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 \(ABD\), \(\sin30^{\circ}=\frac{AB}{AD}\). Given \(AD = 9\) ft. Since \(\sin30^{\circ}=\frac{1}{2}\), then \(AB=AD\times\sin30^{\circ}=9\times\frac{1}{2}=4.5\) ft.

Step2: Find the length of \(BC\)

In right - triangle \(ACD\), \(\sin60^{\circ}=\frac{CD}{AD}\). Since \(AB = CD\) (opposite sides of a rectangle) and \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), \(BC = AD\times\cos60^{\circ}\) (also, in right - triangle \(ACD\), \(\cos60^{\circ}=\frac{AC}{AD}\), and \(AC = BC\)). \(BC=9\times\frac{1}{2}=4.5\) ft, \(CD = 9\times\frac{\sqrt{3}}{2}\approx9\times0.866 = 7.794\) ft.

Step3: Calculate the area of rectangle \(ABCD\)

The area of a rectangle \(A = l\times w\). Here \(l = CD\approx7.794\) ft and \(w = BC = 4.5\) ft. \(A=AB\times BC\). Using another approach:
The area of rectangle \(ABCD\) can also be calculated as \(2\times\) area of \(\triangle ACD\).
The area of \(\triangle ACD=\frac{1}{2}\times AC\times CD\).
Since \(AC = 9\times\cos60^{\circ}=4.5\) ft and \(CD = 9\times\sin60^{\circ}\approx7.794\) ft.
\(A = 2\times\frac{1}{2}\times AC\times CD=AC\times CD\)
\(A=(9\times\cos60^{\circ})\times(9\times\sin60^{\circ})\)
\(A = 81\times\frac{1}{2}\times\frac{\sqrt{3}}{2}=\frac{81\sqrt{3}}{4}\approx\frac{81\times1.732}{4}=\frac{140.292}{4}=35.073\approx35.1\) square feet

Answer:

C. 35.1 square feet