Question

9. In the figure, AD is the bisector of angle BAC. If AB=15cm (3) and AC-20cm, find (a) BD:DC (b) Write the ratio of areas of triangle ABD and triangle ADC (c) If area of triangle ABD is 21cm*,then find area of triangle ADC​

Answer 1

Step-by-step explanation:

Hope this helps you thank you.

I am giving example so you can do that yourself.

Answer 2

Internal Angle Bisector Theorem

Given:

A triangle ABC  is given in which AB=15cm and AC= 20cm.

To find:

BD:DC = ?

$\frac{ar(\triangle ABD)}{ar(\triangle ADC)}=?$

If area of ΔABD is $21cm^{2}$, then find $ar(\triangle ADC)=?$

Explanation:

Internal Angle Bisector Theorem:

This theorem states that the angle bisector of a triangle divides the opposite side into two parts proportional to the other two sides of the triangle.

$\frac{CD}{DB}=\frac{AC}{AB}$

So,

$\frac{CD}{BC-CD} =\frac{15}{20} =\frac{3}{4}\\\\\frac{CD}{CD(\frac{BC}{CD} -1)} =\frac{3}{4}\\\\ \frac{BC}{CD}-1=\frac{4}{3} \\\\\frac{BC}{CD} =\frac{7}{3}$

Hence , $BD:DC=\frac{7}{3}$

From area theorem , we know that, ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

So, we can do construction in attached figure to and prove

$\triangle ABD \\\&\\\triangle ADC$ as similar

So,

$\triangle ABD\sim\triangle ADC$

Then,

$\frac{ar(ABD}{ar(ADC)} =\frac{AB^2}{AC^2} =\frac{15^2}{20^2} =\frac{9}{16}$

Hence the required ratio is  $\frac{9}{16}$.

if area of $\triangle ABD$ is $21 cm^2$, which means,

$\frac{9}{16} =\frac{21}{ar(ADC)} \\\\ar(ADC)=37.3cm^2$

Hence the area is $37.3 cm^2$ .