Question

4 points
Q28) Find the area of the
given quadrilateral PQRS, here
angle SPQ = 90 degree.
IN
0 141 sq. m.
O 276 sq. m.
267 sq. m.
0 114 sq. m.​

Answer 1

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