Q1. - Algebra 2

Question

1. In the figure,  $\angle B\text{ and }\angle BCD$ are right angles. $\overline{BC}$ is 9 units, $\overline{AB}$ is 12 units and $\overline{CD}$ is 8 units. What is the area, in square units of $\Delta ACD$?

$\text{A}.\text{ 36 B. 60 C.72 D.135}$

Step-by-Step Solution

Answer

The area of  $\Delta ACD$ is 36 square units.

Step-1 – Apply the concept of area of a triangle

The area of the triangle is given by $A=\frac{1}{2}bh$, where b is the base and h is the height of the triangle.

For example, suppose a triangle has its base as $16cm$ and 10cm as its height then the area of the triangle is given by

$\begin{array}{c}\text{Area of triangle=}\frac{1}{2}bh\\ =\frac{1}{2}\left(16\right)\left(10\right)\\ =80\end{array}$

Step-2 – Construct a square

Construct a line AM perpendicular to side $DC$ from the point A.

Therefore, the following is obtained:

Since a square is formed, $\overline{AM}=9cm$ and  $\overline{MC}=12cm$. Therefore,

$\begin{array}{c}\overline{MD}=\overline{MC}-\overline{DC}\\ =12-8\\ =4\end{array}$

Step-3 – Find the area of the triangle AMC

In $\Delta AMC$, the base is 12 units and height is 9 units.

Therefore the area is given by

$\begin{array}{c}\text{Area of }\Delta \text{AMC=}\frac{1}{2}bh\\ =\frac{1}{2}\left(12\right)\left(9\right)\\ =54\text{ square units}\end{array}$

Step-4 – Find the area of the triangle AMD

In $\Delta AMD$, the base is 4 units and height is 9 units.

Therefore the area is given by

$\begin{array}{c}\text{Area of }\Delta \text{AMC=}\frac{1}{2}bh\\ =\frac{1}{2}\left(4\right)\left(9\right)\\ =18\text{ square units}\end{array}$

Step-5 – Find the area of the triangle ACD

Area of $\Delta ACD$ is found using the area of $\Delta AMC\text{ and }\Delta AMD$ as follows:

$\begin{array}{c}\text{Area of }\Delta \text{ACD=Area of }\Delta \text{AMC}-\text{Area of }\Delta \text{AMD}\\ =54-18\\ =36\text{ square units}\end{array}$

Hence option A is correct.