NCERT Solutions for Chapter: Areas of Parallelograms and Triangles, Exercise 4: Exercise

The medians $BE$ and $CF$ of a triangle $ABC$ intersect at  $G$. Prove that the area of $\Delta \mathrm{GBC}=$ area of the quadrilateral $AFGE$.

 In given figure, $\text{CD}\left|\right|\text{AE }$and $\text{CY}\left|\right|\text{BA}$. Prove that $\mathrm{ar}\left(\Delta \mathrm{CBX}\right)=\mathrm{ar}\left(\Delta AXY\right)$

$ABCD$is a trapezium in which $AB\left|\right|DC, DC=30 \mathrm{cm}$ and $AB=50 \mathrm{cm}$. If $X$ and $Y$ are respectively the mid-points of $AD$ and $\mathrm{BC}$, prove that: $ar\left(DCYX\right)=\frac{7}{9}ar\left(XYBA\right)$.

In $\Delta \mathrm{ABC}$, if  $L$ and $M$ are the points on $\mathrm{AB}$ and $AC$, respectively such that $LM\left|\right|BC.$ Prove that $ar\left(\text{LOB}\right)=\mathrm{ar}\left(\text{MOC}\right)$.

In the given figure, $ABCDE$ is any pentagon. $BP$ drawn parallel to$AC$ meets $DC$ produced at $P$and $EQ$drawn parallel to $AD$ meets $CD$ produced at $Q.$ Prove that

$\text{ ar }\left(\mathrm{ABCDE}\right)=ar\left(\mathrm{APQ}\right)$

If the medians of a $\Delta \mathrm{ABC}$ intersect at $G,$show that $\text{ ar }\left(\mathrm{AGB}\right)=ar\left(\mathrm{AGC}\right)=ar\left(\mathrm{BGC}\right)=\frac{1}{3}ar\left(ABC\right)$

 In given figure, $X$ and $Y$ are the midpoints of $AC$ and $AB$ respectively, $QP\left|\right|BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $ar\left(\text{ABP}\right)=ar\left(\text{ACQ}\right)$.

In the given figure, $ABCD$and $AEFD$ are two parallelograms. Prove that $$$ar\left(∆PEA\right)=ar\left(QFD\right)$. [ Hint: join $PD$]