Question

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

• A. $61.22$ cm
• B. $42.22$ cm
• C. $12.22$ cm
• D. $24.22$ cm

Answer 1

The correct option is D $24.22$ cm

Given that two triangle, first having sides $35$ cm, $53$ cm, $66$ cm and second having sides $56$ cm, $33$ cm, $65$ cm.
Let us take first triangle with sides $35$ cm, $53$ cm, $66$ cm.
Semiperimeter of triangle, $S$ $=\frac{a+b+c}{2}$, where $a,b,c$ are sides of triangle.
$\therefore S=\frac{35+53+66}{2}=77$
Using Heron's formula,
Area of triangle, $A$ $=\sqrt{s(s-a)(s-b)(s-c)}$
$A$ $=\sqrt{77(77-35)(77-53)(77-66)}$

$=\sqrt{77\times 42\times 24\times 11}$

$=924{cm}^2$
Taking second triangle with sides $56$ cm, $33$ cm, $65$ cm.
Semiperimeter of triangle, $S$ $=\frac{a+b+c}{2}$, where $a,b,c$ are sides of triangle.
$\therefore S=\frac{56+33+65}{2}=77$
Using Heron's formula,
Area of triangle, $A$ $=\sqrt{s(s-a)(s-b)(s-c)}$
$A$ $=\sqrt{77(77-56)(77-33)(77-65)}$

$=\sqrt{77\times 21\times 44\times 12}$

$=924{cm}^2$
Given, Sum of Area of two triangle $=$ Area of circle with radius '$r$'
$\Rightarrow 924+924$ $=$ $\pi r^2$
$\Rightarrow 1848=\frac{22}{7}\times r^2$

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

$\Rightarrow r=24.22$ cm

Radius of circle $=24.22$ cm