If an isosceles triangle has base b and equal sides a, then its area w

1.	If an isosceles triangle has base b and equal sides a, then its area will be
Answer» $\underline {\blue {Given :}}$ $ABC \:is \:an \: isosceles \: triangle.$ $AB = AC = a , \:and \: BC = b$ $\underline {\blue {Construction :}}$ $Draw \: AD \: perpendicular \:to \: BC$ $\underline {\blue {Solution:}}$ $In \: \triangle ABD , \\\angle {ADB} = 90\degree ,\\BD = \frac{BC}{2} = \frac{b}{2} \: ( Altitude \:bisects \: base )$ $AB^{2} = BD^{2} + AD^{2} \\(By \: Pythagoras \: theorem )$ $\implies a^{2} = \big(\frac{b}{2}\big)^{2} + h^{2}$ $\implies h^{2} = a^{2} - \big(\frac{b}{2}\big)^{2}$ $\implies h^{2} = \frac{(4a^{2} - b^{2})}{4}$ $\implies h = \sqrt{\frac{(4a^{2} - b^{2})}{4}}$ $\implies h = \frac{\sqrt{(4a^{2} - b^{2})}}{2}$ $\red { Area \: of \:\triangle ABC} = \frac{1}{2} \times base \times corresponding \:height\\= \frac{1}{2} \times BC \times AD$ $= \frac{1}{2} \times b \times \frac{\sqrt{(4a^{2} - b^{2})}}{2}[From \:(1) ]$ $= \frac{b\sqrt{(4a^{2} - b^{2})}}{4}$ THEREFORE., $\red { Area \: of \:\triangle ABC}$ $\green { = \frac{b\sqrt{(4a^{2} - b^{2})}}{4}}$ •••♪