Question

Find the area of an isosceles triangle with two equal sides as 5 cm each and the third side as 8 cm.

Answer 1

Area of isosceles traingle is 12 cm².

Given:

• An isosceles traingle.
• With one side 8 cm and two equal sides, are each of 5 cm.

To find:

• Find area of isosceles triangle.

Solution:

Formula to be used:

Apply Heron's formula: If a,b and c are the length of sides of a triangle, then area of triangle is given by

$\begin{gathered}\bf Ar.(\triangle)= \sqrt{s(s - a)(s - b)(s - c)} \\ \end{gathered}$

where, s is semi perimeter of triangle.

Step 1:

Find the semi perimeter of triangle.

Let sides are say

$\begin{gathered}a = 8 \\ b = 5 \\ c = 5 \\ \end{gathered}$

So,

$\begin{gathered}s = \frac{8+5 + 5 }{2} \\ \end{gathered}$

or

s=9 cm

Step 2:

Find area of triangle.

Apply Heron's formula.

$\begin{gathered}Ar.(\triangle)= \sqrt{9(9 - 8)(9 - 5)(9 - 5)} \\ \end{gathered}$

or

$\begin{gathered}Ar.(\triangle) = \sqrt{9 \times 1 \times 4 \times 4} \\ \end{gathered}$

or

$\begin{gathered}Ar.(\triangle) = 3 \times 4 \\ \end{gathered}$

or

$\begin{gathered}\bf Ar.(\triangle) = 12 \: {cm}^{2} \\ \end{gathered}$

Thus,

Area of isosceles triangle is 12 cm².

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Answer 2

Area of the given  isosceles traingle is $12 \mathrm{~cm}^{2} .$

Isosceles triangle

An isosceles triangle has both two equal sides and two equal angles

Area of triangle $=\sqrt{\mathbf{s}(\mathbf{s}-\mathbf{a})(\mathbf{s}-\mathbf{b})(\mathbf{s}-\mathbf{c})}$

Given:

An isosceles traingle

One side $8 \mathrm{~cm}$ and two remaining equal sides, are each of$5 \mathrm{~cm} .$

Step by step solution

Step1: Area of isosceles triangle is given by:

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

where,

a,b and c are the side length of the triangle,

and $s$ is semi perimeter of triangle.

$S = (a+b+c)/2$

Step 2:Calculate  the semi perimeter of the  triangle.

Consider

$\begin{aligned}&a=8 \\&b=5 \\&c=5\end{aligned}$

$s=\frac{8+5+5}{2}$

$\mathrm{s}=9 \mathrm{~cm}$

Step 2: Substitute the value in the given formula

$Area of triangle =\sqrt{9(9-8)(9-5)(9-5)}$

$=\sqrt{9 \times 1 \times 4 \times 4}$

Area of triangle = $12 \mathrm{~cm}^{2}$

Hence Area of isosceles triangle is$12 \mathrm{~cm}^{2} .$

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