Area and Volume of Similar Shapes: past paper questions & worked solutions

The marked corresponding bases give the linear scale factor from A to B.

The 5 cm measurement is the perpendicular height.

Areas scale by the square of the linear scale factor.

$$\\text{ASF} = \\frac{\\text{Area of Q}}{\\text{Area of P}} = \\frac{450}{200} = \\frac{9}{4} = 2.25$$

$$\\text{LSF} = \\sqrt{\\text{ASF}} = \\sqrt{\\frac{9}{4}} = \\frac{3}{2} = 1.5$$

$$\\text{VSF} = (\\text{LSF})^3 = \\left(\\frac{3}{2}\\right)^3 = \\frac{27}{8} = 3.375$$

$$\\text{Volume of Q} = \\text{Volume of P} \\times \\text{VSF} = 672 \\times \\frac{27}{8}$$  = 84 × 27 = 2268

Surface area ratio Q:P:

So the linear ratio Q:P is

Therefore the volume ratio Q:P is

Shape P has volume \(672\text{ cm}^3\).

Volume of Q:

The surface area ratio of the small block to the large block is

So the linear ratio is

Therefore the volume ratio is

The small block has volume \(300\text{ cm}^3\).

Large block volume:

The volume ratio of solid A to solid B is

Simplify:

So the linear ratio is

Surface area ratio:

Solid B has surface area \(1120\text{ cm}^2\), so solid A has surface area

Shape R has surface area \(108\text{ cm}^2\) and volume \(135\text{ cm}^3\).

Shape S has surface area \(300\text{ cm}^2\).

Surface area ratio S:R:

So the linear ratio S:R is

Therefore the volume ratio S:R is

Volume of S:

The volume of vase \(S\) is \(72.8\%\) more than vase \(R\), so

\(\dfrac{V_S}{V_R}=1.728\)

For similar solids, volume scale factor is the cube of the linear scale factor:

\(\text{linear scale factor from }R\text{ to }S=\sqrt[3]{1.728}=1.2\)

The height of vase \(T\) is \(6h\), so the linear scale factor from \(R\) to \(T\) is \(6\).

Therefore the linear scale factor from \(S\) to \(T\) is

\(\dfrac{6}{1.2}=5\)

Surface area scale factor is the square of the linear scale factor:

\(k=5^2=25\)

The heights of solids A and B are \(31\) cm and \(18.6\) cm.

So the volumes are in the ratio

parts.

This difference is \(735\text{ cm}^3\), so one part is

Volume of solid A:

Linear scale factor from solid \(A\) to solid \(B\):

So the volume scale factor is

Let the volume of \(A\) be \(V\). Then

The linear scale factor from vase \(B\) to vase \(A\) is \[ \frac{30}{12}=2.5 \]

So the surface area scale factor is \[ 2.5^2=6.25 \]

Let the surface area of vase \(B\) be \(S\).

Therefore the surface area of vase \(A\) is \[ 6.25\times 34=212.5 \]

The volume of statue \(B\) is \(20\%\) less than statue \(A\), so the volume scale factor is \(0.8\).

For similar solids, the linear scale factor is the cube root of the volume scale factor:

\(\text{linear scale factor}=\sqrt[3]{0.8}\)

Surface area scale factor is the square of the linear scale factor:

\(\text{surface area scale factor}=(\sqrt[3]{0.8})^2=0.8^{2/3}=0.861773...\)

So statue \(B\) has \(86.1773...\%\) of the surface area of statue \(A\).

\(k=100-86.1773...=13.8226...\)

The linear scale factor from vase \(A\) to vase \(B\) is \[ \frac{15}{10}=1.5 \]

So the volume scale factor is \[ 1.5^3=3.375 \]

Let the volume of vase \(A\) be \(V\).

Surface area ratio \(A:B=4:9\), so the length ratio is

Volume ratio \(B:C=125:343\), so the length ratio is

Therefore

The two vases are similar.

Their heights are in the ratio

So their surface areas are in the ratio

The total surface area is \(1800\text{ cm}^2\), so there are

parts altogether.

Surface area of vase A:

The volume of statue \(B\) is \(80\%\) of the volume of statue \(A\), so the volume scale factor is

For similar solids:

The surface area scale factor is therefore

So the percentage decrease in surface area is

Correct to 3 significant figures,

Corresponding lengths have the cube-root ratio of volumes.

Corresponding lengths have the square-root ratio of surface areas.

Take \(h_K=10\) units. Then \(h_L=12\) and \(h_M=15\), so \[y=\frac{15-12}{12}\times100=25.\]

Align the shared K-length and compare the increase relative to L.

The straight generators shown through \(A\) give equal vertex angles, so the cones are similar. The area ratio \(4:9\) gives length ratio \(\sqrt4:\sqrt9=2:3\).

For similar solids, surface areas scale as the square of corresponding lengths.

The heights total \(30\text{ cm}\), so \[h_1=\frac25(30)=12\text{ cm},\qquad h_2=\frac35(30)=18\text{ cm}.\]

Split the total in the common length ratio 2:3.

Use the volume of the first cone and choose a positive radius.

Radii scale by the same factor as the heights.

The hemisphere has radius \(x\), so its height is \(x\).

The total height of solid S is \(4x\), so the cone height is

Volume of the cone:

Volume of the hemisphere:

So the volume of solid S is

The sphere has radius \(kx\), so its volume is

This is \(12.5\) times the volume of S:

Cancel \(\frac13\pi x^3\):

For solid T, the volume scale factor is \(512\).

So the linear scale factor is \(8\), and the surface area scale factor is

The area ratio of solid A to solid B is

So the linear ratio A:B is

Therefore the volume ratio A:B is

The volumes are \(8^x\) and \(2^{x+4}\), so

Write \(8^x\) as a power of 2:

So

The height of solid A is

Since the linear ratio A:B is \(4:1\), the height of solid B is

The large cone has height \(15\) cm and radius \(6\) cm.

The small cone has radius \(x\). Since the cones are similar,

So for the small cone,

Volume of the large cone:

Volume of the small cone:

So

Cancel \(\pi\):

The small and large square-based pyramids are similar.

So the side length of the top square of the frustum is

The vertical height of the frustum is

For one trapezium face, the difference between the half-side lengths is

So the slant height of the trapezium face is

Area of one trapezium face:

Area of four trapezium faces:

Add the top and bottom square faces:

Total surface area:

Correct to the nearest whole number:

The two cones are similar.

The large cone has height \(2h\) and base radius \(r\).

The small cone has height \(h\), so its linear scale factor compared with the large cone is

So the small cone has radius

Volume of the large cone:

Volume of the small cone:

So the volume of the frustum is

The sphere has radius \(r\), so its volume is

The two volumes are equal:

Divide by \(\pi r^2\):

Small cone volume:

The larger similar cone has radius \(kr\) and height \(kh\), so its volume is

So the frustum volume is

The volume of one hemisphere of radius \(r\) is

Solid A is the frustum plus one hemisphere:

Solid B is the small cone plus one hemisphere:

Given \(V_A=6V_B\):

Divide by \(\frac13\pi r^2\):