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Chapter 10 : Polygons I

Subjective Questions

Solution for Question 6

(a) x = 180º − 2

80180 °−°

= 180º − 2

100°

= 180º − 50º

= 130º

(b) x + 30º = 150º

x = 150º − 30º

= 120º

Solution for Question 8

∠PQR = ∠PSR = 120º

x + 45º = 120º

x = 120º − 45º

= 75º

Solution for Question 10

In quadrilateral PTSU,

a + b + 80º + 30º = 360º

a + b = 360º − 110º

= 250º

In ∆PQR,

30º + c + d = 180º

c + d = 180º − 30º

= 150º

Therefore, a + b + c + d = 250º + 150º

= 400º

NEXUS VISTA PMR MATHEMATICS FORM 1, 2 & 3

Page 96 (Step-by-step Workings)

Supplementary Materials for

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Objective Questions

1. In the diagram, JLN and KLM are straight lines.

The value of y is

A 51º

B 58º

C 65º

D 74º

2. In the diagram, EFG and EGH are isosceles triangles.

Find the value of x.

A 25°

B 30°

C 45°

D 60°

N

M

L

K

y

32º

55º

E

F

G

H

x

120º

J

NEXUS VISTA PMR MATHEMATICS FORM 1, 2 & 3

Page 96 (Extra Practice)

Supplementary Materials for

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3. In the diagram, RST is an isosceles triangle and RTU is a straight line.

The value of x + y is

A 170°

B 200°

C 230°

D 260°

4. In the diagram, PQRS is a quadrilateral and RST is a straight line.

Find the value of x.

A 40°

B 50°

C 70°

D 80°

P

Q

R

S x

75º

85º

T

U

y

P

Q

R S

x

30º

40º

T

110º

40º

NEXUS VISTA PMR MATHEMATICS FORM 1, 2 & 3

Page 96 (Extra Practice)

Supplementary Materials for

©Sasbadi Sdn. Bhd. iv

5. The diagram shows an isosceles triangle PQS with PQ = PS. PTS and QRS are

straight lines.

Find the value of x.

A 30°

B 40°

C 50°

D 80°

Subjective Questions

6. In the diagram, PQRS is a quadrilateral and QRT is a straight line.

Find the value of x + y.

R

Q

S

T

y

140º x

P

155º

NEXUS VISTA PMR MATHEMATICS FORM 1, 2 & 3

Page 96 (Extra Practice)

Supplementary Materials for

P

Q R S

x

100º T

110º