Checkpoint Quiz 1

Maths Standard Year 11 · Module 2: Measurement

Lesson 1, Formulas & Units Lesson 2, Area of Basic Shapes Lesson 3, Pythagoras' Theorem Lesson 4, Intro to Trigonometry Lesson 5, Perimeter & Arc Length
10 MC 3 SA ~25 min
0/10
Multiple Choice Score
Answer questions to see your score.
Part A, Multiple Choice (1 mark each)
1 Which of the following correctly converts $4.5 \text{ m}^2$ into $\text{cm}^2$? L1
A $4\,500 \text{ cm}^2$
B $45\,000 \text{ cm}^2$
C $45 \text{ cm}^2$
D $450\,000 \text{ cm}^2$
D, $450\,000 \text{ cm}^2$. Area units square the linear conversion: $1\text{ m} = 100\text{ cm}$, so $1\text{ m}^2 = 100^2 = 10\,000\text{ cm}^2$. Therefore $4.5 \times 10\,000 = 450\,000\text{ cm}^2$.
2 A triangle has base $12\text{ cm}$ and perpendicular height $7\text{ cm}$. What is its area? L2
A $84 \text{ cm}^2$
B $42 \text{ cm}^2$
C $19 \text{ cm}^2$
D $38 \text{ cm}^2$
B, $42 \text{ cm}^2$. $A = \tfrac{1}{2}bh = \tfrac{1}{2} \times 12 \times 7 = 42\text{ cm}^2$. Remember to halve the base × height product for a triangle.
3 A trapezium has parallel sides of $5\text{ m}$ and $9\text{ m}$, and a perpendicular height of $4\text{ m}$. Find its area. L2
A $28 \text{ m}^2$
B $56 \text{ m}^2$
C $14 \text{ m}^2$
D $36 \text{ m}^2$
A, $28 \text{ m}^2$. $A = \tfrac{1}{2}(a+b)h = \tfrac{1}{2}(5+9) \times 4 = \tfrac{1}{2} \times 14 \times 4 = 28\text{ m}^2$.
4 A rectangle is $10\text{ m}$ by $6\text{ m}$ with a semicircle of radius $3\text{ m}$ removed from one end. What is the remaining area, to the nearest $0.01\text{ m}^2$? L2
A $45.86 \text{ m}^2$
B $74.57 \text{ m}^2$
C $31.73 \text{ m}^2$
D $88.27 \text{ m}^2$
A, $45.86 \text{ m}^2$. Rectangle area $= 10 \times 6 = 60\text{ m}^2$. Semicircle area $= \tfrac{1}{2}\pi r^2 = \tfrac{1}{2}\pi(3)^2 = \tfrac{9\pi}{2} \approx 14.14\text{ m}^2$. Remaining $= 60 - 14.14 \approx 45.86\text{ m}^2$.
5 A right-angled triangle has legs of $9\text{ cm}$ and $40\text{ cm}$. What is the length of the hypotenuse? L3
A $41 \text{ cm}$
B $49 \text{ cm}$
C $31 \text{ cm}$
D $37.4 \text{ cm}$
A, $41 \text{ cm}$. $c = \sqrt{9^2 + 40^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41\text{ cm}$. This is a Pythagorean triple: $9, 40, 41$.
6 In a right-angled triangle, the hypotenuse is $13\text{ cm}$ and one leg is $5\text{ cm}$. Find the length of the other leg. L3
A $8 \text{ cm}$
B $14 \text{ cm}$
C $12 \text{ cm}$
D $18 \text{ cm}$
C, $12 \text{ cm}$. $b = \sqrt{c^2 - a^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$. The triple $5, 12, 13$ is worth remembering.
7 In a right-angled triangle, the hypotenuse is $20\text{ m}$ and one angle is $40°$. What is the length of the side opposite that angle, to 2 decimal places? L4
A $12.86 \text{ m}$
B $15.32 \text{ m}$
C $26.08 \text{ m}$
D $23.94 \text{ m}$
A, $12.86 \text{ m}$. $\sin 40° = \dfrac{\text{opp}}{\text{hyp}} \Rightarrow \text{opp} = 20 \sin 40° \approx 20 \times 0.6428 \approx 12.86\text{ m}$.
8 The side opposite a $55°$ angle measures $14\text{ cm}$. What is the hypotenuse, to 2 decimal places? L4
A $11.47 \text{ cm}$
B $20.00 \text{ cm}$
C $17.09 \text{ cm}$
D $8.03 \text{ cm}$
C, $17.09 \text{ cm}$. $\sin 55° = \dfrac{14}{\text{hyp}} \Rightarrow \text{hyp} = \dfrac{14}{\sin 55°} \approx \dfrac{14}{0.8192} \approx 17.09\text{ cm}$.
9 An arc subtends an angle of $80°$ at the centre of a circle with radius $9\text{ cm}$. What is the arc length, to 2 decimal places? L5
A $25.13 \text{ cm}$
B $6.28 \text{ cm}$
C $12.57 \text{ cm}$
D $56.55 \text{ cm}$
C, $12.57 \text{ cm}$. $l = \dfrac{\theta}{360°} \times 2\pi r = \dfrac{80}{360} \times 2\pi \times 9 = \dfrac{2}{9} \times 18\pi = 4\pi \approx 12.57\text{ cm}$.
10 A sector has radius $6\text{ cm}$ and arc length $8\text{ cm}$. What is the perimeter of the sector? L5
A $14 \text{ cm}$
B $20 \text{ cm}$
C $8 \text{ cm}$
D $24 \text{ cm}$
B, $20 \text{ cm}$. Perimeter of a sector $= 2r + l = 2(6) + 8 = 12 + 8 = 20\text{ cm}$. The two radii form the straight sides.
Part B, Short Answer (show all working)
1 L1
Convert a speed of $2.4\text{ km/h}$ into metres per second. Show each conversion step.
Convert km → m: $2.4\text{ km} = 2.4 \times 1000 = 2400\text{ m}$
Convert h → s: $1\text{ h} = 60 \times 60 = 3600\text{ s}$
Divide: $\dfrac{2400}{3600} = \dfrac{2}{3} \approx 0.667\text{ m/s}$
Answer: $\approx 0.67\text{ m/s}$
2 L2 & L3
A right-angled triangle has legs of $6\text{ cm}$ and $8\text{ cm}$.
(a) Find the length of the hypotenuse.
(b) Find the area of the triangle.
(a) $c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ cm}$
(b) $A = \tfrac{1}{2} \times 6 \times 8 = 24\text{ cm}^2$
3 L4 & L5
A sector has radius $12\text{ cm}$ and central angle $120°$.
(a) Find the arc length. Give your answer in exact form and as a decimal (2 d.p.).
(b) Find the perimeter of the sector, to 2 decimal places.
(a) $l = \dfrac{120}{360} \times 2\pi \times 12 = \dfrac{1}{3} \times 24\pi = 8\pi \approx 25.13\text{ cm}$
(b) $P = 2r + l = 2(12) + 8\pi = 24 + 8\pi \approx 24 + 25.13 = 49.13\text{ cm}$

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