V K Raman and S N Sharma Solutions for Exercise 3: EXERCISE 31(C)

The parallel sides of an isosceles trapezium are $24 \mathrm{cm}$ and $40 \mathrm{cm}$. The distance between those parallel sides is $15 \mathrm{cm}$.If its area is $k {\mathrm{cm}}^2$.

Then find the value of $k$.

The figure consists of an isosceles trapezium and two rectangles with measures in $\mathrm{cm}$ as shown in the diagram. If the area is $k {\mathrm{cm}}^2$. 

Then find the value of $k$.

$\mathrm{ABCD}$ is a trapezium with parallel sides $\mathrm{AD}$ and  $\mathrm{BC}$. Given that $\mathrm{AD}=24 \mathrm{cm}, \mathrm{BD}=29 \mathrm{cm}, \mathrm{BC}=36 \mathrm{cm}, \mathrm{and} \mathrm{DC}=25 \mathrm{cm}$. If the area of $∆\mathrm{BDC}$ is $k {\mathrm{cm}}^2$

$\mathrm{ABCD}$ is a trapezium with parallel sides $\mathrm{AD}$ and  $\mathrm{BC}$. Given that $\mathrm{AD}=24 \mathrm{cm}, \mathrm{BD}=29 \mathrm{cm}, \mathrm{BC}=36 \mathrm{cm}, \mathrm{and} \mathrm{DC}=25 \mathrm{cm}$ find the altitude from $\mathrm{D}$ to $\mathrm{BC}$.

$\mathrm{ABCD}$ is a trapezium with parallel sides $\mathrm{AD}$ and  $\mathrm{BC}$. Given that $\mathrm{AD}=24 \mathrm{cm}, \mathrm{BD}=29 \mathrm{cm}, \mathrm{BC}=36 \mathrm{cm}, \mathrm{and} \mathrm{DC}=25 \mathrm{cm}$. If the area of $\mathrm{ABCD}$ is $k {\mathrm{cm}}^2$.

Then find the value of $k$.

$\mathrm{ABCD}$ is a trapezium with parallel sides $\mathrm{AD}$ and  $\mathrm{BC}$. Given that $\mathrm{AD}=24 \mathrm{cm}, \mathrm{BD}=29 \mathrm{cm}, \mathrm{BC}=36 \mathrm{cm}, \mathrm{and} \mathrm{DC}=25 \mathrm{cm}$. If the area of $∆\mathrm{ABD}$ is $k {\mathrm{cm}}^2$.

Then find the value of $k$.

The perimeter of the isosceles trapezium given in the figure is $84 \mathrm{cm}$. Find height $\text{'}y\text{'}$

The perimeter of the isosceles trapezium given in the figure is $84 \mathrm{cm}$. If the area of the isosceles trapezium is $k {\mathrm{cm}}^2$.

Then find the value of $k$.