Question

In a right-angled triangle ABC, D is the foot of the perpendicular from B on the hypotenuse AC. If AB = 3 cm and BC = 4 cm, what is the area of ΔABD?

Answer 1

Area of ΔABC = \(\frac{1}{2}\) x 3 x 4 cm² = 6 cm². Also, AC = \(\sqrt{3^2+4^2}\) = 5 cm.

∴ Area of ΔABC = \(\frac{1}{2}\) x BD x AC ⇒ 6 = \(\frac{1}{2}\) BD x 5 ⇒ BD = \(\frac{12}{5}\) cm.

Now in ΔABD, AD = \(\sqrt{AB^2+BD^2}\) = \(\sqrt{3^2-\big(\frac{12}{5}\big)^2}\) = \(\sqrt{9-\frac{144}{25}}\)

= \(\sqrt{\frac{225-144}{25}}\) = \(\sqrt{\frac{81}{25}}\) = \(\frac{9}{5}\) cm

∴ Area of ΔABD = \(\frac{1}{2}\)x AD x BD = \(\frac{1}{2}\) x \(\frac{9}{5}\) x \(\frac{12}{5}\) = \(\frac{54}{25}\) cm².