Two allotment plots are fenced separately, shown in the diagrams. Plot A is a rectangle with length \((x+6)\) m and width \(x\) m. Plot B is a square with s...
Question 1 Report
Two allotment plots are fenced separately, shown in the diagrams. Plot A is a rectangle with length \((x+6)\) m and width \(x\) m. Plot B is a square with side \((x+2)\) m. The total fencing needed for both plots is \(100\) m.
- Write an expression, in terms of \(x\), for the total length of fencing needed. (2)
- Use the total of \(100\) m to form and solve an equation for \(x\). (2)
- Hence write down the perimeter of Plot A. (1)
Answer Details
Plot A is a rectangle with length \((x+6)\) and width \(x\), so its fencing is twice the sum of these two sides; Plot B is a square of side \((x+2)\), so its fencing is four times that side. Adding both perimeters and setting the total to \(100\) m gives an equation for \(x\).
(a) Plot A's fencing is \(2[(x+6)+x]=2(2x+6)=4x+12\). Plot B's fencing is \(4(x+2)=4x+8\). The total is \((4x+12)+(4x+8)=8x+20\). [2 marks]
(b) Setting the total to \(100\): \(8x+20=100\). Subtracting 20: \(8x=80\). Dividing by 8: \(x=10\). [2 marks]
(c) With \(x=10\), Plot A measures \(16\) m by \(10\) m, so its perimeter is \(2[(10+6)+10]=2(26)=52\) m, shown below. [1 mark]
Building the combined expression \(8x+20\) before substituting the total of 100 m keeps the two shapes' perimeters clearly separated, which is why part (a) is worth showing as \(4x+12\) plus \(4x+8\) rather than jumping straight to the simplified total.
Download The App On Google Playstore
Everything you need to excel in your exams
