area problem solving exam questions

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Calculating Areas

Concept quizzes.

• Area Warmup
• Area From Rectangles to Triangles
• Area From Rectangles to Parallelograms and Trapezoids
• Length and Area Warmup
• Area Applications

Challenge Quizzes

• Length and Area: Level 1 Challenges
• Length and Area: Level 2 Challenges
• Length and Area: Level 3 Challenges

Understanding and applying area concepts allows us create robots, airplanes, stadiums, pools, houses, and more.

Expect to see and learn how to solve questions like this one:

Area describes how much two-dimensional space something occupies, or how many unit squares fit into a given figure. Area helps us determine how much much paint, or carpet, or soil we need. From areas of lawns and streets to dimensions of i-phones and airplane wings, the applications of area are limitless and used in nearly every aspect of everyday life.

Examining areas of different shapes helps us understand how common mathematical formulas are created and how various shapes relate to one another. Our study of area extends the foundational knowledge of rectangular areas to develop a deep understanding of areas of triangles, parallelograms, trapezoids, and circles.

Area concepts set the essential foundation for examining other geometric properties such as volume and surface area.

• Area of a Triangle
• Area of a Rectangle
• Length and Area Problem Solving

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Area Problem Solving

When you know the simple area formulas for shapes like rectangles , triangles , and circles , you can often use these in combination to find the area of more complicated shapes. You can add or subtract parts of shapes in order to build up to more complex shapes.

Subtracting to find the area of a shape

What is the area of the shaded region of the following rectangle?

The area of the shaded region will be the difference between the entire outer rectangle and the area of the square that is cut out of the middle.

The dimensions of the rectangle are 6 by 8, making the area L × W = 6 × 8 , which is 48 square units.

Each side of the inner square measures 2 square units, making the area S 2 = 2 × 2 square units, or 4 square units.

So, the shaded part of the rectangle has an area of 48 - 4 = 44 square units.

Adding on to find the area of a shape

Find the total area of the shaded region. In the figure, all angles are right angles.

The given figure can be divided into 3 squares and one rectangle, as shown.

The total area will be the sum of the area of the rectangle and the squares.

The dimensions of the rectangle are 3 by 12, so the area is 3 × 12 . That is 36 square units.

All three squares have sides that are 3 units long, so the area of each of them is 3 × 3 , or 9 square units.

Therefore, the total area of the shape is 36 + ( 3 × 9 ) = 63 square units.

There are other ways to split up the figure into known shapes, so if your first instinct was to slice it vertically into 3 rectangles and one square, that would work just fine too!

If ∆ A B C is a right triangle and BC is a semicircle, find the total area of the figure.

The total area is the sum of the areas of the triangle and the semicircle.

We know the length and altitude of the triangle, so we know the area of the triangle is

1 2 × 4 × 3 square units.

Since arc BC is a semicircle, BC is a diameter. But BC is also the hypotenuse of the right triangle ∆ A B C .

Use the Pythagorean theorem to find the length BC.

BC = 3 2 + 4 2

So the diameter of the semicircle is 5 units, making the radius 2.5 units.

The area of the semicircle with a radius of 2.5 units is 1 2 × π × 2.5 2 is approximately 9.82 square units.

Therefore, the total area of the shape is about 6 + 9.82 , or 15.82 square units.

Topics related to the Area Problem Solving

Perimeter, Area and Volume

Surface Area

Flashcards covering the Area Problem Solving

5th Grade Math Flashcards

Common Core: 5th Grade Math Flashcards

Practice tests covering the Area Problem Solving

MAP 5th Grade Math Practice Tests

Common Core: 5th Grade Math Diagnostic Tests

Get help learning about area problem solving

Considering the different formulas your student needs to know and the creative applications they need to use to find the areas of oddly shaped objects, it can be kind of tricky to master all the skills alone. If your student needs help figuring out how to find the area of objects other than plain rectangles, circles, and triangles, the help of an expert tutor can be useful. To learn more about how tutoring can help your student understand area problem solving, contact the Educational Directors at Varsity Tutors today.

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• Maths Questions

Area Questions

Students can easily access a variety of area questions with complete explanations, which are provided below. One of the most fundamental concepts taught in elementary and secondary schools is the concept of areas of different shapes. NCERT curriculum is used to frame the questions. Students can use these questions to get a fast overview of the topics and practise them so that they will become more knowledgeable about the concept. To cross-verify your answers, look over the complete explanations for each question. To learn more about areas of different shapes, click here .

Go through the different types of area questions and practise them to learn the concept well.

Area Questions with Solutions

1. Find the area of a circle if its diameter is 42 cm. (Use π = 22/7 )

Given: Diameter, d = 42 cm.

Hence, Radius, r = d/2 = 42/2 = 21 cm

We know that the area of a circle = πr 2 square units.

A = (22/7) × 21 × 21

A = 22 × 3 × 21

A = 1386 cm 2 .

Therefore, the area of the circle = 1386 cm 2 .

2. Calculate the radius of a circle if its area is 25π m 2 .

Given: Area = 25π m 2

We know that area of a circle = πr 2

Hence, we can write,

Hence, r = 5 m

Therefore, the radius of the circle is 5 m.

3. Find the area of a square whose side length is 7 cm.

Given, Side, a = 7 cm.

As we know, the area of square = a 2 square units

A = 7 2 cm 2

A = 49 cm 2

Hence, the area of the square is 49 cm 2 .

4. Compute the side length of a square, if its area is 121 cm 2 .

Given: Area of a square = 121 cm 2 .

We know that, A = side 2 square units

121 = side 2

Side = 11 cm

Therefore, the side length of the square is 11 cm, if its area is 121 cm 2 .

5. Determine the area of a rectangle, if its length is 11 cm and breadth is 9 cm.

Given: Length = 11 cm

Breadth = 9 cm.

We know that the formula to find the area of a rectangle is:

Area = Length × Breadth square units

Area = 11 × 9 cm 2

Area = 99 cm 2

Therefore, the area of the rectangle is 99 cm 2 .

6. Compute the breadth of a rectangle if its area is 84m 2 and length is 12 m.

Area of a rectangle = 84 m 2

Length = 12 m

As we know,

Rectangle’s area = Length × Breadth

84 = 12 × Breadth

Therefore, Breadth = 84/12

Breadth = 7 m.

Hence, the breadth of the rectangle is 7 m if its area is 84 m 2 and length is 12 m.

7. Find the area of a parallelogram, if its base length is 9 cm and height is 5 cm.

Given: Base length = 9 cm

Height = 5 cm.

The formula to calculate the area of a parallelogram is:

Area = Base × Height square units.

On substituting the given values, we get

Area = 9 × 5 cm 2

Area = 45 cm 2

Therefore, the area of the parallelogram is 45 cm 2 .

8. Compute the area of a triangle, if its base measurement is 6 cm and height is 10 cm.

Given: Base, b = 6 cm

Height, h = 10 cm.

We know that the area of a triangle = (½) × b × h square units.

Now, substitute the given values, we get

A = ½ × 6 × 10 cm 2

A = 3 × 10 cm 2

A = 30 cm 2

Therefore, the area of the triangle is 30 cm 2 .

9. Determine the height of a triangle, if its base length is 8 cm and its area is 52 cm 2 .

Given: Base length, b = 8 cm

Area = 52 cm 2

As we know, the area of a triangle is ½ bh square units

52 = (½) × 8 × h

52 × 2 = 8 × h

104 = 8 × h

Hence, the height of the triangle is 13 cm if its base is 8 cm and its area is 52 cm 2 .

10. Determine the area of a rhombus if its diagonals are 7 cm and 10 cm.

Diagonal 1 = 7 cm

Diagonal 2 = 10 cm

The area of a rhombus = ½ × diagonal 1 × diagonal 2

Now, substitute the values, we get;

Area = ½ × 7 × 10 cm 2

Area = 7 × 5 cm 2

Area = 35 cm 2

Therefore, the area of the rhombus is 35 cm 2 , if its diagonals are 7 cm and 10 cm.

Explore More Articles:

• Area of Triangle
• Area of Circle
• Area of Rhombus
• Rectangle Questions
• Area of Parallelogram Questions
• Quadrilaterals Questions
• Geometry Questions

Practice Questions

• Compute the area of a triangle, if its base is 14 cm and height is 10 cm.
• Determine the area of a rectangle if its length is 17 cm and breadth is 15 cm.
• Find the area of a square whose side measures 19 cm.

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Functional Skills: Area Revision

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Revision Products

Functional skills maths level 2 pocket revision guide, functional skills maths level 2 mini tests, functional skills maths level 2 revision cards, functional skills maths level 2 practice papers, functional skills maths level 2 practice papers & revision cards, filter by level, filter by exam board, functional skills: area.

The area of 2D shape is the amount of surface it occupies.

Area is calculated by multiplying lengths together, so the metric units for area are squared – cm ^2 , m ^2 , mm ^2 etc. For imperial units for area, we usually say square inches (sq. in) instead of in ^2 , for example.

You will need to know 4 skills to calculate the area of a square , a rectangle , a triangle  and compound shapes .

Would you pass functional skills maths level 2?

Why not test your knowledge now and see if you would pass. A score of around 60% means you would probably pass your functional skills maths level 2. 

Skill 1: Area of a Square

The formula for the area of a square is:

\text{Area} =a \times a = a^2

where a is the length of the sides of the square.

Written in words, this is:

\text{Area} = \text{length} \times \text{length} = \text{length}^2

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Skill 2: Area of a Rectangle

The formula for the area of a rectangle is:

\text{Area} = a\times b

where a is the width of the rectangle and b is the length .

\text{Area} = \text{width} \times \text{length}

Skill 3: Area of a Triangle

The formula for the area of a triangle is:

\text{Area} = \dfrac{1}{2}\times b \times h

where b is the base width of the triangle and h is the vertical height as shown in the diagram on the right.

\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}

Skill 4: Area of Compound Shapes

Sometimes, you may need to split up a compound shape into easier shapes to find the area.

Example: Find the area of the shape to the right.

For this shape you could either:

• Split the shape up into two rectangles, calculate the area of each rectangle and add them up.
• Notice that the shape is a large rectangle, with a smaller rectangle cut out of it. So, we could calculate the area of the smaller rectangle and minus it from the larger rectangle. This would be easier in this situation:

\text{Area of larger rectangle} = 7 \times 12.5 = \textcolor{red}{87.5 \text{ m}^2}

\text{Area of smaller rectangle} = 5 \times 10 = \textcolor{blue}{50 \text{ m}^2}

\text{Area of shape} = \textcolor{red}{87.5} - \textcolor{blue}{50} = \textcolor{limegreen}{37.5 \text{ m}^2}

Example 1: Area of a Square

Calculate the area of a square with sides of length 6 \text{ cm} .

\text{Area of square} = 6 \times 6 = 6^2 = 36 \text{ cm}^2

Example 2: Area of a Rectangle

Calculate the area of the rectangle shown to the right.

\text{Area of rectangle} = 12 \times 5 = 60 \text{ cm}^2

Pass your level 2 maths exam on the first attempt!

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Example 3: Area of a Triangle

Calculate the area of the triangle shown to the right.

\text{Area of triangle} = \dfrac{1}{2} \times 9 \times 6 = 27 \text{ cm}^2

Functional Skills: Area Example Questions

Question 1:  Calculate the area of the triangle ABC shown below, which has a base of 9.8 cm and a height of 11.2 cm.

Using the formula for the area of a triangle:

Area = \dfrac{1}{2}\times base \times height

Area = \dfrac{1}{2}\times 9.8 \times 11.2=54.88 cm ^2

Question 2: Calculate the area of the rectangle shown below.

Area of a rectangle = width \times length

Area = 5.1\times7.6=38.76 cm ^2

Question 3: Below is a right-angled triangle on top of a square. Using the measurements given, calculate the area of the whole shape.

Area of the Square = 6.3\times 6.3=39.69 cm ^2

To work out the area of the triangle we need to know the height. This can be found by subtracting the side length of the square from the total height of the shape: 10.5-6.3=4.2 cm

Area of the triangle = \dfrac{1}{2}\times6.3\times4.2=13.23 cm ^2

Total area = 39.69 + 13.23 = 52.92 cm ^2

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Exam tips cheat sheet, formula booklet, functional skills: area worksheet and example questions.

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This 5 set of Functional Skills Maths Level 2 practice papers are a great way to revise for your Functional Skills Maths Level 2 exam. These practice papers have been specially tailored to match the format, structure, and question types used by each of the main exam boards for functional skills Maths. Each of the 5 papers also comes with a comprehensive mark scheme, so you can see how well you did, and identify areas to improve on.

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Unit 10: Area

About this unit.

Everything around us has a measurable area from the floor we walk on to the walls of our rooms. In this unit, we'll be exploring area! We'll learn some handy ways to figure out just how much space a shape covers--from counting squares, to multiplying, to breaking shapes down into smaller pieces. Let's get measuring!

Area introduction

• No videos or articles available in this lesson
• Understand area Get 3 of 4 questions to level up!

Count unit squares to find area

• Intro to area and unit squares (Opens a modal)
• Measuring rectangles with different unit squares (Opens a modal)
• Creating rectangles with a given area 1 (Opens a modal)
• Creating rectangles with a given area 2 (Opens a modal)
• Find area by counting unit squares Get 5 of 7 questions to level up!
• Compare area with unit squares Get 5 of 7 questions to level up!
• Create rectangles with a given area Get 3 of 4 questions to level up!

Area formula intuition

• Counting unit squares to find area formula (Opens a modal)
• Transitioning from unit squares to area formula (Opens a modal)
• Area with partial grids (Opens a modal)
• Area of rectangles with partial arrays Get 5 of 7 questions to level up!
• Transition from unit squares to area formula Get 5 of 7 questions to level up!

Multiply to find area

• Finding missing side when given area (Opens a modal)
• Comparing areas of plots of land (Opens a modal)
• Area of rectangles review (Opens a modal)
• Area of rectangles Get 5 of 7 questions to level up!
• Find a missing side length when given area Get 5 of 7 questions to level up!
• Compare areas by multiplying Get 3 of 4 questions to level up!

Area and the distributive property

• Area and the distributive property (Opens a modal)
• Area and the distributive property Get 3 of 4 questions to level up!

Decompose figures to find area

• Decomposing shapes to find area: grids (Opens a modal)
• Decomposing shapes to find area: add (Opens a modal)
• Decomposing shapes to find area: subtract (Opens a modal)
• Area: FAQ (Opens a modal)
• Understand decomposing figures to find area Get 3 of 4 questions to level up!
• Decompose figures to find area Get 3 of 4 questions to level up!

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Area and perimeter problem solving

Subject: Mathematics

Age range: 5-7

Resource type: Lesson (complete)

Last updated

25 January 2018

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Analysis of multi representation ability in solving physics problems for class XI students of SMAN 6 Wajo

• Khatimah, U.
• Amin, B. D.
• Palloan, P.

This research is a descriptive study that aims to understand the description of Multi representations Ability in Solving Physics Problems for Class XI Students of SMAN 6 Wajo. The samples used in this study were students of class XI MIPA SMAN 6 Wajo for the 2021/2022 academic year which amounted to 79 students obtained randomly. The research instrument used was in the form of a multi representation ability test on simple harmonic motion material made in the form of essay questions as many as 10 items. The results obtained indicate that the multi-representation ability of students is in the medium category with a percentage of 81,01%. The percentage of scores and categories for each indicator is obtained for the indicator "change graphic representation to verbal" of 72,15%, on the indicator "change verbal representation to mathematical" the percentage is 53,16%, and for the indicator "changing the representation of the image to mathematical" is at a percentage of 86.08%, each indicator is in a medium category. Based on the results of the study it was concluded that some students have the ability In answering the questions, representing the representation in question is in the medium category.

A novel honey badger algorithm with golden sinusoidal survival rate selection for solving optimal power flow problem

• Original Paper
• Published: 28 April 2024

Cite this article

• Fengxian Wang 1 ,
• Senlin Bi 1 ,
• Shaozhi Feng 1 &
• Huanlong Zhang 1  

The original honey badger algorithm (HBA), as one of the newest meta-heuristic techniques, has a better convergence speed. However, HBA has the potential disadvantages of poor convergence accuracy, insufficient balance between exploration and exploitation, and the tendency to slip into local optimization. In this paper, a novel golden sinusoidal survival honey badger algorithm is proposed. Firstly, tent chaotic opposition learning is applied to the initial individual generation so that they can be distributed throughout the entire search area, which improves the precision of initial populations. Secondly, in the position update phase, we use a nonlinear convergence strategy to balance the weight of prey in the next walk and to increase the global search ability. Then, the quality of honey badger is evaluated by the golden sinusoidal survival rate strategy and precocious individuals are updated by Levy flight, which can avoid the premature convergence of the algorithm. Finally, 23 benchmark functions, CEC2019 tests functions and optimal power flow problems are used to evaluate the effectiveness of the improved algorithm. Test results indicate that the algorithm’s ability to evolve, to extract the local optimal and to detect the global optimal placements are improved.

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This work is partly supported by NSFC under Grant No. 62006213, Henan Youth Talent Promotion Project No. 2022HYTP005 and Young Backbone Teacher Training Object Funding Plan of Zhengzhou University of Light Industry.

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School of Electrical and Information Engineering, Zhengzhou University of Light Industry, Zhengzhou, 450000, People’s Republic of China

Fengxian Wang, Senlin Bi, Shaozhi Feng & Huanlong Zhang

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FW contributed to conceptualization, methodology, analysis, resources, writing—review and editing, investigation and supervision. SB contributed to data curation investigation, software, validation and writing—original draft and editing. SF contributed to conceptualization, methodology, validation analysis, review and editing and visualization. HZ contributed to conceptualization, methodology, analysis, resources, investigation and supervision.

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Correspondence to Huanlong Zhang .

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Wang, F., Bi, S., Feng, S. et al. A novel honey badger algorithm with golden sinusoidal survival rate selection for solving optimal power flow problem. Electr Eng (2024). https://doi.org/10.1007/s00202-024-02402-y

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Received : 07 July 2023

Accepted : 01 April 2024

Published : 28 April 2024

DOI : https://doi.org/10.1007/s00202-024-02402-y

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