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

First,let the third side be x.

It is given that the length of the equal sides us 12 cm and it's perimeter is 30 cm.

So,

$30=12+12+x$

$⇒ 30 = 24 + x$

$⇒24 + x = 30$

$⇒ x= 30 - 24$

$⇒ x = 6$

So,the length of the third side is 6 cm.

Thus,the semi perimeter of the isosceles triangle (s) = 30/2 cm =15 cm

By using Heron's Formula,

Area of the triangle,

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

$= \sqrt{15(15 - 12)(15 - 12)(15 - 6)} \: {cm}^{2}$

$= \sqrt{15 \times 3 \times 3 \times 9} \: {cm}^{2}$

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

Answer 2

Given :

• Perimeter of Isosceles triangle = 30 cm

• Equal sides = 12 cm

To Find :

• Area of Triangle = ?

Solution :

Each equal sides of isosceles triangle is 12 cm.

Let's Find 3rd Side ;

Perimeter of Isosceles triangle = a + b + c

Here , 2a + b = 30

• a = 12

• b = ?

$\sf \to \: 2a \: + b \: = 30 \\ \\ \sf \to \: 2 \times 12 + b = 30 \\ \\ \sf \to \: 24 + b = 30 \\ \\ \sf \to \: b = 30 - 24 \\ \\ \bf\to \: b = 6$

So, Third side is 6 cm.

Let's Find Area ;

${\underline{\frak{ area \: of \: Isosceles \:triangle = \sqrt{s(s - a)(s - b)(s - c)} }}}$

• S = semiperimetre

$\sf\to \: semiperimetre \: = \frac{perimetre}{2} \\ \\ \to \sf \: \frac{30}{2} = 15$

Semiperimetre = 15 cm

$\sf \to area = \large \: \sqrt{s(s - a)(s - a)(s - b)}$

$\\ \\ \sf \to \: \sqrt{15(15 - 12)(15 - 12)(15 - 6)}$

$\\ \\ \sf\to \: \sqrt{15(3)(3)(9)}$

$\\ \\ \sf \to \: \sqrt{15 \times 3 \times 3 \times 9}$

$\\ \\ \sf \to \: \sqrt{15 \times 9 \times 9}$

$\\ \\ \bf \to \: 9 \sqrt{15} \: \: cm²$

Area Of Triangle = 9√15 cm²

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