# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 4  and the triangle has an area of 42 . What are the lengths of sides A and B?

Length of sides A & B are $21.1 \left(1 \mathrm{dp}\right)$ unit
Let Altitude (h) be drawn from junction of sides A & B on Side $C$. We know area of the triangle is $\frac{1}{2} \cdot C \cdot h = 42 \mathmr{and} \frac{1}{2} \cdot 4 \cdot h = 42 \therefore h = 21$The altitude meets side C at midpoint and perpendicular to C.$\therefore {h}^{2} + {\left(\frac{C}{2}\right)}^{2} = {A}^{2} \therefore {A}^{2} = {21}^{2} + {\left(\frac{4}{2}\right)}^{2} = 441 + 4 = 445 \therefore A = B = \sqrt{445} = 21.1 \left(1 \mathrm{dp}\right)$unit