R S Aggarwal Solutions for Chapter: Area of a Trapezium and a Polygon, Exercise 1: EXERCISE

A field is in the form of a trapezium. Its area is $1586 {\mathrm{m}}^2$ and the distance between its parallel sides is $26 \mathrm{m}$. If one of the parallel sides is $84 \mathrm{m}$, find the other.

The area of a trapezium is $405 {\mathrm{cm}}^2$. Its parallel sides are in the ratio $4:5$ and the distance between them is $18 \mathrm{cm}$. Find the length of each of the parallel sides.

The area of a trapezium is $180 {\mathrm{cm}}^2$ and its height is $9 \mathrm{cm}$. If one of the parallel sides is longer than the other by $6 \mathrm{cm}$, find the two parallel sides.

In a trapezium-shaped field, one of the parallel sides is twice the other. If the area of the field is $9450 {\mathrm{m}}^2$ and the perpendicular distance between the two parallel sides is $84 \mathrm{m}$, find the length of the longer of the parallel sides.

The length of the fence of a trapezium-shaped field $\mathrm{ABCD}$ is $130 \mathrm{m}$ and side $\mathrm{AB}$ is perpendicular to each of the parallel sides $\mathrm{AD}$ and $\mathrm{BC}$. If $\mathrm{BC}=54 \mathrm{m}$, $\mathrm{CD}=19 \mathrm{m}$ and $\mathrm{AD}=42 \mathrm{m}$, find the area of the field.

In the given figure, $\mathrm{ABCD}$ is a trapezium in which $\mathrm{AD}\parallel \mathrm{BC}$, $\angle \mathrm{ABC}=90°$, $\mathrm{AD}=16 \mathrm{cm}$, $\mathrm{AC}=41 \mathrm{cm}$ and $\mathrm{BC}=40 \mathrm{cm}$. Find the area of the trapezium.

The parallel sides of a trapezium are $20 \mathrm{cm}$ and $10 \mathrm{cm}$. Its nonparallel sides are both equal, each being $13 \mathrm{cm}$. Find the area of the trapezium.

The parallel sides of a trapezium are $25 \mathrm{cm}$ and $11 \mathrm{cm}$, while its nonparallel sides are $15 \mathrm{cm}$ and $13 \mathrm{cm}$. find the area of the trapezium.