Question

In a traingle PQR, point S is on the side PQ and point T is on side PR, such that QRTS is a trapezium. ST : QR = 4 : 9. Calculate the ratio of the area of traingle PST and the trapeziumQRTS?

Answer 1

7.5 cm

Since the lines ST and QR are parallel, PQR & PST are similar triangles

Therefore SQ/PQ =TR/PR

Answer 2

Answer: 16:65

Step-by-step explanation:

Now as given QRTS is a trapezium

∴ ST║QR

Consider ΔPST and ΔPQR

∠PST= ∠PQR         (ST║QR ∴ corresponding angles are equal)

∠PTS= ∠PRQ         (ST║QR ∴ corresponding angles are equal)

∴ Δ PST ≈ Δ PQR   (Angle Angle property)

Now as we know that the ratio of the area of similar triangle is proportional to square of the ratio of corresponding sides.

Therefore

$\dfrac{ar(\triangle PST)}{ar(\triangle PQR)} =(\dfrac{ST}{QR} )^2= (\dfrac{4}{9}) ^2= \dfrac{16}{81}$

Let ar(Δ PST) =16k and ar(Δ PQR )= 81k where k is common ratio

Now

$\dfrac{ar(\triangle PST)}{ar(QRTS)} =\dfrac{ar(\triangle PST)}{ar(\triangle PQR)-ar(\triangle PST)} =\dfrac{16k}{81k-16k} =\dfrac{16k}{65k} =\dfrac{16}{65}$

Hence, the ratio is 16:65