Category: Class 7

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles InText Questions

Check Your Progress [Page No. 92]

Question 1.
Classify the following triangles according to the length of their sides:

Answer:

[Page No. 93]

Question 2.
Classify the following triangles according to the measure of their angles.

Answer:

Let’s Think [Page No. 97]

Question 1.
Can you find a triangle in which each angle is less than 60°?
Answer:
Sum of angles in a triangle is 180°.
If each angle is less than 60°, then the sum of angles of a triangle is < 180°. So, triangle cannot form. So, we cannot find a triangle in which each angle is less than 60°.

Check Your Progress [Page No. 99]

Question 1.
Find the value of ‘x’ in the given figure.

Answer:
In the given figure,
∠HAC = ∠TAE = a (∵ Vertically opposite angles)
In ∆AHC,
we know, ∠H + ∠C + ∠HAC 180°
⇒ 60° + 80° + a = 180°
⇒ 140° + a – 140° = 180°- 140° … a = 40°.
∴ ∠HAC = ∠TAE = a = 40°

In ∆ATE, we know,
∠T + ∠TAE + ∠E = 180°
⇒ x + a + 70° = 180°
⇒ x + 40° + 70° = 180°
⇒ x + 110°- 110° = 180°- 110°
∴ x = 70°

Let’s Do Activity [Page No. 101]

Question 1.
Measure the length of the sides of ∆ABC and fill the following table.

Answer:

From the above table, we can conclude that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Question 2.
Take the same measurements of the previous activity and note the results as given below.
Answer:

From the above table we can conclude that the difference of the lengths of any two sides of a triangle is less than the length of the third side.

Question 3.

Fill the following table as per measurements given in the above triangles:

What do you observe from the last four columns in above table ?

• In any triangle the opposite side to the biggest angle is bigger than the other two sides.
• In any triangle the opposite side to the smallest angle is smaller than the other two sides.

Answer:

Let’s Think [Page No. 102]

Question 1.
What are the measurements of angles of an equilateral triangle?
Answer:

In ∆ADI
AD = DI – AI = 6 cm (All angles are equal)
∠A = ∠D = ∠I = 60° (All angles are equal)
So, in an equilateral triangle,

All Sides are equal in length. Each angle is equal to 60°.

Puzzle Time [Page No. 104]

Question 1.
Find the interior angles of all triangles by using given dues.

Answer:

Hints:
Each angle in an equilateral triangle is 60°.
Straight angle = 180°
In an isosceles triangle, two angles are equal.
In a scalene triangle all angles are unequal.
Right angle = 90°.
Obtuse angle > 90°.

Examples:

Question 1.
In a triangle, two angles are 43° and 57°. Find the third angle.
Answer:
Given two angles of a triangle are 43° and 57°.
Sum of these two angles = 43° + 57° = 100°
In a triangle the sum of the interior angles is 180°.
∴ Third angle = 180° – 100° = 80°

Question 2.
Find the value of ‘x’ in the following triangles :

Answer:
From the figure
∠A + ∠B + ∠C = 180° (v Sum of three angles in triangle is 180°)
⇒ 65° + 60° + x = 180°
⇒ 125° + x = 180°
∴ x =180°- 125° = 55°

Answer:
From the figure
∠P + ∠Q + ∠R = 180°
x + x + 80° = 180°
⇒ 2x + 80° = 180°
⇒ 2x = 180°-80°
⇒ 2x = 100°
⇒ x = \(\frac{100^{\circ}}{2}\) = 50°
∴ x = 50°

Answer:
From the figure
∠X + ∠Y + ∠Z – 180°
⇒ x + 90° + 42° = 180°
⇒ x + 132° = 180°
⇒ x = 180°- 132°
∴ x = 48°

Question 3.
Find the values of x and y in the given triangle.

Answer:
In the ∆ABC
∠ACD + ∠ACB = 180° (linear pair of angles)
110° + y = 180°
⇒ y = 180°- 110°
∴ y = 70° ……………….(i)

∠BAC + ∠ACB + ∠CBA = 180°
⇒ x + y + 60° = 180°
⇒ x + 70° + 60° = 180° (From (i))
⇒ x + 130° = 180°
∴ x = 50°

Question 4.
In a right angled triangle one acute angle is 44°, then find the other acute angle.
Answer:
We know that, sum of the two acute angles in a right angled triangle is 90°.
Given that, in the right angled tri-angle one of the acute angles = 44°
Other acute angle in right angled triangle = 90° – 44° = 46°

Question 5.
The angles of a triangle are (x + 10)°, (x – 20)° and (x + 40)°. Find the value of x and the measure of the angles.
Answer:
Given that the angles of the triangle are
(x + 10)°, (x – 20)° and (x + 40)°
(x + 10)° + (x – 20)° + (x + 40)° = 180°
⇒ x + 10° + x – 20° + x + 40° = 180°
⇒ 3x + 30° = 180°
⇒ 3x – 180° – 30°
⇒ 3x = 150°
⇒ x = \(\frac{150^{\circ}}{3}\) = 50°

The angles are,
x + 10° = 50° + 10° = 60°
x – 20° = 50° – 20° = 30°
x + 40° = 50° + 40° = 90°
∴ Measure of the angles are 60°, 30° and 90°.

Question 6.
In the ∆ABC, exterior angle at ∠C = 105° and ∠A=65°. Find the other interior opposite angle.
Answer:
Given that, exterior angle at ∠C = 105°.
One of the interior opposite angle ∠A = 65°.

The other interior opposite angle is ∠B .
∠A + ∠B =105°(∵ Exterior angle property of a triangle)
65° + ∠B = 105°
⇒ ∠B = 105° – 65° = 40°

Question 7.
Find the exterior angle of the given triangle.
Answer:

From the figure, ∠P = 30°, ∠Q = 35°
Exterior angle at R = ∠P + ∠Q (∵ Exterior angle property of a triangle)
= 30° + 35° = 65°

Question 8.
Find the values of ‘x’ and ‘y’ in the following figure.

Answer:
∠LMN + 110° = 180° (Linear pair angles)
∠LNM = 180° -110° = 70°
∠LMN = ∠LNM (Angles opposite to equal sides)
y = 70°
⇒ x + y – 110° (Exterior angle property)
⇒ x + 70° = 110°
∴ x = 110°-70° – 40°

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Unit Exercise

Question 1.
How many right angles exist in a triangle?
Answer:
Only one right angle can exist in a triangle.

Question 2.
Which is the longest side in ∆XYZ having right angle at ‘Z’?
Answer:

In ∆XYZ, z = 90°
In triangle, the side opposite to the largest angle is longest side. .
So, XY is the longest side in ∆XYZ.

Question 3.
Is the sum of any two angles of a tri-angle always greater than the third angle? Give examples to justify your answer.
Answer:

In ∆ADI, ∠A = 70°, ∠D = 50°, ∠I = 60°
∠A + ∠D = 70° + 50° =120° > 60° ∴ ∠A + ∠D > ∠I.
∠D + ∠I = 50° + 60° = 110° > 70° ∴ ∠D + ∠I > ∠A.
∠A + ∠I = 70° + 60° = 130° > 50° ∴ ∠A + ∠I > ∠D.
Yes, sum of any two angles of a triangle always greater than the third angle.

Question 4.
Choose any three measures from the following to make three different triangular wooden frames.
11 m, 9 m, 3 m, 7 m and 5 m.
Answer:
We know that in a triangle sum of any two sides is greater than the third side.
(i) 3m, 7m, 5m
(ii) 3m, 9m, 11m
(iii) 3m, 7m, 9m

Question 5.
Write any two possible measurements to be suitable for the following triangles.
(i) Right angled triangle.
Answer:

In ∆SRI,
∠S = 40°, ∠R = 90°, ∠I = 50°
SR = 4 cm, RI = 3 cm and SI = 5 cm

(ii) Obtuse angled triangle.
Answer:

In ∆MDI, .
∠A = 200, ∠D = 1200, ∠I = 40°
AD = 4 cm, DI = 5 cm and AI = 7cm

(iii) Acute angled triangle.
Answer:

In ∆KIS,
∠K = 50°, ∠I – 60°, ∠S = 70°
KI = 4 cm, IS = 6 cm, KS = 8 cm

Question 6.
Find the value of ‘x’ and ‘y’ in the adjacent figure.

Answer:
In ∆ACL, AC = CL
So, ∆ACL is an isosceles triangle.

In isosceles triangle, the angles which are opposite to equal sides are also equal.
So, ∠A = ∠L = x°
∠C = 56° (Vertically opposite angle)

In ∆ACL,
we know ∠A + ∠C + ∠L = 180°
⇒ x + 56° + x = 180°
⇒ 2x + 56° – 56° = 180° – 56°
⇒ 2x =124°
⇒ \(\frac{2 x}{2}=\frac{124^{\circ}}{2}\)
∴ x = 62°
Exterior angle at A = ∠C + ∠L
⇒ y = 56 + 62
⇒ y = 118°

Question 7.
In ∆ABC, ∠A is four times to ∠B and ∠C is five times to ∠B. Find the three angles.
Answer:

In ∆ABC, let ∠B = x°
then ∠A 4 times of ∠B = 4x°
∠C = 5 times of ∠B = 5x°

In ∆ABC, we know
∠A + ∠B + ∠C = 180°
⇒ 4x + x + 5x = 180°
⇒ 10x = 180°

x = 18°
∠B = 18°,
∠A = 4x° = 4 × 18° = 72°,
∠C = 5x° = 5 × 18° = 90°
∴ Three angles are 18°, 72° and 90°.

Question 8.
Ladder was faced to a wall, One end of the ladder was making 70° with the floor. Find the angle of the other end of the ladder with the wall.
Answer:

Let A be the one end of the ladder.
C is other end of the ladder.
B is the foot of the wall.

So, ∆ABC is formed.
In ∆ABC, ∠A = 70°, ∠B = 90°

We know, in ∆ABC,
∠A + ∠B + ∠C = 180° ‘
70° + 90° + ∠C = 180°
160° + ∠C = 180°
160° + ∠C – 160° = 180° – 160°
∠C = 20°
So, angle of the other end of the ladder is 20°.

Question 9.
Write the possible measurements of angles in the following table. One example is given for you.
Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.4

Question 1.
Two sides of a triangle are 5cm and 4cm respectively. Write any three possible measurements that suit for the third side.
Answer:
Given two sides of a triangle are 5 cm and 4 cm.
We know that, in triangle sum of lengths of any two sides is always greater than the third side and in the triangle difference of lengths of any two sides is always less than the third side.
So, 5 cm, 4 cm, 2 cm
5 cm, 4 cm, 3 cm
5 cm, 4 cm, 4 cm
5 cm, 4 cm, 5 cm
5 cm, 4 cm, 6 cm
5 cm, 4 cm, 7 cm
Measurements that suit for third side are 2 cm, 3 cm, 4 cm, 5 cm, 6 cm, 7 cm.

Question 2.
The lengths of line segments are 3 cm, 5 cm, 6 cm and 9 cm.
(i) From the above measurements which group of the line segments can form a triangle.
Answer:
3 cm, 5 cm, 6 cm and 5 cm, 6 cm, 9 cm groups form triangle.
Because, sum of any two sides is greater than the third side in this group.

(ii) Which group of line segments cannot form a triangle, give the reason?
Answer:
We know that, sum of the lengths of any two sides is always greater than the third side.
3 cm, 6 cm, 9 cm does not form a triangle.
Because sum of 3 cm and 6 cm is not greater than 9 cm.

Question 3.
Find the value of ‘x’ in the following triangles :

Answer:
Given, in ∆PEN, PE = 4 cm, PN = 4 cm ∠E = 65°, ∠N = x°
Two sides are equal in length.
So, ∆PEN is an isosceles triangle.

In isosceles triangle, the angles which are opposite to equal sides are also equal.
So, ∠N = ∠E
∴ x = 65°

Answer:
Given, in ∆BGA,
∠B = 56°, ∠G = x, ∠A = 56° and BG = x cm, GA = 4.3 cm
We know that, sum of angles of a triangle is 180°.
In ∆BGA, ∠B + ∠G + ∠A = 180°
⇒ 56° + x + 56° = 180°
⇒ 112° + x = 180° :
⇒ 112° + x – 112° =180°- 112°
∴ x = 68°

Two angles are equal in ∆BGA.
So, ∆BGA is an isosceles triangle.
In isosceles triangle the sides which are opposite to equal angles are also equal.
So, BG = GA
∴ x = 4.3 cm

Question 4.
∆ABC is an isosceles triangle in which AB = AC. If ∠A = 80°, then find ∠B and ∠C.
Answer:

In isosceles triangle the angles which are opposite to equal sides are also equal.
So, ∠B = ∠C = x°
In ∆ABC, we know
∠A + ∠B + ∠C — 180°
⇒ 80° + x + x = 180°
⇒ 80° + 2x – 80° = 180° – 80°
⇒ 2x = 100°
⇒ \(\frac{2 x}{2}=\frac{100^{\circ}}{2}\)
∴ x = 50°
∴ ∠B = ∠C = 50°

Question 5.
Find the values of ‘x’ in each of the below triangles.

Answer:
In ∆PQR, PQ = QR and ∠P = 53°
∆PQR is an isosceles triangle.

In isosceles triangle, the angles which are opposite to equal sides are also equal.
So, ∠P = ∠R = 53°
In ∆PQR, we know
∠P + ∠Q + ∠R = 180°
⇒ 53° + x + 53° = 180°
⇒ 106° + x = 180°
⇒ 106° + x- 106° = 180°- 106°
∴ x = 74°

Answer:
In ∆LMN, LM = LN and exterior angle at L = 110° .
∆LMN is an isosceles triangle.
In isosceles triangle, the angles which are opposite to equal sides are also equal.

So, ∠M = ∠N = x° .
We know that an exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at
L = ∠M + ∠N = 110°
⇒ x + x = 110°
⇒ 2x = 110°
⇒ \(\frac{2 x}{2}=\frac{110}{2}\)
∴ x = 55°

Question 6.
Which of the following statements are true in the following diagram.

(i) OY < OT
Answer:

In ∆TOY, ∠T + ∠O + ∠Y = 180°
⇒ ∠T + 90° + 50° = 180°
⇒ ∠T + 140° = 180°
⇒ ∠T + 140°- 140° = 180° – 140°
∴ ∠T – 40°
We know that side opposite to the largest angle is longest.
OY< OT – True

(ii) TY < TO
Answer:
TY < TO False

(iii) ∠Y < ∠T
Answer:
∠Y < ∠T
50° < 40° – False

(iv) TY < OY
Answer:
TY < OY – False

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.3

Question 1.
Write the exterior angles of ∆XYZ.

Answer:
Exterior angles of ∆XYZ are ∠PXY, ∠ZYQ and ∠XZR.
Exterior angle of X is ∠PXY. Exterior angle of Y is ∠QYZ.
Exterior angle of Z is ∠XZR.

Question 2.
Find the exterior angles in each of the following triangles :

Answer:
From the figure
∠A = 60°, ∠B = 73°
Exterior angle of a triangle is equal to the sum of its opposite interior angles.

Exterior angle at
C = ∠A + ∠B
= 600 + 730
= 133°

Answer:
From the figure,
∠D = 90°, ∠E = 30°
Exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at
F = ∠D + ∠E
= 90° +.30°
= 120°

Question 3.
Find the value of ‘x’ in the following figures.

Answer:
From the figure, ∠A = 35°, ∠B = x° and exterior angle at C = 70°
Exterior angle of a triangle is equal to the sum of its opposite interior angles. Exterior anjpNtt
C = ∠A + ∠B = 70°
⇒ 35° + x= 70°
⇒ 35° + x – 35° = 70° – 35°
∴ x = 35°

Answer:
From the figure, ∠P = 4x, ∠Q = 3x and exterior angle at R = 119°
Exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at R = ∠P + ∠Q = 119°
⇒ 4x + 3x = 119°
⇒ 7x = 119°
⇒ \(\frac{7 x}{7}=\frac{119^{\circ}}{7}\)
∴ x = 17°

Question 4.
If the exterior angle of a triangle is 110° and it’s interior opposite angles are x° and (x + 10)°, then find the value of ‘x’.
Answer:
Given interior opposite angles are x° and (x + 10)°.
Exterior angle =110° .
Exterior angle of a triangle is equal to the sum of its opposite interior angles.
x + x + 10° = 110°
⇒ 2x + 10°- 10° = 110°- 10°
⇒ 2x = 100°
⇒ \(\frac{2 x}{2}=\frac{100^{\circ}}{2}\)
∴ x = 50°

Question 5.
Find the values of V and ‘y’ in each of the following figures.

Answer:
From the figure, ∠A = 40°, ∠B = 60°, ∠D = 45°
Exterior angle at C = x°
Exterior angle at E = y°

Exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at C = ∠A + ∠B
x = 40° + 60°
∴ x = 100°

Exterior angle at E = ∠C + ∠D
y = x + 45°
y = 100° + 45°
∴ y = 145°

Answer:
From the figure, ∠M = y, ∠N = 70°
Exterior angle at M = x°
Exterior angle at L = 120°

Exterior angle of a triangle is equal to the sum of its opposite interior angles.
Exterior angle at L = ∠M + ∠N = 120°
⇒ y + 70° = 120°
⇒ y + 70 – 70 = 120 – 70
∴ y = 50°

∠QLN + ∠MLN = 180° (linear pair of angles)
120° + ∠MLN = 180°
120° + ∠MLN – 120° = 180°- 120°
∠MLN = 60°
∴ ∠L = 60°

Exterior angle at M = ∠L + ∠N
x = 60° + 70° = 130°
∴ x = 130°
∴ x = 130° and y = 50°

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.2

Question 1.
Which of the following angles form a triangle?
(a) 60°, 70°, 80°
Answer:
Given angles are 60°, 70°, 80°.
Sum of the angles = 60° + 70° + 80°
= 210° >180°
So, 60°, 70°, 80° cannot form a triangle.

(b) 65°, 45°, 70°
Answer:
Given angles are 65°, 45°, 70°.
Sum of the angles = 65° + 45° + 70° = 180°
So, 65°, 45°, 70° can form.a triangle.

(c) 40°, 50°, 60°
Answer:
Given angles are 40°, 50°, 60°
Sum of the angles = 40° + 50° + 60°
= 150° <180°
So, 40°, 50°, 60° cannot form a triangle.

(d) 60°, 30°, 90°
Answer:
Given angles are 60°, 30°, 90°.
Sum of the angles = 60° + 30° + 90° = 180°
So, 60°, 30°, 90° can form a triangle.

(e) 38°, 102°, 40°
Answer:
Given angles are 38°, 102°, 40°
Sum of the angles = 38° + 102° + 40° = 180°
So, 38°, 102°, 40° can form a triangle.

(f) 100°, 30°, 45°
Sol. Given angles are 100°, 30°, 45°
Sum of the angles = 100° + 30° + 45° = 175° < 180°
So, 100°, 30°, 45° cannot form a triangle.

Question 2.
Sum of two interior angles of a triangle is 105°. Find the third angle.
Answer:

In ∆ABC given sum of two angles is 105°.
Let ∠A + ∠B = 105°
We know that in ∆ABC,
∠A + ∠B + ∠C = 180°
⇒ 105° + ∠C = 180°
⇒ 105° + ∠C – 105° – 180° – 105°
⇒ ∠C = 75°
∴ Third angle is 75°.

Question 3.
In ∆PQR, if ∠P=65° and ∠Q = 50°, then find ∠R.
Answer:

Given in ∆PQR, ∠P = 65°, ∠Q = 50°
We know that in ∆PQR,.
∠P + ∠Q + ∠R = 180°
⇒ 65° + 50° + ∠R =180°
⇒ 115° + ∠R = 180°
⇒ 115° + ∠R – 115°
⇒ 180° – 115°
⇒∠R = 65°
∴ ∠R = 65°

Question 4.
Find the missing angles in each of the following triangles.

Answer:
Given in ∆SKV, ∠K = 60°, ∠V = 70°
We know that in ∆SKV,
∠S + ∠K + ∠V = 180°
⇒ ∠S + 60° + 70° = 180°
⇒ ∠S + 130° = 180°
⇒∠S + 130°- 130° = 180°- 130°
⇒ ∠S = 50°
∴ ∠S = 50°

Answer:
Given in ∆BUN, ∠B = 105°, ∠U – 55°
We know that in ABUN,
∠B + ∠U + ∠N = 180°
⇒ 105° + 55° + ∠N = 180°
⇒ 160° + ∠N = 180°
⇒ 160° + ∠N – 160° = 180° – 160°
∴ ∠N = 20°

Answer:
Given in ∆PAT, ∠A = 90°, ∠T = 38°
We know that in ∆PAT,
∠P + ∠A + ∠T = 180°
⇒ ∠P + 90° + 38° = 180°
⇒ ∠P + 128° = 180°
⇒ ∠P + 128°- 128° = 180° – 128°
∴ ∠P = 52°

Question 5.
Find the value of ‘x’ in each of the given triangles.

Answer:
In ∆CUT, ∠C = 64°, ∠U = 46° and exterior angle ∠CTE = x =?
∠C + ∠U + ∠T= 180°
⇒ 64° + 46° + ∠T = 180°
⇒ 110° + ∠T= 180°
⇒ 110° + ∠T – 110° = 180°- 110°
∠UTC = ∠T = 70°
∠UTC + ∠CTE = 180° (linear pair of angles)
⇒ 70° + x° = 180°
⇒ 70° + x° – 70° = 180°-70°
∴ x = 110°

Answer:
In ∆NTE, ∠N = 78°, ∠T = x, ∠E = x,
We know that in ANTE, ‘
∠N + ∠T + ∠E = 180°
⇒ 78° + x + x = 180°
⇒ 78° + 2x – 78° = 180° – 78°
⇒ 2x = 102°
⇒ \(\frac{2 x}{2}=\frac{102^{\circ}}{2}\)
∴ x = 51°

Question 6.
Find the value of ‘x’ and ‘y’ in each of the following triangles.

Answer:
In ∆TOP given ∠T = 6O, ∠O = y°
∠OPT = x° and ∠RPQ = 68°
∠OFT = ∠RPQ (Vertically, opposite angles are equal)
x = 68°

We know that in ∆TOP
∠T + ∠O + ∠P= 1800
⇒ 60° + y° + x° = 180°
⇒ 60° + y° + 68° = 180°
⇒ 128° + y – 128° = 180°- 128°
∴ y = 52° ‘
∴ x = 68°and y = 52°

Answer:
In ∆EFG, ∠E = 70°, ∠F = 74° and ∠EGF – x° and∠EGH = y°
We know that in ∆EFG,
∠E + ∠F + ∠G = 180°
⇒ 70° + 74° + ∠EGF – 180°
⇒ 144° + x – 144° = 180° – 144°
⇒ x = 36°
∴ ∠EGF .= 36°

∠EGF + ∠EGH = 180° (linear pair of angles)
⇒ 36° + ∠EGH = 180°
⇒ 36° + y° – 36° = 180° – 36°
∴ y = 144°
∴ x = 36° and y = 144°

Question 7.
In a right angled triangle one acute angle is 37°. Find the other acute angle.
Answer:

Given in ∆ADI, ∠A = 37°, ∠D = 90°, ∠I =?
We know that in ∆ADI.
∠A + ∠D + ∠I = 180°
⇒ 37° + 90° + ∠I = 180°
⇒ 127° + ∠I – 127° = 180°- 127°
∴ ∠I – 53°
∴ Other acute angle is 53°.

Question 8.
If the three angles of a triangular sign-board are 2x°, (x – 10)° and (x + 30)° respectively. Then find it’s angles.
Answer:
Given the three angles of a triangular signboard are 2x°, (x – 10)° and (x +30)°.

We know that in a triangle,
2x + (x – 10) + (x + 30) = 180°
⇒ 2x + x- 10° + x + 30 = 180°
⇒ 4x + 20 = 180°
⇒ 4x + 20 – 20 = 180° – 20°
⇒ 4x = 160°
⇒ \(\frac{4 x}{4}=\frac{160^{\circ}}{4}\)
∴ x = 40°
Angles are 2x°, (x- 10)°, (x + 30)° 2(40°), 40° – 10, 40 + 30
Angles of signboard are : 80°, 30°, 70°.

Question 9.
If one angle of a triangle is 80°, find the other two angles which are equal.
Answer:

Given one angle of a triangle is 80°,
In ∆SRI, ZS = 80° and ∠R = ∠I = x

We know that in ∆SRI,
∠S + ∠R + ∠I = 180°
⇒ 80° + x + x = 180°
⇒ 80° + 2x-80° = 180°-80°
⇒ 2x = 100°
⇒ \(\frac{2 x}{2}=\frac{100^{\circ}}{2}\)
∴ x = 50°
Therefore angles of triangle are 80°, 50° and 50°.

Question 10.
State TRUE or FALSE for each of the following statements and write the reasons for the FALSE statement.
(i) A triangle can have two right angles.
Answer:
FALSE.

In triangle sum of three angles is 180°. In triangle, if two angles are two right angles (90° + 90° = 180°).
Then, sum of three angles is greater than 180°.

(ii) A triangle can have two acute angles.
Answer:
TRUE.

(iii) A triangle can have two obtuse angles.
Answer:
FALSE.

In triangle sum of three angles is 180°. In traingle, if two angles are two obtuse angles, then sum of three angles is greater than 180°.

Question 11.
The angles of a triangle are in the ratio 2 : 4 : 3, then find the angles.
Answer:
Given the ratio of the angles of a ’ triangle are 2 : 4 : 3.
2x : 4x : 3x

Sum of the angles of a triangle is 180°.
⇒ 2x + 4x + 3x = 180°
⇒ 9x = 180°
⇒ \(\frac{9 x}{9}=\frac{180^{\circ}}{9}\)
∴ x = 20°
Angles are ⇒ 2x : 4x :,3x
2(20°) : 4(20°): 3(20°)
40°: 80° : 60°
∴ Angles of a triangle are 40°, 80°, 60°.

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Ex 5.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Exercise 5.1

Question 1.
Classify the following triangles based on the length of their sides.

Answer:
In ∆ABC,
AB ≠ BC ≠ AC
So, ∆ABC is scalene traingle.

Answer:
In ∆DEF,
DE = EF = FD = 2 cm
So, ∆DEF is an equilateral triangle.

Answer:
In ∆XYZ,
XZ = YZ = 3.5 cm
So, ∆XYZ is an isosceles triangle.

Answer:
Isosceles triangle (∵ Two sides are equal)

Answer:
Equilateral triangle (∵ Three sides are equal)

Answer:
Scalene triangle (∵ NO two sides are equal).

Question 2.
Classify the following triangles based on the measure of angles.

Answer:
In ∆TUV, ∠T = 90°
So, ∆TUV is Right-angled triangle.

Answer:
In ∆QRS, ∠Q = 108°
So, ∆QRS is an Obtuse angled triangle.

Answer:
In ∆IJK, all are acute angles.
So, ∆IJK is an acute angled triangle.

Question 3.
Classify the following triangles based on their sides and also on their angles.

Answer:
In ∆IJK, IJ ≠ KJ ≠ IK
So, ∆IJK is a scalene triangle.

Answer:
In ∆LMN, ∠M = 90
So, ∆LMN is a right triangle.

Answer:
In ∆TUV, TU = UV = TV = 4.5 cm
So, ∆TUV is an equilateral triangle.

SCERT AP 7th Class Maths Solutions Pdf Chapter 5 Triangles Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangles Review Exercise

Question 1.
Mark any three non collinear points A, B and C in your note book, join them to make a triangle and name it.
Answer:

Question 2.
Observe the given triangle and answer the following :

(i) Write the interior points of the triangle.
Answer:
A, C, D and J.

(ii) Write the points marked on the triangle.
Answer:
B, E, G, P, Q and R.

(iii) Write the exterior points of the triangle.
Answer:
F, H and I.

Question 3.
Observe the given triangle and answer the following:

(i) The opposite side to vertex L is __________
Answer:
\(\overline{\mathrm{KM}}\)

(ii) The opposite side to ZK is __________
Answer:
\(\overline{\mathrm{LM}}\)

(iii) The opposite angle to \(\overline{\mathrm{KL}}\) is __________
Answer:
∠M.

(iv) The opposite vertex to \(\overline{\mathrm{LM}}\) is __________
Answer:
K.

Question 4.
Classify the following angles into acute, obtuse and right angles :
20°, 50°, 102°, 47°, 125°, 65°, 36°, 90°, 95° and 110°.
Answer:
Acute angles : 20°, 50°, 47°, 65°, 36°.
Right angle : 90°.
Obtuse angles : 102°, 125°, 95°, 110°.

Question 5.
Write the intersecting point and concurrent point in the adjust figure.

Answer:
Intersecting point: P
Concurrent point: Q

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.2

Question 1.
Give the steps you will use to separate the variables and then solve the equation.
(i) \(\frac{5 m}{3}\) = 10
Answer:
Given \(\frac{5 m}{3}\) = 10
⇒ \(\frac{5 m}{3}\) × 3 = 10 × 3 (Multiply with 3 on both sides)
⇒ 5m = 30
⇒ \(\frac{5 m}{5}\) = \(\frac{30}{5}\) (Divide both sides by 5)
⇒ m = 6

Check:
Substitute m = 6 in the given equation,
LHS = \(\frac{5 m}{3}\) = \(\frac{5(6)}{3}\) = \(\frac{30}{3}\) = 10 = RHS.
Hence verified.

(ii) 4M – 23 = 13
Answer:
Given 4n – 23 = 13

⇒ 4n = 36
⇒ \(\frac{4 n}{4}\) = \(\frac{36}{4}\) (Divide both sides by 4)
⇒ n = 9

Check:
Substitute n = 9 in the given equation.
LHS = 4n – 23
= 4(9) – 23
= 36 – 23 = 13 = RHS
Hence verified.

(iii) – 5 + 3x = 16
Answer:
Given – 5 + 3x = 16
⇒ – 5 + 3x + 5 = 16 + 5 (Add 5 on both sides)
⇒ 3x = 21
⇒ \(\frac{3 x}{3}\) = \(\frac{21}{3}\) (Divide by 3 on both sides)
⇒ x = 7

Check:
Substitute x = 7 in the given equation
LHS = – 5 + 3x
= – 5 + 3(7)
= – 5 + 21 = 16 = RHS
Hence verified.

(iv) 2(y – 1) =8
Answer:
Given 2(y – 1) = 8
⇒ 2y – 2 = 8 (Distributive property)
⇒ 2y – 2 + 2 = 8 + 2 (Add 2 on both sides)
⇒ 2y = 10
⇒ \(\frac{2 y}{2}\) = \(\frac{10}{2}\) (Divide by 2 on both sides)
⇒ y = 5

Check: Substitute y = 5 in the given equation
LHS = 2(y – 1)
= 2(5 – 1)
= 2 × 4 = 8 = RHS
Hence verified.

Question 2.
Solve the following simple equations and check the results.
(i) 3x = 18
Answer:
Given 3x = 18
⇒ \(\frac{3 x}{3}\) = \(\frac{18}{3}\) (Divide by 3 on both sides)
⇒ x = 6

Check: Substitute x = 6 in 3x = 18
LHS ⇒ 3x = 3(6) = 18 = RHS
Hence verified.

(ii) \(\frac{b}{7}\) = – 2
Sol.
Given \(\frac{b}{7}\) = – 2
⇒ \(\frac{b}{7}\) × 7 = – 2 × 7 (Multiply by 7 on both sides)
⇒ b = – 14

Check: Substitute b = – 14 in
\(\frac{b}{7}\) = – 2
LHS = \(\frac{b}{7}\) = \(\frac{- 14}{7}\) = – 2 = RHS
Hence verified

(iii) – 2x = – 10
Answer:
Given – 2x = – 10
⇒ \(\frac{-2 x}{-2}\) = \(\frac{-10}{-2}\) (Divide by – 2 on both sides)
⇒ x = 5

Check:
Substitute x = 5 in
– 2x = – 10
LHS = – 2x
= – 2(5) = – 10 = RHS
Hence verified.

(iv) 10 + 6a = 40
Answer:
Given 10 + 6a = 40
⇒ 10 + 6a – 10 = 40 – 10 (Subtract 10 on both sides)
⇒ 6a = 30
⇒ \(\frac{6a}{6}\) = \(\frac{30}{6}\) (Divide by 6 on both sides)
⇒ a = 5

Check: Substitute a = 5 in 10 + 6a = 40
LHS = 10 + 6a
= 10 + 6(5)
= 10 + 30
= 40 = RHS
Hence verified.

(v) – 7m = 21
Answer:
Given – 7m = 21
⇒ \(\frac{-7 m}{-7}\) = \(\frac{21}{-7}\) (Divide by – 7 on both sides)
⇒ m = – 3

Check:
Substitute m = in – 7m = 21
LHS = – 7m
= – 7 (- 3)
= 21 = RHS
Hence verified.

(iv) 4p + 7 = – 21
Answer:
Given 4p + 7 = – 21
⇒ 4p + 7 – 7 = – 21 – 7 (Subtract 7 on both sides)
⇒ 4p = – 28
⇒ \(\frac{4p}{4}\) = \(\frac{-28}{4}\) (Divide by 4 on both sides)
⇒ p = – 7

Check:
Substitute p = – 7 in 4p + 7 = – 21
LHS = 4p + 7
= 4(- 7) + 7
= – 28 + 7
= – 21 = RHS
Hence verified.

(vii) 3x – \(\frac{1}{3}\) = 5
Answer:

Check: Substitute x = \(\frac{16}{9}\) in 3x – \(\frac{1}{3}\) = 5
LHS = 3x – \(\frac{1}{3}\)
= 3\(\left(\frac{16}{9}\right)\) – \(\frac{1}{3}\)
= \(\frac{16}{3}-\frac{1}{3}\)
= \(\frac{16-1}{3}\) = \(\frac{15}{3}\)
= 5 = RHS
Hence verified.

(viii) 18 – 7n = – 3
Answer:
Given 18 – 7n = – 3
⇒ 18 – 7n – 18 = – 3 – 18 (Subtract 18 on both sides)
⇒ – 7n = – 21
⇒ \(\frac{-7 n}{-7 n}\) = \(\frac{-21}{-7}\) (Divide by – 7 on both sides)
⇒ n = 3

Check:
Substitute n = 3 in 18 – 7n = – 3
LHS = 18 – 7n
= 18 – 7(3)
= 18 – 21
= – 3 = RHS
Hence verified.

(ix) 3(k + 4) = 21
Sol.
Given 3(k + 4) = 21
⇒ 3k + 12 = 21 (Distributive property)
⇒ 3k + 12 – 12 = 21 – 12 (Subtract 12 on both sides)
⇒ 3k = 9
⇒ \(\frac{3 \mathrm{k}}{3}\) = \(\frac{9}{3}\) (Divide by 3 on both sides)
⇒ k = 3

Check:
Substitute k = 3 in 3(k + 4) = 21
LHS = 3(k + 4)
= 3(3 + 4)
= 3 × 7 = 21= RHS
Hence verified.

(x) 9 (a + 1) + 2 = 11
Answer:
Given 9(a + 1) + 2 = 11
⇒ 9(a + 1) + 2 – 2 = 11 – 2 (Subtract 2 on both sides)
⇒ 9(a + 1) = 9
⇒ \(\frac{9(a+1)}{9}\) = \(\frac{9}{9}\) (Divide by 9 on both sides)
⇒ a + 1 = 1
⇒ a + 1 – 1 = 1 – 1 (Subtract 1 on both sides)
⇒ a = 0

Check:
Substitute a = 0 in 9(a + 1) + 2 = 11
LHS = 9(a + 1) + 2
= 9(0 + 1) + 2
= 9(1) + 2
= 9 + 2 = 11 = RHS
Hence verified.

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Unit Exercise

Question 1.
Choose the correct answer.
(i) Which of the following numbers satisfy the equation – 6 + m = – 10 ?
(a) 2
(b) 4
(c) – 4
(d) – 2
Answer:
(c) – 4

Explaination:
– 6 + m = -10
⇒ – 6 + m + 6 = – 10 + 6
(Add 6 on both sides)
⇒ m = – 4

(ii) The equation having – 2 as solution is
(a) x + 2 = 5
(b) 7 + 3x = 1
(c) 2x + 3 = 7
(d) 2(x + 1) = 4
Answer:
(b) 7 + 3x = 1

Explaination
(a) x + 2 = 5
⇒ x + 2 – 2 = 5 – 2 (Subtract 2 on both sides)
⇒ x = 3 it is not – 2

(b) 7 + 3x = 1
⇒ 7 + 3x – 7 = 1 – 7 (Subtract 7. on both sides)
⇒ 3x = – 6
⇒ \(\frac{3 x}{3}\) = \(\frac{-6}{3}\) (Divide by 3 on both sides)
∴ x = – 2
so, solution is – 2
Similarly, check (c) and (d) also.

(iii) If a and b are positive integers, then the solution of the equation ax = b will always be a ____________ .
(a) positive number
(b) negative number
(c) 1
(d) 0
Answer:
(a) positive number

Explaination:
If a and b are positive integers and b < a, then ax ± b also positive integer.

(iv) The equation which can’t be solved in integers is
(a) 2(x – 3) = 10
(b) \(\frac{x}{3}\) = 5
(c) 5 – 3m = 1
(d) 2k + 1 = 1
Answer:
(c) 5 – 3m = 1

Explaination:
(a) 2(x – 3) = 10
⇒ 2x – 6 = 10 (Distributive property)
⇒ 2x – 6 + 6 = 10 + 6 (Add 6 on both sides)
⇒ 2x = 16
⇒ \(\frac{2x}{2}\) = \(\frac{16}{2}\) (Divide by 2 on both sides)
⇒ x = 8 is an integer.

(b) \(\frac{x}{3}\) = 5
⇒ \(\frac{x}{3}\) × 3 = 5 × 3 (Multiply by 3 on both sides)
⇒ x = 15 is an integer.

(c) 5 – 3m = 1
⇒ 5 – 3m – 5 = 1 – 5 (Subtract 5 on both sides)
⇒ – 3m = – 4
⇒ \(\frac{-3 m}{-3}\) = \(\frac{-4}{-3}\)
⇒ m = \(\frac{4}{3}\) is not an integer.

(d) 2k + 1 = 1
⇒ 2k + 1 – 1 = 1 – 1 (Subtract 1 on both sides)
⇒ 2k = 0
⇒ \(\frac{2 k}{2}\) = \(\frac{0}{2}\) (Divide by 2 on both sides)
⇒ k = 0 is an integer
So, our answer is c.

(v) Which of the following is not allowed in a given equation?
(a) Adding the same number to both sides of the equation.
(b) Subtracting the same number from both sides of the equation.
(c) Multiplying both sides of the equation by the same non-zero number.
(d) Dividing both sides of the equation by the same number.
Ans.
(d) Dividing both sides of the equation by the same number.

Question 2.
Fill in the blanks.
(i) If 2y – 1 = 5, then value of 5y + 3 is ____________
Answer:
Given 2y – 1 = 5
⇒ 2y – 1 + 1 = 5 + 1 (Add 1 on both sides)
⇒ 2y = 6
⇒ \(\frac{2 y}{2}\) = \(\frac{6}{2}\) (Divide by 2 on both sides)
∴ y = 3
then 5y + 3 = 5(3) + 3
= 15 + 3 = 18

(ii) Changing of term from one side of equation to other side is called __________
Answer:
Transposition.

(iii) If the sum of two numbers is 60. One is thrice the other, then the equation formed is ___________.
Answer:
Let one number be x.
Other number = thrice of first number = 3(x) = 3x
Sum of two numbers = 60
⇒ x + 3x = 60
⇒ 4x = 60

(iv) If ‘x’ is a natural number, then the solution of x – 8 = – 8 is _________ .
Answer:
x – 8 = – 8
x – 8 + 8 = – 8 + 8 (Add 8 on both sides)
x = 0 is not a natural number.
So, given equation has no solution.

(v) 13 subtracted from twice of a number gives 3, then the number
Answer:
Let the number be x.
Twice the number = 2x
13 is subtracted from twice the number
⇒ 2x- 13 = 3
⇒ 2x – 13 + 13 = 3 + 13 (Add 13 on both sides)
⇒ 2x = 16
⇒ \(\frac{2 x}{2}\) = \(\frac{16}{2}\) (Divide 2 on both sides)
⇒ x = 8
∴ Number is 8.

Question 3.
Check whether the value given in the brackets is a solution to the given equation or not.
(a) 2n + 5 = 19 (n = 7)
Answer:
Given 2n + 5 = 19
Substitute n = 7 in the given equation
2(7) + 5 = 19
⇒ 14 + 5 = 19
⇒ 19 = 19
∴ LHS = RHS
So, n = 7 is the solution of the given equation.

(b) \(\frac{3 m}{5}\) – 7 = 1 (m = 10)
Answer:
Given \(\frac{3 m}{5}\) – 7 = 1
Substitute m = 10 in the given equation
\(\frac{3(10)}{5}\) – 7 = 1
⇒ 3 × 2 – 7 = 1
⇒ 6 – 7 = 1
∴ – 1 = 1
LHS ≠ RHS
So, m = 10 is not the solution of given equation.

Question 4.
Solve 5 – 2k = – 3 using trial and error method.
Answer:

For k = 4, LHS = RHS
So, k = 4 is the solution of given equation.

Question 5.
Write the following equations in mathe-matical statement form.
(a) 2m + 7 = 21
Answer:
7 more than twice the m is 21.

(b) \(\frac{n}{7}\) = 4
Answer:
One seventh of n is 4.

Question 6.
Give the steps you will use to separate the variables and then solve the equation.
(a) 7(x – 3) = 28
Answer:
Given 7(x – 3) = 28
⇒ 7x – 21 = 28 (Distributive property)
⇒ 7x – 21 + 21 = 28 + 21 (Add 21 on both sides)
⇒ 7x = 49
⇒ \(\frac{7 x}{7}\) = \(\frac{49}{7}\) (Divide by 7 on both sides)
⇒ x = 7

(b) 8y – 9 = 15
Answer:
Given 8y – 9 = 15
⇒ 8y – 9 + 9 = 15 + 9 (Add 9 on both sides)
⇒ 8y = 24
⇒ \(\frac{8 y}{8}\) = \(\frac{24}{8}\) (Divide by 8 on both sides)
⇒ y = 3

Question 7.
Solve the following equations and check the result (Method of Transposition)
(a) 9(a + 3) + 7 = 22
Given 9(a + 3) + 7 = 22
⇒ 9(a + 3) + 7 – 7 = 22 – 7 (Subtract 7 on both sides)
⇒ 9(a + 3) = 15
⇒ 9a + 27 = 15 (Distributive property)
⇒ 9a + 27 – 27 = 15 – 27 (Subtract 27 on both sides)
⇒ 9a = – 12
⇒ \(\frac{9 a}{9}\) = \(\frac{-12}{9}\) (Divide by 9 on both sides0
⇒ a = \(\frac{-4}{3}\)

Check:
Substitute a = \(\frac{-4}{3}\) in the given equation.
LHS = 9(a + 3) + 7
= \(9\left(\frac{-4}{3}+\frac{3}{1}\right)\) + 7
= \(9\left(\frac{-4+9}{3}\right)\) + 7

= 15 + 7 = 22 = RHS
Hence verified.

(b) 25 = 18 – 7(b – 6)
Answer:
Given 25 = 18 – 7(b – 6)
⇒ 25 – 18 = l8 – 7(b – 6) – 18 (Subtract 18 on both sides)
⇒ 7 = – 7(b – 6)
⇒ \(\frac{7}{-7}\) = \(\frac{-7(b-6)}{-7}\) (Divide by – 7 on both sides)
⇒ – 1 = b – 6
⇒ – 1 + 6 = b – 6 + 6 (Add 6 on both sides)
⇒ 5 = b
∴ b = 5 (By transposition)

Check: Substitute b = 5 in the
given equation.
RHS = 18- 7(b – 6) – = 18-7(5-6)
= 18 – 7(- 1)
= 18 + 7 = 25 = LHS
Hence verified.

Question 8.
Six times a number is 72. Find the number.
Answer:
Let the number be x.
Given, six times of number = 72
⇒ 6x = 72
⇒ \(\frac{6 x}{6}\) = \(\frac{72}{6}\)
⇒ x = 12
∴ The number is 12.

Question 9.
Three-fourth of a number is more than one-fourth of same number by 2. Find the number.
Answer:
Let the number be x.
Three fourth of number = \(\frac{3}{4}\) of x = \(\frac{3 x}{4}\)
One-fourth of number = \(\frac{1}{4}\) of x = \(\frac{x}{4}\)
Three-fourth of number
= One-fourth of number + 2
⇒ \(\frac{3 x}{4}=\frac{x}{4}+\frac{2}{1}\)
⇒ \(\frac{3 x}{4}=\frac{x+8}{4}\)
⇒ \(\frac{3 x}{4} \times 4=\frac{x+8}{4} \times 4\) (Multiply by 4 on both sides)
⇒ 3x = x + 8
⇒ 3x – x = x + 8 – x (Subtract x on both sides)
⇒ 2x = 8
⇒ \(\frac{2 x}{2}\) = \(\frac{8}{2}\) (Divide by 2 on both sides)
⇒ x = 4

Question 10.
In the given figure the perimeter of the square is 40m. Then find x.

Answer:
From the above figure,
Side of square = (3x – 5) m
Perimeter of square = 40 m
⇒4 × side = 40
⇒ 4 × (3x – 5) = 40
⇒ \(\frac{4(3 x-5)}{4}\) = \(\frac{40}{4}\) (Divide by 4 on both sides)
⇒ 3x – 5 = 10
⇒ 3x – 5 + 5 = 10 + 5 (Add 5 on both sides)
⇒ 3x = 15
⇒ \(\frac{3 x}{3}\) = \(\frac{15}{3}\) (Divide by 3 on both sides)
∴ x = 5 m

Question 11.
Jeevan is 3 years younger than his brother Sasi. If the sum of their present ages is 19 years. What are their present ages ?
Answer:
Let the present age of Sasi is x years.
The present age of Jeevan = (x – 3) years
Sum of their present ages = 19
⇒x + x – 3 = 19
⇒ 2x – 3 = 19
⇒ 2x – 3 + 3 = 19 + 3 (Add 3 on both sides)
⇒ 2x = 22
⇒ \(\frac{2 x}{2}\) = \(\frac{22}{2}\) (Divide by 2 on both sides)
⇒ x = 11 years
∴ Present age of Sasi x = 11 years
Present age of Jeevan = x – 3
= 11 – 3 = 8 years

Question 12.
The length of a rectangle is 20m more than its width. If the perimeter of the rectangle is 100m, then find the length and breadth of rectangle.
Answer:
Let the width of a rectangle (b) = x m
Then the length of a rectangle (l)
= 20 more than its width
= (x + 20) m
Perimeter of the rectangle = 100 m
⇒ 2(1 + b) = 100
⇒ 2(x + 20 + x) = 100
⇒ 2(2x + 20) = 100
⇒ 4x + 40 = 100
⇒ 4x + 40 – 40 = 100 – 40 (Subtract 40 on both sides)
⇒ 4x = 60 4x 60
⇒ \(\frac{4 x}{4}\) = \(\frac{60}{2}\) (Divide by 4 on both sides)
⇒ x = 15 m
∴ Width x = 15 m
Length = x + 20 = 15 + 20 = 35 m

Question 13.
In a family, the consumption of rice is 4 times that of wheat. The total consumption of the two cereals in a month is 30kg. Find the quantities of rice and wheat consumed in the family.
Answer:
Let the quantity of wheat consumed in the month x kg. .
Quantity of rice = 4 times of wheat = 4x kg
Quantity of rice + Quantity of wheat = 30 kg
⇒ 4x + x = 30
⇒ 5x = 30
⇒ \(\frac{5 x}{5}\) = \(\frac{30}{5}\) (Divide by 5 on both sides)
⇒ x = 6 kg
∴ Quantity of wheat = 6 kg
Quantity of rice = 4x = 4 × 6 = 24 kg

Question 14.
The teacher tells the students in the class that the highest marks obtained by a student in the class is 7 more than twice the lowest marks. If the highest mark is 93, then what is the lowest mark ?
Answer:
Let the lowest mark = x then the highest mark
twice the lowest mark + 7 = 93
⇒ 2x + 7 = 93
⇒ 2x + 7 – 7 = 93 – 7 (Subtract 7 on both side)
⇒ 2x = 86
⇒ \(\frac{2 x}{2}\) = \(\frac{86}{2}\) (Divide by 2 on both sides)
⇒ x = 43
∴ Lowest mark = 43

Question 15.
A man travelled \(\frac{4}{5}\) of his journey by train, \(\frac{1}{7}\) by bus and the remaining 16 km by auto. What is the length of his total journey ?
Answer:
Let the length of total journey = x km
Travelled by train = \(\frac{4}{5}\) of x = \(\frac{4x}{5}\) km
Travelled by bus = \(\frac{1}{7}\) of x = \(\frac{x}{7}\) km
Travelled by auto =16 km
Total length of journey = \(\frac{4x}{5}\) + \(\frac{x}{7}\) + 16
⇒ x = \(\frac{4 x}{5}+\frac{x}{7}+\frac{16}{1}\)
⇒ x = \(\frac{28 x+5 x+16 \times 35}{35}\)
⇒ x = \(\frac{33 x+560}{35}\)

(Multiply 35 on both sides)
⇒ 35x = 33x + 560
⇒ 35x – 33x = 33x + 560 – 33x (Subtract 33x on both sides)
⇒ 2x = 560
⇒ \(\frac{2 x}{2}\) = \(\frac{560}{2}\) (Divide by 2 on both sides)
⇒ x = 280 km
Therefore, length of total journey = 280 km

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations InText Questions

Check your Progress (Page No: 43)

Question 1.
Write simple equations for the following verbal statements.
(i) The sum of five times of x and 3 is 28.
Answer:
Given number = x
Five times of a number = 5 ∙ x
By adding 3 the result = 5x + 3
∴ 5x + 3 = 28

(ii) Taking away 7 from p gives 21.
Answer:
Given number = p
By subtracting 7 from the number = p – 7
∴ p – 7 = 21

(iii) If you add one third to m, then you get 25.
Answer:
Given number = m
By adding \(\frac{1}{3}\) to m = m + \(\frac{1}{3}\)
then, m + \(\frac{1}{3}\) = 25.

(iv) Sum of angles x, (x + 20) is a straight angle.
Answer:
Given angles are x, x + 20
Sum of angles = x + x + 20
Sum of angles is straight angle (180°).
⇒ x + x + 20 = 180°
∴ 2x + 20 = 180°

(v) Perimeter of a rectangle whose length is 2 cm more than its width is 16 cm.
Answer:
Let width of rectangle = x
Length of rectangle
= 2 more than width = x + 2
Perimeter = 2 (length + width)
= 2(x + 2 + x) = 2(2x + 2)
Given Perimeter = 16 cm
∴ 4x + 4 = 16

Question 2.
Write the following equations in state-ment form.
(i) x + 4 = 9
Answer:
4 is added to a number is 9.

(ii) 2y = 15
Answer:
Two times (twice) a number is 15.

(iii) 3m – 13 = 25
Answer:
13 is taking away from 3 times of m is 25.

(iv) \(\frac{n}{4}\) = 5
Answer:
One fourth of n is 5.

Let’s Do Activity (Page No: 45)

Describe the balances given below in two ways.
(a) Using words (b) Using mathematical symbols.
Answer:

Lets Explore (Page No: 46)

Develop equations from the pictures given below and solve pictorially.

(i)

Answer:
x + x + x = 5 + 5 + 1 + 1
3x = 12

(ii)

Answer:
y + y + 1 = 5 + 5 + 5
2y + 1 = 15

Let’s Think (Page No: 47)

An equation is multiplied or divided by two different numbers on either side What will happen to the equality?
Answer:
Consider an equation 7x = 14
If we multiply by two different numbers 5 and 8 on either side.
7x × 5 = 14 × 8 ⇒ 35x = 112
So, the equation 7x = 14 is not equal to 35x = 112

(ii) Consider an equation 3x = 15
If we divide by two different numbers 3 and 5 on either side,
3x ÷ 3 = 15 ÷ 5
\(\frac{3 x}{3}\) = \(\frac{15}{5}\) ⇒ x = 3
So, the equation 3x = 15 is not equal to x = 3 either side, the equality will be changes.

Puzzle Time (Page No: 48)

Question 1.
Solve the puzzle using simple equations.
(i)

Answer:

(ii)

Answer:

(iii)

Answer:

(iv)

Answer:

Check Your Progress (Page No: 51)

Find three equivalent equations having the same solutions x = – 1.
Answer:

Here x + 5 = 4, 8 – x = 9 and 4x + 5 = 1 are equivalent equations because they have the same solution x = – 1

Puzzle Time (Page No. 52)

Question 1.
Here is a puzzle to reveal the name of great mathematician. Solve the equations and fill the below box with the letter opposite to the solution of the equation and finally it reveals the name of mathematician.

Answer:

Check Your Progress (Page No. 53)

Find the value of variable x from the below diagrams.

(i)

Angle at ‘O’ is straight angle
Answer:
Straight angle = 180°
∠ AOB = 180°
∠ AOB + ∠ BOD = 180°
x + 30° + x = 180°
⇒ 2x + 30° = 180°
⇒ 2x + 30° – 30° – 180°- 30° (Subtract 30° on both sides)
⇒ 2x = 150°
⇒ \(\frac{2 x}{2}\) = \(\frac{150^{\circ}}{2}\) (Divide by 2 on both sides)
∴ x = 75°

(ii)

Perimeter = 36 cm
Answer:
Perimeter of rectangle = 36 cm
2(length + breadth) = 36
⇒ 2(x + 4 + x) = 36
⇒ 2(2x + 4) = 36
⇒ \(\frac{2(2 x+4)}{2}\) = \(\frac{36}{2}\) (Divide by 2 on both sides)
⇒ 2x + 4 = 18
⇒ 2x + 4 – 4 = 18 – 4 (Subtract 4 on both sides)
⇒ 2x = 14 .
⇒ \(\frac{2x}{2}\) = \(\frac{14}{2}\) (Divide by 2 on both sides)
∴ x = 7

Puzzle Time (Page No: 59)

Collect the minimum and maximum temperature (in degrees) of any 5 cities of India of any month and convert the temperature into Fahrenheit and Kelvin Scale.

Examples

Question 1.
Convert the mathematical statements into simple equations.
(i) 5 added to a number is 9.
Answer:
Let the number = x
By adding 5 to the number = 5 + x
∴ 5 + x = 9

(ii) 4 times a number decreased by 3 is 5.
Answer:
Let the number = m
4 times the number = 4m
By decreasing 3 the result = 4m – 3
∴ 4m – 3 = 5

(iii) The sum of 3 times of n and 7 is 13.
Answer:
Let the number = n
3 times the number = 3n
By adding 7 the result = 3n + 7
∴ 3n + 7 = 13

(iv) Length of rectangle is 3 m more than its breadth and its perimeter is 24 m.
Answer:
Breadth of rectangle = x
Length of rectangle = x + 3
Perimeter = 2(x + 3 + x) = 4x + 6
Given Perimeter = 24 m
∴ 4x + 6 = 24

Question 2.
Convert simple equations into statements.
(i) y – 7 = 11
Answer:
Taking away 7 from ‘y’ is 11.

(ii) 8m = 24
Answer:
8 times of number ‘m’ is 24.

(iii) 2x + 13 = 25
Answer:
If you add 13 to 2 times of number ‘x’ is 25.

(iv) \(\frac{y}{4}\) – 7 = 1
Answer:
7 is subtracted from one fourth of y is 1.

Question 3.
Check whether the value given in the brackets is a solution to the given equation or not.
2k – 11 = 5 (k = 7)
Answer:
When k = 7
LHS: 2k – 11 = 2(7) – 11 = 14 – 11 = 3
RHS: 5
Here LHS ≠ RHS,
So k = 7 is not a solution.

Question 4.
Solve 6n – 1 = 29 trial and error method.
Answer:

For n = 5, LHS = RHS. So, n = 5 is the solution of given equation.

Question 5.
Solve x + 5 = 8 by using common balance idea.
Answer:

Question 6.
Solve 2x – 5 = 9 pictorially.
Answer:

Question 7.
Solve: 3k + 4 = 28
Answer:
3k + 4 = 28
⇒ 3k + 4 – 4 = 28 – 4 (Add both sides ‘- 4’)
⇒ 3k = 24
⇒ \(\frac{3 \mathrm{k}}{3}\) = \(\frac{24}{3}\) (Divide both sides by ‘3’)
⇒ k = 8

Check: Substitute k = 8 in the given equation.
LHS = 3k + 4
= 3(8) + 4
= 24 + 4 = 28 = RHS
Hence verified.

Question 8.
Solve: – 4(x – 1) = 16
Answer:
– 4(x – 1) = 16
⇒ 4x + 4 = 16 (Distributive property)
⇒ – 4x + 4 – 4 = 16 – 4 (Subtract both sides ‘4’)
⇒ – 4x = 12
⇒ (- 4x) × (- 1) = 12 × (- 1)
(Multiply with ‘- 1 ‘on both sides)
⇒ 4x = – 12
⇒ \(\frac{4 x}{4}\) = \(\frac{-12}{4}\) (Divide both sides by ‘4’)
⇒ x = – 3

Check: Substitute x = – 3 in the given equation.
LHS = – 4(x – 1)
= – 4(- 3 – 1)
= – 4 (- 4)
= 16
= RHS
Hence verified.

Question 9.
Solve: 2(b + 3) + 13 = 27
Answer:
2(b + 3) + 13 = 27
⇒ 2b + 6 + 13 = 27 (distributive law)
⇒ 2b + 19 = 27
⇒ 2b = 27 – 19 (∵+ 19 transposed and becomes – 19)
⇒ 2b = 8
⇒ b = \(\frac{8}{2}\) (∵ × 2 transposed and becomes ÷ 2)
⇒ b = 4

Check:
Substitute b = 4
LHS = 2(b + 3) + 13
= 2(4 + 3) + 13
= 2(7) + 13
= 14 + 13
= 27 = RHS
Hence verified.

Question 10.
Solve : 5(x + 1) – 2(x – 7) = 13
Answer:
5(x + 1) – 2(x – 7) = 13
⇒ 5x + 5 – 2x + 14 = 13 (distributive law)
⇒ (5x – 2x) + (5 + 14) = 13 (regrouping like terms)
⇒ 3x +19 = 13
⇒ 3x = 13 – 19 (∵ +19 transposed arid becomes
⇒ 3x = – 6
⇒ x = \(\frac{-6}{3}\) (∵ × 3 transposed and becomes ÷ 3)
⇒ x = – 2

Question 11.
Solve: 12 = 13 + 7 (y-6)
Answer:
12 = 13 + 7 (y – 6)
An equation remains the same,
when LHS, RHS are interchanged.
⇒ 13 + 7(y – 6) = 12 .
⇒ 3 + 7y – 42 = 12 (distributive law)
⇒ 7y – 29 = 12
⇒ 7y = 12 + 29 (∵ – 29 transposed and becomes + 29)
⇒ 7y = 41
⇒ y = \(\frac{41}{7}\) (∵ × 7 transposed and becomes ÷ 7)

Check:
Substitute y = \(\frac{41}{7}\)
RHS = 13 + 7(y – 6)
= 13 + 7(\(\frac{41}{7}\) – 6)
= 13 + 7\(\left(\frac{41-42}{7}\right)\)
= 13 + 7\(\left(\frac{-1}{7}\right)\)
= 13 – 1
= 12 = LHS
Hence verified.

Question 12.
Solve \(\frac{m}{2}-\frac{2 m}{7}-\frac{m}{10}=\frac{8}{5}\)
Answer:

Check:
Substitute m = 14
LHS = \(\frac{\mathrm{m}}{2}-\frac{2 \mathrm{~m}}{7}-\frac{\mathrm{m}}{10}\)
= 7 – 4 – \(\frac{7}{5}\)
= 3 – \(\frac{7}{5}\)
= \(\frac{15-7}{5}\)
= \(\frac{8}{5}\)
= RHS
Hence verified.

Question 13.
Find the value of variable ‘x’ from the following.
(i) x + 20% of x = 60
Answer:
x + 20% of x = 60
⇒ x + \(\frac{20 x}{100}\) = 60
⇒ x + \(\frac{x}{5}\) = 60
⇒ \(\frac{6x}{5}\) = 60
⇒ 6x = 60 × 5
⇒ x = \(\frac{(60 \times 5)}{6}\)
⇒ x = 50

(ii) If two numbers are in the ratio 2 : 3 and their difference is 5, then find largest number.
Answer:
Let largest number = 3x and smallest number = 2x
Difference = 3x – 2x = x
But as per the problem difference = 5
⇒ x = 5
∴ Largest number = 3x = 3 × 5 = 15

(iii) Find the value of x in the following picture.

Answer:
Sum of two angles is 90
2x + 25 + x – 10 = 90
⇒ 3x + 15 = 90
⇒ 3x = 90 – 15
⇒ 3x = 75
⇒ x = \(\frac{75}{3}\)
⇒ x = 25

(iv) If the perimeter of the following tri-angle is 15 cm, then find the value of x.

Answer:
Perimeter of a triangle = 15 cm
Perimeter = x + x + 2 + 2x – 3 = 4x – 1
∴ 4x – 1 = 15
⇒ 4x = 15 + 1
⇒ 4x = 16
⇒ x = 4

Question 14.
Find two consecutive natural numbers whose sum is 125.
Answer:
Let the number = x
Consecutive number of x = x + 1
Given that, x + (x + 1) = 125
⇒ 2x + 1 = 125
⇒ 2x = 125 – 1
⇒ 2x = 124
⇒ x = \(\frac{124}{2}\)
x = 62 and x + 1 = 62 + 1 = 63
∴ Required numbers are 62 and 63.

Check:
Numbers: 62, 63
Sum = 62 + 63 = 125
Hence verified.

Question 15.
The sum of two numbers is 35. One of the numbers exceeds the other by 7. Find the numbers.
Answer:
Let the first number = x
Then the second number = x + 7 (exceeds first number by 7)
Sum of two numbers = 35
According to problem, x + x + 7 = 35
⇒ 2x + 7 = 35
⇒ 2x = 35 – 7
⇒ 2x = 28
⇒ x = \(\frac{28}{2}\)
⇒ x = 14
x + 7 = 14 + 7 = 21
∴ The two numbers are 14 and 21

Check:
Numbers: 14, 21
Sum = 14 + 21
= 35
Hence verified.

Question 16.
A person has ₹ 1400 in denominations of ₹20, ₹10 and ₹5 notes. The number of notes in each denomination is equal, then find the number of notes of each denomination ?

Answer:
Let the number of notes in each denomination be x.
₹20 denomination amount = 20x
₹10 denomination amount = 10x
₹5 denomination amount = 5x
Total amount = ₹ 1400
⇒ 20x + 10x + 5x = 1400
⇒ 35x = 1400
⇒ x = \(\frac{1400}{35}\)
⇒ x = 40
∴ Number of notes in each denomination is 40.

Check:
Number of notes = 40
Amount = (20 × 40 + 10 × 40 + 5 × 40)
= 800 + 400 + 200
= ₹ 1400
Hence Verified.

Question 17.
The length of a rectangle is five meters more than twice of its breadth. If the perimeter is 148 meter, then find the length and breadth of the rectangle.

Answer:
Let the breadth of the rectangle = x,
length of a rectangle is five more than twice of its breadth.
Then, length of the rectangle = 2x + 5
Perimeter of the rectangle = 148m
2 (Length + Breadth) = 148
⇒ 2(2x + 5 + x) = 148
⇒ 2(3x + 5) = 148
⇒ 6x + 10 = 148
⇒ 6x = 148 – 10
⇒ 6x = 138
⇒ x = \(\frac{138}{6}\)
⇒ x = 23
∴ Breadth of rectangle = 23 m
Length of the rectangle = 2x + 5 = (2 × 23) + 5
= 46 + 5 = 51 m

Check:
Length = 51 m, Breadth = 23 m
Perimeter = 2(l + b)
= 2(51 + 23)
= 2(74)
= 148 m
Hence Verified.

Question 18.
On his birthday Yakshith’s grand father has given ₹ 2000. He used some amount for purchasing books for needy children and thrice of that amount for purchasing food items for orphanage children and the remaining ₹ 200 used for purchasing chocolates . for his friends. Find the amount spent for purchasing books and food items for orphanage children.
Answer:
Let the amount spent for purchasing books for needy children = ₹ x
Amount spent for purchasing food items for orphanage children = ₹ 3x
Amount spent for purchasing chocolates for friends = ₹ 200
Total amount spent = ₹ 2000

⇒ x + 3x + 200 = 2000
⇒ 4x + 200 = 2000
⇒ 4x = 2000 – 200
⇒ 4x = 1800
⇒ x = \(\frac{1800}{4}\)
⇒ x = 450
∴ Amount spent for purchasing books for needy children = ₹ 450
Amount spent for purchasing food items for orphanage children = ₹ 3x
= 3 × 450 = ₹ 1350

Check:
Amount = ₹ 450 + ₹ 1350 + ₹ 200
= ₹ 2000
Hence Verified.

Question 19.
A school bus starts with full strength of 40 students. It drops some students at the first bus stop. At the second bus stop, twice the number of students get down from the bus. 8 students get down at the third bus stop and the number of students remaining in the bus is only 5. How many students got down at the first stop and second stop?

Answer:
Let us take the number of students get down at first stop = x
No.of students get down at the second bus stop = 2x
No. of students get down at the third bus stop = 8
Remaining students in the bus = 5
⇒ x + 2x +8 + 5 = 40
⇒ 3x + 13 = 40
⇒ 3x = 40 – 13
⇒ 3x = 27
⇒ x = \(\frac{27}{3}\)
⇒ x = 9
∴ The number of students got down in the first bus stop = 9.
No. of students got down in second stop = 2x = 2 × 9 = 18.

Practice Questions (Page No: 62)

In the questions given below, equations have become wrong due to wrong order of signs (+, -, ×, ÷, =). Choose the correct order of signs from the alternatives given under each question so that the equation becomes right.

Question 1.
7 + 2 = 2 × 3
(a) = × +
(b) = + ×
(c) = + ÷
(d) + × =
Answer:
(a) = × +

Explaination
Given 7 + 2 = 2 × 3 ,
⇒ 7 = 2 × 2 + 3
⇒ 7 = 4 + 3
∴ 7 = 7

Question 2.
7 + 2 × 6 = 20
(a) = × +
(b) × – =
(c) × + =
(d) ÷ + =
Answer:
(c) × + =

Explaination:
Given 7 + 2 × 6 = 20
⇒ 7 × 2 + 6 = 20
⇒ 14 + 6 = 20
∴ 20 = 20

Question 3.
15 ÷ 5 = 2 × 1
(a) ÷ × =
(b) ÷ = ×
(c) × = +
(d) ÷ = +
Answer:
(d) ÷ = +

Explaination:
Given 15 ÷ 5 = 2 × 1
⇒ 15 ÷ 5 = 2 + 1
∴ 3 = 3

Question 4.
6 = 3 – 6 ÷ 12
(a) = × +
(b) ÷ = ×
(c) + = –
(d) ÷ × =
Sol.
(d) ÷ × =

Explaination:
Given 6 = 3 – 6 ÷ 12
⇒ 6 ÷ 3 × 6 =12
⇒ 2 × 6 = 12
∴ 12 = 12

Question 5.
3 + 1 ÷ 4 = 16
(a) – = ×
(b) × + =
(c) + × =
(d) = × +
Answer:
(c) + × =

Explaination:
Given 3 + 1 ÷ 4 = 16
⇒3 + 1 × 4 = 16
⇒ 4 × 4 = 16
∴ 16 = 16

Question 6.
8 ÷ 4 = 2 + 1
(a) ÷ = +
(b) ÷ = ×
(c) ÷ × =
(d) = ÷ ×
Answer:
(b) ÷ = ×

Explaination:
Given 8 ÷ 4 = 2 + 1
⇒ 8 ÷ 4 = 2 × 1
∴ 2 = 2

Question 7.
2 × 2 + 2 = 2
(a) × ÷ =
(b) × = ÷
(c) + × =
(d) × + =
Answer:
(a) × ÷ =

Explaination:
Given 2 × 2 + 2 = 2
⇒ 2 × 2 ÷ 2 = 2
⇒ 2 × 1 = 2
∴ 2 = 2

Question 8.
5 – 6 + 8 = 3
(a) + – =
(b) + = –
(c) – = ×
(d) ÷ × =
Answer:
(a) + – =

Explaination:
Given 5 – 6 + 8 = 3
⇒ 5 + 6 – 8 = 3
⇒ 11 – 8 = 3
∴ 3 = 3

Question 9.
8 ÷ 2 = 2 × 8
(a) + – =
(b) + ×
(c) ÷ = ×
(d) × = ÷
Answer:
(b) + ×

Explaination:
Given 8 ÷ 2 = 2 × 8
⇒ 8 ÷ 2 × 2 = 8
⇒ 4 × 2 – 8
∴8 = 8

Question 10.
3 = 3 – 7 + 0
(a) – + =
(b) + × =
(c) – × =
(d) = × –
Answer:
(c) – × =

Explaination:
Given 3 = 3 – 7 + 0
⇒ 3 – 3 × 7 = 0
⇒ 0 × 7 = 0
∴ 0 = 0

SCERT AP 7th Class Maths Solutions Pdf Chapter 4 Lines and Angles InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 4th Lesson Lines and Angles InText Questions

Check Your Progress [Page No. 66]

Question 1.
Find the complementary angles of
(i) 27°
Answer:
If the sum of any two angles is 90°, then the angles are called as complementary angles.
Complementary angle of 27° is
(90 – 27) = 63°

(ii) 43°
Answer:
Complementary angle of 43° is
(90 – 43) = 47°

(iii) k°
Answer:
Complementary angle of k° is (90 – k)°

(iv) 2°
Answer:
Complementary angle of 2° is
(90 – 2) = 88°

Question 2.
Find the supplementary angles of
(i) 13°
Answer:
If the sum of any two angles is 180°, then the angles are called as supplementary angles.
Supplementary angle of 13° is
(180 – 13) = 167°

(ii) 97°
Answer:
Supplementary angle of 97° is
(180 – 97) = 83°

(iii) a°
Answer:
Supplementary angle of a° is
(180 – a)°

(iv) 46°
Answer:
Supplementary angle of 46° is
(180 – 46) = 134°

Question 3.
Find the conjugate angles of
(i) 74°
Answer:
If the sum of any two angles is 360°, then the angles are called as conjugate angles.
Conjugate angle of 74° is
(360 – 74) = 286°

(ii) 180°
Answer:
Conjugate angle of 180° is
(360- 180) = 180°

(iii) m°
Answer:
Conjugate angle of m° is (360 – m)°

(iv) 300°
Answer:
Conjugate angle of 300° is
(360.-300) = 60°

[Page No. 66]

Question 1.
Umesh said, “Two acute angles cannot form a pair of supplementary angles.” Do you agree ? Give reason.
Answer:
Yes, acute angle is always less than 90°. So, sum of two acute angles is always less than 180°.
Therefore, two acute angles cannot form a pair of supplementary angles (180°).

Question 2.
Lokesh said, “Each angle in any pair of complementary angles is always acute.” Do you agree? Justify your answer.
Answer:
Yes, sum of any two acute angles is 90°, then they are complementary angles. If they are not acute means they may . be right angle (90°) or obtuse angle (> 90°) or etc.

So, its impossible.
Therefore, each angle in any pair of complementary angles is always acute.

Let’s Think [Page No. 69]

Question 1.
In the figure, ∠AOB and ∠BPC are not adjacent angles. Why? Give reason.

Answer:
In the given figure, ∠AOB and ∠BPC are not adjacent angles. Because, they have no common vertex and no common arm.

Question 2.
In the figure, ∠AOB and ∠COD have common vertex O. But ∠AOB, ∠COD are not adjacent angles. Why? Give reason.

Answer:
In the given figure, ∠AOB and ∠COD are not adjacent angles. Because, they have common vertex. But they have no common arm.

Question 3.
In the figure, ∠POQ and ∠POR have common vertex O and common arm OP but ∠POQ and ∠POR are not adjacent angles. Why? Give reason.

Answer:
∠POQ and ∠POR have common vertex O and common arm OP. But they are not lie either side of the common arm. That’s why they are not adjacent angles.

Check Your Progress [Page No. 70]

Question 1.
In the adjacent figure \(\overrightarrow{\mathbf{P R}}\) is a straight line and O is a point on the line. \(\overrightarrow{\mathbf{O Q}}\) is a ray.

(i) If ∠QOR= 50°, then what is ∠POQ?
Answer:
Given ∠QOR= 50°
∠POQ, ∠QOR are linear pair.
⇒ ∠POQ + ∠QOR = 180°
⇒ ∠POQ + 50° = 180°
⇒ ∠POQ — 180° – 50° = 130°
.-. ∠POQ =130°

(ii) If ∠QOP = 102°, then what is ∠QOR?
. Sol. Given ∠QOP = 102°
∠QOP and ∠QOR are linear pair.
⇒ ∠QOP + ∠QOR = 180°
⇒ 102° + ∠QOR = 180°
⇒ 102°- 102° + ∠QOR = 180°- 102°
⇒ ∠QOR = 78°

Let’s Explore [Page No. 70]

Question 1.
A linear pair of angles must be adja-cent but adjacent angles need not be linear pair. Do you agree? Draw a figure to support your answer.
Answer:
Yes.

Question 2.
Mahesh said that the sum of two angles 30° and 150° is 180°, hence they are linear pair. Do you agree? Justify your answer.
Answer:
No, sum of two angles is 180°, then they are said to be supplementary angles.
If the two angles are on the same straight line and they are adjacent they are said to be linear pair.
So, the two angles 30° and 150° are need not be linear pair.

Let’s Think [Page No. 70]

Question 1.
In the adjacent figure, AB is a straight line, O is a point on AB. OC is a ray. Take a point D in the interior of ∠AOC, join OD.
Find ∠AOD + ∠DOC + ∠COB.

Answer:
Given ∠AOC and ∠COB are linear pair. But ∠AOC = ∠AOD + ∠DOC

⇒ ∠AOC + ∠COB = 180° (linear pair)
⇒ ∠AOD + ∠DOC + ZCOB = 180°

Question 2.
In the given figure, AG is a straight line. Find the value of ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6.

Answer:
Given ∠AOC and ∠COG are linear pair. ∠AOC + ∠COG = 180° (linear pair)
But ∠AOC = ∠AOB + ∠BOC
= ∠1 + ∠2
∠COG – ∠COD + ∠DOE + ∠EOF + ∠FOG – ∠3 + ∠4 + ∠5 + ∠6
=> (∠AOB + ∠BOC) + (∠COD + ∠DOE + ∠EOF + ∠FOG)
= 180°
=> ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6
= 180°
Therefore, the sum of angles at a point on the same side of the line is 180°.

Let’s DO Activity [Page No. 72]

Take a white paper. Draw 3 distinct pairs of intersecting lines on this paper. Measure the angles so formed and fill the table.

From the above table, we observe that “vertically opposite angles are equal.

Check Your Progress [Page No. 73]

In the figure three lines p, q and r inter-sect at a point O. Observe the angles in the figure. Write answers to the following.
Question 1.
What is the vertically opposite angle to ∠1?
Answer:
Vertically opposite angle to ∠1 is ∠4.

Question 2.
What is the vertically opposite angle to ∠6?
Answer:
Vertically opposite angle to ∠6 is ∠3.

Question 3.
If ∠2 = 50°, then what is ∠5?
Answer:
Vertically opposite angle of ∠2 is ∠5. So, ∠5 = ∠2 = 50° ∠5 = 50°

Let’s Think [Page No. 75]

Question 1.
In the figure, the line l intersects other two lines m and n at A and B respectively. Hence l is a transversal. Is there

Give reason.
Answer:
Yes.
1) The line m intersects other two lines / and n at two distinct points A and C respectively. Hence m is a transversal line.
2) The line n intersects other two lines / and m at two distinct points B and C respectively. Hence n is a transversal line.

Question 2.
How many transversals can be drawn for the given pair of lines?
Answer:
One and only one transversal line can be drawn for the given pair of lines.

Check Your Progress [Page No. 76]

Observe the figures (i) and (ii) then fill the table.


Answer:

Check Your Progress [Page No. 78]

In the figure, p ∥ q and t is a transversal. Observe the angles formed.

Question 1.
If ∠1 = 100°, then what is ∠5?
Answer:
In the given figure ∠5 = Z1 (corresponding angles)
Given ∠1 = 100°
So, ∠5 = ∠1 = 100°
∴ ∠5 – 100°

Question 2.
If ∠8 = 80°, then what is ∠4?
Answer:
In the given figure ∠4 = ∠8 (corresponding angles)
Given ∠8 = 80°
So, ∠4 — ∠8 = 80°
∴ ∠4 = 80°

Question 3.
If ∠3 = 145°, then what is ∠7?
Answer:
In the given figure ∠7 = ∠3 (corresponding angles)
Given ∠3 = 145°
So, ∠7 = ∠3 = 145°
∴ ∠7 – 145°

Question 4.
If ∠6 = 30°, then what is ∠2?
Answer:
In the given figure ∠2 = ∠6 (corresponding angles)

Given ∠6 = 30°
So, ∠2 = ∠6 = 30°
∴ ∠2 = 30°

Let’s Think [Page No. 78]

Question 1.
What is the relation between alternate exterior angles formed by a transversal on parallel lines?
Answer:

Alternate exterior angles are equal. That is ∠1 = ∠7 and ∠2 = ∠8

Check Your Progress [Page No. 80]

In the figure, m∥ n and l is a transversal.

Question 1.
If ∠3 = 116°, then what is ∠5?
Answer:
In the figure ∠5 = ∠3 (Alternate interior angles)
Given ∠3 = 116°
So, ∠5 = ∠3 = 116°
∴ ∠5 = 116°

Question 2.
If ∠4 = 51°, then what is ∠5?
Answer:
In the figure the interior angles in the same side of transversal are supple-mentary’.
So, ∠5 + ∠4 = 180° we know ∠4 = 51°
∠5 + 51° = 180°
∠5 + 51° – 51° = 180° – 51°
=> ∠5 = 129°

Question 3.
If ∠1 = 123° then what is ∠7?
Answer:
In the given figure
∠7 = ∠1 (Alternative exterior angles) Given ∠1 = 123°
So, ∠7 = ∠1 = 123°
∴ ∠7 = 123° .

Question 4.
If ∠2 = 66° then what is ∠7?
Answer:
In the given figure,
sum of the exterior angles are the same side of transversal are supplementary. So, ∠2 + ∠7 = 180°
=> 66° + ∠7 = 180° (Given ∠2 = 66°)
=> 66° + ∠7 – 66° = 180° – 66°
=> ∠7 = 114°
∴ ∠7 = 114°

Let’s Think [Page No. 80]

Question 1.
What is the relation between co-exterior angles, when a transversal cuts a pair of parallel lines?

Answer:
Co-exterior angles are supplementary. That is ∠2 + ∠7 = 180° and ∠1 + ∠8 = 180°

Let’s Do Activity [Page N0. 80]

Take a white paper and draw a pair of non-parallel lines p and q and a transversal shown in the fIgure 1. Measure the corresponding angles and fill the table. Measure the pair óf corresponding angles and fill the table.

Check Your Progress [Page No. 81].

From the figure, state which property that is used in each of the following.

Question 1.
If ∠3 = ∠5, then p ∥ q.
Answer:
Given ∠3 = ∠5
Alternative interior angles are equal.

Question 2.
If ∠3 + ∠6 = 180°, then p ∥ q.
Answer:
Given ∠3 + ∠6 = 180°
Interior angles on the same Side of transversal are supplementary.

Question 3.
If ∠3 = ∠8, then p∥q.
Answer:
Given ∠3 = = ∠8 .
∠3 and ∠8 are not corresponding angles and not alternate interior angles.
So, ∠3 ≠ ∠8.

Let’s Explore [Page No. 81]

Question 1.
When a transversal intersects two lines and a pair of alternate exterior angles are equal, what can you say about the two lines?
Answer:
If a pair of alternate exterior angles are equal, then the two lines are parallel to each other.

Question 2.
When a transversal intersects two lines and a pair of co-exterior angles are supplementary, what can you say about the two-lines?
Answer:
If a pair of co-exterior angles are supplementary’ then the two lines are parallel to each other.

Examples:

Question 1.
In the given figure, ∠B and ∠E are complementary angles. Find the value of x.

Answer:
From the figure,
∠B = x + 10°and ∠E =35°
Since ∠B and ∠E are complementary angles,
∠B + ∠E =90°
⇒ x + 10° + 35° = 90°
⇒ x + 45° = 90°
⇒ x = 90°- 45°
x = 45°

Question 2.
If the ratio of supplementary angles is 4 : 5, then find the two angles.
Answer:
Given ratio of supplementary angles = 4:5
Sum of the parts in the ratio = 4 + 5 = 9
Sum of the supplementary angles = 180°
First angle = \(\frac{4}{9}\) × 180° = 80°
Second angle = \(\frac{5}{9}\) × 180 °= 100°

Question 3.
Find the linear pair of angles which are equal to each other?
Answer:
Let the equal linear pair of angles are x° and x°.
⇒ x° + x° = 180° .
⇒ 2x° = 180°
⇒ x° = \(\frac{180^{\circ}}{2}\)
∴ x° = 90°
Hence, each angle = 90°

Question 4.
In the given figure, PS is a straight line, find x°.

Answer:
From the given figure, ∠POQ = 60°
∠QOR = x°
∠ROS = 47°
But ∠POQ + ∠QOR + ∠ROS = 180°
⇒ 60° + x° + 47° = 180°
⇒ x° + 107° = 180°
⇒ x° = 180°- 107°
∴ x° = 73°

Question 5.
Observe the figure, then find x, y and z.

Answer:
From the figure, x = 110° (vertically opposite angles are equal)
y + 110° = 180°
y = 180°- 110° = 70°
z = y ⇒ z = 70°
Hence x = 110°, y = 70° and z = 70°

Question 6.
In the given figure AB ∥ CD and AE is transversal. If ∠BAC =120°, then find x and y.

Answer:
In the given figure, AB ∥ CD and AE is transversal.
∠BAC = 120
∠ACD = x
∠DCE = y
∠BAC = ∠DCE (correpsonding angles are equal)
y = 120°
x + y = 180 ° (Linear pair of angles are supplementary)
x + 120° – 180°
x = 180°- 120°
∴ x = 60°
Hence x = 60°, y =120°.

Question 7.
In the given figure, BA ∥ CD and BC is transversal. Find x.

Answer:
In the given figure, BA ∥ CD and BC is transversal.
∠C = x + 35° and ∠B = 60°
∠C = ∠B (∵ alternate interior angles are equal)
x + 35° = 60° .
x – 60° – 35°
∴ x = 25°

Question 8.
In the figure \(\overrightarrow{\mathbf{M N}} \| \overrightarrow{\mathbf{K L}}\) and \(\overline{\mathrm{MK}}\) is transversal. Find x.

Answer:
From the figure, \(\overrightarrow{\mathbf{M N}} \| \overrightarrow{\mathbf{K L}}\) and \(\overline{\mathrm{MK}}\) is transversal.
∠M = 2x and ∠K = x + 30°
∠M + ∠K = 180° (Q co-interior angles are supplementary)
⇒ 2x + x + 30° = 180°
⇒ 3x + 30° = 180°
⇒ 3x = 180° – 30°
⇒ 3x = 150° ⇒ x = \(\frac{150^{\circ}}{3}\)
∴ x = 50°

Question 9.
In the figure \(\overline{\mathrm{AB}} \| \overline{\mathrm{DE}}\) and C is a point in between them. Observe the figure, then find x, y and ∠BCD.

Answer:
In the figure \(\overline{\mathrm{AB}} \| \overline{\mathrm{DE}}\) and C is a point in between them.
Draw a parallel line CF to \(\overline{\mathrm{AB}}\) through C.
\(\overline{\mathrm{AB}} \| \overline{\mathrm{CF}}\) and \(\overline{\mathrm{BC}}\) is a transversal,
x + 103° = 180°
x = 180°- 103°
x = 77°

From the figure,
\(\overline{\mathrm{DC}} \| \overline{\mathrm{CF}}\) and \(\overline{\mathrm{CD}}\) is a transversal,
y + 103° = 180°
y = 180°- 103°
y = 77°
and ∠BCD = x + y = 77° + 77° = 154°

Question 10.
In the figure transversal p intersects two lines m and n. Observe the figure, check whether m ∥ n or not.

Answer:
In the given figure, it is given that each angle in the pair of corresponding angles is 45°. So they are equal. Since a pair of corresponding angles are equal the lines are parallel. Hence, m ∥ n.

Practice Questions [Page No. 87]
Indicate the group (a, b, c, d, e, f) to which given below belongs to

Question 1.
State, District, Mandai
Answer:

Group: b

Question 2.
Boys, Girls, Artistists
Answer:

Group: c

Question 3.
Hours, Days, Minutes
Answer:

Group: b

Question 4.
Women, Teacher, Doctor
Answer:

Group: c

Question 5.
Food, Curd, Spoon
Answer:

Group: f

Question 6.
Humans, Dancer. Player
Answer:

Group: f

Question 7.
Building, Brick, Bridge
Answer:

Group: c

Question 8.
Tree, Branch, Leaf
Answer:

Group: b

Question 9.
Gold. Silver, Jewellary
Answer:

Group: f

Question 10.
Bulbs, Mixtures, Electricals
Answer:

Group: f

Question 11.
Women, Illiteracy, Men
Answer:

Group: c

Question 12.
Medicine, Tablets. Syrup
Answer:

Group: f

Question 13.
Carrots, Oranges, Vegetables
Answer:

Group: e

Question 14.
Female, Mothers. Sisters
Answer:

Group: f

Question 15.
Table. Furniture, Chair
Answer:

Group: f

Question 16.
Fruits, Mango. Onions
Answer:

Group: e

Question 17.
School, Teacher, Students
Answer:

Group: f

Question 18.
Rivers. Oceans. Springs
Answer:

Group: d

Question 19.
India. Andhra Pradesh, Visakhapatnam
Answer:

Group: b

Question 20.
Animals. Cows. Horses
Answer:

Group: f

Question 21.
Fish. Tiger, snakes
Answer:

Group: a

Question 22.
Flowers. Jasmine. Banana
Answer:

Group: e

Question 23.
Authors, teachers, Men
Answer:

Group: c

Question 24.
Dog. Fish, Parrot
Answer:

Group: a

Question 25.
Rose, Flower, Apple
Answer:

Group: e

Question 26.
School, Benches. Class Room
Answer:

Group: b

Question 27.
Pen, Stationary, Powder
Answer:

Group: e

Question 28.
Crow, Pigeon. Bird
Answer:

Group: f

Question 29.
Mammals, Elephants, Dinosaurs
Answer:

Group: f

Question 30.
Writers. Teachers, Researchers
Answer:

Group: d

REASONING [Practice Questions]

Question 1.
Musician, Instrumentalist, Violinist.
Answer:

Group: b

Question 2.
Officer, Woman, Doctor
Answer:

Group: c

Question 3.
Girls, Students, 7” class girls
Answer:

Group: d

Question 4.
Pencil, Stationary, Toothpaste
Answer:

Group: e

Question 5.
Food, Curd, Fruits
Answer:

Group: f

Question 6.
Bird Pigeon, Cow
Answer:

Group: e

Question 7.
Notes, Pad, Pen
Answer:

Group: a

Question 8.
Banana, Guava, Apple
Answer:

Group: a

Question 9.
Tomato, Food, Vegetables
Answer:

Group: b

Question 10.
Cow, Dog Pet
Answer:

Group: c

SCERT AP 7th Class Maths Solutions Pdf Chapter 3 Simple Equations Ex 3.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 3rd Lesson Simple Equations Exercise 3.1

Question 1.
Write the equations of the following mathematical statements.
(i) A number x decreased by 5 is 14.
Answer:
Given number = x
Number decreased by 5 = x – 5
∴ x – 15 = 14.

(ii) Eight times of y plus 3 is -5.
Answer:
Given number = y
Eight times of y = 8 ∙ y
Eight times of y plus 3 = 8y + 3
∴ 8y + 3 = – 5

(iii) If you add one fourth of z to 3 you get 7.
Answer:
Given number = z
One fourth of z = \(\frac{1}{4}\) ∙ z = \(\frac{z}{4}\)
One fourth of z is added to 3 = \(\frac{z}{4}\) + 3
Result is 7
∴ \(\frac{z}{4}\) + 3 = 7.

(iv) If you take away 5 from 3 times of m, you get 11.
Answer:
Given number = m
3 times of m = 3m
Take away 5 from 3 times of m = 3m – 5
Result is 11
∴ 3m – 5 = 11

(v) Sum of angles 2x, (x – 30) is a right angle.
Answer:
Given angles 2x, (x – 30)
Sum of angles 2x, (x – 30)
= 2x + x – 30 = 3x – 30
Sum of angles is right angle (90°).
∴ 3x – 30 = 90°

(vi) The perimeter of a square of side ‘a’ is 14 m.
Answer:
Given side of a square = a
Perimeter = 4 ∙ side = 4 ∙ a
Given perimeter = 14 m
∴ 4a = 14 m

Question 2.
Write the following equations in statement form.
(i) m – 5 = 12
Answer:
A number m is decreased by 5 is 12.

(ii) \(\frac{\mathbf{a}}{\mathbf{3}}\) = 4
Answer:
One third of a is 4.

(iii) 4x + 7 = 15
Answer:
Sum of 4 times of x and 7 is 15.
(or)
7 is added to 4 times of x is 15.

(iv) 2 – 3y = 11
Answer:
2 is decreased by 3 times of y is 11.
(or)
3 times of y is subtracted from 2 is 11.

Question 3.
Check whether the value given in the brackets is a solution to the given equation or not.
(i) 5n – 7 = 23 (n = 6)
Answer:
Given 5n – 7 = 23
When n = 6
L.H.S = 5n – 7
= 5(6) – 7
= 30 – 7
= 23
R.H.S =23
Here, L.H.S = R.H.S
So, n = 6 is a solution of the given equation.

(ii) \(\frac{p}{4}\) – 7 = 5 (p = 8)
Answer:
Given \(\frac{p}{4}\) – 7 = 5; When p = 8
LHS = \(\frac{p}{4}\) – 7

RHS = 5
– 5 ≠ 5
Here, LHS ≠ RHS
So, p = 8 is not a solution of the given equation.

(iii) 5 – 2x = 19 -7
Answer:
Given 5 – 2x = 19
When x = – 7
LHS = 5 – 2x
= 5 – 2(- 7) = 5 + 14 = 19
RHS = 19
19 = 19
Here, LHS = RHS
So, x = – 7 is a solution of the given equation.

(iv) 2 + 3(m – 1) = 5 (m = -2)
Answer:
Given 2 + 3(m – 1) = 5
When m = – 2
LHS = 2 + 3(m – 1)
= 2 + 3(- 2 – 1)
= 2 + 3(- 3) = 2 – 9 = – 7
RHS = 5
– 7 ≠ 5
Here, LHS ≠ RHS
So, m = – 2 is not a solution of given equation.

Question 4.
Solve the following equations using trial and error method,
(i) 3x – 7 = 5
Answer:

For x = 4, LHS = RHS
So, x = 4 is the solution of given equation.

(ii) 5 – y = – 1
Answer:

For y = 6, LHS = RHS.
So, y = 6 is the solution of given equation.