MAT Exam 2022 Paper

Concept:Use the trisection of the base, isosceles triangle properties, exterior angle theorem, and area formula to find the equal side.Explanation:Let $QT = TZ = ZR = n$.So $QR = 3n$.Since $PQ = PR$, $\triangle PQT \cong \triangle PRZ$ by SAS, hence $PT = PZ$.Therefore, $\triangle PTZ$ is isosceles, giving $\angle PTZ = \angle PZT$.In $\triangle PQT$, by the exterior angle theorem,$\angle PTZ = \angle PQT + \angle QPT$.Given $\angle TPZ = \angle PQR = \angle PQT$,$\angle PTZ = \angle TPZ + \angle QPT = \angle QPZ$.Thus $\angle PZT = \angle QPZ$.Since $ZT$ and $ZQ$ lie on the same straight line, $\angle PZT = \angle PZQ$.Hence, in $\triangle PQZ$, $\angle PZQ = \angle QPZ$, so $PQ = QZ$.Now $QZ = QT + TZ = n + n = 2n$.Therefore, $PQ = PR = 2n$.Height of $\triangle PQR$ is$h = \sqrt{PR^2 - \left(\frac{QR}{2}\right)^2}$$= \sqrt{(2n)^2 - \left(\frac{3n}{2}\right)^2}$$= \frac{\sqrt{7}\,n}{2}$.Area of $\triangle PQR$:$\frac{27\sqrt{7}}{4} = \frac{1}{2} \times 3n \times \frac{\sqrt{7}\,n}{2}$.Solving, $n^2 = 9$, so $n = 3$.Thus $PR = 2n = 6$.Answer:$\mathrm{PR} = 6$, i.e. Option D.