Question

ΔPQR is isosceles with PQ=PR=7.5 cm and QR=9 cm. The height PS from P to QR, is 6 cm. Find the area of ΔPQR. What will be the height from R to PQ i.e. RT?

Answer 1

length of PQ = PR = 7.5 cm
length of QR = 9 cm
length of PS = 6cm

area of triangle = 1/2 × base × altitude
so, area of triangle PQR = 1/2 × length of QR × length of PS
= 1/2 × 9cm × 6cm = 9cm × 3cm = 27cm²
hence, area of ∆PQR = 27 cm²

area of triangle PQR = 1/2 length of QR × length of PS = 1/2 × length of PQ × length of RT
so, length of QR × length of PS = length of PQ × length of RT
=> 9cm × 6cm = 7.5 cm × length of RT
=> 54 cm² = 7.5cm × length of RT
length of RT = 54/7.5 cm = 7.2 cm

hence, length of RT = 7.2 cm

Answer 2

Acc to the question, ΔPQR is isosceles with PQ=PR=7.5 cm and QR=9 cm.

area of ΔPQR = 1/2 x base x height = 1/2 x 9 x 6 = 27 cm².

Since ∠PTR = 90°, 1/2 x PQ x RT = 27 cm². [the area will be the same since it's the same triangle]

or, RT = (27 x 2) / PQ = (27 x 2) / 7.5 = 7.2 cm. [Ans]

Hope the solution is clear to everyone.