In Delta PQR,PM is a median and N is the midp | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $1$. In $\Delta PQR$,$PM$ is a median,so it divides the triangle into two triangles of equal area: $	ext{ar}(PQM) = 	ext{ar}(PRM) = \frac{1}{2}

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(A) $1$. In $\Delta PQR$,$PM$ is a median,so it divides the triangle into two triangles of equal area: $	ext{ar}(PQM) = 	ext{ar}(PRM) = \frac{1}{2} 	ext{ar}(PQR)$.$2$. In $\Delta PQM$,$QN$ is a median because $N$ is the midpoint of $PM$. Therefore,$	ext{ar}(PQN) = 	ext{ar}(QNM) = \frac{1}{2} 	ext{ar}(PQM)$.$3$. Given $	ext{ar}(PQN) = 36 	ext{ cm}^2$,we have $	ext{ar}(PQM) = 2 	imes 	ext{ar}(PQN) = 2 	imes 36 = 72 	ext{ cm}^2$.$4$. Since $	ext{ar}(PQM) = \frac{1}{2} 	ext{ar}(PQR)$,it follows that $	ext{ar}(PQR) = 2 	imes 	ext{ar}(PQM) = 2 	imes 72 = 144 	ext{ cm}^2$.

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

Similar Questions

In the figure,$ABCDE$ is any pentagon. $BP$ is drawn parallel to $AC$ and meets $DC$ produced at $P$,and $EQ$ is drawn parallel to $AD$ and meets $CD$ produced at $Q$. Prove that $\operatorname{ar}(ABCDE) = \operatorname{ar}(APQ)$.

In $\Delta PQR$,$\angle Q = 90^{\circ}$,$QR = 21 \text{ cm}$ and $PR = 29 \text{ cm}$,then find the area of $\Delta PQR$ in $\text{cm}^2$.

Write True or False and justify your answer:
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.

In $\Delta ABC$,point $D$ lies on side $BC$. $E$ is the midpoint of $AD$. Prove that,$ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.

The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8 \, cm$ and $6 \, cm$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module