M L Aggarwal Solutions for Exercise 4: Multiple Choice Questions

In the adjoining figure, $\mathrm{△}\mathrm{ABC}~\mathrm{△}\mathrm{QPR}$. The value of $x$ is

In the adjoining figure, two line segments AC and BD intersect each other at the the point P such that $\mathrm{PA} = 6 \mathrm{cm}, \mathrm{PB}=3 \mathrm{cm}, \mathrm{PC}=2.5 \mathrm{cm}$, $\mathrm{PD}=5\mathrm{cm},\mathrm{\angle }\mathrm{APB}={50}^{\circ }\text{ and }\mathrm{\angle }\mathrm{CDP}={30}^{\circ }$. Then $\mathrm{\angle }\mathrm{PBA}$ is equal to 

In the adjoining figure, $\mathrm{DE}\parallel \mathrm{BC}$ and all measurements are in centimetres. This length of AE is

In the adjoining figure, $\mathrm{PQ}\parallel \mathrm{CA}$ and all lengths are given in centimetres. This length BC is 

In the adjoining figure, $\mathrm{MN}\parallel \mathrm{QR}.$ If $\mathrm{PN}=3.6\mathrm{cm}$ $\mathrm{NR}=2.4\mathrm{cm}\text{ and }\mathrm{PQ}=5\mathrm{cm}$, then $\mathrm{PM}$ is 

It is given that $\mathrm{△}\mathrm{ABC}~\mathrm{△}\mathrm{PQR}\text{ with }\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{1}{3}\text{, then }\frac{\text{ area of }\mathrm{△}\mathrm{PQR}}{\text{ area of }\mathrm{△}\mathrm{ABC}}$ is equal to 

If $\mathrm{△}\mathrm{ABC}~\mathrm{△}\mathrm{PQR},$ area $\mathrm{△}\mathrm{ABC}=81{\mathrm{cm}}^2$ area of $\mathrm{△}\mathrm{PQR}=144{\mathrm{cm}}^2\text{ and }\mathrm{QR}=6\mathrm{cm}$ then length of $\mathrm{BC}$ is

In the adjoining figure, $\mathrm{DE}\parallel \mathrm{CA}\text{ and D}$ is a point on $\mathrm{BC}$ such that $\mathrm{BD}:\mathrm{DC}=2:1.$ The ratio of the area of $\mathrm{△}\mathrm{ABC}$ to area of $\mathrm{△}\mathrm{BDE}$ is