Question

in triangle ABC, if AD divides BC in the ratio m : n, then find the ratio of the areas of triangle ABD and triangle ADC
a) m + n : n
b) m : m + n
c) m : n
$d) {m}^{2} is \: to \: {n}^{2}$
WITH EXPLAINATION

Answer 1

Given in ΔABC, D is a point on BC such that BD: DC = 3:5

⇒ BD: BC = 3:8

That is BD/BC = 3/8 - (1)

Draw AE⊥BC

ar(ΔABC) = 1/2 x BC x AE -- (2)

ar(ΔABD) = 1/2 x BD x AE -- (3)

Now we will divide 2 by 3

it will give us = BD/BC

= 3/8 (from 1)

Therefore, ar(ADC): ar (ABC) = 3:8

If there is any confusion please leave a comment below.

Answer 2

I GUESS THE ASWER IS m:n