Sec 1 Maths: Perimeter, Area & Volume (Mensuration) — practice questions & worked solutions

(a) Area of the shaded region

Radius $= 21$ cm. Length of $DC$:

$$\begin{aligned}DC &= \frac{1}{3} \times 42\\ &= 14 \text{ cm}\end{aligned}$$

The unshaded region is trapezium $ABCD$ with parallel sides $AB$ and $DC$ and height $DE = 20$ cm:

$$\begin{aligned}\text{area of trapezium} &= \frac{1}{2}(42 + 14)(20)\\ &= 560 \text{ cm}^2\end{aligned}$$

Area of circle:

$$\begin{aligned}\text{area of circle} &= 3.142 \times 21^2\\ &= 1385.622 \text{ cm}^2\end{aligned}$$

Shaded area $=$ circle $–$ trapezium:

$$\begin{aligned}\text{shaded area} &= 1385.622 – 560\\ &= 825.622\\ &= 826 \text{ cm}^2 \;(3\text{ s.f.})\end{aligned}$$

(b)(i) Height of the parallelogram

Area $=$ base $\times$ height, base $CD = 10$ cm:

$$\begin{aligned}60 &= 10 \times h\\ h &= 6 \text{ cm}\end{aligned}$$

(b)(ii) Area of triangle $CMB$

$AB = CD = 10$ cm, so $MB = \tfrac{1}{2}(10) = 5$ cm; the height between $AB$ and $DC$ is $6$ cm.

$$\begin{aligned}\text{area of } CMB &= \frac{1}{2} \times 5 \times 6\\ &= 15 \text{ cm}^2\end{aligned}$$

(a) Volume of the block

Convert to cm: $0.22$ m $= 22$ cm, $0.18$ m $= 18$ cm, $0.2$ m $= 20$ cm.

$$\begin{aligned}\text{volume} &= 22 \times 18 \times 20\\ &= 7920 \text{ cm}^3\end{aligned}$$

(b) Number of cylinders

Volume of one cylinder:

$$\begin{aligned}\text{volume} &= 3.142 \times 4^2 \times 10\\ &= 502.72 \text{ cm}^3\end{aligned}$$

Number of cylinders:

$$\begin{aligned}\frac{7920}{502.72} &= 15.75\ldots\end{aligned}$$

Number of full cylinders $= 15$.

Let the common height be $h$ and $DC = 2k$, so $AB = k$. Area of triangle $ECD$:

$$\begin{aligned}\frac{1}{2}\times DC\times h&=53\\ \frac{1}{2}\times 2k\times h&=53\\ kh&=53\end{aligned}$$

Area of trapezium $ABCD$:

$$\begin{aligned}\frac{1}{2}(AB+DC)\times h&=\frac{1}{2}(k+2k)\times h\\ &=\frac{3}{2}kh\\ &=\frac{3}{2}\times 53\\ &=79.5\ \text{cm}^{2}\end{aligned}$$

(i) Surface area to be painted

Open container: curved surface $+$ one circular base.

$$\begin{aligned}\text{Area} &= 2\pi r h+\pi r^{2}\\ &= 2\pi(12)(20)+\pi(12)^{2}\\ &= 480\pi+144\pi\\ &= 624\pi \text{ cm}^{2}\end{aligned}$$

(ii) Cost of 15 containers

Area of $15$ containers:

$$\begin{aligned}15\times 624\pi &= 9360\pi \text{ cm}^{2}\end{aligned}$$

Convert to m$^2$ $(1\text{ m}^{2}=10000\text{ cm}^{2})$:

$$\begin{aligned}\frac{9360\pi}{10000} &= 0.936\pi \text{ m}^{2}\end{aligned}$$

Cost:

$$\begin{aligned}0.936\pi\times 9.50 &= 27.94\end{aligned}$$

The cost is $\$$27.94.

(i) Arc length of minor arc $AB$

Area $=\pi r^{2}=144\pi$, so

$$\begin{aligned}r^{2} &= 144\\ r &= 12 \text{ cm}\end{aligned}$$

Arc length is $\dfrac{9}{24}$ of the circumference:

$$\begin{aligned}\text{arc } AB &= \frac{9}{24}\times 2\pi(12)\\ &= \frac{9}{24}\times 24\pi\\ &= 9\pi \text{ cm}\end{aligned}$$

(ii) Radius of the cone’s base

The arc becomes the circumference of the base of the cone:

$$\begin{aligned}2\pi R &= 9\pi\\ R &= 4.5 \text{ cm}\end{aligned}$$

Triangular prism: cross-section is triangle $ADE$ with base $DE = 19$ cm and height $12$ cm, length $AB = 25$ cm.

$$\begin{aligned}V_{\text{prism}} &= \frac{1}{2}\times 19\times 12\times 25\\ &= 2850 \text{ cm}^{3}\end{aligned}$$

Pyramid: rectangular base $19\times 25$, height $= 48 – 12 = 36$ cm.

$$\begin{aligned}V_{\text{pyramid}} &= \frac{1}{3}\times (19\times 25)\times 36\\ &= 5700 \text{ cm}^{3}\end{aligned}$$

Total volume:

$$\begin{aligned}V &= 2850+5700\\ &= 8550 \text{ cm}^{3}\end{aligned}$$

(i) Length of $HG$

Area of trapezium $= \dfrac{1}{2}(EF + HG) \times FG$:

$$\begin{aligned}26 &= \frac{1}{2}(5 + HG)(4) \\ 26 &= 2(5 + HG) \\ 13 &= 5 + HG \\ HG &= 8 \text{ m}\end{aligned}$$

(ii) Perimeter of the garden

Each rhombus side $= AB + 2 = 5 + 2 = 7$ m, and a sector of radius $2$ m is removed at corners $C$ and $I$. The angle at $E$ (and at $A$) is $53.13^\circ$, so the angle at $C$ (and at $I$) is

$$\begin{aligned}180^\circ – 53.13^\circ &= 126.87^\circ\end{aligned}$$

Arc length of one sector (radius $2$ m):

$$\begin{aligned}\frac{126.87}{360} \times 2 \times \pi \times 2 &= 4.4286\end{aligned}$$

The perimeter consists of $AB$, the arc at $C$, $DE$, $EF$, $FG$, $GH$, the arc at $I$ and $JA$:

$$\begin{aligned}\text{Perimeter} &= 5 + 4.4286 + 5 + 5 + 4 + 8 + 4.4286 + 5 \\ &= 40.9 \text{ m}\end{aligned}$$

(i) Area of $ABCDEF$

The full rectangle has area $70 \times 50 = 3500$ cm$^2$. The carved trapezium has parallel sides $FC = 70$ cm and $ED = 40$ cm with height $50 – 30 = 20$ cm.

$$\begin{aligned}\text{Carved area} &= \frac{1}{2}(70 + 40)(20) \\ &= 1100 \\ \text{Area of } ABCDEF &= 3500 – 1100 \\ &= 2400 \text{ cm}^2\end{aligned}$$

(ii) Form and solve an equation in $x$

The perimeter of the cross-section is $AB + BC + CD + DE + EF + FA = 70 + 50 + x + 40 + x + 50 = 210 + 2x$. Total surface area:

$$\begin{aligned}2(2400) + (210 + 2x)(180) &= 51\,600 \\ 4800 + 180(210 + 2x) &= 51\,600 \\ 180(210 + 2x) &= 46\,800 \\ 210 + 2x &= 260 \\ 2x &= 50 \\ x &= 25\end{aligned}$$

(iii) Volume of cement

$$\begin{aligned}\text{Volume} &= 2400 \times 180 \\ &= 432\,000 \text{ cm}^3\end{aligned}$$

(iv) Cost of cement

Convert to cubic metres: $432\,000 \text{ cm}^3 = 0.432$ m$^3$.

$$\begin{aligned}\text{Cost} &= 0.432 \times 86.90 \\ &= 37.54\end{aligned}$$

The cost of cement is $\$$37.54.

(i)(a) Length

$$\begin{aligned}\text{Length} &= (3x – 4) \text{ cm}\end{aligned}$$

(i)(b) Perimeter of the rectangle

$$\begin{aligned}\text{Perimeter} &= 2(x + 3x – 4) = (8x – 8) \text{ cm}\end{aligned}$$

(ii) Perimeter of the square

Side of square $= \sqrt{100} = 10$ cm.

$$\begin{aligned}\text{Perimeter} &= 4 \times 10 = 40 \text{ cm}\end{aligned}$$

(iii) Form and solve an equation in $x$

Same wire, so perimeters are equal.

$$\begin{aligned}8x – 8 &= 40\\ 8x &= 48\\ x &= 6\end{aligned}$$

(iv) Area of the original rectangle

Breadth $= 6$ cm, length $= 3(6) – 4 = 14$ cm.

$$\begin{aligned}\text{Area} &= 6 \times 14 = 84 \text{ cm}^{2}\end{aligned}$$

Volume of cuboid:

$$\begin{aligned}1.1\times0.2\times2.5&=0.55\text{ m}^3\end{aligned}$$

Volume of half-cylinder (radius $1.5$ m, length $0.4$ m):

$$\begin{aligned}\frac{1}{2}\times\pi\times1.5^2\times0.4&=1.4137\ldots\text{ m}^3\end{aligned}$$

Total volume:

$$\begin{aligned}0.55+1.4137&=1.96\text{ m}^3\end{aligned}$$

Surface area of the half-cylinder (two semicircular ends, the curved surface, and the flat rectangular face), then add the exposed cuboid faces.

Two semicircular ends:

$$\begin{aligned}2\times\frac{1}{2}\pi(1.5)^2&=\pi(2.25)=7.0686\text{ m}^2\end{aligned}$$

Curved (half of cylinder lateral) surface:

$$\begin{aligned}\frac{1}{2}\times2\pi(1.5)(0.4)&=0.6\pi=1.8850\text{ m}^2\end{aligned}$$

Flat rectangular base of the half-cylinder (length $=$ diameter $3$ m, breadth $0.4$ m), with the cuboid top removed where they join. Top of cuboid $= 1.1\times0.2 = 0.22$ m$^2$:

$$\begin{aligned}3\times0.4-0.22&=0.98\text{ m}^2\end{aligned}$$

Exposed cuboid faces (four sides $+$ base):

$$\begin{aligned}2(1.1\times2.5)+2(0.2\times2.5)+(1.1\times0.2)&=5.5+1.0+0.22=6.72\text{ m}^2\end{aligned}$$

Total surface area:

$$\begin{aligned}7.0686+1.8850+0.98+6.72&=16.7\text{ m}^2\end{aligned}$$

(a) Cross-sectional area

Semicircle: radius $= 15$ cm.

$$\begin{aligned}\text{semicircle area} &= \frac{1}{2} \pi (15)^2 = 112.5\pi \text{ cm}^2 \\ \text{trapezium area} &= \frac{1}{2}(8 + 14)(4) = 44 \text{ cm}^2 \\ \text{cross-section} &= (112.5\pi – 44) \text{ cm}^2\end{aligned}$$

(b) Volume

$$\begin{aligned}\text{volume} &= (112.5\pi – 44) \times 25 \\ &= 7735.7 \text{ cm}^3\end{aligned}$$

(c) Total surface area

Perimeter of the cross-section: semicircle arc $+$ two base segments $+$ two slant sides $+$ top.

$$\begin{aligned}\text{arc} &= \pi (15) = 15\pi \text{ cm} \\ \text{base segments} &= \frac{30 – 14}{2} \times 2 = 16 \text{ cm} \\ \text{slant} + \text{top} &= 5 + 5 + 8 = 18 \text{ cm} \\ \text{perimeter} &= (15\pi + 34) \text{ cm}\end{aligned}$$
$$\begin{aligned}\text{total surface area} &= 2(112.5\pi – 44) + (15\pi + 34)(25) \\ &= 2647.0 \text{ cm}^2\end{aligned}$$

The trapezium cross-section has parallel sides $3$ cm and $5$ cm, height $6$ cm; the slant sides are $7$ cm; the prism length is $13$ cm; the cylindrical hole has radius $1$ cm and length $13$ cm.

(i) Volume

Area of trapezium:

$$\begin{aligned}A &= \frac{1}{2}(3+5)(6)\\ &= 24 \text{ cm}^2\end{aligned}$$

Volume of prism minus cylindrical hole:

$$\begin{aligned}V &= 24 \times 13 – \pi (1)^2 (13)\\ &= 312 – 13\pi\\ &= 271 \text{ cm}^3\end{aligned}$$

(ii) Total surface area

Two trapezium faces (less the two circular hole ends), the four rectangular outer faces, and the inner curved surface of the hole. Two trapezium ends:

$$\begin{aligned}2 \times 24 – 2 \times \pi (1)^2 &= 48 – 2\pi\end{aligned}$$

Four rectangular side faces (widths $3$, $5$, $7$, $7$, length $13$):

$$\begin{aligned}(3+5+7+7) \times 13 &= 22 \times 13\\ &= 286\end{aligned}$$

Curved surface of the hole:

$$\begin{aligned}2\pi (1)(13) &= 26\pi\end{aligned}$$

Total surface area:

$$\begin{aligned}&= (48 – 2\pi) + 286 + 26\pi\\ &= 334 + 24\pi\\ &= 409 \text{ cm}^2\end{aligned}$$

(a)(i) Cross-sectional area

Area $=$ trapezium $+$ semicircle, with semicircle radius $0.1$ m.

$$\begin{aligned}\text{trapezium} &= \frac{1}{2}(0.6 + 0.2)(0.5) \\ &= 0.2\end{aligned}$$
$$\begin{aligned}\text{semicircle} &= \frac{1}{2}\pi(0.1)^2 \\ &= 0.015707\end{aligned}$$
$$\begin{aligned}\text{area} &= 0.2 + 0.015707 \\ &= 0.2157 \text{ m}^2\end{aligned}$$

(a)(ii) Volume

$$\begin{aligned}\text{volume} &= 0.2157 \times 3 \\ &= 0.647 \text{ m}^3\end{aligned}$$

(b) Number of structures from 3 tins

Painted area $=$ two cross-section ends $+$ two sloping rectangles $+$ curved semicircular surface (base excluded).

$$\begin{aligned}\text{two ends} &= 2 \times 0.2157 = 0.43142\\ \text{two slopes} &= 2 \times (0.7 \times 3) = 4.2\\ \text{curved} &= \pi (0.1)(3) = 0.94248\\ \text{total} &= 0.43142 + 4.2 + 0.94248 = 5.5739 \text{ m}^2\end{aligned}$$

With $3$ tins covering $60$ m$^2$:

$$\begin{aligned}\frac{60}{5.5739} &= 10.76\end{aligned}$$

Maximum number of structures $= 10$.

(c) Is it safe to lift?

Convert the volume to cm$^3$: $0.647 \text{ m}^3 = 647000 \text{ cm}^3$.

$$\begin{aligned}\text{mass} &= 647000 \times 2.35 \\ &= 1520450 \text{ g} \\ &= 1520.685 \text{ kg}\end{aligned}$$

(Using the unrounded volume gives $1520.685$ kg.) Since $1520.685$ kg $> 900$ kg, the weight exceeds the maximum load, so it is not safe for the lifting gear to lift the barrier.

(a) Name of the quadrilateral

Rhombus.

(b) Total area of the 4 semicircles

Four identical semicircles make $2$ full circles. Radius $= \dfrac{5}{2} = 2.5$ cm.

$$\begin{aligned}\text{Total area} &= 2\times\pi\times2.5^{2}\\ &= 12.5\pi \text{ cm}^{2}\end{aligned}$$

(c)(i) Perpendicular length from $P$ to $QR$

Each side of the rhombus equals a diameter, so $QR = 5$ cm. The diagonal $PR$ splits the rhombus into two equal triangles, so the area of triangle $PQR$ is $12$ cm$^2$. Taking $QR$ as the base and $h$ as the perpendicular distance from $P$ to $QR$:

$$\begin{aligned}\frac{1}{2}\times QR\times h &= 12\\ \frac{1}{2}\times5\times h &= 12\\ h &= 4.8\text{ cm}\end{aligned}$$

(c)(ii) Length of the diagonal $QS$

Area of a rhombus $= \dfrac{1}{2}\times d_1\times d_2$:

$$\begin{aligned}\frac{1}{2}\times PR\times QS &= 24\\ \frac{1}{2}\times6\times QS &= 24\\ QS &= 8\text{ cm}\end{aligned}$$

(a) Area of triangle $ABC$

$$\begin{aligned}\text{Area of triangle } ABC &= \frac{1}{2} \times 8.66 \times 10 \\ &= 43.3 \text{ cm}^2\end{aligned}$$

(b) Volume of the prism

$$\begin{aligned}\text{Base area} &= 43.3 \times 6 \\ &= 259.8 \text{ cm}^2\end{aligned}$$
$$\begin{aligned}\text{Volume} &= 259.8 \times 63 \\ &= 16367.4 \text{ cm}^3\end{aligned}$$

(c) Total surface area

$$\begin{aligned}\text{Surface area of one side} &= 63 \times 10 \\ &= 630 \text{ cm}^2\end{aligned}$$
$$\begin{aligned}\text{Total surface area} &= 259.8 \times 2 + 630 \times 6 \\ &= 4299.6 \text{ cm}^2\end{aligned}$$

(d) Cost of wrapping

$$\begin{aligned}1 \text{ cm}^2 &= 1.12 \div 500 \\ &= 0.00224\end{aligned}$$
$$\begin{aligned}\text{Total cost} &= 0.00224 \times 4299.6 \\ &= \$9.63\end{aligned}$$

(e) Can Elias create 20 cylinders?

$$\begin{aligned}\text{Volume of cylinder} &= \pi r^2 h \\ &= \pi (8^2)(4.5) \\ &= 904.77868 \text{ cm}^3\end{aligned}$$
$$\begin{aligned}\text{Number of cylinders} &= 16367.4 \div 904.77868 \\ &= 18.0899 \\ &\approx 18\end{aligned}$$

Elias will not be able to create $20$ cylinders, as he can only make $18$ such cylinders with the available material.

Original area $= \dfrac{1}{2}\times 3h\times h = \dfrac{3h^2}{2}$.

$$\begin{aligned}1.2\times\frac{3h^2}{2} &= 115.2 \\ 1.8h^2 &= 115.2 \\ h^2 &= 64 \\ h &= 8\end{aligned}$$

(a) Height of the parallelogram

Height of trapezium $= AG = 4$ cm.

$$\begin{aligned}\frac{\text{height of parallelogram}}{4} &= \frac{3}{2} \\ \text{height of parallelogram} &= 6 \text{ cm}\end{aligned}$$

(b) Find $DE$

Area of trapezium $= \dfrac{1}{2}(AB+FC)\times 4 = 2(AB+FC)$. Area of parallelogram $= FC\times 6$ (base $FC$ and the stated height). Area of trapezium $= \dfrac{1}{2}\times$ area of parallelogram:

$$\begin{aligned}2(5+FC) &= \frac{1}{2}(6\,FC) \\ 2(5+FC) &= 3\,FC \\ 10+2\,FC &= 3\,FC \\ FC &= 10\end{aligned}$$

Since $CDEF$ is a parallelogram, $DE = FC$:

$$DE = 10 \text{ cm}$$

(a) Depth of paint in the tin

$1.65$ litres $= 1650$ cm$^3$. Radius $= 5$ cm.

$$\begin{aligned}\pi r^2 h &= 1650 \\ \pi(5)^2 h &= 1650 \\ 25\pi h &= 1650 \\ h &= 21.0 \text{ cm}\end{aligned}$$

(b) Show that $EF = 7$ cm

The trough is a trapezoidal prism of length $HE = 25$ cm. Its cross-section $ABGH$ is a right trapezium with parallel sides $AB = 15$ cm and $EF$, and perpendicular height $AH = 6$ cm. The paint fills it completely, so the trough volume $= 1650$ cm$^3$.

$$\begin{aligned}\text{(area of trapezium)}\times 25 &= 1650 \\ \text{area of trapezium} &= 66 \\ \frac{1}{2}(15+EF)\times 6 &= 66 \\ 3(15+EF) &= 66 \\ 15+EF &= 22 \\ EF &= 7 \text{ cm}\end{aligned}$$

(c) Wetted surface area

The slant side of the trapezium (Pythagoras, horizontal run $15 – 7 = 8$, height $6$):

$$\begin{aligned}\text{slant} &= \sqrt{8^2+6^2} \\ &= 10 \text{ cm}\end{aligned}$$

The top is open, so the wetted surfaces are the two trapezium ends, the bottom rectangle, the vertical side and the slant side.

$$\begin{aligned}\text{two ends} &= 2\times 66=132 \\ \text{bottom}\ (EF\times 25) &= 7\times 25=175 \\ \text{vertical side}\ (6\times 25) &= 150 \\ \text{slant side}\ (10\times 25) &= 250\end{aligned}$$
$$\begin{aligned}\text{Total surface area} &= 132+175+150+250 \\ &= 707 \text{ cm}^2\end{aligned}$$

(a) Value of $h$

Area using base $10$ cm and height $6$ cm:

$$\text{Area} = 10 \times 6 = 60 \text{ cm}^2$$

Same area using base $8$ cm and height $h$:

$$8h = 60$$
$$h = 7.5$$

(b) Length of $CD$

Trapezium area with parallel sides $AB$ and $DC$, height $AE = 12$:

$$\tfrac{1}{2}(14 + CD)(12) = 240$$
$$14 + CD = 40$$
$$CD = 26 \text{ cm}$$

(a) Time to fill

Volume $= 12.6$ litres $= 12600$ cm$^3$. Pipe radius $= 1.5$ cm, rate $= 225$ cm/s. Flow rate:

$$\pi (1.5)^2 (225) = 506.25\pi \text{ cm}^3/\text{s}$$
$$\text{time} = \frac{12600}{506.25\pi}$$
$$= 7.92\ldots$$
$$= 8 \text{ s}$$

(b) Show that $AB = 42$ cm

$FG$ is the perpendicular height to $AB$, so $G$ is the midpoint of $AB$. By Pythagoras’ theorem in $\triangle AGF$:

$$AG = \sqrt{29^2 – 20^2}$$
$$AG = \sqrt{441} = 21$$
$$AB = 2 \times 21 = 42 \text{ cm}$$

(c) Wetted surface area

The prism is open at the top triangular face (the water surface). The water touches the three rectangular faces and the bottom triangular face. Rectangular faces (each length $30$ cm):

$$42 \times 30 + 29 \times 30 + 29 \times 30 = 3000 \text{ cm}^2$$

Bottom triangular face:

$$\tfrac{1}{2}(42)(20) = 420 \text{ cm}^2$$
$$\text{total} = 3000 + 420 = 3420 \text{ cm}^2$$