(a)  We are being asked to find a length in the larger triangle, so use LSF \(=\large\frac{\textsf{big}}{\textsf{small}}\normalsize\)

AB and DE are corresponding lengths, so LSF \(=\large\frac{10}{5}\normalsize =2\)

So DF \(= 2\times\) AC \(= 2\times 4=8\) cm

(b)  Now we are finding a length in the smaller triangle, so use LSF \(=\large\frac{\textsf{small}}{\textsf{big}}\normalsize\)

Again using the corresponding lengths AB and DE, LSF \(=\large\frac{5}{10}\normalsize =\large\frac{1}{2}\normalsize\)

So BC \(= \frac{1}{2}\times\) EF \(=\frac{1}{2}\times 6=3\) cm

Note: Part (b) may also be done using the \(\large\frac{\textsf{big}}{\textsf{small}}\normalsize\) scale factor from part (a), but dividing instead of multiplying. Both methods work perfectly.

(a)  We know both the length and price of the small frame, so consider the enlargement from small to medium.

LSF \(=\large\frac{30}{20}\normalsize=1.5\)

ASF \(=\) LSF² \(=1.5^2=2.25\)

The price of the medium frame is therefore \(2.25\,\times\) £\(4.80=\) £\(10.80\)

(b)  Now let's consider the enlargement from the small frame to the large frame.

ASF \(= \large\frac{24.3}{4.8}\normalsize =5.0625\)

LSF \(= \sqrt{\small\textsf{ASF}}\normalsize =\sqrt{5.0625}=2.25\)

So the length of the large frame is \(2.25\times 20=45\) cm

We are being asked to find the smaller volume, so use LSF \(=\) \(\large\frac{\textsf{small}}{\textsf{big}}\normalsize=\large\frac{7}{9}\normalsize\)

VSF \(=\large(\frac{7}{9})^3\normalsize =\large\frac{343}{729}\small.\) Note that the decimal equivalent of this fraction is an ugly recurring decimal, and rounding it would introduce a rounding error, so we should stick with the exact fraction.

The smaller volume is therefore \(\large\frac{343}{729}\normalsize\times 1458 = 686\) ml

The rectangles are mathematically similar if ASF \(=\) LSF². We need to check whether or not this is true.

LSF \(=\) \(\large\frac{20}{12}\normalsize\) \(=\) \(\large\frac{5}{3}\normalsize\)

ASF \(=\) \(\large\frac{500}{180}\normalsize\) \(=\) \(\large\frac{25}{9}\normalsize\) \(=\) \(\large(\frac{5}{3})^2\normalsize\)

So we can conclude that the two rectangles are mathematically similar.

LSF \(=\) \(\large\frac{24}{15}\normalsize\) \(=\) \(\large\frac{8}{5}\normalsize\)

VSF \(=\) \(\large(\frac{8}{5})^3\normalsize\) \(=\) \(\large\frac{512}{125}\normalsize\)

So the volume of the larger jar is \(\large\frac{512}{125}\normalsize\times 750=3072\) cm³

We are given the area of the smaller triangle. PQTS is part of the larger triangle.

So we use LSF \(= \large\frac{\textsf{big}}{\textsf{small}}\normalsize = \large\frac{30}{24}\normalsize = 1.25\)

ASF \(=\) LSF² \(= 1.25^2= 1.5625\)

So the area of triangle PRS is \(1.5625 \times 400 = 625\) cm²

The area of PQTS is therefore \(625-400=225\) cm²

We are finding the larger volume, so LSF \(= \large\frac{\textsf{big}}{\textsf{small}}\normalsize = \large\frac{36}{15}\normalsize = 2.4\)

VSF \(=\) LSF³ \(= 2.4^3= 13.824\)

So the volume of the larger bottle is \(13.824 \times 250 = 3\small\,\normalsize 456\) ml

(a)  The cartons are mathematically similar if VSF \(=\) LSF³. We need to check whether or not this is true.

LSF \(=\) \(\large\frac{24}{16}\normalsize\) \(=\) \(\large\frac{3}{2}\normalsize\)

VSF \(=\) \(\large\frac{1125}{576}\normalsize\) \(=\) \(\large\frac{125}{64}\normalsize\) \(\neq\) \(\large(\frac{3}{2})^3\normalsize\)

So the two cartons are not mathematically similar.

Note: There are other valid methods for part (a). See marking instructions for details.

(b)  With the redesigned carton, VSF \(= \large\frac{1500}{576}\normalsize = \large\frac{125}{48}\normalsize\)

We need to take care not to approximate this number in the middle of our working, so it is better to keep it as a fraction and work with exact values, only approximating at the very end of our working.

LSF = \( \sqrt[\leftroot{-1}\uproot{3}\scriptstyle 3]{\small\textsf{VSF}\normalsize} = \sqrt[\leftroot{-1}\uproot{3}\scriptstyle 3]{\large\frac{125}{48}}\normalsize\)

So the depth of the redesigned large carton is \(\sqrt[\leftroot{-1}\uproot{3}\scriptstyle 3]{\large\frac{125}{48}}\normalsize\,\times 16 \approx 22.0\) cm

We are finding the width of the larger photograph, so ASF \(= \large\frac{\textsf{big}}{\textsf{small}}\normalsize = \large\frac{500}{80}\normalsize = 6.25\)

LSF \(= \sqrt{\small\textsf{ASF}}\normalsize =\sqrt{6.25}=2.5\)

So the width of the large photograph is \(2.5\times 12=30\) cm