Length and Area Problem Solving
To solve problems on this page, you should be familiar with the following:
You should also know the following formulas:
- Square with side length \(L\): Area is \(L^2\), and perimeter is \(4L \).
- Rectangle with side length \(L\) and breadth \(B\): Area is \(L\times B\), and perimeter is \(2(L+B) \).
- Equilateral triangle with side length \(s\): Area is \( \frac {\sqrt3}4 s^2 \), and perimeter is \(3s\).
If the area of a square is 144, what is the perimeter of the square?
Let \(L\) denote the area of the square, then \(L^2 = 144 = 12^2 \) or \(L =12\). Note that we are only taking the positive root because \(L\) represents a physical dimension. Thus the perimeter of the square is \(4L = 4\times 12= 48. \ _\square\)
The ancients talked about "squaring the circle," by creating a square and a circle which have the same area. If the radius of the circle is 1, what would be the side length of the square?
Let the side length of the square be \( S \). Then, we are given that \( S^2 = \pi \times 1 ^2 = \pi \). Taking square roots, we obtain \( S = \sqrt{\pi } \). \(_\square\)
If the ratio of the area of a square to the area of an equilateral triangle is \( 4 : 3 \), what is the ratio of the side length of the square to the side length of the equilateral triangle?
Let the side length of the square be \(S\). Then the area of the square is \(S^2 \).
Let the side length of the equilateral triangle be \(T \). Then the area of the equilaterial triangle is \( \frac{ \sqrt{3}} {4} T^2 \).We are given that \( \frac{ S^2 } { \frac{ \sqrt{3}} {4} T^2 } = \frac{4}{3} \), or that \( \frac{ S^2 }{ T^2 } = \frac{ \sqrt3 } { 3 } \).
Taking square roots on both sides, \( \frac{ S } { T} = \frac1{3^{1/4}} \). \(_\square\)
If a square and a circle have the same perimeter, which of them will have a greater area?
Let the radius of the circle be \(r\) and the side length of the square be \(s\). Then\[2\pi r=4s \Rightarrow r=\dfrac{2s}{\pi}.\]
Now the area of the square is \(s^2,\) and the area of the circle is \(\pi r^2 = \pi \times \left(\dfrac{2s}{\pi}\right)^2=\dfrac{4s^2}{\pi}.\)
Then the ratio of the area of the square to the area of the circle is \(\dfrac{s^2}{\hspace{2mm} \frac{4s^2}{\pi}\hspace{2mm} } = \dfrac{\pi}{4} <1.\)
Hence, the circle has a larger area. \(_\square\)
