A farmer extends an orchard by adding a rectangular section measuring \(x\) m by \((x+5)\) m next to a triangular section with base \(x\) m and height \(8\)...
Question 1 Report
A farmer extends an orchard by adding a rectangular section measuring \(x\) m by \((x+5)\) m next to a triangular section with base \(x\) m and height \(8\) m. The total extra area is \(220\) m\(^2\).
- Write an expression for the total extra area in terms of \(x\), and show it simplifies to \(x^2+9x=220\). (2)
- Show that this can be written as \(x^2+9x-220=0\), then solve it by factorising. (3)
- Give the value of \(x\), rejecting any solution that is not sensible, with a reason. (1)
Answer Details
(a) The rectangular section has area \(x(x+5) = x^{2}+5x\), and the triangular section has area \(\dfrac{1}{2}(x)(8) = 4x\). Adding these and setting the total equal to \(220\) m\(^{2}\) gives \(x^{2}+5x+4x = x^{2}+9x = 220\). [2 marks]
(b) Moving every term to one side gives \(x^{2}+9x-220=0\). Factorising requires two numbers that multiply to \(-220\) and add to \(9\): these are \(20\) and \(-11\), so \((x+20)(x-11)=0\), giving \(x=-20\) or \(x=11\). [3 marks]
(c) Since \(x\) is a length, it cannot be negative, so \(x=-20\) is rejected and \(x=11\) m. [1 mark]
Quadratic equations arising from area or length problems almost always produce one negative and one positive root; only the positive root has physical meaning, which is why every solution of this type should end with a check against the context.
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