Question

select the correct answer.
rounded to the nearest tenth, what is the area of rectangle abcd?
a
b
c
d
a. 70.1 square feet
b. 40.5 square feet
c. 35.1 square feet
d. 25.5 square feet
e. 24.6 square feet

Explanation:

Step1: Find the length of \(AB\)

In right - triangle \(ABD\), \(\sin60^{\circ}=\frac{AB}{AD}\). Given \(AD = 9\) ft. Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), we have \(AB = AD\times\sin60^{\circ}=9\times\frac{\sqrt{3}}{2}\approx9\times0.866 = 7.794\) ft.

Step2: Find the length of \(BC\)

In right - triangle \(ACD\), \(\sin30^{\circ}=\frac{CD}{AD}\). Since \(AD = 9\) ft and \(\sin30^{\circ}=\frac{1}{2}\), \(CD = 4.5\) ft. (In a rectangle \(AB = CD\) and \(BC=AD\sin60^{\circ}\), but using the other right - triangle relation: In right - triangle \(ACD\), \(\cos60^{\circ}=\frac{AC}{AD}\), \(AC = 4.5\) ft. Also, using the other right - triangle \(ABD\), \(\cos60^{\circ}=\frac{BD}{AD}\), \(BD = 4.5\) ft. The area of rectangle \(A = l\times w\). Using trigonometry in \(\triangle ACD\), \(AC=AD\cos60^{\circ}=9\times\frac{1}{2}=4.5\) ft and \(CD = AD\sin60^{\circ}\approx7.794\) ft. Area \(A=AC\times CD\))
The area of rectangle \(A=\text{length}\times\text{width}\). Let's use another approach. In right - triangle \(ACD\), \(AC = 9\cos60^{\circ}=9\times\frac{1}{2}=4.5\) ft and \(CD = 9\sin60^{\circ}=9\times\frac{\sqrt{3}}{2}\approx7.794\) ft.

$$A=4.5\times\frac{9\sqrt{3}}{2}=\frac{40.5\sqrt{3}}{2}\approx\frac{40.5\times1.732}{2}=\frac{69.94}{2}\approx35.0$$

Answer:

C. 35.1 square feet