Mathematics questions, chapter 10

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Perimeter and area

40 marks. Use Practice to check each answer as you go, or Exam to answer everything first. You may use a calculator. Show your working for Section B.

Section A: objective questions

12 marks

Choose the correct answer.

Question 1

What is the perimeter of a shape?

The total length around the boundary of the shape
The amount of surface covered by the shape
The length of the longest side of the shape
The length multiplied by the width of the shape

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Perimeter is the distance all the way round, so we add all the sides. The surface covered is the area, and length × width is the area of a rectangle.

For the teacher: Remembering, low difficulty, standard 10.1.1

Textbook p. 226

Question 2

Diagram 1 shows a shape with right-angled corners. Find its perimeter.

Staircase shape: top 5 cm, middle step 5 cm, lower step 4 cm, right side 4 cm, bottom 14 cm, left side 10 cm. Two short vertical sides are not labelled. — Diagram 1

42 cm
45 cm
48 cm
140 cm

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The right-hand vertical pieces must add up to the left side, so the two unlabelled vertical sides add up to 10 − 4 = 6 cm. Perimeter = 5 + 5 + 4 + 4 + 14 + 10 + 6 = 48 cm. 42 cm adds only the labelled sides and 45 cm counts only one of the two short sides (3 cm). 140 is 14 × 10, an area idea.

For the teacher: Applying, low difficulty, standard 10.1.1

Textbook p. 226

Question 3

Aina estimates the perimeter of a leaf as 30 cm. She measures it with a thread and gets 32 cm. What is the percentage error of her estimate?

2%
6.25%
6.67%
93.75%

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Percentage error = (32 − 30) ÷ 32 × 100% = 6.25%. 6.67% divides by the estimate (30) instead of the actual value. 2% is only the difference, and 93.75% is 30 ÷ 32 × 100%.

For the teacher: Understanding, medium difficulty, standard 10.1.2

Textbook p. 228

Question 4

A rectangular field is 45 m long and 30 m wide. The whole field is fenced except for a gate 5 m wide. Fencing costs RM12 per metre. What is the cost of the fencing?

RM1 680
RM1 800
RM16 200
RM1 740

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Perimeter = 2 × (45 + 30) = 150 m. Fence = 150 − 5 = 145 m. Cost = 145 × RM12 = RM1 740. RM1 800 forgets the gate, RM1 680 subtracts the gate twice, and RM16 200 uses the area (1 350 m²).

For the teacher: Applying, medium difficulty, standard 10.1.3

Textbook p. 228

Question 5

A trapezium has parallel sides of lengths a and b, and height t. Which formula gives its area?

½(a + b)t
(a + b)t
½abt
½(a + b) + t

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Two identical trapeziums make a parallelogram with base (a + b) and height t, so one trapezium is ½(a + b)t. (a + b)t is the area of the whole parallelogram.

For the teacher: Remembering, low difficulty, standard 10.2.2

Textbook p. 235

Question 6

Find the area of the parallelogram in Diagram 2.

Parallelogram with base 9 cm, perpendicular height 4 cm and slanted side 5 cm — Diagram 2

45 cm²
36 cm²
18 cm²
28 cm²

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Area = base × height = 9 × 4 = 36 cm². 45 cm² uses the slanted side (9 × 5), 18 cm² wrongly halves it like a triangle, and 28 is the perimeter.

For the teacher: Applying, low difficulty, standard 10.2.3

Textbook p. 236

Question 7

Diagram 3 shows a triangle with a base of 6 cm. Its height, 8 cm, is drawn outside the triangle. Find the area of the triangle.

Obtuse triangle with base 6 cm, longest side 17 cm, and a height of 8 cm drawn outside the triangle at right angles to the extended base — Diagram 3

48 cm²
51 cm²
68 cm²
24 cm²

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Area = ½ × base × height = ½ × 6 × 8 = 24 cm². The height is still perpendicular to the base line, even when it is outside. 48 cm² forgets the ½; 51 cm² and 68 cm² use the 17 cm side, which is not perpendicular to the base or the height.

For the teacher: Applying, low difficulty, standard 10.2.3

Textbook p. 236

Question 8

Diagram 4 shows a kite. Find its area.

Kite with a horizontal diagonal of 8 cm. The vertical diagonal is split into 3 cm above and 9 cm below the crossing point. — Diagram 4

96 cm²
24 cm²
48 cm²
36 cm²

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The diagonals are 3 + 9 = 12 cm and 8 cm. Area = ½ × 12 × 8 = 48 cm². 96 cm² forgets the ½, 24 cm² uses half of the 8 cm diagonal, and 36 cm² uses only the 9 cm part.

For the teacher: Applying, medium difficulty, standard 10.2.3

Textbook p. 237

Question 9

Diagram 5 shows a shape on a grid of 1 unit squares. 15 squares are full, 12 squares are half full or more, and 9 squares are less than half full. Estimate the area of the shape.

Oval-shaped region drawn on a grid of 1 unit squares — Diagram 5

15 unit²
27 unit²
36 unit²
12 unit²

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Count full squares plus squares that are half full or more: 15 + 12 = 27 unit². Squares less than half full are ignored, so 36 (15 + 12 + 9) is too big, and 15 leaves out the half-full squares.

For the teacher: Understanding, medium difficulty, standard 10.2.1

Textbook p. 231

Question 10

Kassim has 20 m of fencing to make a rectangular plot. Which length and width give the largest area?

9 m × 1 m
7 m × 3 m
6 m × 4 m
5 m × 5 m

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All four have a perimeter of 20 m. The areas are 9, 21, 24 and 25 m². With a fixed perimeter, the area is largest when the rectangle is a square.

For the teacher: Understanding, medium difficulty, standard 10.3.1

Textbook p. 238

Question 11

The length of a rectangle is 3 cm more than its width. Its perimeter is 26 cm. Find the area of the rectangle.

40 cm²
42.25 cm²
166.75 cm²
65 cm²

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2w + 2(w + 3) = 26, so 4w + 6 = 26 and w = 5 cm; length = 8 cm. Area = 8 × 5 = 40 cm². 42.25 cm² treats it as a square of side 6.5 cm; 166.75 cm² uses w + (w + 3) = 26; 65 cm² multiplies 5 by half the perimeter (13).

For the teacher: Analysing, high difficulty, standard 10.3.2

Textbook p. 241

Question 12

Diagram 6 shows a rectangle 10 cm by 6 cm with a triangle on top. The height of the triangle is 4 cm. Find the area of the whole shape.

House shape: rectangle 10 cm wide and 6 cm high with a triangle of height 4 cm on top — Diagram 6

100 cm²
60 cm²
80 cm²
90 cm²

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Rectangle = 10 × 6 = 60 cm². Triangle = ½ × 10 × 4 = 20 cm². Total = 80 cm². 100 cm² forgets the ½ for the triangle, 60 cm² leaves the triangle out, and 90 cm² uses 6 cm as the height of the triangle.

For the teacher: Analysing, high difficulty, standard 10.2.3

Textbook p. 237

Section B: structured questions

28 marks

Answer all questions. Show your working.

Question 13

Diagram 7 shows a shape drawn on a grid of 1 cm squares.

Pentagon on a 1 cm grid: top 5 cm, right side 2 cm, a slanted side, bottom 2 cm, left side 4 cm — Diagram 7

[1 mark] (a)Estimate the perimeter of the shape.

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5 + 2 + 3.5 + 2 + 4 ≈ 16.5 cm (accept 16 cm to 17 cm) [1]1

The slanted side crosses 3 squares across and 2 squares down, so it is a little more than 3.5 cm.

Tick each point you got.

[2 marks] (b)Ali estimated the perimeter as 16 cm. He measured it with a ruler and got 16.6 cm. Calculate the percentage error of Ali's estimate.

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16.6 − 16 = 0.6 [1]1
0.6 ÷ 16.6 × 100% = 3.61% [1]1

Tick each point you got.

For the teacher: Applying, low difficulty, standard 10.1.2

Textbook p. 228

Question 14

Diagram 8 shows (a) a parallelogram with base p and height q, and (b) a kite with diagonals m and n.

(a) Parallelogram with base p and perpendicular height q. (b) Kite with diagonals m and n crossing at right angles. — Diagram 8

[1 mark] (a)Write an expression for the area of the parallelogram.

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pq [1]1

Tick each point you got.

[1 mark] (b)Write an expression for the area of the kite.

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½mn [1]1

Tick each point you got.

For the teacher: Remembering, low difficulty, standard 10.2.2

Textbook p. 234

Question 15

Diagram 9 shows the plan of a school garden. All corners are right angles. The garden is fenced all round except for a gate 6 m wide. Fencing costs RM25 per metre.

U-shaped garden: bottom 40 m, both outer sides 25 m, top left 12 m, top right 12 m, the gap is 10 m deep and its bottom side is x m. A 6 m gate is on the bottom side. — Diagram 9

[1 mark] (a)Find the value of x.

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x = 40 − 12 − 12 = 16 [1]1

Tick each point you got.

[1 mark] (b)Find the perimeter of the garden.

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12 + 10 + 16 + 10 + 12 + 25 + 40 + 25 = 150 m [1]1

Tick each point you got.

[2 marks] (c)Calculate the cost of the fencing.

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150 − 6 = 144 m [1]1
144 × RM25 = RM3 600 [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 10.1.3

Textbook p. 229

Question 16

Diagram 10 shows a rectangular plot of land PRTV. Triangle PQW is planted with pandan and trapezium UTSX is a fish pond. The shaded region is planted with vegetables.

Rectangle PRTV. PQ = 6 m, QR = 12 m, PW = 4 m, WV = 6 m, RS = 5 m, ST = 5 m, XS = 3 m, UT = 7 m. Triangle PQW at corner P and trapezium UTSX at corner T are not shaded. — Diagram 10

[1 mark] (a)Find the area of the rectangle PRTV.

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(6 + 12) × (4 + 6) = 18 × 10 = 180 m² [1]1

Tick each point you got.

[3 marks] (b)Find the area of the shaded region.

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Triangle PQW = ½ × 6 × 4 = 12 m² [1]1
Trapezium UTSX = ½ × (3 + 7) × 5 = 25 m² [1]1
Shaded = 180 − 12 − 25 = 143 m² [1]1

Tick each point you got.

For the teacher: Applying, medium difficulty, standard 10.2.3

Textbook p. 237

Question 17

In Diagram 11, ABC is an equilateral triangle, BCFG is a square with sides of 7 cm and CDEF is a rectangle. GFE is a straight line. The perimeter of the whole diagram is 65 cm.

Equilateral triangle ABC on top of square BCFG of side 7 cm; rectangle CDEF joined to the right of the square; G, F and E lie on a straight line — Diagram 11

[2 marks] (a)Find the length of CD.

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AB + AC + DE + GB = 4 × 7 = 28, and GE = 7 + CD, so 28 + CD + 7 + CD = 65 or 35 + 2CD = 65 [1]1
CD = 15 cm [1]1

Tick each point you got.

[1 mark] (b)Hence, find the length of GE.

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GE = 7 + 15 = 22 cm [1]1

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 10.1.3

Textbook p. 229

Question 18

Diagram 12 shows a trapezium. Siti calculated its area like this:
Area = ½ × (6 + 10) × 5
= ½ × 16 × 5
= 40 cm²

Trapezium with parallel sides 6 cm (top) and 10 cm (bottom), perpendicular height 4 cm shown dashed, and slanted side 5 cm — Diagram 12

[2 marks] (a)Siti's answer is wrong. Explain her mistake.

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She used the slanted side, 5 cm, as the height [1]1
The height must be perpendicular to the parallel sides, so it is 4 cm [1]1

Tick each point you got.

[2 marks] (b)Calculate the correct area.

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½ × (6 + 10) × 4 [1]1
= 32 cm² [1]1

Tick each point you got.

For the teacher: Evaluating, high difficulty, standard 10.2.3

Textbook p. 236

Question 19

Diagram 13 shows triangle PQS inside rectangle PQRT. S lies on TR. The perimeter of the rectangle is 60 cm and its length PQ is twice its width.

Triangle PQS drawn inside rectangle PQRT, with S on the top side TR — Diagram 13

[2 marks] (a)Find the width and the length of the rectangle.

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2w + 2(2w) = 60, so 6w = 60 [1]1
Width = 10 cm and length = 20 cm [1]1

Tick each point you got.

[2 marks] (b)Find the area of triangle PQS.

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½ × 20 × 10 [1]1
= 100 cm² [1]1

The height of the triangle is the width of the rectangle, wherever S is on TR.

Tick each point you got.

For the teacher: Analysing, high difficulty, standard 10.3.2

Textbook p. 241

Question 20

Puan Lim has 24 m of fencing. She wants to fence a rectangular herb garden using all the fencing. The side lengths must be whole numbers of metres.

[2 marks] (a)List every possible length and width, with the area of each garden.

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Length + width = 12 m [1]1
11 × 1 = 11, 10 × 2 = 20, 9 × 3 = 27, 8 × 4 = 32, 7 × 5 = 35, 6 × 6 = 36 (m²), all six correct [1]1

Tick each point you got.

[1 mark] (b)Puan Lim wants a garden that is not a square and has an area of at least 32 m². Design one such garden. State its length and width and show that it meets both conditions.

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8 m × 4 m: 2(8 + 4) = 24 m, area 32 m², length ≠ width; or 7 m × 5 m: 2(7 + 5) = 24 m, area 35 m², length ≠ width [1]1

Tick each point you got.

[1 mark] (c)Which garden from (a) has the largest area? Explain why, using the pattern in your list.

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6 m × 6 m (a square), 36 m²: for the same perimeter, the area increases as the difference between length and width gets smaller, and is largest when the rectangle is a square [1]1

Tick each point you got.

For the teacher: Creating, high difficulty, standard 10.3.1

Textbook p. 240

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