86.1k views
2 votes
Rounded to the nearest tenth, what is the area of rectangle ABCD?

Rounded to the nearest tenth, what is the area of rectangle ABCD?-example-1

1 Answer

2 votes

Given the rectangle ABCD, you can identify that it is divided into two Right Triangles:


\begin{gathered} \Delta ACD \\ \Delta ABD \end{gathered}

• You can find the length of the rectangle by applying the following Trigonometric Function:


\cos \alpha=(adjacent)/(hypotenuse)

In this case, you can set up that:


\begin{gathered} \alpha=30\degree \\ adjacent=CD=AB=l \\ hypotenuse=AD=9 \end{gathered}

Then, substituting values and solving for "l", you get:


\begin{gathered} \cos (30\text{\degree})=(l)/(9) \\ \\ 9\cdot\cos (30\text{\degree})=l \\ \\ l=(9)/(2)\sqrt[]{3}ft \end{gathered}

• In order to find the width of the rectangle, you can use this Trigonometric Function:


\sin \alpha=(opposite)/(hypotenuse)

In this case, you can say that:


\begin{gathered} \alpha=30\degree \\ opposite=AC=BD=w \\ hypotenuse=AD=9 \end{gathered}

Therefore, substituting values and solving for "w", you get:


\begin{gathered} \sin (30\degree)=(w)/(9) \\ \\ 9\cdot\sin (30\degree)=w \\ \\ w=(9)/(2)ft \end{gathered}

• Now you need to use the following formula for calculating the area of a rectangle:


A=lw

Where "l" is the length and "w" is the width.

Substituting the length and the width of the given rectangle into the formula and evaluating, you get:


A=((9)/(2)\sqrt[]{3}ft)((9)/(2)ft)
A\approx35.1ft^2

Hence, the answer is: Option C.

User Daniel Silva
by
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories