Volume of Prisms

Solution

A has straight upright sides, and every cut across it, parallel to its top and base, is the same hexagon. It is a prism: a hexagonal prism standing on one of its ends. Its cross-section is the hexagon on its top and base.

B has the same circle all the way up, so it has a uniform cross-section. But its ends are circles, not polygons, so it is not a prism. It is a cylinder.

C is a prism. Do not look at the face on the ground or the face towards you: look for the two identical ends. They are the triangles at its left and right, so C is a triangular prism lying on one of its rectangular faces.

D is not a prism. Slice it from front to back: a cut through the middle passes through both blocks and gives an upside-down T, but a cut near the front misses the cube and gives a rectangle. Slicing from side to side gives the same two shapes, and cutting it level gives a large square low down and a small square high up. No way of cutting it gives the same shape all along.

Solution

The cross-section is the triangle at the end. Its base is 6 cm and its height is 4 cm; the 5 cm sides are not needed for the volume.

Area of cross-section $= \frac{1}{2} \times 6 \times 4 = 12\text{ cm}^{2}$

Volume $=$ area of cross-section $\times$ length

$= 12 \times 10$

$= 120\text{ cm}^{3}$

Solution

The prism lies on a rectangular face, but its cross-section is the trapezium at the end: parallel sides 12 cm and 6 cm, height 4 cm. The slant sides are not needed.

Area of cross-section $= \frac{1}{2} \times (12 + 6) \times 4$

$= \frac{1}{2} \times 18 \times 4 = 36\text{ cm}^{2}$

Volume $= 36 \times 15 = 540\text{ cm}^{3}$

Solution

The cross-section is the L-shape. The two lengths not marked are $10 - 4 = 6$ cm (the top of the foot) and $8 - 3 = 5$ cm (the inner upright side).

Cut the L-shape into two rectangles, P and Q:

Area of P $= 4 \times 8 = 32\text{ cm}^{2}$; area of Q $= 6 \times 3 = 18\text{ cm}^{2}$

Area of cross-section $= 32 + 18 = 50\text{ cm}^{2}$

Check by cutting the other way: $10 \times 3 + 4 \times 5 = 30 + 20 = 50\text{ cm}^{2}$.

Volume $= 50 \times 12 = 600\text{ cm}^{3}$

Solution

The paperweight stands on its end, so its cross-section is its triangular base and its length is its height, 5 cm.

Base area $= \frac{1}{2} \times 8 \times 6 = 24\text{ cm}^{2}$ (the two sides about the right angle are a base and a height)

Volume $=$ base area $\times$ height

$= 24 \times 5 = 120\text{ cm}^{3}$

Solution

(a) The lengths are in cm and m. Change them all to metres first: $100\text{ cm} = 1\text{ m}$ and $40\text{ cm} = 0.4\text{ m}$.

Area of cross-section $= \frac{1}{2} \times (1 + 0.4) \times 0.4$

$= \frac{1}{2} \times 1.4 \times 0.4 = 0.28\text{ m}^{2}$

Volume $= 0.28 \times 6 = 1.68\text{ m}^{3}$

Check in centimetres: the area is $\frac{1}{2} \times (100 + 40) \times 40 = 2800\text{ cm}^{2}$, the length is 600 cm, and $2800 \times 600 = 1\,680\,000\text{ cm}^{3}$. Dividing by $1\,000\,000$ gives $1.68\text{ m}^{3}$, as before.

(b) $1\text{ m}^{3} = 1000\text{ L}$, so the capacity is $1.68 \times 1000 = 1680\text{ L}$.

Solution

The water is the same shape all the way across the pool: its cross-section is the trapezium on the side of the pool, and the prism lies on its side with a length of 10 m (the width of the pool). The top and the floor are not cross-sections.

In the trapezium the parallel sides are the two depths, 1 m and 3 m, and the distance between them is 25 m.

(a) Area of cross-section $= \frac{1}{2} \times (1 + 3) \times 25$

$= \frac{1}{2} \times 4 \times 25 = 50\text{ m}^{2}$

Volume $= 50 \times 10 = 500\text{ m}^{3}$

(b) $500 \times 1000 = 500\,000\text{ L}$

Solution

Area of cross-section $= \frac{1}{2} \times 8 \times 3 = 12\text{ cm}^{2}$

Let the length be $x$ cm. Volume $=$ area of cross-section $\times$ length:

$12 \times x = 300$

$x = 25$ (divide both sides by 12)

Check: $12 \times 25 = 300$.

Solution

Volume $=$ area of cross-section $\times$ length, with the numbers put in:

$\frac{1}{2} \times (x + 10) \times 6 \times 20 = 960$

$60(x + 10) = 960$ (since $\frac{1}{2} \times 6 \times 20 = 60$)

$x + 10 = 16$ (divide both sides by 60)

$x = 6$ (subtract 10 from both sides)

Check: $\frac{1}{2} \times (6 + 10) \times 6 = 48\text{ cm}^{2}$, and $48 \times 20 = 960\text{ cm}^{3}$.

Solution

(a) Let the edge be $x$ cm. Volume of a cube $=$ edge $\times$ edge $\times$ edge:

$x^{3} = 343$

$x = \sqrt[3]{343} = 7$ (take the cube root of both sides)

Check: $7 \times 7 \times 7 = 49 \times 7 = 343$.

(b) The edge is wanted in metres, so change the capacity to $\text{m}^{3}$: $1728\text{ L} = 1728 \div 1000 = 1.728\text{ m}^{3}$.

Let the edge be $x$ m. Then $x^{3} = 1.728$

$x = \sqrt[3]{1.728} = 1.2$

Check: $1.2 \times 1.2 \times 1.2 = 1.44 \times 1.2 = 1.728$.

Solution

Each square takes 4 cm from both ends of each side of the card.

Length of the box $= 30 - 2 \times 4 = 22$ cm

Breadth of the box $= 20 - 2 \times 4 = 12$ cm

Height of the box $= 4$ cm (the side of a square)

Volume $= 22 \times 12 \times 4 = 1056\text{ cm}^{3}$

Capacity: $1000\text{ cm}^{3} = 1\text{ L}$, so $1056\text{ cm}^{3} = 1056 \div 1000 = 1.056\text{ L}$.