R K Bansal Solutions for Exercise 1: EXERCISE

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find :

$\frac{\mathrm{AD}}{\mathrm{AB}}$

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find:

$\frac{\mathrm{AE}}{\mathrm{AC}}$

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find:

$\mathrm{DE}=2.4\mathrm{cm},$ find the length of $\mathrm{BC}.$

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find:

$\mathrm{BC}=4.8\mathrm{cm},$ find the length of $\mathrm{DE}.$

In the given figure, $\mathrm{AX}:\mathrm{XB}=3:5$

If the length of $\mathrm{BC}$ is $k \mathrm{cm}$ and the length $\mathrm{XY}$ is $18 \mathrm{cm}$, then find the value of $k$.

In the given figure, $\mathrm{AX}:\mathrm{XB}=3:5$

Find the ratio between the areas of trapezium $\mathrm{XBCY}$ and triangle $\mathrm{ABC}.$

$\mathrm{ABC}$ is a triangle. $\mathrm{PQ}$ is a line segment intersecting $\mathrm{AB}$ in $\mathrm{P}$ and $\mathrm{AC}$ in $\mathrm{Q}$ such that $\mathrm{PQ}\mathit{\parallel }\mathrm{BC}$ and divides  triangle $\mathrm{ABC}$ into two parts equal in area. The value of ratio $\mathrm{BP}:\mathrm{AB}$ is $\frac{2-\sqrt{a}}{b}$. Find $a+b$.

A triangle $\mathrm{ABC}$ is enlarged, about the point $\mathrm{O}$ as centre of enlargement, and the scale factor is $3.$ Find :

$\mathrm{A}\text{'}\mathrm{B}\text{'},$ if $\mathrm{AB}=4\mathrm{cm}.$