Question

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

Answer 1

Answer:

Given -

QPS = SQR = 90°, PQ = 12cm, PS = 9cm, QR = 8cm and SR = 17cm.

To Find -

Area of Quadrilateral PQRS

Solution -

by Pythagoras theorem in ΔSPQ we get,

$= > {sp}^{2} + {pq}^{2} = {sq}^{2}$

$= > {9}^{2} + {12}^{2} = {sq}^{2}$

$= > 81 + 144 = {sq}^{2}$

$= > {sq}^{2} = 225$

$= > sq = \sqrt{225}$

$= > sq = 15cm$

Area of parallelogram = Area of triangle spq + Area of triangle SQR

$= > area \: of \: triangle \: spq \: = \frac{1}{2} \times base \times height$

$= > \frac{1}{2} \times pq \times sp$

$\frac{1}{2} \times 12 \times 9$

Area of triangle SPQ =

${54}^{2}$

$= > area \: of \: triangle \: sqr = \frac{1}{2} \times base \times height$

$= > \frac{1}{2} \times qr \times sq$

$= > \frac{1}{2} \times 8 \times 15$