Question

The sides of two triangles are $35cm,53cm,66cm$ and $56cm,33cm,65cm$ respectively. Area of circle is equal to sum of the areas of these two triangles. Find the radius of the circle $(\sqrt{3}=1.73)$.

Answer 1

Let the sides of the first triangle be $a,b$ and $c$.
$a=35cm$
$b=53cm$
$c=66cm$

Perimeter of triangle $2s=35cm+53cm+66cm$

$\Rightarrow 2s=154cm$

$\Rightarrow s=\frac{154}{2}=77cm$

$\Rightarrow s-a=77cm-35cm=42cm$

$s-b=77cm-53cm=24cm$

$s-a=77cm-66cm=11cm$

By Herons formula,

Area of first triangle,

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

$\Rightarrow {\Delta }_1=\sqrt{77cm\times 42cm\times 24cm\times 11cm}$

$\Rightarrow {\Delta }_1=\sqrt{77\times 42\times 24\times 11}c{m}^2$

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

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

$\Rightarrow {\Delta }_1=924c{m}^2$ ---(2)

Let the sides of the first triangle be

Let the sides of the second triangle be $x,y$ and $z$_.
$x=56cm$

$y=33cm$

$z=65cm$
Perimeter of triangle $2s=56cm+33cm+65cm$

$\Rightarrow 2s=154cm$

$\Rightarrow s=77cm$

$\Rightarrow s-x=77cm-56cm=21cm$

$\Rightarrow s-a=77cm-33cm=44cm$

$\Rightarrow s-a=77cm-65cm=12cm$

By Herons formula,

Area of first triangle,

${\Delta }_2=\sqrt{s(s-x)(s-y)(s-z)}$

$\Rightarrow {\Delta }_2=\sqrt{77cm\times 21cm\times 44cm\times 12cm}$

$\Rightarrow {\Delta }_2=\sqrt{77\times 21\times 44\times 12}c{m}^2$

$\Rightarrow {\Delta }_2=\sqrt{7\times 11\times 7\times 3\times 11\times 4\times 4\times 3}c{m}^2$

$\Rightarrow {\Delta }_2=11\times 7\times 4\times 3c{m}^2$

$\Rightarrow {\Delta }_2=11\times 7\times 4\times 3c{m}^2$

$\Rightarrow {\Delta }_2=924c{m}^2$---(2)

Let the radius of required circle be $r$.

$\Rightarrow$ Area of circle is $=\pi r^2$ sq. units
It is given that,

Area of circle is equal to sum of the areas of triangles.

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

$\Rightarrow \pi r^2=924c{m}^2+924c{m}^2$

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

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

$\Rightarrow r^2=7\times 84c{m}^2$

$\Rightarrow r^2=7\times 4\times 3\times 7c{m}^2$

$\Rightarrow r=7\times 2\times \sqrt{3}cm$

$\Rightarrow r=14\times 1.73cm$ [Since, $\sqrt{3}=1.73$]

$\Rightarrow r=24.22cm$

Hence, the radius of required circle is $24.22cm$.