M L Aggarwal Solutions for Exercise 1: EXERCISE 14

Any point $D$ is taken on the side $BC$ of a triangle $ABC$ and $AD$ is produced to $E$ such that $AD=DE$. Prove that area of triangle $BCE=$ area of triangle $ABC$.

$ABCD$ is a rectangle and $P$ is the mid-point of $AB. DP$ is produced to meet $CB$ at $Q$. Prove that area of rectangle $ABCD=$ area of the triangle $DQC$.

In the figure given below, the perimeter of the parallelogram is $42 \mathrm{cm}$. Calculate the lengths of the sides of the parallelogram.

in the given figure perimeter of a triangle is $37 \mathrm{cm}$. if the lengths of the altitude $AM, BN$ and $CL$ are $5x,6x \mathrm{and} 4x$, respectively, calculate the length of the sides of triangle $ABC$.

In the figure given below $ABCD$ is parallelogram. $P$ is the point on the side $DC$, such that the area of the triangle $DAP=25 {\mathrm{cm}}^2$ and the area of the triangle$BCP=15 {\mathrm{cm}}^2$. Find the $DP:PC$.

In the figure given below, $\mathrm{BC}\Vert \mathrm{AE}$ and $\mathrm{CD}\Vert \mathrm{BE}$. Prove that area of $∆\mathrm{ABC}=$ area of $∆\mathrm{EBD}$.

In the figure given below, $\mathrm{ABC}$ is right angled triangle at $A$. $AGFB$ is a square on the side $AB$ and $BCDE$ is a square on the hypotenuse $BC$. If $AN\perp ED$, prove that $∆\mathrm{BCF}\cong ∆\mathrm{ABE}$.

In the figure given below, $\mathrm{ABC}$ is right angled triangle at $A$. $AGFB$ is a square on the side $AB$ and $BCDE$ is a square on the hypotenuse $BC$. If $AN\perp ED$, prove that area of square $\mathrm{ABFG}=$ area of rectangle $BENM$.