Area and Perimeter Word Problems

Area and Perimeter Word Problems

8th Grade

15 Qs

quiz-placeholder

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Area and Perimeter Word Problems

Area and Perimeter Word Problems

Assessment

Quiz

Mathematics

8th Grade

Hard

CCSS
4.MD.A.3, 6.G.A.1, 3.MD.D.8

+3

Standards-aligned

Created by

Anthony Clark

FREE Resource

15 questions

Show all answers

1.

FILL IN THE BLANK QUESTION

1 min • 1 pt

Media Image

A triangular wall needs to be painted. The triangle is an equilateral triangle with 35 metre sides and the height of the wall is 30 metres. Nina wants to make a gold border around the wall. What is the length of the gold border that is needed?

Answer explanation

An equilateral triangle means that all 3 sides are equal. This means you need to add up all 3 sides to get the length of the gold border.

Tags

CCSS.3.MD.D.8

2.

OPEN ENDED QUESTION

1 min • 2 pts

Media Image

The garden has two sections: a triangle section for vegetables and a circular section for flowers, with the rest grass. The vegetables section is a right-angle with sides of 3.45m and 2.1m. The flower section is 5.7m from one side to the other. What is the total area of each section of the garden? If grass seed costs $1.77/m², how much will it cost to buy enough seed for the grass area?

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Answer explanation

Vegetable Section (Triangle)

Area of triangle = (base × height) ÷ 2

Area = (3.45 × 2.1) ÷ 2 = 3.6225 m²

Flower Section (Circle)

Area of circle = π × r²

Where r is the radius (half the diameter) - Radius = 5.7 ÷ 2 = 2.85m

Area = π × 2.85² ≈ 25.5173 m²

Grass Section

subtract vegetable and flower areas from the total garden area.

TOTAL GARDEN AREA: 52.46m² (8.6 x 6.1)

VEGETABLE & FLOWER GARDENS:

3.6225 m² + 25.5173 m² = 29.1398m²

GRASS AREA = 52.46 - 29.1398 = 23.32m²

Grass Seed Cost = Grass area × Price per m²

         = 23.32m² × $1.77/m²  = 41.2764

         = $41.30

3.

OPEN ENDED QUESTION

1 min • 1 pt

Media Image

For a school event, circular pizzas with a radius of 12cm are ordered. If each student should get 50cm² of pizza, how many students can one pizza feed?

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Answer explanation

Area of pizza = πr² = π × 12² =  452.39 cm²

Number of students = 452.39  ÷ 50 ≈ 9.05

= 9 students

Tags

CCSS.HSG.C.B.5

4.

OPEN ENDED QUESTION

1 min • 2 pts

Media Image

In Hobbiton, hobbit houses have circular windows. A hobbit needs to replace a window which has a circumference of 4.8 metres. What is the area of the window?

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Answer explanation

We need to find the radius first using the circumference formula: 

C = 2πr

4.8 = 2πr

4.8 = 2 × 3.14 × r

 r = 4.8 ÷ (2 × 3.14) 

 r = 0.764 metres

Calculate the area

A = πr²

A  = 3.14 × (0.764)²

       = 3.14 × 0.584 

Tags

CCSS.7.G.B.4

5.

OPEN ENDED QUESTION

1 min • 2 pts

Media Image

The school is painting a mural on a wall shaped like a trapezium. The parallel sides of the wall are 8 metres and 12 metres long, and the height between them is 5 metres. How much paint will they need if 1L covers 10 square metres?

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Answer explanation

Area of trapezium

= ½(a+b)h

= ½(8+12) × 5

= 50m²

Paint needed

= 50 ÷ 10

Tags

CCSS.6.G.A.1

6.

OPEN ENDED QUESTION

1 min • 3 pts

Media Image

A church is commissioning a circular stained glass window with a smaller circular design in its centre. The outer circle has a radius of 2.4 metres, while the inner circle has an area that is 1/9 of the total window area. What is the radius of the inner circle?

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Answer explanation

Area of outer circle

= πr² = π(2.4)² = 18.1 m² (rounded to one dp)

Area of inner circle

=The inner circle's area is 1/9 of the total area

   =  18.1 ÷ 9 = 2.01 m²

use the area formula for a circle to find the radius of the inner circle:

   2.01 = πr²

   r² = 2.01 ÷ π

   r = √(2.01 ÷ π) ≈ 0.8 m

7.

OPEN ENDED QUESTION

1 min • 4 pts

Media Image

A farmer notices a mysterious crop circle in his field. It's composed of two circles: a larger outer circle and a smaller inner circle. The area between the two circles is 200 square metres. If the radius of the larger circle is 10 metres, what is the radius of the smaller circle?

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Answer explanation

 Area between circles

= π(R² - r²) = 200

=  π(10² - r²) = 200

= 100π - πr² = 200

=  πr² = 100π - 200

= r² = 100 - 200/π ≈ 36.34

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