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Problem of the Week
Problem C and Solution
Just Outside

Problem

In square \(OABC\), points \(A\) and \(C\) lie on the circumference of a circle with centre \(O\), and \(B\) lies outside of the circle. Square \(OABC\) has an area of \(36\,\text{m}^2\).

Determine the area of the shaded region inside square \(OABC\) and outside the circle with centre \(O\), rounded to two decimal places.

Solution

Since \(OABC\) is a square with an area of \(36\,\text{m}^2\), its side length must be \(6\) m. That is, \(OA = OC = 6\) m.

Since \(A\) lies on the circumference of the circle with centre \(O\), the radius of the circle is \(r = OA = 6\) m.

Therefore, the area of the circle is \(\pi \times r^2 = \pi \times 6^2 = 36\pi\,\text{m}^2\).

Since \(OABC\) is a square, \(\angle AOC = 90^{\circ}\).

Therefore, the area of sector \(OAC\) is \(\frac{90^{\circ}}{360^{\circ}} = \frac{1}{4}\) of the area of the circle.

In other words, the area of the sector \(OAC\) is \(\frac{1}{4}\times 36\pi = 9\pi\,\text{m}^2\).

Therefore, \[\begin{aligned} \text{Area of shaded region} &= \text{Area of square }OABC-\text{Area of sector }OAC\\ &= 36 - 9\pi\\ &\approx 7.73\, \text{m}^2 \end{aligned}\]

Note: In the problem you were asked to give your answer rounded to two decimal places. However, many times in mathematics we are actually interested in the exact answer. In this case, the exact answer would be \((36 - 9\pi)\,\text{m}^2\).