Question

2. PQRS in which PQ = QR = 3.5 cm, PS = RS = 5.2 cm and angle POR=120°.​

Answer 1

Answer:

Given:

PQ

is parallel to

SR

, PQ = 8 cm, ∠ PQR = 70

o

QR = 6 cm and PS = 6 cm.

Steps for construction

Step 1 : Draw a rough diagram and mark the given measurements.

Step 2 : Draw a line segment PQ = 8 cm.

Step 3 : At Q on

PQ

make ∠ PQX whose measure is 70

o

.

Step 4 : With Q as centre and 6 cm as radius draw an arc. This cuts $$\overline{QX}$$ at R.

Step 5 : Draw

RY

parallel to

OP

.

Step 6 : With Q as centre and radius 6 cm draw an arc cutting

RY

at S.

Step 7 : Join

PS

. PQRS is the required trapezium.

Step 8 : From S draw

ST

⊥

PQ

and measure the length of ST. ST = h = 5.6 cm, RS = b = 6 cm, PQ = a = 8 cm.

Calculation of area:

In the trapezium PQRS, a = 8 cm, b = 3.9 cm and h = 5.6 cm.

Area of the trapezium ABCD =

2

1

h(a+b)

=

2

1

(5.6)(8+3.9)

=

2

1

× 5.6×11.9

=33.32 cm

2