In a Δ ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of �

1.	In a Δ ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ΔABC and hence its altitude on AC
Answer» Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by: A = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}{2}\)[Heron’s Formula] a = 15,b = 13, c = 14 a = \(\frac{a+b+c}{2}\) = \(\frac{15+12+14}{2}\) = 21 A = \(\sqrt{21(21-15)(21-13)(21-14)}\) A = \(\sqrt{21\times6 \times8 \times7}\) = 84 cm² Area of triangle = \(\frac{1}2\)(Base x Altitude) 84 = \(\frac{1}2\) (14 x Altitude) Altitude = 12 cm

In a Δ ABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ΔABC and hence its altitude on AC

Answer»

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}{2}\)[Heron’s Formula]

a = 15,b = 13, c = 14

a = \(\frac{a+b+c}{2}\) = \(\frac{15+12+14}{2}\) = 21

A = \(\sqrt{21(21-15)(21-13)(21-14)}\)

A = \(\sqrt{21\times6 \times8 \times7}\) = 84 cm²

Area of triangle = \(\frac{1}2\)(Base x Altitude)

84 = \(\frac{1}2\) (14 x Altitude)

Altitude = 12 cm