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

Answer: Option D

Solution

Answer: Option D

Solution:

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