Question

Using Heron's Formula, find the area of an isosceles triangle, the measure of one of its equal sides being $a$ units and the third side $2b$ units.

• A. $b\sqrt{(a^2-b^2)}$ $sq.$ units
• B. $a\sqrt{(a^2+b^2)}$ $sq.$ units
• C. $a\sqrt{(a^2-b^2)}$ $sq.$ units
• D. $b\sqrt{(a^2+b^2)}$ $sq.$ units

Answer 1

Answer: (A)

$b\sqrt{(a^2-b^2)}$ $sq.$ units

In an isosceles triangle, the height of the triangle divides it into two right angled triangles
${Equalside}^2={Height}^2+{\left(\frac{Base}{2}\right)}^2$

$\Rightarrow a^2=h^2+{\left(b\right)}^2$
$\Rightarrow h^2=a^2-b^2$
$\Rightarrow h=\sqrt{(a^2-b^2)}$
Area of triangle $=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times \left(2b\right)\times \sqrt{(a^2-b^2)}$

$=b\sqrt{(a^2-b^2)}$ sq. units