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find the area of the figure. round to the nearest hundredths when neces…

Question

find the area of the figure. round to the nearest hundredths when necessary. 16.9 ft 11.9 ft answer attempt 1 out of 2

Explanation:

Step1: Find the height of the right - triangle (also the radius of the semicircle)

Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\). Let the height \(h\), \(c = 16.9\), \(a=11.9\). Then \(h=\sqrt{16.9^{2}-11.9^{2}}=\sqrt{(16.9 + 11.9)(16.9 - 11.9)}=\sqrt{28.8\times5}=\sqrt{144}=12\) ft.

Step2: Calculate the area of the right - triangle

The area formula for a triangle is \(A_{\triangle}=\frac{1}{2}\times base\times height\). Here, base \(b = 11.9\) ft and height \(h = 12\) ft. So \(A_{\triangle}=\frac{1}{2}\times11.9\times12=71.4\) \(ft^{2}\).

Step3: Calculate the area of the semicircle

The area formula for a circle is \(A=\pi r^{2}\), for a semicircle \(A_{semicircle}=\frac{1}{2}\pi r^{2}\). Since \(r = 12\) ft, \(A_{semicircle}=\frac{1}{2}\times\pi\times12^{2}=\frac{1}{2}\times\pi\times144 = 72\pi\approx72\times3.14 = 226.08\) \(ft^{2}\).

Step4: Calculate the total area of the figure

\(A = A_{\triangle}+A_{semicircle}\). So \(A=71.4 + 226.08=297.48\) \(ft^{2}\).

Answer:

\(297.48\) \(ft^{2}\)