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Question

Find the area of a quadrilateral PQRS in which QPS=SQR=90o,PQ=12 cm,PS=9 cm,QR=8 cm and SR=17 cm
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Solution

Given : QPS=90 SQR=90
PQ=12cm , PS=9cm , QR=8cm and SR=17cm

By Pythagoras theorem in ΔSPQ we get :

SP2+PQ2=SQ2

92+122=SQ2

81+144=SQ2

SQ2=225

SQ=225

SQ=15cm

Area of the parallelogram = Area of triangle SPQ + area of triangle SQR.

Area of Triangle SPQ = 12×base×height

12×PQ×SP

12×12×9

Area of triangle SPQ = 54cm2

Area of Triangle SQR = 12×base×height

12×QR×SQ

12×8×15

Area of triangle SPQ = 60cm2

Total area = area of triangle SPQ + area of trinagle SQR

54+60=114cm2

Area of quadrilateral PQRS is 114cm2

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