1. Find the following. i. Find the area of a quadrilateral PQRS if: a. ZQ = 90°, PQ = 20 cm, QR= 21 cm, SR=15 cm and PS = 16 cm b. PQ = 12 cm, QR = 16 cm, RS = 20 cm, PS = 24 cm and PR= 20 cm .
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Answer:
Given : ∠QPS=90 ∠ SQR=90
PQ=12cm , PS=9cm , QR=8cm and SR=17cm
By Pythagoras theorem in ΔSPQ we get :
⇒SP
2
+PQ
2
=SQ
2
⇒9
2
+12
2
=SQ
2
⇒81+144=SQ
2
⇒SQ
2
=225
⇒SQ=
225
⇒SQ=15cm
Area of the parallelogram = Area of triangle SPQ + area of triangle SQR.
⇒ Area of Triangle SPQ =
2
1
×base×height
⇒
2
1
×PQ×SP
⇒
2
1
×12×9
⇒ Area of triangle SPQ = 54cm
2
⇒ Area of Triangle SQR =
2
1
×base×height
⇒
2
1
×QR×SQ
⇒
2
1
×8×15
⇒ Area of triangle SPQ = 60cm
2
Total area = area of triangle SPQ + area of trinagle SQR
⇒54+60=114cm
2
Area of quadrilateral PQRS is 114cm
2
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