AP State Syllabus AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 3 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 3

Question 1.

Find the value of the unknown ‘x’ in the following triangles.

Solution:

i) In ΔABC

∠A + ∠B + ∠C = 180° (angle sum property

x° + 50° + 60° = 180°

x° + 110° = 180°

x° = 180° – 110°

ii) In ΔPQR,

∠P + ∠Q + ∠R = 180° (angle – sum property)

90° + 30° + x° = 180°

120° + x° – 180°

x° = 180° – 120°

∴ x° = 60°

iii) In ΔXYZ,

∠X + ∠Y + ∠Z = 180° (angle – sum property)

30°+110° + x° =180°

140° + x° = 180°

x° = 180° – 140°

∴ x° = 40°

Question 2.

Find the values of the unknowns ‘x ‘and ‘y ‘in the following diagrams.

Solution:

i) In ΔPQR,

x° + 50° = 120° (exterior angle property)

x°= 120°- 50°

x°= 70°

Also

∠P + ∠Q +∠R = 180° (angle – sum property)

70° + 50° + y° = 180°

120° + y° = 180°

y° = 180° – 120°

y° = 60°

(OR)

y° + 120° = 1800 (linear pair of angles)

y° = 180°- 120°

∴ y° = 60°

ii) In the figure ΔRST,

x° = 80° (vertically opposite angles)

also ∠R + ∠S + ∠T = 180° (angle – sum property)

80° + 50°+ y°= 180°

130° + y° = 180°

y° = 180° – 130°

∴ y° = 50°

iii) m ΔMAN,

x° = ∠M + ∠A (exterior angle property)

x° = 50° + 60°

x° = 110°

Also x° + y° = 180°

110°+ y°= 180°

y° = 180° – 110°

y° = 70°

(OR) in ΔMAN,

∠M + ∠A + ∠N = 180° (angle – sum property )

50° + 60° + y° = 180°

110° + y° = 180°

y° = 180°- 110°

∴ y° = 70°

iv) In the figure ΔABC,

x° = 60° (vertically opposite angles)

∠A + ∠B + ∠ACB = 180° (angle – sum property)

y° + 30° + 60° = 180°

y° + 90° = 180°

y° = 180° – 90°

∴ y° = 90°

v) In the figure ΔEFG,

y° = 90° (vertically opposite angles)

Also in ΔEFG;

∠F + ∠E + ∠G = 180° (angle – sum property)

∴ x° + x° + y° = 180

2x° + 90° = 180°

2x° = 180°- 90°

2x° = 90°

x° = \(\frac{90^{\circ}}{2}\)

∴ x° = 45°

vi) In the figure ΔLET,

∠L = ∠T = ∠E = x° (vertically opposite angles)

Also in ΔLET

∠L + ∠E + ∠T = 180° (angle – sum property)

x° + x° + x° = 180°

3x° = 180°

x° = \(\frac{180^{\circ}}{3}\)

x° = 60°

Question 3.

Find the measure of the third angle of triangles whose two angles are given below:

(i) 38° , 102°

(ii) 116°, 30°

(iii) 40°, 80°

Solution:

(i) 38° , 102°

Let the third angle be x° then

38° + 102° + x° = 180° (angle – sum property)

140° + x° = 180°

x° = 180° – 140° = 40°

(ii) 116°, 30°

Let the third angle be x°

then 116° + 300 + x° = 180° (angle – sum property)

146° + x = 180°

x= 180°- 146° = 34°

(iii) 40°, 80°

Let the third angle be x° then

40C + 80° + x° = 180° (angle-sum property)

120°+x° = 180°

x° = 180 – 120° – 60°.

Question 4.

In a right-angled triangle, one acute angle is 30°. Find the other acute angle.

Solution:

Given triangle is right angled triangle.

Let the third angle be x°

then 90° +30° + x° = 180° (angle – sum property)

120° + x° – 180°

x° = 180° – 120° = 60°

Question 5.

State true or false for each of the following statements.

(i) A triangle can have two right angles.

(ii) A triangle can have two acute angles.

(iii) A triangle can have two obtuse angles.

(iv) Each angle of a triangle can be less than 60°.

Solution:

(i) A triangle can have two right angles. – False

(ii) A triangle can have two acute angles. – True

(iii) A triangle can have two obtuse angles. – False

(iv) Each angle of a triangle can be less than 60°. – False

Question 6.

The angles of a triangle are in the ratio 1 : 2 : 3. Find the angles.

Solution:

Given that ratio of the angle = 1: 2: 3

Sum of the terms of the ratio – 1 + 2 + 3 = 6

Sum of the angles of a triangle = 1800

∴ 1st angle = \(\frac { 1 }{ 6 }\) x 180° = 30° (angle-sum property)

∴ 2nd angle = \(\frac { 2 }{ 6 }\) x 180° = 60° (angle – sum property)

∴ 3rd angle = \(\frac { 3 }{ 6 }\) x 180° = 90° (angIe – sum property)

Question 7.

In the figure, \(\overline{\mathrm{DE}} \| \overline{\mathrm{BC}}\) , ∠A = 300 and ∠B = 50°. Find the values of x, y and z.

Solution:

y° = 50° (correspondIng angles)

x° = z° (corresponding angles)

Also in MBC;

∠A + ∠B +∠C = 180° (angIe – sum property)

30° + 50° + z° = 180°

80° + z° = 180°

z° = 180° – 80°

∴ z° = 100°

∴ x° = 100°; y° = 50°; z° = 100°

Question 8.

In the figure, ∠ABD = 3 ∠DAB and ∠BDC = 96°. Find ∠ABD.

Solution:

In the figure

∠ABD + ∠DAB = 96° (exterior angle property)

3∠DAB + ∠DAB = 96° (given)

4∠DAB = 96°

∠DAB = \(\frac{96^{\circ}}{4}\) = 24°

∠ABD = 3 x 24° = 72°

Question 9.

In ΔPQR ∠P= 2 ∠Q and 2 ∠R =3 ∠Q , calculate the angles of ΔPQR.

Solution:

In ΔPQR

∠P + ∠Q + ∠R = 180° also

∠P:∠Q:∠R = 2∠Q:∠Q: \(\frac { 3 }{ 2 }\)∠Q = 4 : 2 : 3

Sum of the term of the ratio = 4 + 2 + 3 = 9

∠P = \(\frac { 4 }{ 9 }\) x 180° = 80°

∠Q = \(\frac { 2 }{ 9 }\) x 180° = 40°

∠R = \(\frac { 3 }{ 9 }\) x 180° = 60°

Question 10.

If the angles of a triangle are in the ratio 1:4: 5, find the angles.

Solution:

Given that ratio of the angles = 1 : 4 : 5

Sum of the terms of the ratio = 1 + 4 + 5 = 10

Sum of the angles 180°

∴ 1st angIe = \(\frac { 1 }{ 10 }\) x 180° = 18°

2nd angle \(\frac { 4 }{ 10 }\) x 180° = 72°

3rd angle = \(\frac { 5 }{ 10 }\) x 180°= 90°

Question 11.

The acute angles of a right triangle are in the ratio 2 : 3. Find the angles of the triangle.

Solution:

Given that ratio of acute angles = 2 : 3

Sum of the terms of the ratio = 2 + 3 = 5

Sum of the acute angles = 90°

∴ 1st acute angle x 900 = 36°

2nd acute angle = x 900 = 540

∴ Angles of the triangle = 36°, 54° and 90°

Question 12.

In the figure, ∆PQR is right angled at Q, \(\overline{\mathrm{ML}} \| \overrightarrow{\mathrm{RQ}}\) and ∠LMR = 130°. Find ∠LPM, ∠PML and ∠PRQ.

Solution:

∠PRQ + ∠LMR = 180° (int, angles on the same side of the transversal)

∠PRQ + 130° = 18 0°

∴ ∠PRQ = 180° – 130°= 50°

In ∆PRQ,

∠P + ∠R +∠Q = 180° (angle – sum property)

∠LPM + 50° + 90° = 180°

∠LPM + 140° = 180°

∴ ∠LPM = 180° – 140° = 40°

Also ∠PML + ∠LMR = 180° (Linear pair of angles)

∴ ∠PML + 130° = 180°

∠PML = 180° – 130° = 50°

(or) ∠PML = ∠PRQ = 50° (corresponding angles)

Question 13.

In Figure ABCDE, find ∠1 +∠2 + ∠3 + ∠4 + ∠5.

Solution:

In ∆ABC, ∠BAC + ∠ACB + ∠ABC = 180° ………….(1)

In ∆ACD, ∠CAD + ∠ADC + ∠ACD = 180° ……………. (2)

In ∆ADE, ∠DAE + ∠ADE + ∠AED = 180° …………(3)

Adding (1), (2) and (3) we get

(∠BAC + ∠ACB + ∠ABC) + ∠CAD+ (∠ADC + ∠ACD + ∠DAE + ∠ADE + ∠AED = 180°+ 180° 180°

(∠BAC + ∠CAD + ∠DAE) + ∠ABC + (∠ACB + ∠ACD) + ∠ADC + ∠ADE) + ∠AED = 180° + 180° + 180°

Hene

∠1 + ∠2 + ∠3 + ∠4 + ∠5 = 540°