Q. 24.4( 32 Votes )

# Find the area of a quadrilateral PQRS in which ∠QPS = ∠SQR = 90°, PQ = 12 cm, PS = 9 cm, QR = 8 cm and SR = 17 cm (Hint: PQRS has two parts)

Answer :

Area of quadrilateral PQRS = area(ΔSQR) + area(ΔPQS)…(i)

Let us find area(ΔPQS)

Base = PQ = 12 cm

Height = PS = 9 cm

area of triangle = × base × height

⇒ area(ΔPQS) = × PQ × PS

⇒ area(ΔPQS) = × 12 × 9

⇒ area(ΔPQS) = 6 × 9

⇒ area(ΔPQS) = 54 cm^{2}

Using pythagoras theorem

SQ =

⇒ SQ =

⇒ SQ =

⇒ SQ =

⇒ SQ = 15 …(ii)

Now let us find area(ΔSQR)

Base = QR = 8 cm

Height = SQ = 15 cm …from (ii)

area of triangle = × base × height

⇒ area(ΔSQR) = × QR × SQ

⇒ area(ΔSQR) = × 8 × 15

⇒ area(ΔSQR) = 4 × 15

⇒ area(ΔSQR) = 60 cm^{2}

Therefore from (i)

Area of quadrilateral PQRS = area(ΔSQR) + area(ΔPQS)

= 60 + 54

= 114 cm^{2}

Hence area of quadrilateral PQRS = 114 cm^{2}

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