Exercise solution 2, page 345 - AQA GCSE MATHS HIGHER

⭐ What you need to know:

A) The largest circle that fits inside a square is called an inscribed circle. Its diameter is equal to the side length of the square. So if a square has side length \(s\), then the circle inside it has:

\[ \text{diameter} = s \qquad\text{and}\qquad \text{radius} = \frac{s}{2} \]

For example, if a square has side length \(10\text{ cm}\), then the largest possible circle inside it has diameter \(10\text{ cm}\) and radius \(5\text{ cm}\).

B) To find the area of the square, use:

\[ \text{Area of square} = \text{side} \times \text{side} = s^2 \]

For example, if the side length is \(7\text{ cm}\), then:

\[\begin{aligned} \text{Area of square} &= 7 \times 7 \\ &= 49\text{ cm}^2 \end{aligned}\]

C) To find the area of a circle, use:

\[ \text{Area of circle} = \pi r^2 \]

Here, \(r\) is the radius of the circle. Remember that you must square the radius before multiplying by \(\pi\).

For example, if the radius is \(6\text{ cm}\), then:

\[\begin{aligned} \text{Area of circle} &= \pi \times 6^2 \\ &= \pi \times 36 \\ &= 36\pi \text{ cm}^2 \end{aligned}\]

D) The area of the card remaining is found by subtracting the area of the circle from the area of the square:

\[ \text{remaining area} = \text{area of square} - \text{area of circle} \]

This is because the circle is cut out of the square, so the part left is what was there at the start minus what was removed.

For example, if a square has area \(64\text{ cm}^2\) and the circle cut out has area \(16\pi \text{ cm}^2\), then the remaining area is:

\[ 64 - 16\pi \text{ cm}^2 \]

E) An exact answer should be left in terms of \(\pi\), not changed into a decimal approximation. So instead of writing a rounded value such as \(50.3\text{ cm}^2\), you should keep the answer in a form like:

\[ 25 - 9\pi \]

This is more accurate because \(\pi\) is an irrational number.

F) Always check that the units are correct. Since you are finding an area, the final unit must be in square centimetres, written as \(\text{cm}^2\).

Lengths are measured in \(\text{cm}\), but areas are measured in \(\text{cm}^2\).