In Delta ABC,AD is a median. P and Q are the | vedclass

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(C) $1$. Since $AD$ is the median of $\Delta ABC$,it divides the triangle into two triangles of equal area.$\operatorname{ar}(\Delta ABD) = \frac{1}{2

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) $1$. Since $AD$ is the median of $\Delta ABC$,it divides the triangle into two triangles of equal area.$\operatorname{ar}(\Delta ABD) = \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{2} 	imes 72 = 36 \, 	ext{cm}^2$.$2$. In $\Delta ABD$,$Q$ is the midpoint of $AD$. Therefore,$BQ$ is a median of $\Delta ABD$.$\operatorname{ar}(\Delta ABQ) = \frac{1}{2} \operatorname{ar}(\Delta ABD) = \frac{1}{2} 	imes 36 = 18 \, 	ext{cm}^2$.$3$. In $\Delta ABQ$,$P$ is the midpoint of $AB$. Therefore,$QP$ is a median of $\Delta ABQ$.$\operatorname{ar}(\Delta APQ) = \frac{1}{2} \operatorname{ar}(\Delta ABQ) = \frac{1}{2} 	imes 18 = 9 \, 	ext{cm}^2$.

In $\Delta ABC$,$AD$ is a median. $P$ and $Q$ are the midpoints of $AB$ and $AD$ respectively. If $\operatorname{ar}(\Delta ABC) = 72 \, \text{cm}^2$,then $\operatorname{ar}(\Delta APQ) = \dots \text{cm}^2$.

Similar Questions

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

Write True or False and justify your answer:
$PQRS$ is a parallelogram whose area is $180 \, cm^{2}$ and $A$ is any point on the diagonal $QS$. The area of $\triangle ASR = 90 \, cm^{2}$.

In $\Delta ABC$,$\angle B = 90^{\circ}$ and $BM$ is an altitude to the hypotenuse $AC$. If $AB = 12 \, cm$ and $BC = 16 \, cm$,then find the length of $BM$ in $cm$.

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 $\text{ar}(ABC) = 24 \text{ cm}^2$.

$(1)$ If a planar region formed by a figure $T$ is made up of two non-overlapping planar regions formed by figures $P$ and $Q$,then $\operatorname{ar}(T) = \dots$
$(2)$ Area of a parallelogram $= \dots$

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