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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.
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Solution
Draw a perpendicular from Q to line MR. Where it meets the line MR, name it point N.
MN = PQ = 7 cm
In ΔPMS,
PM² + SM² = PS²
⇒ 4² + 3² = PS²
⇒ PS² = 16 + 9
⇒ PS² = 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) × height
=`1/2 xx (7 + 13) xx 4`
= 40 cm²
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