ΔPQR is isosceles with PQ = PR = 7.5 cm and QR = 9 cm. The height PS

1.	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» Given: PQ = PR = 7.5cm, QR = 9cm PS = 6cm Area of ΔPQR = 1/2 bh = 1/2 x 9 x 6 = 27cm² Also area of ΔPQR = 1/2 × PQ × RT 27 = 1/2 × 7.5 × RT [PQ = PR] ∴ RT = \(\frac {2\times27}{7.5} = \frac {2\times27\times2}{15} = \frac {36}{5}\) = 7.2 cm

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»

Given: PQ = PR = 7.5cm, QR = 9cm PS = 6cm

Area of ΔPQR = 1/2 bh = 1/2 x 9 x 6 = 27cm²

Also area of ΔPQR = 1/2 × PQ × RT

27 = 1/2 × 7.5 × RT [PQ = PR]

∴ RT = \(\frac {2\times27}{7.5} = \frac {2\times27\times2}{15} = \frac {36}{5}\) = 7.2 cm