Question

for the given figure, find the area of the rectangle a. the other two labeled areas are squares.
27 ft²
108 ft²
a
the area is
(simplify your answer.)

Explanation:

Step1: Find the side lengths of the squares

Let the side length of the square with area \(108\ ft^{2}\) be \(a\), and the side length of the square with area \(27\ ft^{2}\) be \(b\).
For a square, \(A = s^{2}\) (where \(A\) is the area and \(s\) is the side - length).
If \(A=a^{2}=108\), then \(a=\sqrt{108}=\sqrt{36\times3}=6\sqrt{3}\) (using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\), \(a\geq0,b\geq0\)).
If \(A = b^{2}=27\), then \(b=\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\).

Step2: Calculate the area of rectangle \(A\)

The area of a rectangle is \(A = l\times w\).
The length of rectangle \(A\) is \(a = 6\sqrt{3}\) and the width is \(b=3\sqrt{3}\).
Using the formula \(A=l\times w\), we substitute \(l = 6\sqrt{3}\) and \(w = 3\sqrt{3}\).
\(A=(6\sqrt{3})\times(3\sqrt{3})\).
By the property \(a\sqrt{m}\times b\sqrt{n}=ab\sqrt{mn}\), we have \(A = 6\times3\times\sqrt{3\times3}\).
Since \(\sqrt{3\times3}=3\), then \(A=18\times3\).

Answer:

\(54\ ft^{2}\)