Question

∠B is a right angle in ΔABC and BD is an altitude to hypotenuse. AB = 8, BC = 6. Find the area of ΔBDC.

Answer 1

see attachment, in ∆ABC , angle B = 90°
AB = 8 and BC = 6
we know, area of triangle = 1/2 × height × base
= 1/2 × BC × AB
= 1/2 × 6 × 8
= 24
hence, area of triangle , ABC = 24 ----(1)
In ∆ABC, ∠B is a right angle and BD is an altitude,
In ∆ABC,
∠A + ∠C = 90° and,
In ∆BDC,
∠DBC + ∠C = 90°
So, ∠A = ∠DBC …(2)
In ∆ABD and ∆BDC,
∠DAB =  ∠DBC [from (2)]
∠ADB = ∠BDC [by right angles]
from A - A rule of similarity ,
∆ABD ~ ∆BCD
Areas of similar triangles are proportional to the squares of their corresponding sides.
so, $\bf{\frac{ar(\Delta{ABD})}{ar(\Delta{BCD})}=\frac{AB^2}{BC^2}=\frac{8^2}{6^2}=\frac{16}{9}}$

hence, ar(∆ABD) = 16/9 ar(∆BCD) ----(3)
now, ar(ABD) + ar(BCD) = ar(ABC)
so, 16/9 ar(BCD) + ar(BCD) = 24 [ from (1) and (3)]
=>25/9 ar(BCD) = 24
=> ar(BCD) = 24 × 9/25 = 8.64

hence, area of BCD is 8.64