Question

select the correct answer.
rounded to the nearest tenth, what is the area of rectangle abcd?
a, b, c, d are vertices of the rectangle, with diagonal ad = 9 ft, angles at a and d are 30° and 60° as shown. options: 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 \(AD\)

In right - triangle \(ACD\), \(\cos60^{\circ}=\frac{AD}{AC}\). Given \(AC = 9\) ft. Since \(\cos60^{\circ}=\frac{1}{2}\), then \(AD=AC\times\cos60^{\circ}=9\times\frac{1}{2}=4.5\) ft.

Step2: Find the length of \(CD\)

In right - triangle \(ACD\), \(\sin60^{\circ}=\frac{CD}{AC}\). Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\) and \(AC = 9\) ft, then \(CD=AC\times\sin60^{\circ}=9\times\frac{\sqrt{3}}{2}\approx9\times0.866 = 7.794\) ft.

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

The area of a rectangle \(A = l\times w\), where \(l = CD\) and \(w = AD\). So \(A=AD\times CD\). Substitute \(AD = 4.5\) ft and \(CD\approx7.794\) ft. Then \(A=4.5\times7.794 = 35.073\approx35.1\) square feet.

Answer:

C. 35.1 square feet