# 5 Tricky MCQs on Area and Perimeter

Many students struggle with area and perimeter questions because they are unable to spot the visual clues in the question. However, spatial visualization skills can help your child ace area and perimeter MCQs easily.

Before you read on, you might want to download this entire revision notes in PDF format to print it out for your child, or to read it later.

This will be delivered to your email inbox. In this tutorial, we’re solving area and perimeter examples from 2021 prelim papers, with each tackling different shapes to demonstrate the spatial visualization skills.

Example 1

This question was taken from an Ai Tong School Prelim paper.

Question: The figure below is made up of a semicircle and 2 squares. The length of XY is 10 cm and the length of the larger square is 6 cm. Find the perimeter of the figure. Give your answer in terms of π. Usually, π is given as $\frac{22}{7}$ or 3.14. Sometimes the question requires you to give your answer in terms of π. That means we use π to represent $\frac{22}{7}$ or 3.14.

Step 1: Analyse the shape

The perimeter is made up of the circumference of a semicircle, 2 sides of a small square, 2 sides of a larger square, and a small awkward length between the 2 squares.

The combined length method means that you don’t have to work out every individual length. So, you do not need to know that awkward length between the 2 squares.

Eventually, the student would have to add up all the lengths to find a perimeter. So, it’s simpler to find the combined length right away.

We know that the vertical length from X to Y is 10 cm. So, the 6 cm of the larger square and the length of the small square also make up 10 cm when we transfer over.

Here, we’ve highlighted some of the combined lengths we can use for this question: Step 2: Find the circumference

The circumference of a semicircle is calculated as $\frac{1}{2}$ x π x diameter. The diameter here is 10 cm. So, the circumference is $\frac{1}{2}$ x π x 10 = 5π cm. 5π + 6 + 10 = (5π + 22) cm. So, option 2 is our answer to this question.

Example 2

This question was taken from a Henry Park Primary School Prelim paper.

Question: Maliki cut a square piece of paper measuring 12 cm in length into 2 pieces of squares and 2 pieces of rectangles as shown in Figure 1. He arranged the pieces to form a big rectangle as shown in figure 2. What is the perimeter of the big rectangle in figure 2? This is a rearrangement question, meaning one shape has been rearranged to form another.

In area and perimeter questions like this, many students find the perimeter of figure 1 and take that as the answer. But that would be incorrect. When 1 sheet is rearranged to form another, the perimeter will not necessarily be the same. The only information we are given in this question is that the length of the original square is 12 cm.

Step 1: Analyse the shapes

We need to study the relation among all the different lengths in the shape. So, let’s focus on figure 2. Here, 2 breadths of the small rectangle make up one side of the square. If each breadth is 1u, then one side of the square is 2u.

Now, let’s look at figure 1. 2 sides of the square make up the length of the rectangle. That means 2u + 2u make up the length of the rectangle, which is 4u.

Step 2: Find the perimeter

Now we know that the top center length of figure 2 is 4u. If we focus on the top side of the square in figure 1, we can see clearly that 1u + 2u + 1u makes up 12 cm. This means 4u = 12 cm. So, 1u = 3 cm.

To find the perimeter of figure 2, we have to work out how many units are there in total. We marked all the units, as shown below: We have 6 sets of 2u, which makes up 12u. We then have 2 sets of 4u, making up 8u. So, we have a total of 20u. Since 1u is 3 cm, 20u = 20 × 3 = 60 cm. Therefore, option 2 is our answer to this question.

Example 3

This question was taken from CHIJ St. Nicholas Girls’ School

Question: The figure shows a rectangle ADEG. The area of triangle BCF is 21 cm2. What is the total area of the unshaded parts in ADEG? Now, let’s recap the formula for area of a triangle. The challenging part of this question is that we only know length AD (21 cm), not the individual lengths of the areas between points A and D.

Using the combined length method, we can effectively work out the area of multiple triangles. All the triangles in the given figure share the same perpendicular height (A to G).

Although we don’t know the individual lengths, we know that the combined base is 21 cm. So, we can find the combined area of all the unshaded triangles and BCF in just one step. As we know that BCF is 21 cm2, we can subtract that from the total area and find our answer.

So, $\frac{1}{2}$ × 21 × 14 = 147 cm2. Finally, 147- 21 = 126 cm2. Therefore, option 4 is our answer to this question.

Example 4

This question was taken from the Rosyth School 2020 Prelim paper.

Question: The figure below is made up of 4 identical circles inside a square. The length of the square is 28 cm. Find the perimeter of the shaded part. (Take π = $\frac{22}{7}$) Here, the perimeter we’re supposed to calculate is indicated in red.

Step 1: Analyse the shapes

Most students calculate the individual lengths and then add them up. While this is not wrong, it’s very time consuming.

The combined length method saves a lot of time when solving area and perimeter questions. It also prevents careless mistakes because there are fewer steps needed.

Since we know that 2 semi circles (SC) make up 1 circle (C), 4 semi circles would make up 2 circles. So, we should work out the circumference of 2 full circles.

Step 2: Find the circumference From the horizontal lines, we can see that the length from the beginning to the centre is the radius of 1 circle. This means we have 4 radii.

So, we can work out 4 radii (4r), which is 2 diameters. Since the length of the square is 28 cm, and 28 cm is the length of 2 diameters, we already have the value of 4r. All you need to do now is to find the circumference of 2 full circles. Since 2 diameters make up 28 cm, 1 diameter will make up 14 cm.

So, the circumference of 2 circles will be 2 × $\frac{22}{7}$ × 14 = 88 cm. Now, you can add up the 2 numbers: Therefore, option 3 is our answer to this question.

Example 5

This question was taken from the Rosyth School 2021 Prelim paper.

Question: Aishah cut out 4 identical right-angled triangles. Each right-angled triangle has a perimeter of 36 cm. She formed the shape shown below. What is the perimeter of the figure formed by the 4 right-angled triangles? The only information we have is the perimeter of 1 right-angled triangle (36 cm) and the base length of the triangle (9 cm). The perimeter of the bigger shape is highlighted to show that it doesn’t involve the inner lengths.

We already know that the perimeter of 1 triangle is 36 cm. And inside the figure, we have 4 bases (highlighted in yellow). To work out the length of just the pink portion in the top right triangle, we simply need to subtract 9 cm and 9 cm from 36 cm. So, 36 – 9 – 9 = 18 cm. The figure is made up of 4 sets of 18 cm. So, 18 × 4 = 72 cm. Therefore, option 1 is our answer to this question.

I hope this tutorial has shown you how effective the combined length method is when solving area and perimeter questions. If you have any questions or suggestions, please leave them in the comments.

You can also watch the full area and perimeter MCQs video tutorial here:

Before you go, you might want to download this entire revision notes in PDF format to print it out for your child, or to read it later.

This will be delivered to your email inbox.  Ms Elaine Wee
Math and Science Specialist
Jimmy Maths and Grade Solution Learning Centre

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