Homework corrected (21 Sep) — Maths challenges

The task

For each of the following solids, find:

The base area.
The lateral area.
The total area.
The volume, in cm³ and in litres.

Remember: $1000\ \text{cm}^3 = 1\ \text{dm}^3 = 1$ L. These are the same three solids as the modules in the Survival laboratory, on a different scale.

About rounding. Results are rounded to two decimal places and, as in class, each step uses the rounded result of the previous one. Where rounding only at the end changes a digit, the solution says so.

1. Rectangular prism

Rectangular prism: 8 cm × 5 cm base and 7 cm high.

See the solution

a) Base area

The base is an $8$ cm by $5$ cm rectangle:

A_B = 8 \cdot 5 = 40\text{ cm}^2

$A_B = 40\ \text{cm}^2$

b) Lateral area

Unfolded, the side faces form a rectangle whose length is the perimeter of the base and whose height is the height of the prism:

P_B = 8 + 5 + 8 + 5 = 26 \text{ cm}

A_L = P_B \cdot h = 26 \cdot 7 = 182\text{ cm}^2

$A_L = 182\ \text{cm}^2$

c) Total area

Two bases plus the lateral area:

A_T = 2 \cdot A_B + A_L = 2 \cdot 40 + 182 = 262\text{ cm}^2

$A_T = 262\ \text{cm}^2$

d) Volume

V = A_B \cdot h = 40 \cdot 7 = 280\text{ cm}^3

To convert to litres we divide by $1000$:

280 : 1000 = 0{,}28 \text{ L}

$V = 280\ \text{cm}^3 = 0{,}28$ L

2. Cylinder

Cylinder: radius 4 cm and height 9 cm. Unrolled, the curved face is a rectangle as long as the circumference of the base.

See the solution

a) Base area

The base is a circle of radius $4$ cm:

A_B = \pi \cdot r^2 = \pi \cdot 4^2 \approx 50{,}27\text{ cm}^2

$A_B \approx 50{,}27\ \text{cm}^2$

b) Lateral area

The perimeter of the base is the circumference, $2\pi r$:

A_L = P_B \cdot h = 2 \cdot \pi \cdot 4 \cdot 9 \approx 226{,}19\text{ cm}^2

$A_L \approx 226{,}19\ \text{cm}^2$

c) Total area

A_T = 2 \cdot A_B + A_L = 2 \cdot 50{,}27 + 226{,}19 = 326{,}73\text{ cm}^2

$A_T \approx 326{,}73\ \text{cm}^2$

d) Volume

V = A_B \cdot h = 50{,}27 \cdot 9 = 452{,}43\text{ cm}^3

452{,}43 : 1000 \approx 0{,}45 \text{ L}

Rounding only at the end, $V = \pi \cdot 4^2 \cdot 9 = 144\pi \approx 452{,}39\ \text{cm}^3$: the difference comes from rounding the base area before multiplying by $9$. In litres it is the same, $0{,}45$ L.

$V \approx 452{,}43\ \text{cm}^3 \approx 0{,}45$ L

3. Square-based pyramid

Square-based pyramid: 7 cm side and 8 cm high. The height of each side triangle is the slant height, not the height of the pyramid.

See the solution

a) Base area

A_B = c^2 = 7^2 = 49\text{ cm}^2

$A_B = 49\ \text{cm}^2$

b) Lateral area

The side faces are 4 triangles with a $7$ cm base. Their height is the slant height $h'$, which is not given: we find it with Pythagoras in the right triangle formed by the height of the pyramid ($8$ cm), half the side of the base ($3{,}5$ cm) and the slant height.

h'^2 = 8^2 + 3{,}5^2 = 64 + 12{,}25 = 76{,}25 \;\Longrightarrow\; h' = \sqrt{76{,}25} \approx 8{,}73 \text{ cm}

A_L = 4 \cdot A_{\triangle} = 4 \cdot \dfrac{7 \cdot 8{,}73}{2} = 122{,}22\text{ cm}^2

In class we wrote $A_T$ for the area of each triangle; here we use $A_{\triangle}$ so it is not confused with the total area. Rounding only at the end (with $h' = \sqrt{76{,}25}$) gives $A_L \approx 122{,}25\ \text{cm}^2$.

$h' \approx 8{,}73$ cm · $A_L \approx 122{,}22\ \text{cm}^2$

c) Total area

A pyramid has only one base:

A_T = A_B + A_L = 49 + 122{,}22 = 171{,}22\text{ cm}^2

Rounding only at the end, $A_T \approx 171{,}25\ \text{cm}^2$.

$A_T \approx 171{,}22\ \text{cm}^2$

d) Volume

Here we use the height of the pyramid, $8$ cm, not the slant height:

V = \dfrac{A_B \cdot h}{3} = \dfrac{49 \cdot 8}{3} \approx 130{,}67\text{ cm}^3

130{,}67 : 1000 \approx 0{,}13 \text{ L}

$V \approx 130{,}67\ \text{cm}^3 \approx 0{,}13$ L