# Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm.

Question:

Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm. Find the area of the triangle.

Solution:

Let the height of the triangle be h cm.

Each of the equal sides measures $a=(h+2) \mathrm{cm}$ and $b=12 \mathrm{~cm}$ (base).

Now,
Area of the triangle = Area of the isosceles triangle

$\frac{1}{2} \times$ base $\times$ height $=\frac{1}{4} \times b \sqrt{4 a^{2}-b^{2}}$

$\Rightarrow \frac{1}{2} \times 12 \times h=\frac{1}{4} \times 12 \times \sqrt{4(h+2)^{2}-144}$

$\Rightarrow 6 h=3 \sqrt{4 h^{2}+16 h+16-144}$

$\Rightarrow 2 h=\sqrt{4 h^{2}+16 h+16-144}$

On squaring both the sides, we get:

$\Rightarrow 4 h^{2}=4 h^{2}+16 h+16-144$

$\Rightarrow 16 h-128=0$]

$\Rightarrow h=8$

Area of the triangle $=\frac{1}{2} \times b \times h$

$=\frac{1}{2} \times 12 \times 8$

$=48 \mathrm{~cm}^{2}$

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