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 the triangle = 30 cm
• Equal sides of isosceles triangle = 12 cm

• To find

• Area of the triangle

• Solution

Let the third side of the triangle be x.

Using he formula,

Perimeter = Sum of all sides

Substituting their values,

⟶ 30 = 12 + 12 + x

⟶ 30 = 24 + x

⟶ 30 - 24 = x

⟶ 6 = x

The value of x = 6

Third side of triangle = 6 cm

Using formula,

Heron's formula = √s(s - a)(s - b)(s - c)

To find s,

⟶ s = (a + b + c)/2

Substituting their values,

⟶ (12 + 12 + 6)/2

⟶ 30/2

⟶ 15

s = 15 cm

Substituting the values in heron's formula-

⟶ √15(15 - 12)(15 - 12)(15 - 6)

⟶ √15(3)(3)(9)

⟶ 9√15 cm²

Therefore, area of isosceles triangle = 9√15 cm²