Category: Class 10

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.2

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.2 Textbook Questions and Answers

Question 1.
Choose the correct answer and give justification for each.
(i) The angle between a tangent to a circle and the radius drawn at the point of contact is
a) 60°
b) 30°
c) 45°
d) 90°
Answer: [ d ]
If radius is not perpendicular to the tangent, the tangent must be a secant i.e., 90°.

(ii) From a point Q, the length of the tangent to a circle is 24 cm. and the distance of Q from the centre is 25 cm. The radius of the circle is
a) 7 cm
b) 12 cm
c) 15 cm
d) 24.5 cm
Answer: [ a ]

O – centre of the circle
OP – a circle radius = ?
OQ = 25 cm
PQ = 24 cm
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
25² = OP² + 24²
OP² = 625 – 576
OP² = 49
OP = √49 = 7 cm.

iii) If AP and AQ are the two tangents a circle with centre O, so that ∠POQ = 110°. Then ∠PAQ is equal to

a) 60°
b) 70°
c) 80°
d) 90°
Answer: [ b ]
In □ OPAQ,
∠OPA = ∠OQA = 90°
∠POQ = 110°
∴ ∠O + ∠P + ∠A + ∠Q = 360°
⇒ 90° + 90° + 110° + ∠PAQ – 360°
⇒ ∠PAQ = 360° – 290° = 70°

iv) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
a) 50°
b) 60°
c) 70°
d) 80°
Answer: [None]

If ∠APB = 80°
then ∠AOB = 180° – 80° = 100°
[∴ ∠A + ∠B = 90° + 90° = 180°]

v) In the figure XY and XV are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and XV at B then ∠AOB =
a) 80°
b) 100°
c) 90°
d) 60°

Answer: [ c ]

Question 2.
Two concentric circles of radii 5 cm and 3 cm are drawn. Find the length of the chord of the larger circle which touches the smaller circle.
Answer:
Given: Two circles of radii 3 cm and 5 cm with common centre.

Let AB be a tangent to the inner/small circle and chord to the larger circle.
Let ‘P’ be the point of contact.
Construction: Join OP and OB.
In △OPB ;
∠OPB = 90°
[radius is perpendicular to the tangent]
OP = 3cm OB = 5 cm
Now, OB² = OP² + PB²
[hypotenuse² = Adj. side² + Opp. side², Pythagoras theorem]
5² = 3² + PB²
PB² = 25 – 9 = 16
∴ PB = √l6 = 4cm.
Now, AB = 2 × PB
[∵ The perpendicular drawn from the centre of the circle to a chord, bisects it]
AB = 2 × 4 = 8 cm.
∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.

Question 3.
Prove that the parallelogram circumscribing a circle is a rhombus.
Answer:

Given: A circle with centre ‘O’.
A parallelogram ABCD, circumscribing the given circle.
Let P, Q, R, S be the points of contact.
Required to prove: □ ABCD is a rhombus.
Proof: AP = AS …….. (1)
[∵ tangents drawn from an external point to a circle are equal]
BP = BQ ……. (2)
CR = CQ ……. (3)
DR = DS ……. (4)
Adding (1), (2), (3) and (4) we get
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + DC = AD + BC
AB + AB = AD + AD
[∵ Opposite sides of a parallelogram are equal]
2AB = 2AD
AB = AD
Hence, AB = CD and AD = BC [∵ Opposite sides of a parallelogram]
∴ AB = BC = CD = AD
Thus □ ABCD is a rhombus (Q.E.D.)

Question 4.
A triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm. and 3 cm. respectively (See below figure). Find the sides AB and AC.

Answer:
The given figure can also be drawn as

Given: Let △ABC be the given triangle circumscribing the given circle with centre ‘O’ and radius 3 cm.
i.e., the circle touches the sides BC, CA and AB at D, E, F respectively.
It is given that BD = 9 cm
CD = 3 cm

∵ Lengths of two tangents drawn from an external point to a circle are equal.
∴ BF = BD = 9 cm
CD = CE = 3 cm
AF = AE = x cm say
∴ The sides of die triangle are
12 cm, (9 + x) cm, (3 + x) cm
Perimeter = 2S = 12 + 9 + x + 3 + x
⇒ 2S = 24 + 2x
or S = 12 + x
S – a = 12 + x – 12 = x
S – b = 12 + x – 3 – x = 9
S – c = 12 + x – 9 – x = 3
∴ Area of the triangle

Squaring on both sides we get,
27 (x² + 12x) = (36 + 3x)²
27x² + 324x = 1296 + 9x² + 216x
⇒ 18x² + 108x- 1296 = 0
⇒ x² + 6x – 72 = 0
⇒ x² + 12x – 6x – 72 = 0
⇒ x (x + 12) – 6 (x + 12) = 0
⇒ (x – 6) (x + 12) = 0
⇒ x = 6 or – 12
But ‘x’ can’t be negative hence, x = 6
∴ AB = 9 + 6 = 15 cm
AC = 3 + 6 = 9 cm.

Question 5.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythagoras Theorem.
Answer:
Steps of construction:

1. Draw a circle with centre ‘O’ and radius 6 cm.
2. Take a point P outside the circle such that OP =10 cm. Join OP.
3. Draw the perpendicular bisector to OP which bisects it at M.
4. Taking M as centre and PM or MO as radius draw a circle. Let the circle intersects the given circle at A and B.
5. Join P to A and B.
6. PA and PB are the required tan¬gents of lengths 8 cm each.

Proof: In △OAP
OA² + AP² = 6² + 8²
= 36 + 64 = 100
OP² = 10² = 100
∴ OA² + AP² = OP²
Hence AP is a tangent.
Similarly BP is a tangent.

Question 6.
Construct, a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Answer:
Steps of construction:

1. Draw two concentric circles with centre ‘O’ and radii 4 cm and 6 cm.
2. Take a point ‘P’ on larger circle and join O, P.
3. Draw the perpendicular bisector of OP which intersects it at M.
4. Taking M as centre and PM or MO as radius draw a circle which intersects smaller circle at Q.
5. Join PQ, which is a tangent to the smaller circle.

Question 7.
Draw a circle with the help of a bangle, take a point outside the circle. Con-struct the pair of tangents from this point to the circle measure them. Write conclusion.
Answer:
Steps of construction:

1. Draw a circle with the help of a bangle.
2. Draw two chords AB and AC. Perpendicular bisectors of AB and AC meets at ‘O’ which is the centre of the circle.
3. Taking an outside point P, join OP.
4. Let M be the midpoint of OP. Taking M as centre OM as radius, draw a circle which intersects the given circle at R and S. Join PR, PS which are the required tangents.

Conclusion: Tangents drawn from an external point to a circle are equal.

Question 8.
In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.
Answer:
Let ABC be a right triangle right angled at P.
Consider a circle with diametere AB.
From the figure, the tangent to the circle at B meets BC in Q.
Now QB and QP are two tangents to the circle from the same point P.
QB = QP …….. (1)
Also, ∠QPC = ∠QCP
∴ PQ = QC (2)
From (1) and (2);
QB = QC Hence proved.

Question 9.
Draw a tangent to a given circle with center O from a point ‘R’ outside the circle. How many tangents can be drawn to the circle from that point? [Hint: The distance of two points to the point of contact is the same.
Answer:
Only two tangents can be drawn from a given point outside the circle.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.3

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.3 Textbook Questions and Answers

Question 1.
Find the area of the triangle whose vertices are
i) (2, 3), (-1, 0), (2,-4)
Answer:
Given: A (2, 3), B (- 1, 0) and C (2, – 4) are the vertices of a △ABC.
Area of the triangle ABC = \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|2(0+4)-1(-4-3)+2(3-0)|\)
= \(\frac{1}{2}|8+7+6|\)
= \(\frac{21}{2}\)
= 10\(\frac{1}{2}\) sq.units

ii) (-5, -1), (3, -5), (5, 2)
Answer:
Given: A (- 5, – 1), B (3, – 5) and C (5, 2) are the vertices of △ABC.
Area of the △ABC
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|-5(-5-2)+3(2+1)+5(-1+5)|\)
= \(\frac{1}{2}|35+9+20|\)
= \(\frac{64}{2}\)
= 32 sq.units

iii) (0, 0), (3, 0), (0, 2)
Answer:
Given: O (0, 0), A (3, 0) and B (0, 2) are the vertices of a triangle, △AOB.
Area of the △AOB
= \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(0-2)+3(2-0)+0(0-0)|\)
= \(\frac{1}{2}|6|\)
= 3 sq.units

Or

△AOB = \(\frac{1}{2}\) × OA × OB
= \(\frac{1}{2}\) × 3 × 2
= 3 sq.units

Question 2.
Find the value of ‘K’ for which the points are collinear.
i) (7, -2), (5, 1), (3, K)
Answer:
Given: A (7, – 2), B (5, 1) and C (3, K) are collinear.
∴ Area of △ABC = 0
But area of triangle
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
⇒ \(\frac{1}{2} \mid 7(1-\mathrm{K})+5(\mathrm{~K}+2)+3(-2-1)\) = 0
⇒ \(|7-7 K+5 K+10-9|\) = 0
⇒ \(|-2 \mathrm{~K}+8|\) = 0
⇒ -2K + 8 = 0
⇒ -2K = -8
⇒ K = \(\frac{8}{2}\)
i.e., K = 4

ii) (8, 1), (K,-4), (2,-5)
Answer:
Given: A (8, 1), B(K, – 4) and C (2, – 5) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|8(-4+5)+\mathrm{K}(-5-1)+2(1+4)|\) = 0
⇒ \(|8-6 \mathrm{~K}+10|\) = 0
⇒ \(|18-6 \mathrm{~K}|\) = 0
⇒ 18 – 6K = 0
⇒ 6K = 18
⇒ K = \(\frac{18}{6}\)
i.e., K = 3

iii) (K,K), (2, 3), and (4,-1)
Answer:
A (K, K), B (2, 3) and C (4, – 1) are collinear.
∴ Area of △ABC = 0
⇒ \(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\) = 0
⇒ \(\frac{1}{2}|\mathrm{~K}(3+1)+2(-1-\mathrm{K})+4(\mathrm{~K}-3)|\) = 0
⇒ \(|4K-2-2K+4K-12|\) = 0
⇒ \(|6 \mathrm{~K}-14|\) = 0
⇒ 6K – 14 = 0
⇒ 6K = 14
⇒ K = \(\frac{14}{6}\) = \(\frac{7}{3}\)
∴ K = \(\frac{7}{3}\)

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Answer:

Given: A (0, – 1), B (2, 1) and C (0, 3) are the vertices of △ABC.
Let D, E and F be the midpoints of the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AC}}\).

Area of a triangle ABC =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)
= \(\frac{1}{2}|0(1-3)+2(3+1)+0(-1-1)|\)
= \(\frac{1}{2}|8|\)
= 4 sq.units
Area of △DEF = \(\frac{1}{2}|1(2-1)+1(1-0)+0(0-2)|\)
= \(\frac{1}{2}|1+1|\)
= \(\frac{2}{2}\)
= 1 sq.units
Ratio of areas = △ABC : △DEF = 4 : 1.
△ADF ≅ △BED ≅ △DEF ≅ △CEF
∴ △ABC : △DEF = 4 : 1

Question 4.
Find the area of the quadrilateral whose vertices taken inorder are (-4, -2), (-3, -5),(3, -2) and (2, 3).
Answer:

Given: A (- 4, – 2), B (- 3, – 5), C (3, – 2) and D (2, 3) are the vertices of the quadrilateral ▱ ABCD.
Area of ▱ ABCD = △ABC + △ACD.
Area of a triangle =
\(\frac{1}{2}\left|\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-{\mathrm{y}}_{2}\right)\right|\)

Question 5.
Find the area of the triangle formed by the points by using Heron’s formula.
i) (1, 1), (1, 4) and (5, 1)
ii) (2, 3), (-1,3) and (2, -1)
Answer:
i) (1, 1) (1, 4) and (5, 1)
let A (1, 1) B(l, 4) and C(5, 1) are the vertices then length of sides can be calculated using the formula
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)
now

now formula for area of triangle using Heron’s formula = △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
∴ s = \(\frac{3+4+5}{2}\) = \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

ii) let the vertices of given triangle A (2, 3), B (-l, 3) and C (2, -1)

∴ a = 5, b = 4, c = 3 units
now from using Heron’s formula area of triangle
= △ = \(\sqrt{s(s-a)(s-b)(s-c)}\)
where s = \(\frac{a+b+c}{2}\)
= \(\frac{5+4+3}{2}\)
= \(\frac{12}{2}\) = 6
∴ △ = \(\sqrt{6(6-5)(6-4)(6-3)}\)
= \(\sqrt{6 \times 1 \times 2 \times 3}\)
= \(\sqrt{6 \times 6}\)
= 6 sq. units
∴ area of given triangle = 6 sq units

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.4

10th Class Maths 11th Lesson Trigonometry Ex 11.4 Textbook Questions and Answers

Question 1.
Evaluate the following:
i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
Answer:
Given (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

ii) (sin θ + cos θ)² + (sin θ – cos θ)²
Answer:
Given (sin θ + cos θ)² + (sin θ – cos θ)²
= (sin² θ + cos² θ + 2 sin θ cos θ) + (sin² θ + cos² θ – 2 sin θ cos θ) [∵ (a + b)² = a² + b² + 2ab
(a – b)² = a² + b² – 2ab]
= 1 + 2 sin θ cos θ + 1 – 2 sin θ cos θ [∵ sin² θ + cos² θ = 1]
= 1 + 1
= 2

iii) (sec² θ – 1) (cosec² θ – 1)
Answer:
Given (sec² θ – 1) (cosec² θ – 1)
= tan² θ × cot² θ [∵ sec² θ – tan² θ = 1; cosec² θ – cot² θ = 1]
= tan² θ × \(\frac{1}{\tan ^{2} \theta}\) = 1

Question 2.
Show that (cosec θ – cot θ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:

Question 3.
Show that \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A
Answer:
Given that L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)
Rationalise the denominator, rational factor of 1 – sin A is 1 + sin A.

[∵ (a + b)(a + b) = (a + b)²]; (a – b)(a + b) = a² — b²]
= \(\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)
= \(\frac{1+\sin A}{\cos A}\)
= \(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)
= sec A + tan A = R.H.S.

Question 4.
Show that \(\frac{1-\tan ^{2} A}{\cot ^{2} A-1}\) = tan² A
Answer:

Question 5.
Show that \(\frac{1}{\cos \theta}\) – cos θ = tan θ – sin θ.
Answer:

Question 6.
Simplify sec A (1 – sin A) (sec A + tan A)
Answer:
L.H.S. = sec A (1 – sin A) (sec A + tan A)
= (sec A – sec A . sin A) (sec A + tan A)
= (sec A – \(\frac{1}{\cos A}\) . sin A) (sec A + tan A)
= (sec A – tan A) (sec A + tan A)
= sec² A – tan² A [∵ sec² A – tan² A = 1]
= 1

Question 7.
Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
Answer:
L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= (sin² A + cosec² A + 2 sin A . cosec A) + (cos² A – sec² A + 2 cos A . sec A) [∵ (a + b)² = a² + b² + 2ab]
= (sin² A + cos² A) + cosec² A + 2 sin A . \(\frac{1}{\sin A}\) + sec² A + 2 cos A . \(\frac{1}{\cos A}\)
[∵ \(\frac{1}{\sin A}\) = cosec A; \(\frac{1}{\cos A}\) = sec A]
= 1 +(1 + cot² A) + 2 + (1 + tan² A) + 2
[∵ sin² A + cos² A = 1; cosec² A = 1 + cot² A; sec² A = 1 + tan² A]
= 7 + tan² A + cot² A
= R.H.S.

Question 8.
Simplify (1 – cos θ) (1 + cos θ) (1 + cot² θ)
Answer:
Given that
(1 – cos θ) (1 + cos θ) (1 + cot² θ)
= (1 – cos² θ) (1 + cot² θ)
[∵ (a – b) (a + b) = a² – b²]
= sin² θ. cosec² θ [∵ 1 – cos² θ = sin² θ; 1 + cot² θ = cosec² θ]
= sin² θ . \(\frac{1}{\sin ^{2} \theta}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]
= 1

Question 9.
If sec θ + tan θ = p, then what is the value of sec θ – tan θ?
Answer:
Given that sec θ + tan θ = p ,
We know that sec² θ – tan² θ = 1
sec² θ – tan² θ = (sec θ + tan θ) (sec θ – tan θ)
= p (sec θ – tan θ)
= 1 (from given)
⇒ sec θ – tan θ = \(\frac{1}{p}\)

Question 10.
If cosec θ + cot θ = k, then prove that cos θ = \(\frac{k^{2}-1}{k^{2}+1}\)
Answer:
Method-I:
Given that cosec θ + cot θ = k
R.H.S. = \(\frac{k^{2}-1}{k^{2}+1}\)

Method – II:
Given that cosec θ + cot θ = k ……..(1)
We know that cosec² θ – cot² θ = 1
⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1 [∵ a² – b² = (a -b)(a + b)]
⇒ k (cosec θ – cot θ) = 1
⇒ (cosec θ – cot θ) = \(\frac{1}{k}\)
By solving (1) and (2)

According to identity cos² θ + sin² θ = 1

Hence proved.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 8 Similar Triangles Ex 8.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 8th Lesson Similar Triangles Exercise 8.1

10th Class Maths 8th Lesson Similar Triangles Ex 8.1 Textbook Questions and Answers

Question 1.
In △PQR, ST is a line such that \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and also ∠PST = ∠PRQ.
Prove that △PQR is an isosceles triangle.
Answer:
Given : In △PQR,
\(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\) and ∠PST= ∠PRQ.

R.T.P: △PQR is isosceles.
Proof: \(\frac{PS}{SQ}\) = \(\frac{PT}{TR}\)
Hence, ST || QR (Converse of Basic proportionality theorem)
∠PST = ∠PQR …….. (1)
(Corresponding angles for the lines ST || QR)
Also, ∠PST = ∠PRQ ……… (2) given
From (1) and (2),
∠PQR = ∠PRQ
i.e., PR = PQ
[∵ In a triangle sides opposite to equal angles are equal]
Hence, APQR is an isosceles triangle.

Question 2.
In the given figure, LM || CM and LN || CD. Prove that \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Answer:
Given : LM || CB and LN || CD In △ABC, LM || BC (given) Hence,

Adding ‘1’ on both sides.

From (1) and (2)
∴ \(\frac{AM}{AB}\) = \(\frac{AN}{AD}\).

Question 3.
In the given figure, DE || AC and DF || AE. Prove that \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\).

Answer:
In △ABC, DE || AC
Hence \(\frac{BE}{EC}\) = \(\frac{BD}{DA}\) ………. (1)
[∵ A line drawn parallel to one side of a triangle divides the other two sides in the same ratio – Basic proportionality theorem]
Also in △ABE, DF || AE
Hence \(\frac{BF}{FE}\) = \(\frac{BD}{DA}\) ………. (2)
From (1) and (2), \(\frac{BF}{FE}\) = \(\frac{BE}{AC}\) Hence proved.

Question 4.
Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using Basic proportionality theorem).
Answer:

Given: In △ABC; D is the mid-point of AB.
A line ‘l’ through D, parallel to BC, meeting AC at E.
R.T.P: E is the midpoint of AC.
Proof:
DE || BC (Given)
then
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)(From Basic Proportional theorem)
Also given ‘D’ is mid point of AB.
Then AD = DB.
⇒ \(\frac{AD}{DB}\) = \(\frac{DB}{DB}\) = \(\frac{AE}{EC}\) = 1
⇒ AE = EC
∴ ‘E’ is mid point of AC
∴ The line bisects the third side \(\overline{\mathrm{AC}}\).
Hence proved.

Question 5.
Prove that a line joining the mid points of any two sides of a triangle is parallel to the third side. (Using converse of Basic proportionality theorem)
Answer:
Given: △ABC, D is the midpoint of AB and E is the midpoint of AC.
R.T.P : DE || BC.
Proof:

Since D is the midpoint of AB, we have AD = DB ⇒ \(\frac{AD}{DB}\) = 1 ……. (1)
also ‘E’ is the midpoint of AC, we have AE = EC ⇒ \(\frac{AE}{EC}\) = 1 ……. (2)
From (1) and (2)
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
If a line divides any two sides of a triangle in the same ratio then it is parallel to the third side.
∴ DE || BC by Basic proportionality theorem.
Hence proved.

Question 6.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Answer:
Given: △PQR, DE || OQ; DF || OR
R.T.P: EF || QR
Proof:
In △POQ;
\(\frac{PE}{EQ}\) = \(\frac{PD}{DO}\) ……. (1)
[∵ ED || QO, Basic proportionality theorem]
In △POR; \(\frac{PF}{FR}\) = \(\frac{PD}{DO}\) ……. (2) [∵ DF || OR, Basic Proportionality Theorem]
From (1) and (2),
\(\frac{PE}{EQ}\) = \(\frac{PF}{FR}\)
Thus the line \(\overline{\mathrm{EF}}\) divides the two sides PQ and PR of △PQR in the same ratio.
Hence, EF || QR. [∵ Converse of Basic proportionality theorem]

Question 7.
In the given figure A, B and C are points on OP, OQ and OR respec¬tively such that AB || PQ and AC || PR. Show that BC || QR.

Answer:
Given:
In △PQR, AB || PQ; AC || PR
R.T.P : BC || QR
Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OB}{BQ}\) ……… (1)
(∵ Basic Proportional theorem)
and in △OPR, Proof: In △POQ; AB || PQ
\(\frac{OA}{AP}\) = \(\frac{OC}{CR}\) ……… (2)
From (1) and (2), we can write
\(\frac{OB}{BQ}\) = \(\frac{OC}{CR}\)
Then consider above condition in △OQR then from (3) it is clear.
∴ BC || QR [∵ from converse of Basic Proportionality Theorem]
Hence proved.

Question 8.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point ‘O’. Show that\(\frac{AO}{BO}\) = \(\frac{CO}{DO}\).
Answer:
Given: In trapezium □ ABCD, AB || CD. Diagonals AC, BD intersect at O.
R.T.P: \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\)

Construction:
Draw a line EF passing through the point ‘O’ and parallel to CD and AB.
Proof: In △ACD, EO || CD
∴ \(\frac{AO}{CO}\) = \(\frac{AE}{DE}\) …….. (1)
[∵ line drawn parallel to one side of a triangle divides other two sides in the same ratio by Basic proportionality theorem]
In △ABD, EO || AB
Hence, \(\frac{DE}{AE}\) = \(\frac{DO}{BO}\)
[∵ Basic proportionality theorem]
\(\frac{BO}{DO}\) = \(\frac{AE}{ED}\) …….. (2) [∵ Invertendo]
From (1) and (2),
\(\frac{AO}{CO}\) = \(\frac{BO}{DO}\)
⇒ \(\frac{AO}{BO}\) = \(\frac{CO}{DO}\) [∵ Alternendo]

Question 9.
Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts.
Answer:

Steps of construction:

1. Draw a line segment \(\overline{\mathrm{AB}}\) of length 7.2 cm.
2. Construct an acute angle ∠BAX at A.
3. Mark off 5 + 3 = 8 equal parts (A₁, A₂, …., A₈) on \(\stackrel{\leftrightarrow}{\mathrm{AX}}\) with same radius.
4. Join A₈ and B.
5. Draw a line parallel to \(\stackrel{\leftrightarrow}{\mathrm{A}_{8} \mathrm{~B}}\) at A₅ meeting AB at C.
6. Now the point C divides AB in the ratio 5:3.
7. Measure AC and BC. AC = 4.5 cm and BC = 2.7 cm.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.2

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.2 Textbook Questions and Answers

Question 1.
Find the coordinates of the point which divides the line segment joining the points (-1, 7) and (4, -3) in the ratio 2 :3.
Answer:
Given points P (-1, 7) and Q (4, – 3). Let ‘R’ be the required point which divides \(\overline{\mathrm{PQ}}\) in the ratio 2:3. Then

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Answer:
Given points A (4, – 1) and B (- 2, – 3) Let P and Q be the points of trisection
of \(\overline{\mathrm{AB}}\), then AP = PQ = QB.

∴ P divides \(\overline{\mathrm{AB}}\) internally in the ratio 1 : 2.

Also, Q divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1 internally.

Question 3.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Answer:
Let the point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in a ratio of m : n

⇒ 6m – 3n = -(m + n) = -m – n
⇒ 6m + m = – n + 3n
⇒ 7m = 2n
⇒ \(\frac{m}{n}=\frac{2}{7}\)
⇒ m : n = 2 : 7
∴ The point (-1, 6) divides the given line segment in a ratio of 2 : 7.

Question 4.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Answer:

Given: ▱ ABCD is a parallelogram where A (1, 2), B (4, y), C (x, 6) and D (3, 5).
In a parallelogram, diagonals bisect each other.
i.e., the midpoints of the diagonals coincide with each other.
i.e.,midpoint of AC = midpoint of BD

⇒ 1 + x = 7 and 8 = y + 5
⇒ x = 7 – 1 and y = 8 – 5
∴ x = 6 and y = 3.

Question 5.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).
Answer:

Given:
A circle with centre ‘C’ (2, -3). \(\overline{\mathrm{AB}}\) is a diameter where
B = (1, 4); A = (x, y).
C is the midpoint of AB.
[∵ Centre of a circle is the midpoint of the diameter]

4 = x + 1 and – 6 = y + 4
⇒ x = 4 – 1 = 3 and y = -6 – 4 = -10
A (x, y) = (3, -10)

Question 6.
If A and B are (-2, -2) and (2, -4) respectively. Find the coordinates of P such that AP = \(\frac{3}{7}\) AB and P lies on the segment AB.
Answer:
Given: A (- 2, – 2) and B (2, – 4)
P lies on AB such that AP = latex]\frac{3}{7}[/latex] AB

i.e., P divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 4 By section formula,

Question 7.
Find the coordinates of points which divide the line segment joining A (-4, 0) and B (0, 6) into four equal parts.
Answer:
Given, A (- 4, 0) and B (0, 6).
Let P, Q and R be the points which divide AB into four equal parts.

P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 3, Q → 1 : 1 and R → 3 : 1 Use section formula to find P, Q and R.
Then, Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 8.
Find the coordinates of the points which divides the line segment joining A(-2, 2) and B(2, 8) into four equal parts.
Answer:
Given, A (- 2, 2) and B (2, 8).
Let P, Q and R be the points which divide AB into four equal parts.

Then, Q is the midpoint of \(\overline{\mathrm{AB}}\)
P is the midpoint of \(\overline{\mathrm{AQ}}\)
R is the midpoint of \(\overline{\mathrm{QB}}\)

Question 9.
Find the coordinates of the point which divide the line segment joining the points (a + b, a-b) and (a-b, a + b) in the ratio 3 : 2 internally.
Answer:
Given : A (a + b, a – b) and B (a – b, a + b).
Let P (x, y) divides \(\overline{\mathrm{AB}}\) in the ratio 3 : 2 internally.
Section formula

Question 10.
Find the coordinates of centroid of the triangle with following vertices:
i) (-1, 3), (6, -3) and (-3, 6)
Answer:
Given: △ABC in which- A (- 1, 3), B (6, -3) and C (-3, 6)

ii) (6, 2), (0, 0) and (4, -7)
Answer:
Given: The three vertices of a triangle are A (6, 2), B (0, 0) and C (4, – 7).
Centroid (x, y)

iii) (1,-1), (0, 6) and (-3, 0)
Answer:
Given: (1, -1), (0, 6) and (-3, 0) are the vertices of a triangle.
Centroid (x, y)

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.1

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers

Question 1.
Fill in the blanks.
i) A tangent to a circle intersects it in ——— point(s). (one)
ii) A line intersecting a circle in two points is called a ———. (secant)
iii) The number of tangents drawn at the end of the diameter is ———. (two)
iv) The common point of a tangent to a circle and the circle is called ———. (point of contact)
v) We can draw ——— tangents to a given circle. (infinite)

Question 2.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find length of PQ.
Answer:
Given: A circle with centre O and radius OP = 5 cm
\(\overline{\mathrm{PQ}}\) is a tangent and OQ = 12 cm
We know that ∠OPQ = 90°
Hence in △OPQ
OQ² = OP² + PQ²
[∵ hypotenuse² = Adj. side² + Opp. side²]
12² = 5² + PQ²
∴ PQ² = 144 – 25 .
PQ² = 119
PQ = √119

Question 3.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer:
Steps:

1. Draw a circle with some radius.
2. Draw a chord of the circle.
3. Draw a line parallel to the chord intersecting the circle at two distinct points.
4. This is secant of the circle (l).
5. Draw another line parallel to the chord, just touching the circle at one point (M). This is a tangent of the circle.

Question 4.
Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm.
Answer:
Given: A circle with radius OP = 9 cm
A tangent PQ from a point Q at a distance of 15 cm from the centre, i.e., OQ =15 cm
Now in △POQ, ∠P = 90°
OP² + PQ² – OQ²
9² + PQ² = 15²
PQ² = 15² – 9²
PQ² = 144
∴ PQ = √144 = 12 cm.
Hence the length of the tangent =12 cm.

Question 5.
Prove that the tangents to a circle at the end points of a diameter are parallel.
Answer:
A circle with a diameter AB.
PQ is a tangent drawn at A and RS is a tangent drawn at B.
R.T.P: PQ || RS.
Proof: Let ‘O’ be the centre of the circle then OA is radius and PQ is a tangent.
∴ OA ⊥ PQ ……….(1)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
Similarly, OB ⊥ RS ……….(2)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
But, OA and OB are the parts of AB.
i.e., AB ⊥ PQ and AB ⊥ RS.
∴ PQ || RS.
O is the centre, PQ is a tangent drawn at A.
∠OAQ = 90°
Similarly, ∠OBS = 90°
∠OAQ + ∠OBS = 90° + 90° = 180°
∴ PQ || RS.
[∵ Sum of the consecutive interior angles is 180°, hence lines are parallel]

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 13 Probability Ex 13.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 13th Lesson Probability Exercise 13.1

10th Class Maths 13th Lesson Probability Ex 13.1 Textbook Questions and Answers

Question 1.
Complete the following statements:
i) Probability of an event E + Probability of the event ‘not E’ = 1 .
ii) The probability of an event that cannot happen is zero.
Such an event is called an impossible event.
iii) The probability of an event that is certain to happen is  1  such an event is called sure or certain event.
iv) The sum of the probabilities of all the elementary events of an experiment is 1 .
v) The probability of an event is greater than or equal to zero and less than or equal to 1 .

Question 2.
Which of the following experiments have equally likely outcomes? Explain.
i) A driver attempts to start a car. The car starts or does not start.
Answer:
Equally likely. Since both have the same probability \(\frac{1}{2}\).

ii) A player attempts to shoot a basket-ball. She/he shoots or misses the shot.
Answer:
Equally likely. Since both have the same probability \(\frac{1}{2}\).

iii) A trial is made to answer a true-false question. The answer is right or wrong.
Equally likely. Since both have the same probability \(\frac{1}{2}\).

iv) A baby is born. It is a boy or a girl.
Equally likely. Since both the events have the same probability \(\frac{1}{2}\).

Question 3.
If P(E) = 0.05, what is the probability of not E?
Answer:
Given: P(E) = 0.05
Hence, P(E) + P(\(\overline{\mathrm{E}}\)) = 1, where P(\(\overline{\mathrm{E}}\)) is the probability of ‘not E’
0.05 + P(\(\overline{\mathrm{E}}\)) = 1
∴ P(\(\overline{\mathrm{E}}\)) = 1 -0.05 = 0.95.

Question 4.
A bag contains lemon flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out
i) an orange flavoured candy?
ii) a lemon flavoured candy?
Answer:
Bag contains only lemon flavoured candies.
i) Taking an orange flavoured candy is an impossible event and hence the probability is zero.
ii) Also taking a lemon flavoured candy is a sure event and hence its probability is 1.

Question 5.
Rahim removes all the hearts from the cards. What is the probability of
i. Picking out an ace from the remaining pack.
ii. Picking out a diamond.
iii. Picking out a card that is not a heart.
iv. Picking out the Ace of hearts.
Answer:
Total number of cards in the deck = 52.
Total number of hearts in the deck of cards =13.
When Hearts are removed, remaining cards = 52 – 13 = 39.
i)Picking out an Ace:
Number of outcomes favourable to Ace = 3 [∵ ♦ A, ♥ A, ♠ A, ♣ A]
Total number of possible outcomes from the remaining cards = 39
– after removing Hearts.
Probability = P(A)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{39}\) = \(\frac{1}{13}\)

ii) Picking out a diamond:
Number of favourable outcomes to diamonds (♦) = 13
Total number of possible outcomes = 39
∴ p(♦) = \(\frac{13}{39}\) = \(\frac{1}{3}\)

iii) Picking out a card that is ‘not a heart’:
As all hearts are removed, the remain-ing cards are all non-heart cards. So the picked card will be definitely a non-heart card. So this is a sure event.
Hence its probability is one
P(E) = \(\frac{39}{39}\) = 1

iv) Picking out the Ace of Hearts:
a) As all the heart cards are removed the left over cards will have three suits (i) spades, (ii), clubs, (iii) dia¬monds of each 13.
Hence total outcomes = 3 × 13 = 39 But among them there is no Ace of heart. So number of favourable outcomes for picking Ace of heart = zero.
∴ Probability P(E) = \(\frac{0}{39}\) = 0
So it is an impossible event.

b) If picking from the rest of the cards, it is an impossible event and hence probability is zero.
It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992.

Question 6.
What is the probability that the 2 students have the same birthday?
Answer:
Let P(E) = The probability that two students not having the same birthday = 0.992
Then P(\(\overline{\mathrm{E}}\)) = The complementary event of E, i.e., two students having the same birthday Also, P(E) + p(\(\overline{\mathrm{E}}\)) = 1
∴ The probability that two students have the same birthday P(\(\overline{\mathrm{E}}\)) = 1 – P(E)
= 1 – 0.992 = 0.008

Question 7.
A die is thrown once. Find the probability of getting
(i) a prime number;
(ii) a number lying between 2 and 6;
(iii) an odd number.
Answer:
i) When a die is thrown for one time, total number of outcomes = 6
No. of outcomes favourable to a prime number (2, 3, 5) = 3
∴ Probability of getting a prime = \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

ii) No. of outcomes favourable to a number lying between 2 and 6 (3, 4, 5) = 3
∴ Probability of getting a number between 2 and 6
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

iii) Number of outcomes favourable to an odd number (1, 3, 5) = 3
∴ Probability of getting an odd number P(odd)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{3}{6}\) = \(\frac{1}{2}\)

Question 8.
What is the probability of drawing out a red king from a deck of cards?
Answer: Number of favourable outcomes to red king (♥ K, ♦ K) = 2.
Number of total outcomes = 52
(∵ Number of cards in a deck of cards = 52)
∴ Probability of getting a red king P (Red king)
= \(\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}\)
= \(\frac{2}{52}\) = \(\frac{1}{26}\)

Question 9.
Make 5 more problems getting probability using dice, cards or birthdays and discuss with friends and teacher about their solutions.
Answer:
Class-room activity.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.4

10th Class Maths 14th Lesson Statistics Ex 14.4 Textbook Questions and Answers

Question 1.
The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Answer:
Since the curve is a less than type graph the data changes to

X – axis – upper limits 1 cm = 50 units.
Y – axis – less than c.f. 1 cm = 5 units.

Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Answer:
Given: Upper limits of the classes and less than cumulative frequencies. Therefore required points are (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35)
X – axis – upper limits 1 cm = 2 units.
Y – axis – less than c.f. 1 cm = 4 units.

Number of observations = 35
∴ \(\frac{N}{2}\) = \(\frac{35}{2}\) = 17.5
Locate the point on the ogive whose ordinate is 17.5.
The x – coordinate of this point is the required median.
From the graph, median = 46.5.

Number of observations = n = 35
∴ \(\frac{N}{2}\) = \(\frac{35}{2}\) = 17.5
17.5 belongs to the class 46 – 48
∴ Median class = 46-48
l – lower boundary of class = 46
f – frequency of the median class =14
c.f = 14
Class size = 2
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 46 + \(\frac{17.5-14}{14}\) × 2
= 46 + \(\frac{3.5}{14}\) × 2
= 46 + \(\frac{7}{14}\)
= 46 + \(\frac{1}{2}\)
= 46.5
Here median is 46.5 by either by ways.

Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution, and draw its ogive.
Answer:
The given data is to be changed to more than frequency distribution type.

A graph is plotted by taking the lower limits on the X – axis and respective of Y – axis.
Scale:
X – axis: 1 cm = 5 units
Y – axis: 1 cm = 5 units

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.2

10th Class Maths 11th Lesson Trigonometry Ex 11.2 Textbook Questions and Answers

Question 1.
Evaluate the following.
i) sin 45° + cos 45°
Answer:
sin 45° + cos 45°
= \(\frac{1}{\sqrt{2}}\) + \(\frac{1}{\sqrt{2}}\)
= \(\frac{1+1}{\sqrt{2}}\)
= \(\frac{2}{\sqrt{2}}\)
= \(\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}\)
= √2

ii)

Answer:

iii)

Answer:

iv) 2 tan² 45° + cos² 30° – sin² 60°
Answer:
2 tan² 45° + cos² 30° – sin² 60°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^{2}\) – \(\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= \(\frac{2}{1}\) + \(\frac{3}{4}\) – \(\frac{3}{4}\)
= \(\frac{8+3-3}{4}\)
= \(\frac{8}{4}\)
= 2

v)

Answer:

Question 2.
Choose the right option and justify your choice.
i) \(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) sin 60°
b) cos 60°
c) tan 30°
d) sin 30°
Answer:

ii) \(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\)
a) tan 90°
b) 1
c) sin 45°
d) 0
Answer:
\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) = \(\frac{1-(1)^{2}}{1+(1)^{2}}\)
= \(\frac{0}{1+1}\) = \(\frac{0}{2}\) = 0

iii) \(\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\)
a) cos 60°
b) sin 60°
c) tan 60°
d) sin 30°
Answer:

Question 3.
Evaluate sin 60° cos 30° + sin 30° cos 60°. What is the value of sin (60° + 30°). What can you conclude?
Answer:
Take sin 60°.cos 30° + sin 30°.cos 60°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{(\sqrt{3})^{2}}{4}\) + \(\frac{1}{4}\)
= \(\frac{3}{4}\) + \(\frac{1}{4}\)
= \(\frac{3+1}{4}\)
= \(\frac{4}{4}\) = 1 …… (1)
Now take sin (60° + 30°)
= sin 90° = 1 …….. (2)
From equations (1) and (2), I conclude that
sin (60°+30°) = sin 60° . cos 30° + sin 30° . cos 60°.
i.e., sin (A + B) = sin A . cos B + cos A . sin B

Question 4.
Is it right to say cos (60° + 30°) = cos 60° cos 30° – sin 60° sin 30° ?
Answer:
L.H.S. = cos (60° + 30°)
cos 90° = 0
R.H.S. = cos 60° . cos 30° – sin 60° . sin 30°.
= \(\frac{1}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\) = 0
∴ L.H.S = R.H.S
Yes, it is right to say
cos (60°+30°) = cos 60° . cos 30° – sin 60° . sin 30°.
i.e., cos (A + B) = cos A . cos B – sin A . sin B

Question 5.
In right angle triangle △PQR, right angle is at Q and PQ = 6 cms, ∠RPQ = 60°. Determine the lengths of QR and PR.
Answer:
Given that △PQR is a right angled triangle, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°.

tan 60° = \(\frac{\text { Opposite side to } \angle P}{\text { Adjacent side to } \angle P}\)
√3 = \(\frac{RQ}{6}\)
which gives RQ = 6√3 cm ……. (1)
To find the length of the side RQ, we consider

∴ The length of QR is 6√3 and RP is 12 cm.

Question 6.
In △XYZ, right angle is at Y, YZ = x, and XY = 2x then determine ∠YXZ and ∠YZX.
Answer:
Note: In the problem take
YX = x, and XZ = 2x.

Given that △XYZ is a right angled triangle and right angle at Y, and YX = x and XZ = 2x.
By Pythagoras theorem
XZ² = XY² + YZ²
(2x)² = (x)² + YZ²
4x² = x² + YZ²
YZ² = 4x² – x² = 3x²
YZ = \(\sqrt{3 x^{2}}\) = √3x
Now, from the △XYZ
tan X = \(\frac{XZ}{XY}\) = \(\frac{\sqrt{3} x}{x}\)
tan X = √3 = tan 60°
∴ Angle YXZ is 60°.
tan Z = \(\frac{XY}{YZ}\) = \(\frac{x}{\sqrt{3} x}\)
tan Z = \(\frac{1}{\sqrt{3}}\) = tan 30°
∴ Angle YZX is 30°.
Hence ∠YXZ and ∠YZX are 60° and 30°.

Question 7.
Is it right to say that
sin (A + B) = sin A + sin B? Justify your answer.
Answer:
Let A = 30° and B = 60°
L.H.S = sin (A + B)
= sin (30° + 60°) = sin 90° = 1
R.H.S = sin 30° + sin 60°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
Hence L.H.S ≠ R.H.S
So, it is not right to say that sin (A + B) = sin A + sin B

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 12 Applications of Trigonometry Ex 12.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 12th Lesson Applications of Trigonometry Exercise 12.1

10th Class Maths 12th Lesson Applications of Trigonometry Ex 12.1 Textbook Questions and Answers

Question 1.
A tower stands vertically on the ground. From a point which is 15 meter away from the foot of the tower, the angle of elevation of the top of the tower is 45°. What is the height of the tower?
Answer:
Let the height of the tower = h m
Distance of the point of observation from the foot of the tower =15 cm.
Angle of elevation of the top of the tower = 45°
From the figure tan θ = \(\frac{\text { opp. side }}{\text { adj. side }}\)
tan 45° = \(\frac{h}{15}\)
⇒ 1 = \(\frac{h}{15}\)
∴ h = 1 × 15 = 15 m

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground by making 30° angle with the ground. The distance between the foot of the tree and the top of the tree on the ground is 6 m. Find the height of the tree before falling down.
Answer:
Distance between the foot of tree and the point of contact of the top of the tree on the ground = 6 cm.
Let the length of the remaining part be = h m.
Let the length of the broken part be = x m.
Angle made by the broken part with the ground = 30°.
From the figure
tan 30° = \(\frac{h}{6}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{6}\)
∴ h = \(\frac{6}{\sqrt{3}}=\frac{3 \times 2}{\sqrt{3}}\) = 2√3 m
Also cos 30° = \(\frac{6}{x}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6 \times 2}{\sqrt{3}}\) = \(\frac{3 \times 2 \times 2}{\sqrt{3}}\) = 4√3
∴ Height of the tree = broken part + remaining part
= x + h
= 2√3 + 4√3 = 6√3 m
= 6 × 1.732
≃ 10.392 m.

Question 3.
A contractor wants to set up a slide for the children to play in the park. He wants to set it up at the height of 2 m and by making an angle of 30° with the ground. What should be the length of the slide?
Answer:
Height of slide = 2 m
Let the length of the slide = x m.
Angle made by the slide with the ground = 30°
From the figure
sin 30° = \(\frac{2}{x}\)
⇒ \(\frac{1}{2}\) = \(\frac{2}{x}\)
⇒ x = 2 × 2 = 4 m
Length of the slide = 4 m.

Question 4.
Length of the shadow of a 15 meter high pole is 5√3 meters at 7 o’clock in the morning. Then, what is the angle of elevation of the Sun rays with the ground at the time?
Answer:
Height of the pole = 15 m
Length of the shadow = 5√3 m
Let the angle of elevation be ‘θ’.
Then from the figure
tan θ = \(\frac{15}{5 \sqrt{3}}=\frac{5 \times \sqrt{3} \times \sqrt{3}}{5 \times \sqrt{3}}\) = √3
tan θ = √3 = tan 60°
∴ θ = 60°
∴ Angle of elevation of Sun rays with the ground = 60°.

Question 5.
You want to erect a pole of height 10 m with the support of three ropes. Each rope has to make an angle 30° with the pole. What should be the length of the rope?
Answer:
Height of the pole = 10 m
Let the length of each rope = x
Angle made by the rope with the pole = 30°

From the figure
cos 30° = \(\frac{10}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{10}{x}\)
⇒ x = \(\frac{10 \times 2}{\sqrt{3}}=\frac{20}{\sqrt{3}}\)
∴ Length of each rope = \(\frac{20}{\sqrt{3}}\)m
= 11.546 m.

∴ Total length of the rope = 3 × \(\frac{20}{\sqrt{3}}\)
= 20√3
= 20 × 1.732
≃ 34.64 m.

Question 6.
Suppose you are shooting an arrow from the top of a building at a height of 6 m to a target on the ground at an angle of depression of 60°. What is the distance between you and the object?
Answer:
Height of the building = 6 m
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = \(\frac{6}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6 \times 2}{\sqrt{3}}=\frac{2 \times \sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}\) = 4√3
∴ Distance = 4√3 m or
4 × 1.732 = 6.928 m.

Question 7.
An electrician wants to repair an electric connection on a pole of height 9 m. He needs to reach 1.8 m below the top of the pole to do repair work. What should be the length of the ladder which he should use, when he climbs it at an angle of 60° with the ground? What will be the distance between foot of the ladder and foot of the pole?
Answer:
Height of the pole = 9m
Height of the point from the ground where he reaches the pole = 9 – 1.8 = 7.2 m
Angle of elevation = 60°
Angle of depression = Angle of elevation at the ground = 60°
Let the distance of the target from the shooting point = x m
Then from the figure
sin 60° = \(\frac{7.2}{x}\)
\(\frac{\sqrt{3}}{2}\) = \(\frac{7.2}{x}\)
⇒ x = \(\frac{7.2 \times 2}{\sqrt{3}}=\frac{3 \times 2.4 \times 2}{\sqrt{3}}=\frac{\sqrt{3} \times \sqrt{3} \times 4.8}{\sqrt{3}}\)
⇒ x = 1.732 × 4.8
≃ 8.31 m
Also tan 60° = \(\frac{7.2}{d}\)
√3 = \(\frac{7.2}{d}\)
⇒ d = \(\frac{7.2}{\sqrt{3}}=\frac{2.4 \times 3}{\sqrt{3}}\) = 2.4 × √3 = 2.4 × 1.732
∴ d ≃ 4.1568 m

Question 8.
A boat has to cross a river. It crosses the river by making an angle of 60° with the bank of the river due to the stream of the river and travels a distance of 600 m to reach the another side of the river. What is the width of the river?
Answer:
Let the width of the river = AB = x m
Angle made by the boat = 60°
Distance travelled = AC = 600 m
From the figure
cos 60° = \(\frac{x}{600}\)
\(\frac{1}{2}\) = \(\frac{x}{600}\)
⇒ x = \(\frac{600}{2}\) = 300 m.
In the figure
A = Boat’s place
C = Reach place of another side (or) Point of observation.
AC = Travelling distance of the boat ∠AC = 60°
AB = width of the river AB
In △ABC, sin 60° = \(\frac{AB}{AC}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{AB}{600}\)
⇒ AB = 600 × \(\frac{\sqrt{3}}{2}\) = 300√3

Question 9.
An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is 45°. What is the height of the palm tree?
Answer:
Height of the observer = 1.8 m
Distance of the observer from the palm tree = 13.2 m
From the figure
tan 45° = \(\frac{x}{13.2}\)
⇒ 1 = \(\frac{x}{13.2}\)
⇒ x = 13.2 m
∴ Height of the palm tree = 13.2 + 1.8 = 15 m.

Question 10.
In the given figure, AC = 6 cm, AB = 5 cm and ∠BAC = 30°. Find the area of the triangle.
Answer:
Draw a perpendicular BD to AC
∴ BD ⊥ AC
Now let AD = 6 – x and DC = x
Given AB = 5 cm and ∠BAD = 30° then in △ABD
sin 30° = \(\frac{BD}{AB}\) = \(\frac{BD}{5}\) = \(\frac{1}{2}\)
⇒ BD = \(\frac{5}{2}\) = 2.5 cm
and cos 30° = \(\frac{AD}{AB}\) = \(\frac{6-x}{5}\) = \(\frac{\sqrt{3}}{2}\)
⇒ 6 – x = \(\frac{5 \sqrt{3}}{2}\)
⇒ x = 6 – \(\frac{5 \sqrt{3}}{2}\) = 6 – \(\frac{5(1.732)}{2}\)
∴ x = 1.67
∴ Area of △ABC = \(\frac{1}{2}\)bh
= \(\frac{1}{2}\) × AC × BD
= \(\frac{1}{2}\) × 6 × 2.5
= 7.5 cm²

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.3

10th Class Maths 14th Lesson Statistics Ex 14.3 Textbook Questions and Answers

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Answer:

Sum of the frequencies = 68
∴ \(\frac{n}{2}\) = \(\frac{68}{2}\) = 34
Hence median class = 125 – 145
Lower boundary of the median class, l = 125
cf – cumulative frequency of the class preceding the median class = 22
f – frequency of the median class = 20
h = class size = 20
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 125 + \(\frac{[34-22]}{20}\) × 20
∴ Median = 125 + 12 = 137
Maximum number of consumers lie in the class 125 – 145
Modal class is 125 -145
l – lower limit of the modal class =125
f₁ – frequency of the modal class = 20
f₀ – frequency of the class preceding the modal class =13
f₂ – frequency of the class succeeding the modal class =14
h – size of the class = 20
Mode (Z) = \(l+\frac{f_{1}-f_{0}}{\left(f_{1}-f_{0}\right)+\left(f_{1}-f_{2}\right)} \times h\)
Mode (Z) = 125 + \(\frac{20-13}{(20-13)+(20-14)} \times 20\)
= 125 + \(\frac{7}{7+6}\) × 20
= 125 + \(\frac{140}{13}\)
= 125 + 10.76923
∴ Mode = 135.769
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}} \times \mathrm{h}\)
a = assumed mean = 135
∴ \(\overline{\mathbf{x}}\) = 135 + \(\frac{7}{68}\)
= 135 + 0.102941
≃ 135.1
Mean, Median and Mode are approximately same in this case.

Question 2.
If the median of 60 observations, given below is 28.5, find the values of x and y.

Answer:

Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
It is given that ∑f = n = 60
So, 45 + x + y = 60
x + y = 60 – 45 = 15
x + y = 15 ….. (1)
The median is 28.5 which lies be-tween 20 and 30.
Median class = 20 – 30
Lower boundary of the median class ‘l’ = 20
\(\frac{N}{2}\) = \(\frac{60}{2}\) = 30
cf – cumulative frequency = 5 + x
h = 10
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
⇒ 28.5 = 20 + \(\frac{30-5-x}{20}\) × 10
⇒ 28.5 = 20 + \(\frac{25-x}{2}\)
\(\frac{25-x}{2}\) = 28.5 – 20 = 8.5
25 – x = 2 × 8.5
x = 25- 17 = 8
also from (1); x + y = 15
8 + y = 15
y = 7
∴ x = 8; y = 7.

Question 3.
A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. [Policies are given only to persons having age 18 years onwards but less than 60 years.

Answer:

The given distribution being of the less than type, 25, 30, 35, give the upper limits of corresponding class intervals. So the classes should be 20 – 25, 25 – 30, 30 – 35, ………. 55 – 60.
Observe that from the given distribution 2 persons with age less than 20.
i.e., frequency of the class below 20 is 2.
Now there are 6 persons with age less than 25 and 2 persons with age less than 20.
∴ The number of persons with age in the interval 20 – 25 is 6 – 2 = 4.
Similarly, the frequencies can be calculated as shown in table.
Number of observations = 100
n = 100
\(\frac{n}{2}\) = \(\frac{100}{2}\) = 50, which lies in the class 35-40
∴ 35 – 40 is the median class and lower boundary l = 35
cf = 45;
h = 5;
f = 33
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 35 + \(\frac{50-45}{33}\) × 5
= 35 + \(\frac{5}{33}\) × 5
= 35 + 0.7575
= 35.7575
∴ Median ≃ 35.76

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5,…, 171.5 – 180.5.)
Answer:
Since the formula, Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\) assumes continuous classes assumes continuous class, the data needs to be converted to continuous classes.
The classes then changes to 117.5 – 126.5; 126.5 – 133.5, …… 171.5 – 180.5.

∑f_i = n = 40
\(\frac{n}{2}\) = \(\frac{40}{2}\) = 20
\(\frac{n}{2}\)th observation lie in the class 144.5- 153.5
∴ Median class = 144.5 – 153.5
Lower boundary, l = 144.5
Frequency of the median class, f = 12
c.f. = 17
h = 9
∴ Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 144.5 + \(\frac{20-17}{12}\) × 9
= 144.5 + \(\frac{3}{12}\) × 9
= 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25
∴ Median length = 146.75 mm.

Question 5.
The following table gives the distribution of the life-time of 400 neon lamps.

Find the median life-time of a lamp.
Answer:

Total observations are n = 400
\(\frac{n}{2}\)th observation i.e \(\frac{400}{2}\) = 200
200 lies in the class 3000 – 3500
∴ Median class = 3000 – 3500
Lower boundary l = 3000
frequency of the median class f = 86
c.f = 130
Class size, h = 500
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 3000 + \(\frac{200-130}{86}\) × 500
= 3000 + \(\frac{70}{86}\) × 500
= 3000 + 406.977
= 3406.98
∴ Median life ≃ 3406.98 hours

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows.

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames ? Also, find the modal size of the surnames.
Answer:
Number of letters in the surnames.
Also find the modal size of the surnames.

Total observations are n = 100
\(\frac{n}{2}\) = \(\frac{100}{2}\) = 50
50 lies in the class 7 – 10
∴ Median class = 7 – 10
l – lower boundary = 7
f – frequency of the median class = 40
cf = 36
Class size h = 3
Median:
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 7 + \(\frac{50-36}{40}\) × 3
= 7 + \(\frac{14}{40}\) × 3
= 7 + \(\frac{42}{40}\)
= 7 + 1.05
= 8.05
∴ Median = 8.05.

Mean:
Assumed mean, a = 8.5
Mean \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 8.5 + \(\frac{(-18)}{100}\)
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32.

Mode:
Maximum number of surnames = 40
∴ Modal class = 7-10
l – lower boundary of the modal class = 7
Mode (Z) = \(l+\frac{f_{1}-f_{0}}{\left(f_{1}-f_{0}\right)+\left(f_{1}-f_{2}\right)} \times h\)
l = 7; f₁ = 40, f₀ = 30, f₂ = 16, h = 3
Mode (Z) = 7 + \(\frac{40-30}{(40-30)+(40-16)}\) × 3
= 7 + \(\frac{10}{10+24}\) × 3
= 7 + \(\frac{30}{34}\)
= 7 + 0.882
= 7.882

Median = 8.0.5; Mean = 8.32; Modal size = 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

Number of observations (n) = ∑f_i
\(\frac{n}{2}\) = \(\frac{30}{2}\) = 15
15 lies in the class 50 – 55
∴ Median class = 50-55
l – lower boundary of the median class = 55
f – frequency of the median class = 8
c.f = 5
Class size h = 6
Median = \(l+\frac{\left[\frac{\mathrm{n}}{2}-\mathrm{cf}\right]}{\mathrm{f}} \times \mathrm{h}\)
= 50 + \(\frac{15-5}{8}\) × 6
= 50 + 7.5
= 57.5
= 50 + 7.5 = 57.5
∴ Median weight = 57.5 kg.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 14 Statistics Ex 14.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 14th Lesson Statistics Exercise 14.1

10th Class Maths 14th Lesson Statistics Ex 14.1 Textbook Questions and Answers

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Answer:

Since f_i and x_i are of small values we use direct method.
∴ \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{162}{20}\)
= 8.1

Question 2.
Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:

Here the x_i are of large numerical values.
So we use deviation method then,
\(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
Here the assumed mean is taken as 275.
∴ \(\overline{\mathrm{x}}=\mathrm{a}+\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= 275 + \(\frac{1900}{50}\)
= 275 + 38
= 313.

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f.

Answer:

\(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
x_i = 18 (given)
\(\Rightarrow 18=\frac{752+20 \mathrm{f}}{(44+\mathrm{f})}\)
18 (44 + f) = 752 + 20 f
⇒ 20f- 18f= 792-752
⇒ 2f = 40
∴ f = \(\frac{40}{20}\) = 20.

Question 4.
Thirty women were examined in a hospital by a doctor and their of heart beats per minute were recorded and summarised as shown. Find the mean heart beats per minute for these women, choosing a suitable method.

Answer:

\(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
75.5 + \(\frac{12}{30}\)
= 75.5 + 0.4
= 75.9.

Question 5.
In a retail market, fruit vendors were selling oranges kept in packing baskets. These baskets contained varying number of oranges. The following was the distribution of oranges.

Find the mean number of oranges kept in each basket. Which method of finding the mean did you choose?
Answer:

Here we use step deviation method where a = 135, h = 5,a multiple of all d_i
\(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)
= 22 + \(\frac{25}{400}\) × 5
= 22 + 0.31
= 22.31

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by a suitable method.
Answer:

Here a = 125, h = 50, ∑f_iu_i = 43
Now
\(\overline{\mathrm{x}}=\mathrm{a}+\left(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\right) \times \mathrm{h}\)
= 125 + \(\frac{43}{25}\) × 50
= 125 + (43 × 2)
= 125 + 86
= 211.
NOTE: If we consider first value as “a” then we dont get negative values in u_i, f_iu_i columns. Then it becomes easy for calculation.

Question 7.
To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.
Answer:

∴ \(\overline{\mathrm{x}}=\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
= \(\frac{2.96}{30}\)
= 0.00986666…….
≃ 0.099

Question 8.
A class teacher has the following attendance record of 40 students of a class for the whole term. Find the mean number of days a student was present out of 56 days in the term.

Answer:

Here, a = 51.5
∴ \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{d}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}}\)
= 51.5 – \(\frac{99}{40}\)
= 51.5 – 2.475
= 49.025
≃ 49 days

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Answer:

a = 70; h = 10
∴ \(\overline{\mathbf{x}}=\mathbf{a}+\frac{\Sigma \mathbf{f}_{\mathbf{i}} \mathbf{u}_{\mathbf{i}}}{\Sigma \mathbf{f}_{\mathbf{i}}} \times \mathbf{h}\)
\(\Rightarrow \bar{x}=70-\frac{2}{35} \times 10\)
= 70 – \(\frac{2}{35}\) × 10
= 70 – \(\frac{20}{35}\)
= 70 – 0.57142
= 69.4285
≃ 69.43%