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 $\Delta ACD$?

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

Short answer

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

Step-by-step solution

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

The area of the triangle is given by $A=\frac{1}{2}bh$, where bis the base and his 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

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

Step-2 – Construct a square

Construct a line AMperpendicular 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

$\begin{array}{c}\text{Areaof}\Delta \text{AMC=}\frac{1}{2}bh\\ =\frac{1}{2}\left(12\right)\left(9\right)\\ =54\text{squareunits}\end{array}$

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

$\begin{array}{c}\text{Areaof}\Delta \text{AMC=}\frac{1}{2}bh\\ =\frac{1}{2}\left(4\right)\left(9\right)\\ =18\text{squareunits}\end{array}$

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:

$\begin{array}{c}\text{Areaof}\Delta \text{ACD=Areaof}\Delta \text{AMC}-\text{Areaof}\Delta \text{AMD}\\ =54-18\\ =36\text{squareunits}\end{array}$

Hence option A is correct.