



 
Hi Karan, I drew an isosceles triangle and I will use it to help you get started. $D$ is the midpoint of $BC$ and hence, since triangle $ABC$ is isosceles the triangles $ACD$ and $ADC$ are congruent and angle $ADC$ is a right angle. Suppose that $h = AD,$ the height of the triangle and $b = BC,$ the length of its base. You know the area of the triangle is 4 square centimeters and hence \[\frac12 b \times h = 4 \mbox{ square centimeters.}\] Pythagoras Theorem applied to triangle $ADC$ hives a second equation in $b$ and $h.$ Solve the first equation for $h$ and substitute into the second. Square both sides and simplify. This gives you a fourth degree equation in $b$ but if you look closely it is a quadratic in $b^{2}.$ Solve this quadratic. Can you complete the problem now? Write back if you need more assistance,  


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