Question

If the area of $\Delta \text{ABD}$is 280, state the area of the polygon $\text{ABCD}$.

A. $560$ B. $630$

C. $700$ D. $840$

Short answer

Area of the polygon $ABCD$ is$630$ and the correct option isB.

Step-by-step solution

- Consider the diagram.

From the diagram,

$\angle A=90°$ and $\angle D=90°$.

$AB=16$, $CD=20$.

- Prove that ABCD is a trapezium.

From the diagram,

$\angle A+\angle D=180°$.

So, sum of interior angles on same side of transversal is ${180}^{°}$

So, $AB\parallel CD$.

Thus, $ABCD$is a trapezium.

- Calculate the area of ABCD.

Given,

Area of $\Delta ABD$is 280.

Again area of triangle is given by:

$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$

Area of triangle $ABD$is:

$\begin{array}{c}\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}\\ 280=\frac{1}{2}\times 16\times \text{height}\\ 280\times 2=\frac{1}{2}\times 16\times \text{height}\times 2\\ \frac{560}{16}=\frac{16}{16}\times \text{height}\\ \text{height}=35\end{array}$

Since, $ABCD$is a trapezium.

Area of $ABCD$is:

$\text{Area}=\frac{\left(\text{Sum of paralle sides}\right)}{2}\times \text{height}$

In $ABCD$, $AB$and $CD$are parallel sides.

And $AB=16$, $CD=20$and height is 35.

Then area is:

$\begin{array}{c}\text{Area}=\frac{AB+CD}{2}\times \text{height}\\ =\frac{16+20}{2}\times 35\\ =\frac{36}{2}\times 35\\ =630\end{array}$

- State the conclusion.

Area of $ABCD$ is 630. Therefore, option B is correct.