#### Question

☐ PQRS is an isosceles trapezium l(PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm, Distance between two parallel sides is 4 cm, find the area of ☐ PQRS

#### Solution

Draw a perpendicular from Q to line MR. Where it meets the line MR name it point N.

MN = PQ = 7 cm

In Δ PMS,

PM² + SM² = PS²

⇒ 4² + 3² = PS²

⇒ PS² = 16 + 9 = 25

⇒ PS = 5cm

PQRS is an isosceles trapezium so, PS = QR = 5 cm

PM = QN = 4 cm

So, NR = SM = 3 cm

SR = SM + MN + NR = 3 + 7 + 3 = 13 cm

Area of trapezium PQRS = `1/2 xx` (sum of parallel sides) x height

=`1/2 xx` (7 + 13) x 4

= 40 cm²

Is there an error in this question or solution?

Solution ☐ Pqrs is an Isosceles Trapezium L(Pq) = 7 Cm. Seg Pm ⊥ Seg Sr, L(Sm) = 3 Cm, Distance Between Two Parallel Sides is 4 Cm, Find the Area of ☐ Pqrs Concept: Area of a Trapezium.