B Nirmala Shastry Solutions for Exercise 1: EXERCISE 13

$ABCD$ and $BEFG$ are parallelograms and $CE\parallel AG$. Prove that, $\mathrm{Area}\left(\parallel \mathrm{gm} \mathrm{ABCD}\right)=\mathrm{Area}\left(\parallel \mathrm{gm} \mathrm{BEFG}\right)$

$ABCD$ is a parallelogram, $BC$ is produced to $Q$ so that$\angle DQC=90°$. $AP\perp DC$. If $AP=2.5 \mathrm{cm}$,$BC=4 \mathrm{cm}$ and $DQ=5 \mathrm{cm}$, find the length of $AB$.  

$ABCD$ is a parallelogram, and $AF\parallel DE$. Prove that $\mathrm{area}\left(\parallel \mathrm{gm} \mathrm{DEFH}\right)=\mathrm{area}\left(\parallel \mathrm{gm} \mathrm{ABCD}\right)$

In $∆ABC, M$ is the mid-point of $BC, L$ is a point on $AB$ such that $AL=2LB$. Find the area of $∆ALM$ if area of $∆ABC=72 {\mathrm{cm}}^2$.

$ABCD$ is a parallelogram, prove that $\mathrm{area}\left(∆\mathrm{ABC}\right)=\mathrm{area}\left(∆\mathrm{PAD}\right)$.

$ABCD$ is a parallelogram, prove that $\mathrm{area}\left(∆\mathrm{PAB}\right)=\mathrm{area}\left(\mathrm{quadrilateral} \mathrm{ACPD}\right)$.

$D$ is the mid-point of $AC$. $E$ is on $BC$ such that $BE=\frac{1}{3}BC$. $\mathrm{area}\left(∆\mathrm{ABC}\right)=48 {\mathrm{cm}}^2$. Find the area of $∆ADB$ and $∆BDE$.

Area of $∆PCD=10 {\mathrm{cm}}^2$. Find the area of $∆ADQ$ and $\parallel gm ABCD$.