ΔPQR is an isosceles triangle and PQ = PR = 2a unit, QR = a unit. Draw PX ⊥ QR, and find the length of PX.

$$\eqalign{ & {\text{Given that,}} \cr & PQ = PR = 2a, \cr & QR = a, \cr & QX = \frac{a}{2} \cr & \Rightarrow {\left( {2a} \right)^2} = {\left( {PX} \right)^2} + {\left( {\frac{a}{2}} \right)^2} \cr & \Rightarrow 4{a^2} - \frac{{{a^2}}}{4} = {\left( {PX} \right)^2} \cr & \Rightarrow {\left( {PX} \right)^2} = \frac{{15{a^2}}}{4} \cr & \Rightarrow PX = \frac{{\sqrt {15} a}}{2}{\text{ Answer}} \cr} $$