Question

find the area of a quadrilateral PQRS in which angle QPS+angleSQR=90 , PQ=12cm , PS=9cm , QR=8cm and SR=17cm

Answer 1

 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 = 2