Question

In triangle ABC If D is a point on BC and divides it in the ratio 3:5 i.e.if BD:DC=3:5 then area of (triangle ADC): area of (triangle ABC)

Answer 1

hope it will help you

Answer 2

Given:

In triangle ΔABC If D is a point on BC and divides it in the ratio 3:5

To Find:

What is ratio of area of ΔADC and area of ΔABC ?

Solution:

Let altitude of both triangle is:

Altitude = H                  ...1)

Here it is given that:

$\mathbf{\dfrac{BD}{DC}=\dfrac{3}{5}}$

Then we can write that:

BD = 3 k

DC = 5k            

Then value of BC is sum of BD and DC:

BC = BD + DC

BC = 3 k + 5 k

So

BC = 8 k              

In case of ΔADC:

Base of ΔADC = DC = 5 k          ...2)

In case of ΔABC:

Base of ΔABC = BC = 8 k          ...3)

We know the formula of area of triangle:

$\mathbf{Area\ of\ Triangle =\dfrac{1}{2}\times Base\times Altitude}$

$\mathbf{\dfrac{ar\Delta ADC}{ar\Delta ABC}=\dfrac{\dfrac{1}{2}\times DC\times H}{\dfrac{1}{2}\times BC\times H}}$

On cancel out numerator and denominator in R.H.S of above equation:

$\mathbf{\dfrac{ar\Delta ADC}{ar\Delta ABC}=\dfrac{ DC}{ BC}}$          ...4)

Putting value of DC and BC in equation 4):

$\mathbf{\dfrac{ar\Delta ADC}{ar\Delta ABC}=\dfrac{ 5k}{ 8k}}$

$\mathbf{\dfrac{ar\Delta ADC}{ar\Delta ABC}=\dfrac{ 5}{ 8}}$    

This is answer.

Ratio of area of ΔADC and area of ΔABC is 5:8