## 2012 May June Paper 21/23 Q2

A cuboid has a square base of side \((2+\sqrt{3}) cm\) and a volume of \((16+9 \sqrt{3}) cm ^{3}\). Without using a calculator, find the height of the cuboid in the form \((a+b \sqrt{3}) cm\), where \(a\) and \(b\) are integers.

Volume of cuboid

\(V=(2+\sqrt{3})(2+\sqrt{3}) \times h=16+9 \sqrt{3}\)

\(h=\frac{16+9 \sqrt{3}}{4+4 \sqrt{3}+3}\)

\(=\frac{16+9 \sqrt{3}}{7+4 \sqrt{3}} \times \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}}\)

\(=\frac{112+63 \sqrt{3}-64 \sqrt{3}-108}{49-\left(4\sqrt{3})^{2}\right.}\)

\(=4-\sqrt{3}\)

## 2015 May June Paper 22 Q3

Do not use a calculator in this question.

The diagram shows the right-angled triangle \(A B C\), where \(A B=(6+3 \sqrt{5}) cm\) and angle \(B=90^{\circ} .\) The area of this triangle is \(\left(\frac{36+15 \sqrt{5}}{2}\right) cm ^{2}\).

- Find the length of the side \(B C\) in the form \((a+b \sqrt{5}) cm\), where \(a\) and \(b\) are integers.
- Find \((A C)^{2}\) in the form \((c+d \sqrt{5}) cm ^{2}\), where \(c\) and \(d\) are integers.

## 2016 Oct Nov Paper 23

In this question all lengths are in centimetres.

In the triangle \(A B C\) shown above, \(A C=\sqrt{3}+1, B C=\sqrt{3}-1\) and angle \(A C B=60^{\circ}\).

- Without using a calculator, show that the length of \(A B=\sqrt{6}\).
- Show that angle \(C A B=15^{\circ}\).
- Without using a calculator, find the area of triangle \(A B C\).