Sec 1 G1 Maths Perimeter and area worksheet with answers

A free Sec 1 G1 Maths worksheet on perimeter and area: 9 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 1 G1 Maths worksheet: Perimeter and area

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

In the diagram,

ABCD

is a trapezium with

AB

parallel to

DC

. The dashed line is perpendicular to

AB

. Find the area of

ABCD

Answercm²[2]

A running track is made up of two straight sections and two semicircles, as shown. Find the perimeter of the track, correct to 3 significant figures.

Answerm[2]

The diagram shows a circular garden pond. Its diameter is

16

m. Find its circumference, leaving your answer in terms of

\pi

Answerm[2]

The diagram shows a rectangular carpet measuring

4

m by

135

cm. Find its area, in m².

Answerm²[2]

A poster has an area of

2600

cm². Express this area in m².

Answerm²[2]

The diagram shows a semicircle of diameter

24

m. Find its area, correct to 3 significant figures.

Answerm²[3]

(a)Express

1.75

m² in cm².

Answer (a)cm²[2]

(b)A poster has an area of

25\,000

cm². Express this area in m².

Answer (b)m²[2]

Section B

(a)A notice board has an area of

52\,000

cm². Express this area in m².

Answer (a)m²[2]

(b)The diagram shows a circle of diameter

12.4

cm. Using the

\pi

key on your calculator, find the area of the circle, correct to 3 significant figures.

Answer (b)cm²[3]

(a)The diagram shows a square of side

7

m and a quarter circle with its centre at a corner of the square. Find the area of the shaded region, correct to 3 significant figures.

Answer (a)m²[2]

(b)The diagram shows a semicircle of radius

3

cm. Find its area, correct to 3 significant figures.

Answer (b)cm²[2]

(c)Express

5.2

m² in cm².

Answer (c)cm²[2]

(d)In the diagram,

ABCD

is a trapezium with

AB

parallel to

DC

. The dashed line is perpendicular to

AB

. Find the area of

ABCD

Answer (d)cm²[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 144

cm²[2]

\text{Area of } ABCD = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

= \dfrac{1}{2} \times (8 + 28) \times 8

= 144

cm²

Question 2

Answer:

= 79.4

m (3 s.f.)[2]

\text{Curved parts} = \pi d

= \pi(10)

\text{Perimeter} = 24 + 24 + \pi(10)

= 79.4159\ldots

= 79.4

m (3 s.f.)

Question 3

Answer:

= 16\pi

m[2]

r = \dfrac{16}{2}

= 8

C = 2\pi r

= 2\pi(8)

= 16\pi

Question 4

Answer:

= 5.4

m²[2]

135 \text{ cm} = 135 \div 100 \text{ m}

= 1.35 \text{ m}

\text{Area} = 4 \times 1.35

= 5.4

m²

Question 5

Answer:

= 0.26

m²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

2600 \text{ cm}^2 = \dfrac{2600}{10\,000} \text{ m}^2

= 0.26

m²

Question 6

Answer:

= 226

m² (3 s.f.)[3]

r = \dfrac{24}{2}

= 12

A = \dfrac{1}{2} \times \pi r^2

= \dfrac{1}{2} \times \pi(12)^2

= 226.1946\ldots

= 226

m² (3 s.f.)

Question 7

(a)Answer:

= 17\,500

cm²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

1.75 \text{ m}^2 = 1.75 \times 10\,000 \text{ cm}^2

= 17\,500

cm²

(b)Answer:

= 2.5

m²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

25\,000 \text{ cm}^2 = \dfrac{25\,000}{10\,000} \text{ m}^2

= 2.5

m²

Question 8

(a)Answer:

= 5.2

m²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

52\,000 \text{ cm}^2 = \dfrac{52\,000}{10\,000} \text{ m}^2

= 5.2

m²

(b)Answer:

= 121

cm² (3 s.f.)[3]

r = \dfrac{12.4}{2}

= 6.2

A = \pi r^2

= \pi(6.2)^2

= 120.7628\ldots

= 121

cm² (3 s.f.)

Question 9

(a)Answer:

= 10.5

m² (3 s.f.)[2]

\text{Area} = 7^2 - \dfrac{1}{4} \times \pi(7)^2

= 10.5154\ldots

= 10.5

m² (3 s.f.)

(b)Answer:

= 14.1

cm² (3 s.f.)[2]

A = \dfrac{1}{2} \times \pi r^2

= \dfrac{1}{2} \times \pi(3)^2

= 14.1371\ldots

= 14.1

cm² (3 s.f.)

(c)Answer:

= 52\,000

cm²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

5.2 \text{ m}^2 = 5.2 \times 10\,000 \text{ cm}^2

= 52\,000

cm²

(d)Answer:

= 380

cm²[2]

\text{Area of } ABCD = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

= \dfrac{1}{2} \times (9 + 29) \times 20

= 380

cm²

Sec 1 G1 Maths Perimeter and area worksheet with answers

In the print window, choose Save as PDF to download it.

Or practise it online with step-by-step help when you get one wrong.

Answers and worked solutions

Question 1

Answer:

= 144

cm²[2]

\text{Area of } ABCD = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

= \dfrac{1}{2} \times (8 + 28) \times 8

= 144

cm²

Question 2

Answer:

= 79.4

m (3 s.f.)[2]

\text{Curved parts} = \pi d

= \pi(10)

\text{Perimeter} = 24 + 24 + \pi(10)

= 79.4159\ldots

= 79.4

m (3 s.f.)

Question 3

Answer:

= 16\pi

m[2]

r = \dfrac{16}{2}

= 8

C = 2\pi r

= 2\pi(8)

= 16\pi

Question 4

Answer:

= 5.4

m²[2]

135 \text{ cm} = 135 \div 100 \text{ m}

= 1.35 \text{ m}

\text{Area} = 4 \times 1.35

= 5.4

m²

Question 5

Answer:

= 0.26

m²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

2600 \text{ cm}^2 = \dfrac{2600}{10\,000} \text{ m}^2

= 0.26

m²

Question 6

Answer:

= 226

m² (3 s.f.)[3]

r = \dfrac{24}{2}

= 12

A = \dfrac{1}{2} \times \pi r^2

= \dfrac{1}{2} \times \pi(12)^2

= 226.1946\ldots

= 226

m² (3 s.f.)

Question 7

(a)Answer:

= 17\,500

cm²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

1.75 \text{ m}^2 = 1.75 \times 10\,000 \text{ cm}^2

= 17\,500

cm²

(b)Answer:

= 2.5

m²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

25\,000 \text{ cm}^2 = \dfrac{25\,000}{10\,000} \text{ m}^2

= 2.5

m²

Question 8

(a)Answer:

= 5.2

m²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

52\,000 \text{ cm}^2 = \dfrac{52\,000}{10\,000} \text{ m}^2

= 5.2

m²

(b)Answer:

= 121

cm² (3 s.f.)[3]

r = \dfrac{12.4}{2}

= 6.2

A = \pi r^2

= \pi(6.2)^2

= 120.7628\ldots

= 121

cm² (3 s.f.)

Question 9

(a)Answer:

= 10.5

m² (3 s.f.)[2]

\text{Area} = 7^2 - \dfrac{1}{4} \times \pi(7)^2

= 10.5154\ldots

= 10.5

m² (3 s.f.)

(b)Answer:

= 14.1

cm² (3 s.f.)[2]

A = \dfrac{1}{2} \times \pi r^2

= \dfrac{1}{2} \times \pi(3)^2

= 14.1371\ldots

= 14.1

cm² (3 s.f.)

(c)Answer:

= 52\,000

cm²[2]

1 \text{ m}^2 = 100 \text{ cm} \times 100 \text{ cm}

= 10\,000 \text{ cm}^2

5.2 \text{ m}^2 = 5.2 \times 10\,000 \text{ cm}^2

= 52\,000

cm²

(d)Answer:

= 380

cm²[2]

\text{Area of } ABCD = \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}

= \dfrac{1}{2} \times (9 + 29) \times 20

= 380

cm²