A science lab orders microscope coverslips in two similar circular sizes. The standard coverslip has area \(12.5\) mm\(^2\) and diameter \(4\) mm. An enlarg...
Question 1 Report
A science lab orders microscope coverslips in two similar circular sizes. The standard coverslip has area \(12.5\) mm\(^2\) and diameter \(4\) mm. An enlarged coverslip, similar to the standard one, has area \(32\) mm\(^2\).
- Find the linear scale factor of the enlargement, using the two areas. (2)
- Hence find the diameter of the enlarged coverslip. (2)
- State, with a reason, whether finding the scale factor directly from the diameters would give a different value. (1)
Answer Details
For similar shapes, the ratio of areas equals the square of the linear scale factor, so the scale factor is recovered by taking the square root of the area ratio; that same scale factor then converts any length, such as diameter, from the original shape to the enlarged one.
- \[ \text{area ratio} = 32\div12.5 = 2.56 \] so \[ \text{scale factor} = \sqrt{2.56} = 1.6 \] [2 marks]
- \[ \text{diameter} = 4\times1.6 = 6.4 \text{ mm} \] [2 marks]
- No, finding the scale factor directly from the diameters would give the same value, since a linear scale factor between two similar shapes is the same whichever pair of corresponding lengths is used to calculate it. [1 mark]
The area scaled up by a factor of \(2.56\) while the linear dimensions only scaled up by \(\sqrt{2.56}=1.6\); this is a direct consequence of area being a two-dimensional measure, so it always grows with the square of whatever factor the lengths themselves grow by.
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