In $\Delta PQR$,$PM$ is a median and $N$ is the midpoint of $PM$. If $\text{ar}(PQN) = 36 \text{ cm}^2$,then $\text{ar}(PQR) = \dots \text{ cm}^2$.

  • A
    $144$
  • B
    $9$
  • C
    $72$
  • D
    $18$

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Similar Questions

In $\Delta PQR$,$M$ and $N$ are the midpoints of $PQ$ and $PR$ respectively. $X$ is any point on $QR$. Prove that,$ar(MXN) = \frac{1}{4} ar(PQR)$.

In square $ABCD$,$AC = 16 \text{ cm}$,then $\operatorname{ar}(ABCD) = \dots \text{ cm}^2$.

$AC$ is one of the diagonals of quadrilateral $ABCD$. $BM$ and $DN$ are altitudes on $AC$ from $B$ and $D$ respectively. If $AC = 18 \, cm$,$BM = 10 \, cm$,and $DN = 6 \, cm$,then $ar(ABCD) = \dots \dots \, cm^2$.

In $\Delta ABC$,medians $AD$,$BE$,and $CF$ intersect at point $G$. Prove that,$ar(GAB) = ar(GBC) = ar(GCA) = \frac{1}{3} ar(ABC)$.

$A$ point $E$ is taken on the side $BC$ of a parallelogram $ABCD$. $AE$ and $DC$ are produced to meet at $F$. Prove that $\operatorname{ar}(\triangle ADF) = \operatorname{ar}(ABFC)$.

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