Stuck on a Question?In this section, we compile some of the questions that have been asked by students or parents, and explain the best way to solve them.

PSLE Math Question: Circles
Since PSLE is just around the corner, recently many questions have been asked related to the PSLE past papers. This week, the topic we will be focusing on is Circles. Circle related questions test students' spatial skills. They are challenging because they require students to visualise space and area, and apply creative problem solving skills in order to be able to solve the problem at hand. Here are 3 circle questions that have been asked by students/parents.
Question 1: the shaded petals/leaves/eyes
For the problem above, the key to answering the question is realising that: the inner shaded petals (or "eye", as per the diagram on the left) can be divided into two and spread out to fill the white segments.
In the end, the area of the shaded part is just:
Area of big circle  Area of big square
= (pi x 15 x 15)  (15 x 15 x 2) = 256.86 cm2 (2 d.p.)
Once you can see the trick, the question becomes very easy!
Area of big circle  Area of big square
= (pi x 15 x 15)  (15 x 15 x 2) = 256.86 cm2 (2 d.p.)
Once you can see the trick, the question becomes very easy!
Question 2: the overlapping quadrants
Questions on overlapping quadrants are quite common when it comes to the topic Circles.
For the above question, we can label two of the white parts as A and B (see picture on the left).
The trick to answering the question successfully, is to realise that you do not need to know X and Y's exact area to solve the question. Instead, we are going to make use of part A to help us reach the same objective and solve the problem efficiently.
A+Y = whole square  quadrant
= (14 x 14)  (1/4 x 22/7 x 14 x 14)
= 196  154 = 42 cm2
A+X = quadrant  semicircle B
= (1/4 x 22/7 x 14 x 14)  (1/2 x 22/7 x 7 x 7)
= 154  77 = 77 cm2
Note: the difference between (A+X) and (A+Y) is the same as the difference between X and Y (ie. the A part cancels out).
Hence:
XY = 77  42 = 35 cm2
For the above question, we can label two of the white parts as A and B (see picture on the left).
The trick to answering the question successfully, is to realise that you do not need to know X and Y's exact area to solve the question. Instead, we are going to make use of part A to help us reach the same objective and solve the problem efficiently.
A+Y = whole square  quadrant
= (14 x 14)  (1/4 x 22/7 x 14 x 14)
= 196  154 = 42 cm2
A+X = quadrant  semicircle B
= (1/4 x 22/7 x 14 x 14)  (1/2 x 22/7 x 7 x 7)
= 154  77 = 77 cm2
Note: the difference between (A+X) and (A+Y) is the same as the difference between X and Y (ie. the A part cancels out).
Hence:
XY = 77  42 = 35 cm2
Question 3: the concentric quadrants
Concentric quadrants are when a smaller quadrant is placed inside a bigger quadrant and they share the same centre point. For any circle question, it is important to be very observant about each detail you see in the picture.
In the above picture, we can see that the small shaded quadrant that is jutting out below is the key to finding out the difference in radius. Since the radius written is 4 cm, the difference in radius must be half of that, which is 2 cm.
Radius of big quadrant = 4 + 2 = 6 cm
Perimeter = (0.5 x 2 x 3.14 x 6) + (3/4 x 2 x 3.14 x 4) + 2 + 2 + 4 = 45.68 cm
In the above picture, we can see that the small shaded quadrant that is jutting out below is the key to finding out the difference in radius. Since the radius written is 4 cm, the difference in radius must be half of that, which is 2 cm.
Radius of big quadrant = 4 + 2 = 6 cm
Perimeter = (0.5 x 2 x 3.14 x 6) + (3/4 x 2 x 3.14 x 4) + 2 + 2 + 4 = 45.68 cm
Question 4: the Pincers
As if rugby balls, leaves, petals and quadrants are not confusing enough already, here comes the "killer" of them all  the Pincers. At first glance, you might think it looks "somewhat solvable". Then you try to solve, and then you realise, this may not be as easy as you initially thought.
But it actually is easy! ... If you know the trick that is!
But it actually is easy! ... If you know the trick that is!
For each "pincer", the smaller part of the pincer can be rearranged and rotated around such that it fits nicely under the larger part of the pincer. If you do that, you will get a bottom outline as per the blue line in the picture above.
So the area of 1 pincer is:
Area of large 3u quadrant  area of 4 squares  area of small 1u quadrant
= ( 1/4 x pi x 27 x 27 )  ( 4 x 81 )  ( 1/4 x pi x 9 x 9 )
Since there are 4 pincers, that would equate to:
= ( pi x 27 x 27 )  ( 16 x 81 )  ( pi x 9 x 9 )
= 738.72 cm2
Pinchingly easy! If you need extra help, contact us and our SG Math Tutor will guide you with proper training on how to handle these tricky questions.
So the area of 1 pincer is:
Area of large 3u quadrant  area of 4 squares  area of small 1u quadrant
= ( 1/4 x pi x 27 x 27 )  ( 4 x 81 )  ( 1/4 x pi x 9 x 9 )
Since there are 4 pincers, that would equate to:
= ( pi x 27 x 27 )  ( 16 x 81 )  ( pi x 9 x 9 )
= 738.72 cm2
Pinchingly easy! If you need extra help, contact us and our SG Math Tutor will guide you with proper training on how to handle these tricky questions.