Question

<- The equal aides of the
isosceles triangle are
12 cm, and the perimeter is 30cm. The
area of this triangle is:
a) 9√15 (cm²)
b) 6√15(cm²)
c) 3√15(cm²)
d)√15(cm²)
​

Answer 1

Perimeter of isosceles triangle=30cm
Length of equal sides=12cm
Let third side of triangle=xcm
According to problem,
x+12+12=30
x+24=30
x=30−24
x=6

∴ Third side of triangle=6cm

Using Heron's formula

Area of triangle=
s(s−a)(s−b)(s−c)
​
sq. units

where s = a+b+c/2

s= 30/2

=15

Area of triangle=
/15(15−12)(15−12)(15−6)cm2
=/15×3×3×9
​
=3*3*/15 cm2

=9 /15cm2

∴ Area of triangle=9 /15cm2

Answer 2

Given :

• Equal side of the isosceles triangle = 12 cm

• Perimeter of the isosceles triangle = 30 cm

To find :

The Area of the isosceles triangle.

Solution :

First we have to find the base of the isosceles triangle.

Let the base of the isosceles triangle be x cm.

We know the formula for Perimeter of a triangle I.e,

$\boxed{\bf{P = a + b + c}}$

Where :

• P = Perimeter of the triangle

• a, b and c = Side of the triangle

Using the above equation and substituting the values in it, we get :

$:\implies \bf{P = a + b + c} \\ \\ \\ :\implies \bf{30 = 12 + 12 + x} \\ \\ \\ :\implies \bf{30 - 24 = x} \\ \\ \\ :\implies \bf{6 = x} \\ \\ \\ \boxed{\therefore \bf{x = 6}}$

Hence the case of the triangle is 6.

Area of the isosceles triangle :

We know the formula area of a isosceles triangle i.e,

$\boxed{\bf{A = \dfrac{1}{4}b\sqrt{4a^{2} - b^{2}}}}$

Where :

• A = Area of the isosceles triangle.

• b = Base of the isosceles triangle.

• a = Equal side of the isosceles triangle.

Now by using the above equation and substituting the values in it, we get :

$:\implies \bf{A = \dfrac{1}{4}b\sqrt{4a^{2} - b^{2}}}$

$:\implies \bf{A = \dfrac{1}{4} \times 6 \times \sqrt{(4 \times 12^{2}) - 6^{2}}}$

$:\implies \bf{A = \dfrac{1}{\not{4}^{2}} \times \not{6}^{3} \times \sqrt{(4 \times 12 \times 12) - 6^{2}}}$

$:\implies \bf{A = \dfrac{3}{2} \times \sqrt{(4 \times 144) - 36}}$

$:\implies \bf{A = \dfrac{3}{2} \times \sqrt{576 - 36}}$

$:\implies \bf{A = \dfrac{3}{2} \times \sqrt{576 - 36}}$

$:\implies \bf{A = \dfrac{3}{2} \times \sqrt{540}}$

$:\implies \bf{A = \dfrac{3}{2} \times \sqrt{2 \times 2 \times 3 \times 3 \times 15}}$

$:\implies \bf{A = \dfrac{3}{2} \times 2 \times 3 \sqrt{15}}$

$:\implies \bf{A = \dfrac{3}{2} \times 6\sqrt{15}}$

$:\implies \bf{A = \dfrac{3}{\not{2}} \times \not{6\sqrt{15}}}$

$:\implies \bf{A = 3 \times 3\sqrt{15}}$

$:\implies \bf{A = 9\sqrt{15}}$

$\underline{\boxed{\therefore \bf{A = 9\sqrt{15}\:cm^{2}}}}$

Hence the area of the isosceles triangle is 9√15 cm².