Kesab Chandra Nag Solutions for Exercise 1: EXERCISE-3

$E$ is any point on the side $BC$ of the parallelogram $ABCD$. If $∆ABE$ and $∆DCE$ have the areas $10 \mathrm{sq}. \mathrm{cms}$ and $12 \mathrm{sq}. \mathrm{cms}$ respectively, then the area of the parallelogram $ABCD=$

The ratio of the areas of a triangle and a parallelogram having the Same base and the same height is

The area of a rhombus is $\_\_\_\_$ the area of triangle having same base and between the same parallels

The length of the base $AB$ of the parallelogram $ABCD$ is $10 \mathrm{cm}$. If the area of $ABCD$ be $110 \mathrm{sq}. \mathrm{cm}$, then the height of $ABCD$ with respect to the base $AB$ is

$\mathrm{P}$ is any point on the side $\mathrm{BC}$ of the parallelogram $\mathrm{ABCD}$. Extended $\mathrm{DP}$ intersects $\mathrm{AB}$ at $\mathrm{Q}$. If the area of the $∆\mathrm{ACD}$ is $22 \mathrm{sq}. \mathrm{cm}$, then the area of the $∆\mathrm{DCQ}$ is

$\mathrm{D}$ is the mid-point of the sides $\mathrm{BC}$ of the $△\mathrm{ABC}$ and $\mathrm{P}$ is any point on $\mathrm{BC}$. Let us join $\mathrm{P}, \mathrm{A}$. The line segment drawn through $\mathrm{D}$, parallel to $\mathrm{PA}$ intersects $\mathrm{AB}$ at $\mathrm{Q}$. Prove that $△\mathrm{BPQ}=\frac{1}{2}△\mathrm{ABC}$.

Prove that sum of the perpendiculars drawn from any interior point of an equilateral triangle to its sides equal to its height.

In isosceles triangle $\mathrm{ABC}$. $\mathrm{D}$ and $\mathrm{E}$ are the midpoints of the sides $\mathrm{AB} \mathrm{and} \mathrm{AC}$. If $BE \mathrm{and} \mathrm{CD}$ intersects each other at $\mathrm{F}$, prove that $△\mathrm{BDE}=3△\mathrm{DEF}$.