Question

In a ∆ABC, perpendicular AD from A and BC meets BC at D. If BD = 8 cm, DC = 2 cm and AD = 4 cm, then

(a) ∆ABC is isosceles
(b) ∆ABC is equilateral
(c) AC = 2AB
(d) ∆ABC is right-angled at A

Answer 1

Given: In ΔABC,, BD = 8cm, DC = 2 cm and AD = 4cm.

In ΔADC,

Similarly, in ΔADB

Now, In ΔABC

${\mathrm{BC}}^2={\left(\mathrm{CD}+\mathrm{DB}\right)}^2={\left(2+8\right)}^2={\left(10\right)}^2=100$

and

${\mathrm{AB}}^2+{\mathrm{CA}}^2=80+20=100$

$\therefore {\mathrm{AB}}^2+{\mathrm{CA}}^2={\mathrm{BC}}^2$

Hence, triangle ABC is right angled at A.

We got the result as