Question

Pq perpendicular to pr then find the area of trapezium pqrs in which pq =10,qr=16,ps=25

Answer 1

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GIVEN THAT PQ perpendicular to PR

PQR IS RIGHT TRIANGLE

PYTHAGORAS THEOREM
${10}^{2} + {pr}^{2} = {16}^{2}$
${x}^{2} = 256 - 100 \\ x = 12.5cm$
AR(PQRS)=PR×PQ
=12.5cm×10cm
=125cm^2

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