AP State Syllabus AP Board 7th Class Maths Solutions Chapter 5 Triangle and Its Properties Ex 4 Textbook Questions and Answers.

## AP State Syllabus 7th Class Maths Solutions 5th Lesson Triangle and Its Properties Exercise 4

Question 1.

In ΔABC, name all the interior and exterior angles of the triangle.

Solution:

Interior angles

∠ABC, ∠BAC, ∠CAB

Exterior angles

∠ACZ, ∠BAY and ∠CBX

Question 2.

For ΔABC, find the measure of ∠ACD.

Solution:

In ΔABC,

∠ACD = ∠BAC + ∠ABC (exterior angle property)

=66° + 45° = 111°.

Question 3.

Find the measure of angles x and y.

Solution:

In the figure,

x° + 65° = 180° (lInear pair of an1es)

∴ x°= 180°- 65°= 115°

Also y° + 30° = 65° (exterior angle property)

y° = 65° – 30° = 35°

Question 4.

In the following figures, fmd the values of x and y.

Solution:

In ΔABC,

107° = x° + 57° (exterIor angle property)

∴ x° = 107° – 57° =50°

In ΔADC,

∠CAD + ∠ADC + ∠ACD = 180° (angle – sum property)

40° + 107°+ x°= 180°

∴ x° = 180° – 147° = 33°

Also in ΔABC,

∠A +∠B +∠C = 180°

40° + ∠B + (33° + 65°) = 180°

∠B + 138° = 180°

B = 180° – 138° = 42°

Now y° = ∠A + ∠B for ΔABC (exterior angle property)

=40° + 42° = 82°

Question 5.

In the figure ∠BAD = 3 ∠DBA, find ∠CDB. ∠DBC and ∠ABC.

Solution:

In ΔBCD,

∠DBC + ∠BCD = ∠BDA (exterior angle property)

∠DBC + 65° -104°

∴ ∠DBC = 104° – 65° = 39°

Also ∠BDA + ∠BDC = 180° (linear pair of angles)

104° + ∠BDC = 180°

∠CDB or ∠BDC = 180° – 104° = 76°

Now in ΔABD,

∠BAD + ∠ADB + ∠DHA = 180°

3∠DBA + ∠DBA + 104° = 180° (given)

4∠DBA + 104° – 180°

∴ 4∠DBA = 180° – 104°= 76°

∴∠DBA = \(\frac{76^{\circ}}{4}\) = 19°

Now ∠ABC = ∠DBA + ∠DBC = 19° + 39° = 58°

Question 6.

Find the values of x and y in the following figures.

Solution:

i) In the figure the angles are

70°, x°, x° (angles in an Isosceles triangle)

Also 70° + x° + x° = 180° (angle – sum property)

2x° + 70°= 180°

2x° = 180° – 70° [∵ y° = x°]

2x° = 110°

x°= \(\frac{110^{\circ}}{2}\) = 55°

ii) From the figure,

the Interior opposite angles of x are 50°, 500 (angles in an isosceles triangle )

50° + 50° = x° (exterior angle is equal to sum of the Interior opposite angles)

∴ x = 100°

iii) From the figure,

y° = 30° (vertically opposite angles)

ALso y = x° – 30° (equal angles of an Isosceles triangles)

∴ x° = 30° and y = 30°

iv) From the figure,

a + 110° = 180° (linear pair of angles)

∴ a = 180° – 110°= 70

Also y° = a° = 70° (equal angles of an isosceles triangle)

x + y + a = 180 (sum of interior angles)

x + 70 + 70 = 180

x + 140 = 180

x = 180 – 40 = 40°, ∴ x = 40°

v) From the figure,

30° + y° – 180°

y° = 180° – 30° = 150°

Also x° + a° = y° (exterior angle property)

x° + 90° = 150°

x° = 150° – 90° = 60°

vi) From the figure,

b = 80° (vertically opposite angles)

Also x° = a° (equal angles of an isosceles triangle.)

° 80° + x° + a° = 180

80 + x° + x° = 180

2x° = 180° – 80° =100°

2x° = 100°

∴ x° = \(\frac{100}{2}\) = 50°

Now y = x° b° (exterior angle property)

50° + 80° = 130°

Question 7.

One of the exterior angles of a triangle is 125° and the interior opposite angles are in the

ratio 2 :3. Find the angles of the triangle.

Solution:

Ratio of the interior opposite angles = 2 : 3

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

Sum of the interior angles exterior angle = 125°

∴ First angle = \(\frac{2}{5}\) x 125° = 500

Second angle = \(\frac{3}{5}\) x 125° = 75°

Question 8.

The exterior ∠PRS of ∆PQR is 105°. If Q = 70°. find ∠P. Is ∠PRS > ∠P?

Solution:

∠P+ ∠Q = ∠PRS (exterior angle is equal to sum of the interior opposite angles)

∠P + 70° = 105°

∠P = 105° – 70° = 35°

Now ∠PRS > ∠P.

Question 9.

If an exterior angle of a triangle is 130° and one of the interior opposite angle is 6. Find

the other interior opposite angle.

Solution:

Let the other interi6r opposite angle be = x°

Give interior opposite angle be = 60°

Now sum of the interior opposite angle = exterior angLe

x° + 60° = 130°

x° = 130° – 60° = 70°

∴ The other interior opposite angle = 70°

Question 10.

One of the exterior angle ofa triangle is 105° and the interior opposite angles are in the ratio 2 : 5. Find the angles of the triangle.

Solution:

Ratio of interior opposite angles = 2 : 5

Sum of the terms of the ratio = 2 + 5 = 7

Sum of the angles = 105°

∴ 1st angle = \(\frac { 2 }{ 7 }\) x 105° = 30°

2nd angIe = \(\frac { 5 }{ 7 }\) x 105° = 75°

3rd angle = 180° – (30° + 75°) [∵ angle – sum property]

= 180° – (105°) = 75°

Question 11.

In the figure find the values of x andy.

Solution:

From the figure,

∠y = 50° + 30° = 80° (exterior angle property)

∠x = y° + 55° (exterior angle property)

= 80° + 55°

= 135°