Q10. - Algebra 2

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. $840$

Step-by-Step Solution

Answer

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

Step 1 - Consider the diagram.

From the diagram,

 

$\angle A=90°$ and $\angle D=90°$.

 

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

Step 2 - 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.

Step 3 - 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:

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

Step 4 - State the conclusion.

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