At a festival, a rope of length 34 metres runs round a rectangular first aid tent covering 60 square metres. The tent measures \(a\) metres by \(b\) metres....
Question 1 Report
At a festival, a rope of length 34 metres runs round a rectangular first aid tent covering 60 square metres. The tent measures \(a\) metres by \(b\) metres.
- Write down two equations in \(a\) and \(b\). (2)
- Show that \(a^2 - 17a + 60 = 0\). (1)
- Find the two side lengths of the tent. (2)
Answer Details
Two facts are given about the same rectangle: the rope round it is its perimeter, and the ground it covers is its area.
(a)
- Perimeter: \(2(a + b) = 34\). [1]
- Area: \(ab = 60\). [1]
(b) Divide the perimeter equation by 2 to get \(a + b = 17\), so \(b = 17 - a\). Substituting into the area equation gives \(a(17 - a) = 60\), that is \(17a - a^2 = 60\). Multiplying through by \(-1\) and rearranging gives \(a^2 - 17a + 60 = 0\) as required. [1]
(c)
- Factorise: two numbers multiplying to \(+60\) and adding to \(-17\) are \(-5\) and \(-12\), so \((a-5)(a-12) = 0\) and \(a = 5\) or \(a = 12\). [1]
- The two roots are not two different tents; they are the two sides of the same one, since \(a = 5\) gives \(b = 12\) and \(a = 12\) gives \(b = 5\). The tent measures 5 metres by 12 metres. [1]
Check both conditions: \(2(5 + 12) = 34\) metres of rope and \(5\times 12 = 60\) square metres of ground.
Here both roots are positive, so neither is rejected. Recognising that the pair of roots represents the pair of sides is exactly what the final mark rewards.
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