S.chand publication New Learning Composite mathematics solution of class 7 Chapter 10 Properties of Triangles Exercise 10A

 Exercise 10A

Question 1

Find the unknown marked angle in each triangle

(a) a+63°+70°=180°

Sol :

a=180°-(63°-70°)=47°

(b) b+90°+49°=180°

Sol :

b=180°-(90°+49°)

b=180°-139°=41°

(c) c+125°+30°=180°

Sol :

c=180°-(125°+30°)

c=180°-155°=25°

(d) d+48°+89°=180°

Sol :

d=180°-(48°+84°)

d=180°-132°=48°

(e) e+60°+60°=180°

Sol :

e=180°-(60°+60)

e=180°-120°=60°

(f) f+45°+45°=180°

Sol :

f=180°-(45°+45°)

f=180°-90°=90°

(g) fig to be added

Sol :

a+37°+108°=180°

a=180°-(37°+108°)

∴a=180°-145°=35°

∴∠ACD=180°-108°=72°

∴b+28°+72°=180°

∴b=180°-100°=80°

(h) fig to be added

In ΔABC, P+100°+45°=180°

P=180°-145°=35°

In ΔBCD, q+40°+65°=180°

q=180°-(40°+65°)

q=180°-105°=75°

Question 2

Calculate the value of each variable .Give reasons

(a) fig to be added

Draw BC||AE

∴∠DCE=∠DBF=35°  [AE||BF]

∴∠P=90°-35°=55°

(b) fig to be added

∠ABC=180°-125°=55°

In ΔABC, 

x+55°+65°=180°

x=180°-(55°+65°)

x=180°-120°=60°

(c) fig to be added

∠CAD=∠ACB=3x  [AD||BC]

2x+3x=50°=180°

5x=130°

$x=\frac{130}{5}$

x=26°

∠ABC=2x=26°×2=52°

∠CAD=3x=26°×3=78°

(d) fig to be added

x+x+x=180°

3x=180°

$x=\frac{180}{3}=60$

∴x=60°

(e) fig to be added

2x+x+90°=180°

3x=180°-90°=90°

3x=90°

$x=\frac{90}{3}$

x=30°

2x=2×30°=60°

(f) fig to be added

∠DCE=∠ACB=85°  [vertically opposite ∠s]

In ΔABC, 

y+85°+54°=180°

y=180°-(80°+54°)

y=41°

∠BCP=y=41° [PQ||AB corresponding ∠s]

x=180°-(85°+41°)

=180°-126°=54°

(g) fig to be added

∠ACB=180°-89°=91° 

∠BAC=∠ADE=42° [corresponding ∠s, DE||AB]

In ΔABC, 

x+42°+91°=180°

x=180°-(42°+91°)

x=180°-133°=47°

∴x=47°

(h) fig to be added

∠APR=118° [AB||CD corresponding ∠s]

118°+x+22°=180°

x=180°-(118°+22°)

x=40°

In ΔPQR, 

40°+75°+y=180°

y=180°-(40°+75°)

y=180°-115°=65°

Question 3

In the figure, show that  ∠A+∠B+∠C+∠D+∠E+∠F=360°

fig to be added

In ΔABC, ∠A+∠B+∠C=180°

In ΔDEF, ∠D+∠E+∠F=180°

∴∠A+∠B+∠C+∠D+∠E+∠F=180°+180°=360°

Question 4

In the figure , BO and Co are the bisector of ∠S ,∠B and ∠C respectively. If ∠BAC=80° and ∠ACO=30° , find ∠BOC and ∠OBC

Sol :

Fig to be added

OC is the bisector of ∠ACB

∴∠ACO=∠OCB=30°

∴∠ACB=30°+30°=60°

In ΔABC, ∠B+80°+60°=180°

∠B=180°-(80°+60°)

∠B=180°-140°=40°

OB is the bisector of ∠ABC

∴∠ABO+∠OBC$=\frac{40}{2}$

In ΔOBC, 

∠O+20°+30°=180°

∠O=180°-(20°+30°)

=180°-50°=130°

Ans : ∠BOC=130°

∠OBC=20°

Question 5

The angles of a triangle are in the ratio 2 : 3 : 4. Find the measure of each angle of the triangle.

Sol :

Let , the angles are 2x, 3x, 4x

2x+3x+4x=180°

9x=180°

$x=\frac{180}{9}$

x=20°

The angles are =(2×20), (3×20), (4×20)

=40°, 60°, 80°

Question 6

(6) Which of the following groups of 3 angles cannot be the angles of a triangle?

(a) 60°, 50°, 70°

Sol :

60°+50°+70°=180°

Ans : The groups of angles can be a triangle

(b) 49°, 72°, 63°

Sol :

49°+72°+63°=184°

Ans : The groups of angles can't be the angles of a triangle

Question 7

A triangle has two equal angles, each of which is equal to twice the remaining third angle. Find the angles of the triangle.

Sol :

Let , the third angle be x

ATQ,

2x+2x+x=180°

5x=180°

$x=\frac{180}{5}$

x=36°

2x=(36×2)=72°

Ans : The angles of the triangles are 36°, 72°, 72°,