A bicycle repair shop is cutting a right-angled triangular metal brace, shown below. The two shorter sides are \(3\sqrt{3}\) cm and \(4\sqrt{3}\) cm. 4√3 cm...
Question 1 Report
A bicycle repair shop is cutting a right-angled triangular metal brace, shown below. The two shorter sides are \(3\sqrt{3}\) cm and \(4\sqrt{3}\) cm.
- Work out the exact length of the hypotenuse, giving your answer as a simplified surd. (2)
- Work out the exact perimeter of the brace, giving your answer in the form \(a\sqrt{3}\). (2)
- The shop only stocks metal strip in whole-centimetre lengths. Given \(\sqrt{3} \approx 1.73\), find the least whole number of centimetres of strip needed for the perimeter. (2)
Answer Details
Pythagoras' theorem applies to sides written as surds just as to ordinary numbers, since \((a\sqrt3)^2=3a^2\); the perimeter is then found by adding surds that share the same root, and converting to a decimal for a practical rounding decision must round up, since anything less would fall short of the length needed.
- Hypotenuse\(^2=(3\sqrt3)^2+(4\sqrt3)^2=27+48=75\) [1 mark]; hypotenuse \(=\sqrt{75}=\sqrt{25\times3}=5\sqrt3\) cm. [1 mark]
- Perimeter \(=3\sqrt3+4\sqrt3+5\sqrt3=12\sqrt3\) cm, since all three sides share the same \(\sqrt3\) term and their coefficients simply add. [2 marks]
- \(12\sqrt3\approx12\times1.73=20.76\) cm [1 mark]; since even a whole-centimetre length of strip must be at least this long, the least whole number of centimetres needed is \(21\) (rounding up, not to the nearest whole number). [1 mark]
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