The rectangular photograph in the diagram has length \((x+5)\) cm and width \(8\) cm. Its area is \(96\) cm\(^2\). Find the value of \(x\) and work out the ...
Question 1 Report
The rectangular photograph in the diagram has length \((x+5)\) cm and width \(8\) cm.
Its area is \(96\) cm\(^2\).
Find the value of \(x\) and work out the perimeter of the photograph.
Answer Details
The area of a rectangle is length times width, so when one dimension is written as an algebraic expression the area statement becomes an equation you can solve for the unknown.
The length is \((x+5)\) cm and the width is \(8\) cm, so the area is \(8(x+5)\) cm\(^2\). Setting that equal to the given area of \(96\) cm\(^2\):
\[8(x+5)=96\]
which earns the method mark [M1]. Dividing both sides by \(8\) gives \(x+5=12\), so \(x=7\) [A1].
The perimeter needs the actual dimensions, not \(x\). With \(x=7\) the length is \(7+5=12\) cm and the width is \(8\) cm, so
\[\text{perimeter}=2(12+8)=40\text{ cm}\]
giving the final answer of \(40\) cm [A1]. A quick check: \(12\times 8=96\) cm\(^2\), which matches the stated area.
The common slip is to give the perimeter as \(2(x+5+8)=2x+26\) or to substitute \(x=7\) directly as a side length. Always convert \(x\) back into the real lengths before working out a perimeter or an area.
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