10th Class Maths 5th Lesson Quadratic Equations InText Questions and Answers

Try this

Question 1.
Check whether the following equations are quadratic or not. (Page No. 102)
i) x² – 6x – 4 = 0
ii) x³ – 6x² + 2x – 1 = 0
iii) 7x = 2x²
iv) x² + \(\frac{1}{\mathbf{x}^{2}}\) = 2
v) (2x + 1) (3x + 1) = 6(x – 1) (x – 2)
vi) 3y² = 192
Answer:
i) x² – 6x – 4 = 0
Yes. It’s a quadratic equation.
ii) x³ – 6x² + 2x – 1 =0
No. It is not a quadratic equation. [∵ degree is 3]
iii) 7x = 2x²
Yes. It’s a quadratic equation.
iv) ² + \(\frac{1}{\mathbf{x}^{2}}\) = 2
No. It is not a quadratic equation. [∵ degree is 4]
v) (2x + 1) (3x + 1) = 6(x – 1) (x – 2) No. It is not a quadratic equation, [∵ co-efficient of x² on both sides is same i.e. 6]
vi) 3y² = 192
Yes. It’s a quadratic equation.

Try this

Question 1.
Verify that 1 and \(\frac{3}{2}\) are the roots of the equation 2x² – 5x + 3 = 0. (Page No. 107)
Answer:
Let the given Q.E. be p(x) = 2x² – 5x + 3
Now p(1) = 2(1)² – 5(1) + 3
= 2 – 5 + 3 = 0
∴ 1 is a root of 2x² – 5x + 3 = 0
also p\(\left(\frac{3}{2}\right)\) = 2\(\left(\frac{3}{2}\right)^{2}\) – 5\(\left(\frac{3}{2}\right)\) + 3
= 2 × \(\frac{9}{4}\) – \(\frac{15}{2}\) + 3
= \(\frac{9}{2}\) + 3 – \(\frac{15}{2}\)
= \(\frac{9+6-15}{2}\) = 0
∴ \(\frac{3}{2}\) is also a root of 2x² – 5x + 3 = 0.

Do this

Question 1.
Solve the equations by completing the square.  (Page No. 113)
i) x² – 10x + 9 = 0
Answer:
Given: x² – 10x + 9 = 0
⇒ x² – 10x = -9
⇒ x² – 2.x.5 = -9
⇒ x² – 2.x.5 + 52 = -9 + 52
⇒ (x – 5)² = 16
∴ x – 5 = ± 4
∴ x – 5 = 4 (or) x – 5 = -4
⇒ x = 9 (or) x = 1
⇒ x = 9 (or) 1

ii) x² – 5x + 5 = 0
Answer:
Given: x² – 5x + 5 = 0
⇒ x² – 5x = 5

iii) x² + 7x-6 = 0
Answer:
x² + 7x – 6 = 0
x² + 7x = 6

Think & Discuss

Question 1.
We have three methods to solve a quadratic equation. Among these three, which method would you like to use 7 Why? (Page No. 115)
Answer:
If the Q.E. has distinct and real roots, we use factorisation. If Q.E. has no real roots, we use quadratic formula.

Try these

Question 1.
Explain the benefits of evaluating the discriminant of a quadratic equation before attempting to solve it. What does its value signifies?  (Page No. 122)
Answer:
The discriminant b² – 4ac of a Q.E.
ax² + bx + c = 0 gives the clear idea about the nature of the roots of the Q.E.
If the discriminant D = b² – 4ac > 0, the Q.E. has distinct and real roots.
If b² – 4ac = 0, the Q.E. has equal roots.
If b² – 4ac < 0, the Q.E. has no real roots.
By the value of the discriminant, we can state the nature of the roots of a Q.E. without actually finding them.

Question 2.
Write three quadratic equations one having two distinct real solutions, one having no real solution and one having exactly one real solution.  (Page No. 122)
Answer:

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables InText Questions and Answers.

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables InText Questions and Answers

Question 1.
Solve the following systems of equations: i) x – 2y = 0; 3x + 4y = 20     (Page No. 79)
Answer:
i) x – 2y = 0;                                                                         3x + 4y = 20
-2y = -x                                                                               4y = 20 – 3x
y = \(\frac{x}{2}\)                                                                                      y = \(\frac{20-3x}{4}\)


The two lines meet at (4, 2).
The solution set is {(4, 2)}

ii) x + y = 2
2x + 2y = 4
Answer:
x + y = 2
2x + 2y = 4


These two are coincident lines.
∴ There are infinitely many solutions.

iii) 2x – y = 4
4x – 2y = 6
Answer:
2x – y = 4                                                                                4x – 2y = 6
⇒ y = 2x – 4                                                                           ⇒ 2y = 4x – 6 ⇒ y = 2x – 3

These two are parallel lines.
∴ The pair of linear equations has no solution.

Question 2.
Two rails of a railway track are represented by the equations.
x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation graphically.    (Page No. 79)
Answer:
x + 2y – 4 = 0                                                                     2x + 4y – 12 = 0
2y = 4 – x                                                                         4y = 12 – 2x (or) 4y = 2 (6 – x)
y = \(\frac{4-x}{2}\)                                                                                  y = \(\frac{6-x}{2}\)
x + 2y – 4 = 0                                                                     2x + 4y – 12 = 0


These lines are parallel and hence no solution.

Question 3.
Check each of the given systems of equations to see if it has a unique solution, infinitely many solutions or no solution. Solve them graphically. (Page No. 83)
i) 2x + 3y = 1
3x – y = 7
Answer:
Let a₁x + b₁y – c₁ = 0 ≃ 2x + 3y – 1 = 0
a₂x + b₂y + c₂ = 0 ≃ 3x – y – 7 = 0
Now comparing their coefficients i.e., \(\frac{a_{1}}{a_{2}}\) and \(\frac{b_{1}}{b_{2}}\)
⇒ \(\frac{2}{3}\) ≠ \(\frac{3}{-1}\)
The given lines are intersecting lines.
2x + 3y = 1                                                                     3x – y = 7
3y = 1 – 2x                                                                      y – 3x = 7
y = \(\frac{1-2x}{3}\)


The system of equations has a unique solution (2, – 1).

ii) x + 2y = 6
2x + 4y = 12
Answer:
From the given pair of equations,
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\)
∴ The lines are dependent and have infinitely many solutions.
x + 2y = 6                                                                          2x + 4y = 12
2y = 6 – x                                                                          4y = 12 – 2x (or) 4y = 2(6 – x)
y = \(\frac{6-x}{2}\)                                                                             y = \(\frac{12-2x}{4}\) or y = \(\frac{6-x}{2}\)

iii) 3x + 2y = 6
6x + 4y = 18
Answer:
From the given pair of equations,
\(\frac{a_{1}}{a_{2}}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}\) = \(\frac{6}{18}\) = \(\frac{1}{3}\)
∴ \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\)
∴ The lines are parallel and hence no solution.
3x + 2y = 6                                                                                   6x + 4y = 18
2y = 6 – 3x                                                                                    4y = 18 – 6x
y = \(\frac{6-3x}{2}\)                                                                                    y = \(\frac{18-6x}{4}\)

Try these

(Page No. 75, 76)

Question 1.
Mark the correct option in the following questions:
Which of the following equations is not a linear equation?
a) 5 + 4x = y + 3
b) x + 2y = y – x
c) 3 – x = y² + 4
d) x + y = 0
Answer:
[ c ]

Question 2.
Which of the following is a linear equation in one variable?
a) 2x + 1 = y – 3
b) 2t – 1= 2t + 5
c) 2x – 1 = x²
d) x² – x + 1 =0
Answer:
[ b ]

Question 3.
Which of the following numbers is a solution for the equation 2(x + 3) = 18?
a) 5
b) 6
c) 13
d) 21
Answer:
[b]

Question 4.
The value of x which satisfies the equation 2x – (4 – x) = 5 – x is
a) 4.5
b) 3
c) 2.25
d) 0.5
Answer:
[ c ]

Question 5.
The equation x – 4y = 5 has
a) no solution
b) unique solution
c) two solutions
d) infinitely many solutions
Answer:
[ d ]

Question 6.
In the example given above, can you find the cost of each bat and ball?    (Page No. 79)
Answer:
We can’t find the exact values for the costs of bat and ball as there are infinitely many possibilities.

Question 7.
For what value of ‘p’ the following pair of equations has a unique solution.     (Page No. 83)
2x + py = – 5 and 3x + 3y = – 6
Answer:
Given: 2x + py = – 5
3x 4- 3y = – 6
To have an unique solution we should have

∴ The pair has unique solution when p ≠ 2.

Question 8.
Find the value of ‘k’ for which the pair of equations 2x – ky + 3 = 0, 4x + 6y – 5 = 0 represent parallel lines.   (Page No. 83)
Answer:
Given: 2x – ky + 3 = 0
4x + 6y – 5 = 0
If the above lines are to be parallel, then

∴ k = – 3 is the required value for which lines are parallel.

Question 9.
For what value of ‘k’, the pair of equation 3x + 4y + 2 = 0 and 9x + 12y + k = 0 represent coincident lines.   (Page No. 83)
Answer:
Given: 3x + 4y + 2 = 0
9x + 12y + k = 0
If the lines are to be coincident with each other, then

∴ k = 2 × 3 = 6

Question 10.
For what positive values of ‘p’ the following pair of linear equations have infinitely many solutions?   (Page No. 83)
px + 3y – (p – 3) = 0
12x + py – p = 0
Answer:
Given: px + 3y – (p – 3) = 0
12x + py – p = 0
The above equations to have infinitely many solutions

p.p = 12 × 3
⇒ p² = 36
⇒ p = ±6

Think & Discuss

Question 1.
Two situations are given below:
i) The cost of 1 kg potatoes and 2 kg tomatoes was Rs. 30 on a certain day. After two days, the cost of 2 kg potatoes and 4 kg tomatoes was found to be Rs. 66.
ii) The coach of a cricket team of M.K. Nagar High School buys 3 bats and 6 balls for Rs. 3900. Later he buys one more bat and 2 balls for Rs. 1300.
Identify the unknowns in each situation. We observe that there are two unknowns in each case. (Page No. 73)
Answer:
i) The unknowns in the first problem are
a) cost of 1 kg tomatoes
b) cost of 1 kg potatoes
ii) In the second problem, the unknowns are
a) cost of each bat
b) cost of each ball

Question 2.
Is a dependent pair of linear equations always consistent? Why or why not? (Page No. 79)
Answer:
Reason: \(\frac{a_{1}}{a_{2}}\) = \(\frac{b_{1}}{b_{2}}\) = \(\frac{c_{1}}{c_{2}}\) always holds. In other words, they have infinitely many solutions.

Do these

Solve each pair of equations by using the substitution method. (Page No. 88)
Question 1.
3x – 5y = -1
x – y = -1
Answer:
Given: 3x – 5y = -1 ……. (1)
x – y = -1 …….. (2)
From equation (2), x – y = – 1
x = y – 1
Substituting x = y – 1 in equation (1)
we get
3 (y – 1) – 5y = – 1
⇒ 3y – 3 – 5y = r 1
⇒ – 2y = – 1 + 3
⇒ 2y = – 2
⇒ y = -1
Substituting y = – 1 in equation (1) we get
3x – 5 (- 1) = -1
3x + 5 = – 1
3x = – 1 – 5
x = \(\frac{-6}{3}\) = -2
∴ The solution is (-2, -1)

Question 2.
x + 2y = – 1
2x – 3y = 12
Answer:
Given: x + 2y = -1 ……. (1)
2x – 3y = 12 …….. (2)
From equation (1)x + 2y = -l
⇒ x = – 1 – 2y
Substituting x = – 1 – 2y in equation (2), we get
2 (- 1 – 2y) – 3y = 12
– 2 – 4y – 3y = 12
– 2 – 7y = 12
7y = – 2 – 12
∴ y = \(\frac{-14}{7}\) = -2
Substituting y = – 2 in equation (1), we get
x + 2 (- 2) = – 1
x = – 1 + 4
x = 3
∴ The solution is (3, – 2)

Question 3.
2x + 3y = 9
3x + 4y = 5
Answer:
Given: 2x + 3y – 9 …….. (1)
3x + 4y = 5 ……. (2)
From equation (1);
2x = 9 – 3y
x = \(\frac{9-3y}{2}\)
Substituting x = \(\frac{9-3y}{2}\) in equation (2)
we get

Substituting y = + 17 in equation (1) we get
2x + 3 (+ 17) = 9
⇒ 2x = 9 – 51
⇒ 2x = -42
⇒ x = -21
∴ The solution is (-21, 17)

Question 4.
x + \(\frac{6}{y}\) = 6
3x – \(\frac{8}{y}\) = 5
Answer:
Given:
x + \(\frac{6}{y}\) = 6 …….. (1)
3x – \(\frac{8}{y}\) = 5 …….. (2)
From equation (1) x = 6 – \(\frac{6}{y}\)
Substituting x = 6 – \(\frac{6}{y}\) in equation (2)
we get

Substituting y = 2 in equation (1) we get
x + \(\frac{6}{2}\) = 6 ⇒ x + 3 = 6
∴ x = 3
∴ The solution is (3, 2)

Question 5.
0.2x + 0.3y =1.3
0.4x + 0.5y = 2.3
Answer:
Given:
0.2x + 0.3y = 1.3
⇒ 2x + 3y = 13 …… (1)
0.4x + 0.5y = 2.3
⇒ 4x + 5y = 23 …… (2)
From equation (1)
2x = 13 – 3y
⇒ x = \(\frac{13-3y}{2}\)
Substituting x = \(\frac{13-3y}{2}\) equation (2) we get
\(\frac{13-3y}{2}\) + 5y = 23
⇒ 26 – 6y + 5y = 23
⇒ -y + 26 = 23
⇒ y = 26 — 23 = 3
Substituting y = 3 in equaion (1) we get
2x + 3(3) = 13
⇒ 2x + 9 = 13
⇒ 2x = 13 – 9
⇒ 2x = 4
⇒ x = \(\frac{4}{2}\) = 2
∴ The solution is (2, 3)

Question 6.
√2x + √3y = 0
√3x – √8y = 0
Answer:
Given:
√2x + √3y = 0 ……. (1)
√3x – √8y = 0 ……. (2)
Substitute x = 0 in (1),
√2(0) + √3y = 0
√3y = 0
∴ y = 0
∴ The solution is x = 0, y = 0
Note: a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0, if c₁ = c₂ = 0
then, x = 0, y = 0 is a solution.

Solve each of the following pairs of equations by the elimination method. (Page No. 89)
Question 7.
8x + 5y = 9
3x + 2y = 4
Answer:
Given: 8x + 5y = 9 ……. (1)
3x + 2y = 4 …….. (2)

∴ y = 5
Substituting y = 5 in equation (1) we get
8x + 5 × 5 = 9
⇒ 8x = 9 – 25
x = \(\frac{-16}{8}\) = -2
∴ The solution is (- 2, 5)

Question 8.
2x + 3y = 8
4x + 6y = 7
Answer:
Given: 2x + 3y = 8 ……. (1)
4x + 6y = 7 …….. (2)

The lines are parallel.
∴ The pair of lines has no solution.

Question 9.
3x + 4y = 25
5x – 6y = -9
Answer:
Given: 3x + 4y = 25 ……. (1)
5x – 6y = -9 …….. (2)

Substituting y = 4 in equation (1) we get
3x + 4 × 4 = 25
3x = 25 – 16
⇒ x = \(\frac{9}{3}\) = 3
∴ (3,4) is the solution for given pair of lines.

Question 10.
In a competitive exam, 3 marks are awarded for every correct answer and for every wrong answer, 1 mark will be deducted. Madhu scored 40 marks in this exam. Had 4 marks been awarded for each correct answer and 2 marks deducted for each incorrect answer, Madhu would have scored 50 marks. How many questions were there in the test? (Madhu attempted all the questions) .
Now use the elimination method to solve the above example – 9.
Answer:
The equations formed are
3x – y = 40 ……. (1)
4x – 2y = 50 …….. (2)
Substituting y = 5 in equation (1) we get
3x – 5 = 40
⇒ 3x = 40 + 5
⇒ x = \(\frac{45}{3}\) = 15
Total number of questions = Number of correct questions + Number of wrong answers
= x + y
= 15 + 5 = 20

Question 11.
Mary told her daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Find the present age of Mary and her daughter. Solve example – 10 by the substitution method.
Answer:
The equations formed are
The equations formed are
x – 7y + 42 = 0 ……. (1)
x – 3y – 6 = 0 …….. (2)
From (1), x = – 42 + 7y
Substituting x = – 42 + 7y in equation (2) we get
-42 + 7y – 3y – 6 = 0
⇒ 4y – 48 = 0
⇒ y = \(\frac{48}{4}\) = 12
Substituting y = 12 in equation (2) we get
x – 3 × 12 – 6 = 0
x – 36 – 6 = 0
x = 42

Try this

Question 1.
Solve the given pair of linear equations, (a – b)x + (a + b)y = a² – 2ab – b²
(a + b) (x + y) = a² + b² (Page No. 89)
Answer:

x(-2b) = – 2b(a + b)
⇒ x = (a + b)
Put this value of ‘x’ in eq (1) we get
(a – b) (a + b) + (a + b)y = a² – 2ab – b²
a² – b² + (a + b)y = a² – 2ab – b²
⇒ y = \(\frac{-2ab}{a+b}\)
∴ Solution to given pair of linear equations x = a + b, y = \(\frac{-2ab}{a+b}\)

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials InText Questions and Answers.

10th Class Maths 3rd Lesson Polynomials InText Questions and Answers

Do these

Question 1.
State which of the following are polynomials and which are not? Give reasons.   (Page No. 48)
(i) 2x³
(ii) \(\frac{1}{x-1}\)
(iii) 4z² + \(\frac{1}{7}\)
(iv) m² – √2 m + 2
(v) p^-2 + 1
Answer:
i) 2x³ is a polynomial.
ii) \(\frac{1}{x-1}\) is not a polynomial because its power is negative integer exponent.
iii) 4z² + \(\frac{1}{7}\) is a polynomial.
iv) m² – √2 m + 2 is a polynomial.
v) p^-2 + 1 is not a polynomial because its power is negative integer exponent.

Question 2.
p(x) = x² – 5x – 6, find the values of p(l), p(2), p(3), p(0), p(-l), p(-2), p(-3).    (Page No. 49)
Answer:
Given polynomial p(x) = x² – 5x – 6
p(1) = (1)² – 5(1) – 6 = 1 – 5 – 6 = – 10
p(2) = (2)² – 5(2) – 6 = 4 – 10 – 6 = -12
p(3) = 3² – 5(3) – 6 = 9 – 15 – 6 = – 12
p(0) = 0² – 5(0) – 6 = – 6
p(-1) = (-1)² – 5(-1) – 6 = 1 + 5 – 6 = 0
p(-2) = (-2)² – 5(-2) – 6 = 4 + 10 – 6 = 8
p(-3) = (-3)² – 5(-3) – 6 = 9 + 15 – 6 = 18

Question 3.
p(m) = m² – 3m + 1, find the values of p(1)and p(-1).    (Page No. 49)
Answer:
Given polynomial p(m) = m² – 3m + 1
p(1) = (1)² – 3(1) + 1 = 1 – 3 + 1 = 2 – 3 = – 1
p(-1) = (-1)² – 3(-1) + 1 = 1 + 3 + 1 = 5

Question 4.
Let p(x) = x² – 4x + 3. Find the values of p(0), p(l), p(2), p(3) and obtain zeroes of the polynomial p(x).   (Page No. 50)
Answer:
Given polynomial p(x) = x² – 4x + 3
p(0) = (0)² – 4(0) + 3 = 3
p(1) = (1)² – 4(1) + 3 = 1 – 4 + 3 = 0
p(2) = (2)² – 4(2) + 3 = 4 – 8 + 3 = – 1
p(3) = (3)² – 4(3) + 3 = 9 – 12 + 3 = 0
We see that p(1) and p(3) are zeroes of the polynomial p(x).

Question 5.
Check whether -3 and 3 are the zeroes of the polynomial x² – 9.    (Page No. 50)
Answer:
Given polynomial p(x) = x² – 9
Zero of the polynomial p(x) = 0
x² – 9 = 0
⇒ x² = 9
⇒ x = V9 = ± 3
∴ x = + 3, – 3
∴ Zeroes of the polynomial p(x) are – 3 and 3.

Try these

Question 1.
Write 3 different quadratic, cubic and 2 linear polynomials with different number of terms. (Page No. 48)
Answer:
Quadratic polynomials:

Cubic polynomials :

Linear polynomials :
f(t) = √2 t + 5
g(u) = \(\frac{2}{3}\) u – \(\frac{5}{2}\) and
q(y) = 3y
Yes, we can write polynomials of any degree.

Question 2.
Write a quadratic polynomial and a cubic polynomial in variable x in the general form. (Page No. 49)
Answer:
General form of a quadratic polynomial having variable ‘x’ is
f(x) = ax³ + bx² + c, a ≠ 0
General form of a cubic polynomial having variable ‘x’ is
f(x) = ax³ + bx² + cx + d, a ≠ 0

Question 3.
Write a general polynomial q(z) of degree n with coefficients that are b₀…. b_n. What are the conditions on b₀…. b_n. (Page No. 49)
Answer:
q(z) = b₀zⁿ + b₁z^n-1 + b₂z^n-2 …….. + b_n-1z + b_n is a polynomial of n degree where b₀, b₁, b₂,…… b_n-1, b_n are real coefficients and b₀ ≠ 0.

Do this

Question 1.
Draw the graph of i) y = 2x + 5, ii) y = 2x – 5, iii) y = 2x and find the point of intersection on X – axis. Is the x-coordinates of these points also the zero of the polynomial?   (Page No. 52)
Answer:
i) Given that y = 2x + 5


Result: The graph y = 2x + 5 cuts the X – axis at the point (-2.5, 0).
Hence, the zeroes of the polynomial is -2.5.

ii) Given that y = 2x – 5


Result: The graph of y = 2x – 5 cuts the X – axis at the point (2.5, 0).
The zeroes of the polynomial is 2.5 = \(\frac{5}{2}\)

iii) Given that y = 2x


Result: The graph passes through the origin.
So, the zeroes of the polynomial y = 2x is zero.

Try these

Question 1.
Draw the graphs of (i) y = x² – x – 6 (ii) y = 6 – x – x² and find zeroes in each case. What do you notice? (Page No. 53)
Answer:
i) Given that y = x² – x – 6


Result: From the graph we observe that 3 and -2 are the intersecting points of X – axis.
So, the zeroes of given quadratic polynomial are 3 and -2.

ii) Given that y = 6 – x – x²


Result: From the graph we observe that – 3 and 2 are the intersecting points ol X – axis.
So, the zeroes of given quadratic polynomial are – 3 and 2.

Question 2.
Write three quadratic polynomials that have 2 zeroes each.   (Page No. 55)
Answer:
y = x² – x – 2 having two zeroes, i.e., (2, 0) and (- 1, 0).
y = 3 – 2x – x² having two zeroes i.e., (1,0) and (- 3, 0).
y = x² – 3x – 4 having two zeroes i.e., (-1, 0) and (4, 0)

Question 3.
Write one quadratic polynomial that has one zero. (Page No. 55)
Answer:
Quadratic Polynomial y = x² – 6x + 9 has only one zero i.e., 3.

Question 4.
How will you verify if a quadratic polynomial it has only one zero?  (Page No. 55)
Answer:
If the graph of the given quadratic polynomial touches X – axis at exactly one point, then I can confirm it has only one zero.

Question 5.
Write three quadratic polynomials that have no zeroes for x that are real numbers.  (Page No. 55)
Answer:
The quadratic polynomials y = 2x² – 4x + 5 and y = – 3x² + 2x – 1 and y = x² – 2x + 4 have no zeroes.

Question 6.
Find the zeroes of cubic polynomials
(i) – x³
(ii) x² – x³
(iii) x³ – 5x² + 6x
without drawing the graph of the polynomial.  (Page No. 57)
Answer:
i) Given polynomial is y = – x³
f(x) = -x³ ; f(x) = 0
x³ = 0
x = \(\sqrt[3]{0}\) = 0
∴ Zero of the polynomial f(x) is only one i.e., 0.

ii) Given that y = x² – x³
f(x) = x² (1 – x)
f(x) = 0
⇒ x² (1 – x) = 0
⇒ x² = 0 and 1 – x = 0
⇒ x = 0 and x = 1
∴ The zeroes of the polynomial f(x) are two i.e., 0 and 1.

iii) Given that x³ – 5x² + 6x Let f(x) = x³ – 5x² + 6x
= x(x² – 5x + 6)
= x(x² – 2x – 3x + 6)
= x[x(x – 2) – 3(x – 2)]
= x(x – 2) (x – 3)
∴ The zeroes of the polynomial f(x) are x = 0 and x = 2 and x = 3

Do these

Question 1.
Find the zeroes of the quadratic polynomials given below. Find the sum and product of the zeroes and verify relationship to the coefficients of terms in the polynomial. (Page No. 62)
i) p(x) = x² – x – 6
ii) p(x) = x² – 4x + 3
iii) p(x) = x² – 4
iv) p(x) = x² + 2x + 1
Answer:
i) Given polynomial p(x) = x² – x – 6
We have x² – x – 6 = x² – 3x + 2x – 6
= x(x – 3) + 2(x – 3)
= (x – 3) (x + 2)
So, the value of x² – x – 6 is zero when x – 3 = 0 or x + 2 = 0
i.e., x = 3 or x = -2
So, the zeroes of x² – x – 6 are 3 and – 2.
∴ Sum of the zeroes = 3 – 2 = 1
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-1)}{1}\) = 1
And product of the zeroes = 3 × (-2) = -6
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-6}{1}\) = -6

ii) p(x) = x² – 4x + 3
Answer:
Given polynomial p(x) = x² – 4x + 3
We have, x² – 4x + 3 = x² – 3x – x + 3
= x(x – 3) – 1 (x – 3)
= (x – 3) (x – 1)
So, the value of x² – 4x + 3 is zero when x – 3 = 0 or x – 1 =0, i.e.,
when x = 3 or x = 1 So, the zeroes of x² – 4x + 3 are 3 and 1
∴ Sum of the zeroes = 3 + 1 = 4
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-4)}{1}\) = 4
And product of the zeroes = 3 × 1 = 3
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{3}{1}\) = 3

iii) Given polynomial p(x) = x² – 4
We have, x² – 4 = (x – 2) (x + 2)
So, the value of x² – 4 is zero
when x – 2 = 0 or x + 2 = 0
i.e., x = 2 or x = – 2
So the zeroes of x² – 4 are 2 and – 2
∴ Sum of the zeroes = 2 + (- 2) = 0
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-0}{1}\) = 0
And product of the zeroes = 2 × (-2) = -4
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-4}{1}\) = -4

iv) Given polynomial p(x) = x² + 2x + 1
We have x² + 2x + 1 = x² + x + x + 1
= x(x + 1) + l(x + 1)
= (x + 1) (x + 1)
So, the value of x² + 2x + 1 is zero
when x + 1 = 0 (or) x + 1 = 0, i.e.,
when x = – 1 or – 1
o, the zeroes of x² + 2x + 1 are – 1 and – 1.
∴ Sum of the zeroes = (-1) + (-1) = -2
= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-2}{1}\) = -2
And product of the zeroes = (-1) × (-1) = 1
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{1}{1}\) = 1

Question 2.
If α, β and γ are the zeroes of the given cubic polynomials, find the values as given in the table. (Page No. 66)

Answer:
l) Given polynomial is x³ + 3x² – x – 2.
Comparing given polynomial with ax³ + bx² + cx + d,
the values are a = 1, b = 3, c = -l, d = -2
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-3}{1}\) = -3
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{-1}{1}\) = -1
αβγ = \(\frac{-d}{a}\) = \(\frac{-(-2)}{1}\) = 2

2) Given polynomial is 4x³ + 8x² – 6x – 2
Compare the polynomial with ax³ + bx² + cx + d = 0
Then a = 4, b = 8, c = – 6 and d = – 2
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-8}{4}\) = -2
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{-6}{4}\) = \(\frac{-3}{2}\)
αβγ = \(\frac{-d}{a}\) = \(\frac{-(-2)}{4}\) = \(\frac{1}{2}\)

3) Given polynomial is x³ + 4x² – 5x – 2
Compare the polynomial with ax³ + bx² + cx + d = 0
Then a = 1, b = 4, c = – 5 and d = – 2
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-4}{1}\) = -4
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{-5}{1}\) = -5
αβγ = \(\frac{-d}{a}\) = \(\frac{-(-2)}{1}\) = 2

4) Given polynomial is x³ + 5x² + 4
Compare the polynomial with ax³ + bx² + cx + d = 0
Then a = 1, b = 5, c = 0 and d = 4
α + β + γ = \(\frac{-b}{a}\) = \(\frac{-5}{1}\) = -5
αβ + βγ + γα = \(\frac{c}{a}\) = \(\frac{0}{1}\) = 0
αβγ = \(\frac{-d}{a}\) = \(\frac{-4}{1}\) = -4

Try this

Question 1.
i) Find a quadratic polynomial with zeroes -2 and \(\frac{1}{3}\). (Page No. 64)
Answer:
Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α and β.
Here α = – 2 and β = \(\frac{1}{3}\)
Sum of the zeroes = α + β
= -2 + \(\frac{1}{3}\) = \(\frac{-5}{3}\)
Product of the zeroes = αβ
= \(\frac{1}{3}\) × (-2) = \(\frac{-2}{3}\)
∴ ax² + bx + c is [x² – (α + β)x + αβ]
= [x² – \(\left(\frac{-5}{3}\right)\)x + \(\left(\frac{-2}{3}\right)\)]
the quadratic polynomial will be 3x² + 5x – 2.

ii) What is the quadratic polynomial whose sum of zeroes is \(\frac{-3}{2}\) and the product of zeroes is -1.
Answer:
Let the quadratic polynomial be ax² + bx + c and its zeroes be α and β.
Here α + β = \(\frac{-3}{2}\) and αβ = -1
Thus, the polynomial formed = x² – (α + β)x + αβ
= x² – \(\left(\frac{-3}{2}\right)\)x + (-1)
= x² + \(\frac{3x}{2}\) – 1
The other polynomials are (x² + \(\frac{3x}{2}\) – 1)
then the polynomial is 2x² + 3x – 2.

AP State Syllabus SSC 10th Class Maths Solutions 2nd Lesson Sets InText Questions

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 2 Sets InText Questions and Answers.

10th Class Maths 2nd Lesson Sets InText Questions and Answers

Question 1.
List the teeth under each of the following type (Page No. 25)

i) Incisors
Answer:
Central incisors = 4
Lateral incisors = 4
Total incisors = 8
ii) Canines
Answer:
Total canines = 4
iii) Pre-molars
Answer:
First premolars = 4
Second premolars = 4
Total premolars = 8
iv) Molars
Answer:
First molars = 4
Second molars = 4
Third molars = 4
Total molars = 12

Question 2.
Identify and write the “common property” of the following collections. (Page No. 26)
1) 2, 4, 6, 8, …….
Answer:
All are even numbers {x : x is even}
2) 2, 3, 5, 7, 11, …….
Answer:
All are prime numbers (x : x is a prime}
3) 1, 4, 9, 16, …….
Answer:
All are perfect squares
(x : x is a perfect square}
4) January, February, March, April,…
Answer:
All are English months
{x : x is a month of the year}
5) Thumb, index finger, middle finger, ring finger, pinky.
Answer:
All are fingers of a hand
{x : x is a finger of a hand}

Question 3.
Write the following sets. (Page No. 27)
1) Set of the first five positive integers.
Answer:
{11, 2, 3, 4, 5}
2) Set of multiples of 5 which are more than 100 and less than 125.
Answer:
{105, 110, 115, 120}
3) Set of first five cubic numbers.
Answer:
{1³, 2³, 3³, 4³, 5³}
{1, 8, 27, 64, 125}
4) Set of digits in the Ramanujan number.
Answer:
Ramanujan’s number is 1729
{1, 2, 7, 9}

Question 4.
Some numbers are given below. Decide the numbers to which number sets they belong to and does not belong to and express with correct symbols. (Page No. 28)
i) 1
Answer:
1 ∈ N
ii) 0
Answer:
0 ∈ W and 0 ∉ N
iii) -4
Answer:
– 4 ∈ I and – 4 ∉ N
iv) \(\frac{5}{6}\)
Answer:
\(\frac{5}{6}\) ∈ Q and \(\frac{5}{6}\) ∉ Z
v) \(1 . \overline{3}\)
Answer:
\(1 . \overline{3}\) ∉ N and \(1 . \overline{3}\) ∉ Z
vi) √2
Answer:
√2 ∈ S and √2 ∉ Q
vii) log 2
Answer:
log 2 ∉ N
viii) 0.03
Answer:
0.03 ∉ Q
ix) π
Answer:
π ∉ Z
x) \(\sqrt{-4}\)
Answer:
\(\sqrt{-4}\) ∉ Q and \(\sqrt{-4}\) ∈ C

Question 5.
List the elements of the following sets. (Page No. 29)
i) G = {all the factors of 20}
ii) F = {the multiples of 4 between 17 and 61 which are divisible by 7}
iii) S = {x : x is a letter in the word ‘MADAM’}
iv) P = {x : x is a whole number between 3.5 and 6.7}
Answer:
i) G = {1, 2, 4, 5, 10, 20}
ii) Multiples of 4 between 17 and 61
x = {20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60}
F = {28, 56}
iii) S = {M, D, A}
iv) P = {4, 5, 6}

Question 6.
Write the following sets in the roster form.   (Page No. 29)
i) B is the set of all months in a year having 30 days.
ii) P is the set of all prime numbers smaller than 10.
iii) X is the set of the colours of the rainbow.
Answer:
i) B = {April, June, September, November}
ii) P = {2, 3, 5, 7}
iii) X = {Violet, Indigo, Blue, Green, Yellow, Orange, Red}

Question 7.
A is the set of factors of 12. Which one of the following is not a member of A?   (Page No. 29)
A) 1
B) 4
C) 5
D) 12
Answer:
[C]

Think & Discuss

Question 1.
Observe the following collections and prepare as many as generalized statements you can describing their more properties. (Page No. 26)
i) 2, 4, 6, 8,….
Answer:
a) All even natural numbers
b) All positive even integers
c) Multiples of 2

ii) 1, 4, 9, 16, …..
Answer:
a) Squares of natural numbers
b) All perfect square numbers

Question 2.
Can you write set of rational numbers listing elements in it? (Page No. 28)
Answer:
We can’t list all elements in the set of rational numbers. We know that rational numbers are infinite.

Try this

Question 1.
Write some sets of your choice, involving algebraic and geometrical ideas. (Page No. 29)
Answer:
The set of all natural numbers ‘x’ such that 4x + 9 < 50,
ii) A = {x : x is an integer and -3 ≤ x ≤ 7}
iii) B = {Equilateral triangle, Right angled triangle, Scalene triangle, Obtuse angled triangle, Acute angled triangle)
iv) C = {Rectangle, Square, Parallelogram, Rhombus, Trapezium}

Question 2.
Match roster forms with the set builder form. (Page No. 29)

Answer:
i) d
ii) c
iii) a
iv) b

Do this

Question 1.
A = {1, 2, 3, 4},
B = {2, 4},
C = {1, 2, 3, 4, 7}, ∅ = { }.
Fill in the blanks with ⊂ or ⊄.  (Page No. 33)
i) A …. B
ii) C …. A
iii) B …. A
iv)A …. C
v) B …. C
vi) ∅ …. B
Answer:
i) A ⊄ B
ii) C ⊄ A
iii) B ⊆ A
iv) A ⊆ C
v) B ⊆ C
vi) ∅ ⊆ B

Question 2.
State which of the following statements are true.    (Page No. 33)
i) { } = ∅
ii) ∅ = 0
iii) 0 = { 0 }
Answer:
i) True (T)
ii) False (F)
iii) False (F)

Question 3.
Let A = {1, 3, 7, 8} and B = [2, 4, 7, 9}.
Find A ∩ B.     (Page No. 37)
Answer:
Given sets A = (1, 3, 7, 8} and B = {2,4, 7,9}
A ∩ B = {1, 3, 7, 8} ∩ (2, 4, 7, 9} = {7}

Question 4.
If A = {6,9,11 }; ∅ = {}, find A ∪ ∅, A ∩ ∅).  (Page No. 37)
Answer:
Given sets
A = {6, 9, 11} and ∅ = { }
A ∪ ∅ = {6, 9, 11} ∪ { }
= {6, 9, 11} = A
∴ A ∪ ∅ = A
A ∩ ∅ = {6,9,11} ∩ { } = { } = ∅
∴ A ∩ ∅ = ∅

Question 5.
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
B = {2, 3, 5, 7}. Find A ∩ B and show that A ∩ B = B.    (Page No. 37)
Answer:
Given sets
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}
A ∩ B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∩ {2, 3, 5, 7}
= {2, 3, 5, 7} = B
∴ A ∩ B = B

Question 6.
If A = {4, 5, 6}; B = {7, 8}, then show that A ∪ B = B ∪ A.  (Page No. 37)
Answer:
Given sets are
A = {4, 5, 6} and B = {7, 8}
A ∪ B = {4, 5, 6} ∪ {7, 8}
= {4, 5, 6, 7, 8}.
B ∪ A = {7, 8} ∪ {4, 5, 6}
= {4, 5, 6, 7, 8}
∴ A ∪ B = B ∪ A.

Question 7.
If A = {1, 2, 3, 4, 5 }; B = {4, 5, 6, 7}, then find A – B and B – A. Are they equal?  (Page No. 38)
Answer:
Given sets are
A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}
A – B = {1, 2, 3, 4, 5} – {4, 5, 6, 7}
= {1, 2, 3}
B – A = {4, 5, 6, 7} – {1, 2, 3, 4, 5}
= {6, 7}
∴ No, A – B ≠ B – A.

Question 8.
If V = {a, e, i, o, u} and B = {a, i, k, u}, find V – B and B – V.    (Page No. 38)
Answer:
Given sets are
V = {a, e, i, o, u} and B = {a, i, k, u}
V – B = {a, e, i, o, u} – {a, i, k, u}
= {e, o}
B – V = {a, i, k, u} – {a, e, i, o, u}
= {k}.

Try this

Question 1.
A = {set of quadrilaterals},
B = {square, rectangle, trapezium, rhombus}.
State whether A ⊂ B or B ⊂ A. Justify your answer.     (Page No. 33)
Answer:
A = {set of quadrilaterals} means A = {square, rectangle/trapezium, rhombus, parallelogram}
B = {square, rectangle, trapezium, rhombus}
So, B ⊂ A because A’ is having elements more than ‘B’.

Question 2.
If A = {a, b, c, d}. How many subsets does the set A have?    (Page No. 33)
A) 5 B) 6 C) 16 D) 65
Answer:
Given A = {a, b, c, d}
n(A) = 4
Number of subsets for a set, which is having ‘n’ elements is 2ⁿ.
So n(A) = 4
Number of subsets for A is 2⁴ = 16.
Answer is [C].

Question 3.
P is the set of factors of 5, Q is the set of factors of 25 and R is the set of factors of 125. Which one of the following is false?    (Page No. 33)
A) P ⊂ Q
B) Q ⊂ R
C) R ⊂ P
D) P ⊂ R
Answer: [C]

Question 4.
A is the set of prime numbers smaller than 10, B is the set of odd numbers < 10 and C is the set of even numbers < 10. How many of the following statements are true?    (Page No. 33)
i) A ⊂ B
ii) B ⊂ A
iii) A ⊂ C
iv) C ⊂ A
v) B ⊂ C
vi) C ⊂ B
Answer:
All the statements are false.

Question 5.
List out some sets A and B and choose their elements such that A and B are disjoint.  (Page No. 37)
Answer:
Consider the disjoint sets
A = {1, 2, 3, 4} and B = {a, b, c}

Question 6.
If A = {2, 3, 5}, find A ∪ ∅ and ∅ ∪ A and compare.    (Page No. 37)
Answer:
Given sets A = {2, 3, 5} and ∅ = { }
A ∪ ∅ = {2,3,5} ∪ { } = {2,3,5}
∅ ∪ A = { } ∪ {2, 3, 5} = {2,3,5}
A ∪ v = ∅ ∪ A = A

Question 7.
If A = {1, 2, 3, 4}; B = {1, 2, 3, 4, 5, 6, 7, 8}, then find A ∪ B, A ∩ B. What do you notice about the result?   (Page No. 37)
Answer:
Given sets are
A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6, 7, 8}
A ∪ B = {1, 2, 3, 4} ∪ {1, 2, 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8} = B
A ∩ B = {1, 2, 3, 4} ∩ {1, 2, 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4} = A
If A ⊂ B, then A ∪ B = B and A ∩ B = A

Question 8.
A = {1, 2, 3, 4, 5, 6}; B = {2, 4, 6, 8, 10}. Find the intersection of A and B.     (Page No. 37)
Answer:
Given sets are
A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10}
A ∩ B = {1, 2, 3, 4, 5, 6} ∩ {2, 4, 6, 8, 10} = {2, 4, 6}

Think & Discuss

Question 1.
Is empty set subset to every set?    (Page No. 34)
Answer:
‘Yes’. Empty set is subset to every set
Justify: If A ⊂ B, it means all the elements of set ‘A’ belong to set ‘B’.
In other words we can say no element of ‘A’ missed in set B.
now empty set means which has no elements, now no element of empty set can be missed in any set. So we can write empty set is subset of every set.

Question 2.
Is any set subset to itself?    (Page No. 34)
Answer:
‘Yes’. Every set is subset to itself.
Let ‘A’ is any set.
Now every element of ‘A’ definitely belongs to ‘A’.
So A ⊂ A
Hence every set is a subset to it.

Question 3.
You are given two sets such that a set is not a subset of the other. If you have to prove this, how do you prove?    (Page No. 34)
Answer:
Let the given sets are ‘A’ and ‘B’.
To prove are set (A) is not subset of other (B).
We check if all elements of ‘A’ belong to the set ‘B’ or not.
If any of the element doesn’t belong to ‘B’ then we can say ‘A’ is not subset of ‘B’. So we have to prove at least one element of ‘A’ does not belong to ‘B’.
Hence ‘A’ is not subset of ‘B’.

Question 4.
The intersection of any two disjoint sets is a null set. Justify your answer.    (Page No. 37)
Answer:
Let A and B be any two disjoint sets,
i.e., A and B have no elements in common.
∴ A ∩ B is a null set. (∵ A ∩ B is the set of all elements which are common to both A and B)

Question 5.
The sets A – B, B – A and A ∩ B are mutually disjoint sets. Use examples to observe if this is true.    (Page No. 38)
Answer:
Let the sets are A = {1, 2, 3, 4} and B = {5, 6, 7, 8}
A – B = {1, 2, 3, 4} – {5, 6, 7, 8} = {1, 2, 3, 4}
B – A = {5, 6, 7, 8} – {1, 2, 3, 4} = {5, 6, 7, 8}
A ∩ B = {1, 2, 3, 4} ∩ {5, 6, 7, 8} = { } = ∅
∴ A – B, B – A and A ∩ B are disjoint sets.

Do these

Question 1.
Which of the following are empty sets? Justify your answer.    (Page No. 44)
i) Set of integers which lie between 2 and 3 .
ii) Set of natural numbers that are smaller than 1.
iii) Set of odd numbers that leave remainder zero, when divided by 2.
Answer:
i) This is null set. We know that there is no integer that lie between 2 and 3.
ii) This is also a null set. We know that there is’ no natural number less than ‘1’.
iii) This is a null set. We know that odd numbers do not leave remainder zero when divided by 2.

Question 2.
State which of the following sets are finite and which are infinite. Give reasons for your answer.    (Page No. 44)
i) A = {x : x e N and x < 100}
ii) B = {x : x e N and x ≤ 5}
iii) C = {1² , 2², 3², ……}
iv) D = {1, 2, 3, 4}
v) {x : x is a day of the week}
Answer:
i) A = (1, 2, 3, 4, , 98, 99}
This set is finite, because there are 99 numbers possible to count.
ii) B = {1, 2, 3,  4, 5}
This set is finite because there are 5 numbers possible to count.
iii) C = {1² , 2², 3², ……}
This set is infinite because there are infinite numbers.
iv) D – {1, 2, 3, 4}
This set is finite because there are 4 numbers that are possible to count.
v) E = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
This set is finite, because there are 7 days in a week possible to count.

Question 3.
Tick the set which is infinite.      (Page No. 44)
A) The set of whole numbers < 10
B) The set of prime numbers < 10
C) The set of integers < 10
D) The set of factors of 10
Answer:
[C]
The set of integers < 10
{….., -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Try this

Question 1.
Which of the following sets are empty sets? Justify your answer.    (Page No. 44)
i) A = {x : x² = 4 and 3x = 9}.
ii) The set of all triangles in a plane having the sum of their three angles less than 180.
Answer:
i) x² = 4 ⇒ x = ± 2
3x = 9 ⇒ x = 3
The value of ‘x’ is not same in both cases, so this is a null set.
ii) This is a null set, because the sum of the three angles of a triangle is equal to 180°.

Question 2.
B = {x : x + 5 = 5} is not an empty set. Why?   (Page No. 44)
Answer:
B = {x: x + 5 = 5} is not an empty set
let x ∈ Z or x ∈ W
then for x = 0 ⇒ x + 5 = 0 + 5 = 5
So if x ∈ W, or x ∈ Z then for x = 0,
x + 5 = 5 is true.
Then the set B = {0} which is not an empty set.
Note: But if x ∈ N
We will have no ‘x’ such that x + 5 = 5
then ‘B’ will be an empty set.
But in the textbook it is not given whether x ∈ N (or) x ∈ W (or) x ∈ Z.
Hence we consider first one.

Think & Discuss

Question 1.
An empty set is a finite set. Is this statement true or false? Why?   (Page No. 44)
Answer:
Yes, it is a finite set because there is finite number i.e., ‘0’ elements it consists.

Think & Discuss

Question 1.
What is the relation between n(A), n(B), n(A ∩ B) and n(A ∪ B)?   (Page No. 45)
Answer:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B). This is called Fundamental theorem of sets.

Question 2.
If A and B are disjoint sets, then how can you find n(A ∪ B)?    (Page No. 45)
Answer:
If A and B are disjoint then A ∩ B is a null set.
∴ n(A ∩ B) = 0 and it gives us n(A ∪ B) = n (A) + n(B).

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Optional Exercise Textbook Questions and Answers.

10th Class Maths 13th Lesson Probability Optional Exercise Textbook Questions and Answers

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Answer:
Shyam and Ekta can visit the shop in the following combination:

(T Tu) (W, Tu) (Th, Tu) (F, Tu) (S, Tu) (Tu, W) (W, W) (Th, W) (F, W) (S, W) (Tu, Th) (W, Th) (Th, F) (F, Th) (S, Th) (Tu, F) (W, F) (Th, S) (F, F) (S, F) (Tu, S) (W, S) (Th, Th) (F, S) (S, S)
∴ Number of total outcomes = 5 × 5 = 5² = 25 [also from the above table]
i) Number of favourable outcomes to that of visiting on the same day
(Tu, Tu), (W, W), (Th, Th), (F, F), (S, S) = 5
∴ Probability of visiting the shop on the same day = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{5}{25}\) = \(\frac{1}{5}\)
ii) Number of outcomes favourable to consecutive days
(Tu, W), (W, Th), (Th, F), (F, S), (W, Tu), (Th, W), (F, Th), (S, F) = 8
∴ Probability of visiting the shop on consecutive days = \(\frac{8}{25}\)
iii) If P(E) is the probability of visiting the shop on the same day,
then P(\(\overline{\mathrm{E}}\)) is the probability of visiting the shop not on the same day.
i.e., P(\(\overline{\mathrm{E}}\)) is the probability of visiting the shop on different days.
Such that P(E) + P(\(\overline{\mathrm{E}}\)) = 1
P(\(\overline{\mathrm{E}}\)) = 1 – P(E) = 1 – \(\frac{1}{5}\) = \(\frac{4}{5}\)

Question 2.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Answer:
Number of red balls in the bag = 5 As the probability of blue balls is double the probability of red balls, we have that number of blue balls is double the number of red balls.
∴ Blue balls = 5 × 2 = 10.
[!! Let the number of blue balls = x
Number of red balls = 5
Total no. of balls = x + 5
Total outcomes in drawing a ball at random = x + 5
Number of outcomes favourable to red ball = 5
∴ P(R) = \(\frac{5}{x+5}\)
from the problem,
P(B) = 2 × \(\frac{5}{x+5}\) = \(\frac{10}{x+5}\)
Also \(\frac{5}{x+5}\) + \(\frac{10}{x+5}\) = 1
[∵ P(E) + P(\(\overline{\mathrm{E}}\)) = 1]
⇒ \(\frac{5+10}{x+5}\) = 1
⇒ \(\frac{15}{x+5}\) = 1
⇒ x + 5 = 15]

Question 3.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Answer:
Number of black balls = x
Total number of balls in the box = 12
Probability of drawing a black ball = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{x}{12}\) …….. (1)
When 6 more black balls are placed in the box, number of favourable outcomes to black ball becomes = x + 6.
Total number of balls in the box becomes = 12 + 6 = 18.
Now the probability of drawing a black ball become = \(\frac{x+6}{18}\) …….. (1)
By Problem,
\(\frac{x+6}{18}\) = 2. \(\frac{x}{12}\)
⇒ \(\frac{x+6}{18}\) = \(\frac{x}{6}\)
⇒ 6(x + 6) = 18(x)
⇒ 6x + 36 = 18x
⇒ 18x – 6x = 36
⇒ 12x = 36
⇒ x = \(\frac{36}{12}\) = 3
Check:

and hence proved.

Question 4.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue marbles in the jar.
Answer:
Total number of marbles in the jar = 24.
Let the number of green marbles = x.
Then number of blue marbles = 24 – x.
Probability of drawing a green marbles = \(\frac{\text { No. of favourable outcomes }}{\text { No. of total outcomes }}\) = \(\frac{x}{24}\)
By problem,
\(\frac{x}{24}\) = \(\frac{2}{3}\)
⇒ 3x = 24 × 2
x = \(\frac{24 \times 2}{3}\) = 16
∴ Number of green marbles = 16
Number of blue marbles = 24 – 1 = 8
∴ Probability of picking blue marble = \(\frac{8}{24}\) = \(\frac{1}{3}\)
(OR)
P(B) = P(E) – P(G) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
[!! P(G) = \(\frac{2}{3}\)
P(G) + P(B) = 1
∴ P(B) = 1 – P(G) = 1 – \(\frac{2}{3}\) = \(\frac{1}{3}\)
Number of blue marbles in the jar = \(\frac{1}{3}\) × 24 = 8]

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Optional Exercise Textbook Questions and Answers.

10th Class Maths 12th Lesson Applications of Trigonometry Optional Exercise Textbook Questions and Answers

Question 1.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval.
Answer:
Height of the balloon from the ground = 88.2 m
Height of the girl = 1.2 m
Angles of elevations = 60° and 30°

Let the distance travelled = dm
From the figure
tan 60° = \(\frac{87}{x}\)
√3 = \(\frac{87}{x}\)
⇒ 87 = √3x …….(1)
⇒ x = \(\frac{87}{\sqrt{3}}\) m
Also tan 30° = \(\frac{87}{x+d}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{87}{x+d}\)
⇒ 87 = \(\frac{x+d}{\sqrt{3}}\) ………(2)
From equations (1) and (2)
√3x = \(\frac{x+d}{\sqrt{3}}\)
√3 × √3x = x + d
⇒ 3x = x + d
⇒ 2x = d

Question 2.
The angle of elevation of the top of a tower from the foot of the building is 30° and the angle of elevation of the top of the building from the foot of the tower is 60°. What is the ratio of heights of tower and building?
Answer:

Let the height of the tower = x m
Let the height of the building = y m
Distance between the tower and building = d m.
Angle of elevation of the top of the tower = 30°.
From the figure,

∴ x : y = 1 : 3
∴ The ratio of heights of tower and building = 1 : 3.

Question 3.
The angles of elevation of the top of a lighthouse from 3 boats A, B and C in a straight line of same side of the light- house are a, 2a, 3a respectively. If the distance between the boats A and B is x meters. Find the height of lighthouse.
Answer:
From the figure,

Let PQ be the height of the lighthouse = h m
A = First point of observation
B = Second point of observation
C = Third point of observation Given,
AB = x and BC = y
(Not given in the text)
Exterior angle = Sum of the opposite interior angles
∠PBQ = ∠BQA + ∠BAQ and
∠PCQ = ∠CBQ + ∠CQB
∴ AB = x = OB
By applying the sine rule,

From △PBQ

Question 4.
Inner part of a cupboard is in the cuboidical shape with its length, breadth and height in the ratio 1 : √2 : 1. What is the angle made by the longest stick which can be inserted cupboard with its base inside?
Answer:

The ratio of the length, breadth and height = 1 : √2 : 1
Let its length be = x
breadth = √2x height = x
The longest stick that can be placed on the base is along its hypotenuse

[!! Again, the longest stick that can be inserted in the cup board is along the line join of the bottom corn on with’ its opposite top corner, i.e., along the hypotenuse of the right triangle formed by height of the cup board, hypotenuse of the base and the line join of bottom corner with its opposite top corner.

Length of the largest stick = \(\sqrt{(\sqrt{3} x)^{2}+x^{2}}\)
= \(\sqrt{3 x^{2}+x^{2}}\)
= \(\sqrt{4 x^{2}}\) = 2x]
Now the angle made by the largest stick be = θ

Then tan θ = \(\frac{\text { opp. side }}{\text { adj. side }}\) = \(\frac{x}{\sqrt{3} x}\) = \(\frac{1}{\sqrt{3}}\)
tan θ = tan 30°
∴ θ = 30°.

Question 5.
An iron spherical ball of volume 232848 cm³ has been melted and converted into a cone with the vertical angle of 120°. What are its height and base?
Answer:
Volume of the spherical ball = Volume of the cone

Given that vertical angle = 60°

Let its height be h cm. and radius r cm.
From the figure
Also tan 30° = \(\frac{h}{r}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{r}\)
∴ h = \(\frac{r}{\sqrt{3}}\)
Substituting h = \(\frac{r}{\sqrt{3}}\) equation (1) we get

⇒ r = h√3 = (22.4) (1.732) = 38.79 m
r = 38.79 cm and h = 22.4 cm.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Optional Exercise Textbook Questions and Answers.

10th Class Maths 11th Lesson Trigonometry Optional Exercise Textbook Questions and Answers

Question 1.
Prove that

Answer:

Hence Proved.

Question 2.
Prove that \(\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}\) = \(\frac{1}{\sec \theta-\tan \theta}\) using the identity sec² θ = 1 + tan² θ.
Answer:


From (1) and (2), we get L.H.S. = R.H.S.
Hence Proved.

Question 3.
Prove that (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)
Answer:
L.H.S. = (cosec A – sin A) (sec A – cos A)

Now

From (1) and (2), we get
L.H.S. = R.H.S.
∴ (cosec A – sin A) (sec A – cos A) = \(\frac{1}{\tan A+\cot A}\)

Question 4.
Prove that \(\frac{1+\sec A}{\sec A}\) = \(\frac{\sin ^{2} A}{1-\cos A}\)
Answer:

Question 5.
Prove that \(\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)\) = \(\left(\frac{1+\tan A}{1+\cot A}\right)^{2}\) = tan² A
Answer:

Question 6.
Prove that \(\left(\frac{\sec A-1}{\sec A+1}\right)\) = \(\left(\frac{1-\cos A}{1+\cos A}\right)\)
Answer:

∴ L.H.S. = R.H.S.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Optional Exercise Textbook Questions and Answers.

10th Class Maths 10th Lesson Mensuration Optional Exercise Textbook Questions and Answers

Question 1.
A golf ball has diameter equal to 4.1 cm. Its surface has 150 dimples each of radius 2 mm. Calculate total surface area which is exposed to the surroundings. (Assume that the dimples are all hemispherical) [π = \(\frac{22}{7}\)]
Answer:
Area exposed = surface area of the ball – total area of 150 hemispherical with radius 2 mm
∴ C.S.A of hemisphere = 2πr²

= 52.831 – 37.71 = 15.12 cm².

Question 2.
A cylinder of radius 12 cm. contains water to a depth of 20 cm. A spherical iron ball is dropped into the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. [π = \(\frac{22}{7}\)]
Answer:
Rise in the water level is seen as a cylinder of radius ‘r’ = 12 cm
Height, h = 6.75 cm.
Volume of the rise = Volume of the spherical iron ball dropped

= 9 × 12 × 6.75
= 108 × 6.75
= 729

r³ = 9 × 9 × 9
∴ 729 = (3 × 3) × (3 × 3) × (3 × 3)
∴ Radius of the ball r = 9 cm.

Question 3.
A solid toy is in the form of a right circular cylinder with a hemispheri¬cal shape at one end and a cone at the other end. Their common diameter is 4.2 cm. and height of the cylindrical and conical portion are 12 cm. and 7 cm. respectively. Find the volume of the solid toy. [π = \(\frac{22}{7}\)]
Answer:

Volume of the toy = volume of the hemisphere + volume of the cylinder + volume of the cone.

Hemisphere:
Radius = \(\frac{d}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm

Cylinder:
Radius, r = \(\frac{d}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm
height, h = 12 cm
V = πr²h
= \(\frac{22}{7}\) × 2.1 × 2.1 × 12
= 166.32 cm³
Cone: Radius, r = \(\frac{d}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm
Height, h = 7 cm
Volume = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 2.1 × 2.1 × 7
= 32.34 cm³
∴ Total volume = 19.404 + 166.32 + 32.34 = 218.064 cm³.

Question 4.
Three metal cubes with edges 15 cm., 12 cm. and 9 cm. respectively are melted together and formed into a simple cube. Find the diagonal of this cube.
Answer:
Edges l₁ = 15 cm, l₂ = 12 cm, l₃ = 9 cm.

Volume of the resulting cube = Sum of the volumes of the three given cubes
L³ = l₁³ + l₂³ + l₃³
L³ = 15³ + 12³ + 9³
L³ = 3375 + 1728 + 729
L³ = 5832 = 18 × 18 × 18
∴ Edge of the new cube l = 18 cm

Question 5.
A hemispherical bowl of internal diameter 36 cm. contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm. and height 6 cm. How many bottles are required to empty the bowl?
Answer:
Let the number of bottles required = n
Then total volume of n bottles = volume of the hemispherical bowl
n. πr₁²h = πr₂²h
Bottle:
Radius, r = 3 cm; Height, h = 7 cm
Volume, V = πr²h
= \(\frac{22}{7}\) × 3 × 3 × 7
= 198 cm³
∴ Total volume of n bottles = 198 . n cm³.
Bowl:
Radius, r = \(\frac{d}{2}\) = \(\frac{36}{2}\) = 18 cm

∴ 62 bottles are required to empty the bowl.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise Textbook Questions and Answers.

10th Class Maths 9th Lesson Tangents and Secants to a Circle Optional Exercise Textbook Questions and Answers

Question 1.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line – segment joining the points of contact at the centre.
Answer:

Given: A circle with centre ‘O’.
Two tangents \(\overleftrightarrow{\mathrm{PQ}}\) and \(\overleftrightarrow{\mathrm{PT}}\) from an external point P. Let Q, T be the points of contact.
R.T.P: ∠P and ∠QOT are supplementary.
Proof: OQ ⊥ PQ
[∵ radius is perpendicular to the tangent at the point of contact] also OT ⊥ PT
∴ ∠OQP + ∠OTP = 90° + 90° = 180° Nowin oPQOT,
∠OTP + ∠TPQ + ∠PQO + ∠QOT
= 360° (angle sum property)
180° + ∠P + ∠QOT = 360°
∠P + ∠QOT = 360°- 180° = 180° Hence proved. (Q.E.D.)

Question 2.
PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.

Answer:

Given: PQ = 8
⇒ PR = 4
⇒ PO² = PR² + OR²
⇒ 25 = 16 + OR²
⇒ OR = 3
Now let RT = x and PT in △OPT, ∠P = 90°
∴ OT is hypotenuse.
∴ OT² = OP² + PT²
(Pythagoras theorem)
(3 + x)² = 5² + y² …….. (1)
and in △PRT, ∠R = 90°
∴ \(\overline{\mathrm{PT}}\) is hypotenuse.
∴ PT² = PR² + RT²
y² = 4² + x² …….. (2)
Now putting the value of y² = 4² + x² in equation (1) we got
(3 + x)² = 5² + x² + 4²
9 + x² + 6x = 25 + 16 + x²
6x = 25 + 16 – 9 = 25 + 7 = 32
⇒ x = \(\frac{32}{6}\) = \(\frac{16}{3}\)
Now from equation (2), we get

Question 3.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer:

Given: Let a circle with centre ‘O’ touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively.
R.T.P: ∠AOB + ∠COD = 180°
∠AOD + ∠BOC = 180°
Construction: Join OP, OQ, OR and OS.
Proof: Since the two tangents drawn from an external point of a circle subtend equal angles.
At the centre,
∴ ∠1 = ∠2
∠3 = ∠4 (from figure)
∠5 = ∠6
∠7 = ∠8
Now, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
[∵ Sum of all the angles around a point is 360°]
So, 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°
and 2 (∠1 + ∠8 + ∠4 + ∠5) = 360°
(∠2 + ∠3) + (∠6 + ∠7) = \(\frac{360}{2}\) = 180°
Also, (∠1 + ∠8) + (∠4 + ∠5) = \(\frac{360}{2}\) = 180°
So, ∠AOB + ∠COD = 180°
[∵ ∠2 + ∠3 = ∠AOB;
∠6 + ∠7 = ∠COD
∠1 + ∠8 = ∠AOD
and ∠4 + ∠5 = ∠BOC [from fig.]]
and ∠AOD + ∠BOC = 180°

Question 4.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Answer:

Steps of construction:

1. Draw a line segment AB of length 8 cm.
2. With A and B as centres and 4 cm, 3 cm as radius draw two circles.
3. Draw the perpendicular bisectors \(\stackrel{\leftrightarrow}{\mathrm{XY}}\) of AB. Let \(\stackrel{\leftrightarrow}{\mathrm{XY}}\) and AB meet at M.
4. Taking M as centre and MA or MB as radius draw a circle which cuts the circle with centre A at P and Q and circle with centre B at R, S.
5. Join BP, BQ and AR, AS.

Question 5.
Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Answer:

Steps of construction:

1. Draw AABC such that AB = 6 cm; ∠B = 90° and BC – 8 cm.
2. Drop a perpendicular BD from B on AC.
3. Draw the circumcircle to ABCD. Let ‘E’ be its centre.
4. Join AE and draw its perpendicular bisector \(\stackrel{\leftrightarrow}{\mathrm{XY}}\). Let it meet AE at M.
5. Taking M as centre and MA or ME as radius draw a circle, which’ cuts the circumcircle of △BCD at P and B.
6. Join AP and extend AB, which are the required tangents.

Question 6.
Find the area of the shaded region in the figure, given in which two circles with centres A and B touch each other at the point C. If AC = 8 cm. and AB = 3 cm.

Answer:
Given: Two circles with centres A and B, whose radii are 8 cm and 5 cm.
[∵ AC = 8 cm, AB = 3 cm ⇒ BC = 8 – 3 = 5 cm]
Area of the shaded region = (Area of the larger circle) – (Area of the smaller circle)

Question 7.
ABCD is a rectangle with AB = 14 cm. and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded region.

Answer:
Given AB = 14 cm, AD = BC = 7 cm Area of the shaded and unshaded region
= (2 × Area of the semi-circles with AD as diameter) + Area of the rectangle

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Optional Exercise

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Optional Exercise Textbook Questions and Answers.

10th Class Maths 8th Lesson Coordinate Geometry Optional Exercise Textbook Questions and Answers

Question 1.
In the given figure, \(\frac{QT}{PR}\) = \(\frac{QR}{QS}\) and ∠1 = ∠2. Prove that △PQS ~ △TQR.

Answer:
Given: \(\frac{QT}{PR}\) = \(\frac{QR}{QS}\)
∠1 = ∠2
R.T.P : △PQS ~ △TQR
Proof: In △PQR; ∠1 = ∠2 Thus, PQ = PR
[∵ sides opp. to equal angles are equal]
\(\frac{QT}{PR}\) = \(\frac{QR}{QS}\) ⇒ \(\frac{QT}{PQ}\) = \(\frac{QR}{QS}\)
i.e., the line PS divides the two sides QT and QR of △TQR in the same ratio.
Hence, PS // TR.
[∵ If a line join of any two points on any two sides of triangle divides the two sides in the same ratio, then the line is parallel to the third side]
Hence, PS // TR (converse of B.P.T)
Now in △PQS and △TQR
∠QPS = ∠QTR
[∵ ∠P, ∠T are corresponding angles for PS // TR]
∠QSP = ∠QRT
[∵ ∠S, ∠R are corresponding angles for PS // TR]
∠Q = ∠Q (common)
∴ △PQS ~ △TQR (by AAA similarity)

Question 2.
Ravi is 1.82 m tall. He wants to find the height of a tree in his backyard. From the tree’s base he walked 12.20 m. along the tree’s shadow to a position where the end of his shadow exactly overlaps the end of the tree’s shadow. He is now 6.10 m from the end of the shadow. How tall is the tree?

Answer:
Given:
Height of Ravi ‘BC’ = 1.82 m.
Distance of Ravi from the foot of the tree BD = 12.2 m.
Length of the shadow of Ravi = AB = 6.10 m
Let DE represent the tree.
From the figure, △ABC ~ △ADE.
Thus, \(\frac{AB}{AD}\) = \(\frac{BC}{DE}\) = \(\frac{AC}{AE}\)
Ratio of corresponding sides of two similar triangles are equal]
\(\frac{6.10}{6.10+12.20}\) = \(\frac{1.82}{\mathrm{DE}}\)
DE = \(\frac{1.82 \times 18.30}{6.10}\) = 5.46 m
Thus the height of the tree = 5.46 m.

Question 3.
The diagonal AC of a parallelogram ABCD intersects DP at the point Q, where ‘P’ is any point on side AB. Prove that CQ × PQ = QA × QD.
Answer:

Given: □ ABCD is a parallelogram.
P is a point on AB.
DP and AC intersect at Q.
R.T.P.: CQ . PQ = QA . QD.
Proof: In △CQD, △AQP
∠QCD = ∠QAP
∠CQD = ∠AQP
∴ ∠QDC = ∠QPA
(∵ Angle sum property of triangles)
Thus, △CQD ~ △AQP by AAA similarity condition.
\(\frac{CQ}{AQ}\) = \(\frac{QD}{QP}\) = \(\frac{CD}{AP}\)
[∵ Ratio of corresponding sides of similar triangles are equal]
\(\frac{CQ}{AQ}\) = \(\frac{QD}{QP}\)
CQ . PQ = QA . QD [Q.E.D]

Question 4.
△ABC and △AMP are two right triangles right angled at B and M respectively. Prove that (i) △ABC ~ △AMP (ii) \(\frac{CA}{PA}\) = \(\frac{BC}{MP}\)

Answer:
Given: △ABC; ∠B = 90°
AAMP; ∠M = 90°
R.T.P : i) △ABC ~ △AMP
Proof: In △ABC and △AMP
∠B = ∠M [each 90° given]
∠A = ∠A [common]
Hence, ∠C = ∠P
[∵ Angle sum property of triangles]
∴ △ABC ~ △AMP (by A.A.A. similarity)
ii) △ABC ~ △AMP (already proved)
\(\frac{AB}{AM}\) = \(\frac{BC}{MP}\) = \(\frac{CA}{PA}\)
[∵ Ratio of corresponding sides of similar triangles are equal]
\(\frac{CA}{PA}\) = \(\frac{BC}{MP}\)

Question 5.
An aeroplane leaves an airport and flies due north at a speed of 1000 kmph. At the same time another aeroplane leaves the same airport and flies due west at a speed of 1200 kmph. How far apart will the two planes be after 1\(\frac{1}{2}\) hour?
Answer:

Given: Speed of the first plane due north = 1000 kmph.
Speed of the second plane due west = 1200 kmph.
Distance = Speed × Time
Distance travelled by the first plane in
1\(\frac{1}{2}\) hrs = 1000 × 1\(\frac{1}{2}\) = 1000 × \(\frac{3}{2}\) = 1500 km.
Distance travelled by the second plane
in 1\(\frac{1}{2}\) hrs = 1200 × \(\frac{3}{2}\) = 1800 km.
From the figure, △ABC is a right triangle; ∠A = 90°.
AB² + AC² = BC²
[Pythagoras theorem]
1500² + 1800² = BC²
2250000 + 3240000 = BC²
∴ BC = √5490000
= 100 × √549 m
≅ 100 × 23.43
≅ 2243km.

Question 6.
In a right triangle ABC right angled at C. P and Q are points on sides AC and CB respectively which divide these sides in the ratio of 2:1. Prove that
i) 9AQ² = 9AC² + 4BC²
ii) 9BP² = 9BC² + 4AC²
iii) 9(AQ² + BP²) = 13AB²
Answer:

Given: In △ABC; ∠C = 90°
R.T.P.: i) 9AQ² = 9AC² + 4BC²
Proof: In △ACQ; ∠C = 90°
AC² + CQ² = AQ²
[side² + side² = hypotenuse²]
AQ² = AC² + \(\left(\frac{2}{3} \mathrm{BC}\right)^{2}\)
[∵ Q divides CB in the ratio 2 : 1
CQ = \(\frac{2}{3}\)BC]
AQ² = AC² + \(\frac{4}{9} \mathrm{BC}^{2}\)
AQ² = \(\frac{9 \mathrm{AC}^{2}+4 \mathrm{BC}^{2}}{9}\)
⇒ 9AQ² = 9AC² + 4BC²

ii) 9BP² = 9BC² + 4AC²
Proof: In △PCB,
PB² = PC² + BC² [Pythagoras theorem]

⇒ PB² = AC² + 9BC²
[!! If we take P on CA, in the ratio 2 : 1 then we get
BP² = PC² + BC²
BP² = \(\left(\frac{2}{3} \mathrm{A}\right)^{2}\) + BC²
BP² = \(\frac{4}{9} \mathrm{AC}^{2}\) + BC²
BP² = \(\frac{4 \mathrm{AC}^{2}+9 \mathrm{BC}^{2}}{9}\)
9BP² = 4AC² + 9BC²

iii) 9 (AQ² + BP²) = 13 AB²
Proof: In △ABC,
AC² + BC² = AB²
[Pythagoras theorem]
Also, from (i) and (ii),