S.chand publication New Learning Composite mathematics solution of class 7 Chapter 11 Congruence of Triangles Exercise 11D

 Exercise 11D

Question 1

(a) What will be the other angles of a right angled isosceles triangle ?

Sol :

Figure to be added

In right angled isosceles triangle ABC

∠B=90°

AB=BC

∠A=∠C

∴90°+∠A+∠C=180°

90°+2∠A=180°

∠A$=\frac{90}{2}$

∴∠A=45°

Ans : The other triangle of a right angled isosceles triangle is 45°

(b) Can you draw an obtuse angled isosceles triangle?

Figure to be added

Ans : Yes, I can draw an obtuse angled isosceles triangle

Question 2

(a) The vertical angle of an isosceles triangle is 110° .What must be the size of each of its base angles ?

Sol :

Let, the size of each of base angles=x

∴2x+100°=180°

2x=180°-110°

2x=70°

$x=\frac{70}{2}$

∴x=35°

Ans : The size of each base angle is 35°

(b) What is the size of each exterior angle of an equilateral triangle ?

Sol :

In a equilateral triangle each angle is 60°

∴Each exterior angle of an equilateral triangle is 180°-60°=120°

Question 3

ΔABC is isosceles with AB=AC , if ∠A=80°. What is the measure of ∠B?

Figure to be added

∠B=∠C

∴80°+∠B+∠C=180°

80°+2∠B=180°

2∠B=180°-80°

$=\frac{100}{2}$

∠B=50°

Question 4

In ΔABC, ∠A=∠B=50°. Name the pair of sides which are equal

Figure to be added

Ans : AC and BC are equal

Question 5

In the figure given alongside , AN=AC, ∠BAC=52°, ∠ACK=84° and BCK is a straight line. Prove that NB=NC

Figure to be added

In ΔANC, AN=AC

∴∠ANC=∠ACN

∴52°+∠ANC+∠ACN=180°

2∠ANC=180°-52°

∠ANC$=\frac{128}{2}$

∴∠ANC=64°

In ΔNBC,

∴∠BNC=180°-64°=116°

∴∠NCB=180°-(64°+84°)

=180°-148°=32°

∴∠NCB=180°-(116°+32°)

=180°-32°

∠NCB=∠NBC

∴ΔNBC is an isosceles triangle

∴NB=NC

Question 6

In the figure , AB=AC. Prove that BD=BC

Figure to be added

In ΔABC, ∠ADB=180°-(40°+30°)

=180°-70°=110°

∴∠BDC=180°-110°=70°

In ΔABC, AB=AC

∴∠ABC=∠ACB

40°+∠ABC+∠ACB=180°

40°+2∠ABC=180°

2∠ABC=180°-40°

∠ABC$=\frac{140}{2}$

∴∠ABC=∠ACB=70°

∴∠DBC=70°-30°=40°

In ΔBDC,

∴∠BDC=∠DCB

∴ΔBDC is an isosceles triangle

∴BD=BC