Question

Find to the three places of decimals the radius of the circle whose area is the sum of the areas of two triangles whose sides are $35,53,66$ and $33,56,65$ measured in centimetres. (Use $\pi =\frac{22}{7}$).

Answer 1

For the first triangle, we have

$a=35$ , $b=53$ and $c=66$

$\therefore s=\frac{a+b+c}{2}=\frac{35+53+66}{2}=77cm$

now ,
${\Delta }_1$ = Area of the first triangle

$\Rightarrow {\Delta }_1=\sqrt{s(s-a)(s-b)(s-c)}$

$\Rightarrow {\Delta }_1=\sqrt{77(77-35)(77-53)(77-66)}=\sqrt{77\times 42\times 24\times 11}$

$\sqrt{7\times 11\times 7\times 6\times 6\times 4\times 11}=\sqrt{7^2\times {11}^2\times 6^2\times 2^2}$

$\Rightarrow {\Delta }_1=7\times 11\times 6\times 2=924c{m}^2$

For the second triangle , we have

$a=33,b=56,c=65$

$\therefore s=\frac{a+b+c}{2}=\frac{33+56+65}{2}=77cm$

Area of the second triangle

$\Rightarrow {\Delta }_1=\sqrt{s(s-a)(s-b)(s-c)}$

$\Rightarrow {\Delta }_1=\sqrt{77(77-33)(77-56)(77-65)}=\sqrt{77\times 44\times 21\times 12}$

$\Rightarrow {\Delta }_1=\sqrt{7\times 11\times 4\times 11\times 3\times 7\times 3\times 7}$

$\Rightarrow {\Delta }_1=\sqrt{7^2\times {11}^2\times 4^2\times 3^2}=7\times 11\times 4\times 3=924c{m}^2$

Let $r$ be the radius of the circle . Then
Area of the circle = sum of the area of two triangles
Area related to circles

$\Rightarrow \pi r^2={\Delta }_1+{\Delta }_2$

$\Rightarrow \pi r^2=924+924$

$\Rightarrow \frac{22}{7}\times r^2$ = 1848

$\Rightarrow r^2=1848\times \frac{7}{22}=3\times 4\times 7\times 7$
$\Rightarrow r=\sqrt{3\times 2^2\times 7^2}=2\times 7\times \sqrt{3}=14\sqrt{3}=14\times 1.732=24.248cm$

Therefore, the radius of the circle is $24.248cm$.