Question

An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find
the area of the triangle.​

Answer 1

Answer:

Area of the isosceles triangle = $\sf{9\sqrt{15}{cm}^{2}}$

Step-by-step Explanation:

Given :

Perimeter of the isosceles triangle = 30cm

Equal sides are of = 12 cm

To find :

The area of the triangle

Concept :

First find the third side of the isosceles triangle .

So , use this equation to find the third side -:

$\boxed{\sf{30cm-(12\times2)cm}}$

After that to find the area of the isosceles triangle . And first divide the sum of a,b and c with 2.

After that to find the area of the isosceles triangle use this formula -:

$\boxed{\sf{A=\sqrt{s(s-a)\times(s-b)\times(s-c)}}}$

A= Area of the isosceles triangle

s = Division a,b and c with 2

a = First side

b = Second side

c = Third side

Solution :

Third side :

$:\implies\sf{30cm-(12\times2)cm}$

$:\implies\sf{30cm-24cm}$

$:\implies\sf{6cm}$

Now , area of the isosceles triangle :

First division -:

$:\implies\sf{s=\dfrac{a+b+c}{2}}$

$:\implies\sf{s=\dfrac{12cm+12cm+6cm}{2}}$

$:\implies\sf{s=\dfrac{\cancel{30}cm}{\cancel{2}}}$

$:\implies\sf{s=15cm}$

Area -:

$:\implies\sf{A=\sqrt{15cm(15cm-12cm)\times(15cm-12cm)\times(15cm-6cm)}}$

$:\implies\sf{A=\sqrt{15cm\times3cm\times3cm\times9cm}}$

$:\implies\sf{A=\sqrt{15cm(9)(9)}}$

$:\implies\sf{A=\sqrt{9\times9}\times\sqrt{15}}$

$:\implies\sf{A=\sqrt{{9}^{2}}\times\sqrt{15}}$

$:\implies\sf{A=(9)\times\sqrt{15}}$

$:\implies\sf{A=9\sqrt{15}{cm}^{2}}$

So , the area of the isosceles triangle is $\sf{9\sqrt{15}{cm}^{2}}$