In $\Delta PQR$,$PM$ is a median. If $\text{ar}(\Delta PMQ) = 36 \text{ cm}^2$,then find $\text{ar}(\Delta PMR)$ in $\text{cm}^2$.

  • A
    $53$
  • B
    $50$
  • C
    $36$
  • D
    $75$

Explore More

Similar Questions

In $\Delta ABC$,$AB = 9 \, \text{cm}$ and $BC = 12 \, \text{cm}$. If $x < \text{perimeter of } \Delta ABC < y$,then find the values of $x$ and $y$.

Difficult
View Solution

In $\Delta ABC$,$\angle A = 90^{\circ}$ and $AB = AC$. The bisector of $\angle A$ meets $BC$ at $D$. Prove that $BC = 2 AD$.

In $\Delta ABC$,$AB = AC$ and $\angle B = 75^{\circ}$,then $\angle C = \dots$ (in $^{\circ}$)

In $\Delta ABC$,$AD$ is a median. If $ar(\Delta ABC) = 50 \, cm^2$,then $ar(\Delta ADC) = \dots \dots \dots cm^2$.

In $\Delta PQR$,$\angle P = 42^{\circ}$ and $\angle Q = 75^{\circ}$,then find $\angle R$. (in $^{\circ}$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo