- Numbers
- What does the number stand for?
- Nearest number
- Which digit is in the place?
- Round off to the nearest number
- Number divide
- Find the fraction nearest to 1
- Sort fraction numbers
- Find the largest or smallest fraction
- Fraction to decimal
- Convert decimal to a mixed fraction.
- Fraction - Addition
- Fraction - Substraction
- Fraction - Multiply
- Fraction - Divide
- List all the common factors of two integers
- Chicken and Rabbit problem
- Assumption Method
- Measurement
- Data Analysis
- Geometry
- Speed
- Others
- P5 Maths
- P4 Maths
- Tap to switch levels
- Numbers
- Measurement
- Data Analysis
- Geometry
- Speed
- Others
- P5 Maths
- P4 Maths
SELECT A SUBJECT
- Numbers
Question 1 of 84
The figure shown is made up of three identical small circles and a Larger circle with radius 10cm.Find the ratio of the unshaded area to the shaded area. $ (Take\pi = 3.14)$.
Sorry. Please check the correct answer below.
You are Right
The diagram below shows a rectangle STUV. Area of triangle STW is 30cm$^2$ and the area of triangle SWV is 45cm$^2$. Give that SW = VW and the ratio of the lenght of VX to the length of XU is 2 : 3, Find the area of TUXW.
Sorry. Please check the correct answer below.
You are Right
$30 + 30 = 45 + WTU$
$WTU = 15$
$5u --- 30$
$3u ---- 18$
$18 + 15 = 33$
The figure below is made up of a right –angled triangle and a rhombus overlapping each other. $\frac{2}{9}$ of the triangle and $\frac{1}{2}$ of the rhombus are unshaded. The base and height of the triangle are 18cm and 28cm respectively. What is the area of the rhombus?
Sorry. Please check the correct answer below.
392cm$^2$ $\rightarrow$ 9u $\rightarrow$ $\frac{1}{2} \times 18cm \times 28cm = 252cm^2$
1U $\rightarrow$ 252cm ÷ 9 = 28$cm^2$
14u $\rightarrow$ $28cm^2 \times 14 = 392cm^2$
You are Right
392cm$^2$ $\rightarrow$ 9u $\rightarrow$ $\frac{1}{2} \times 18cm \times 28cm = 252cm^2$
1U $\rightarrow$ 252cm ÷ 9 = 28$cm^2$
14u $\rightarrow$ $28cm^2 \times 14 = 392cm^2$
The figure below is not drawn to scale. BCE is an equilateral triangle. ABC and AFD are straight line. If $\angle$BAF = 22$^\circ$, what is the difference between the marked angles, $\angle$EDF and $\angle$BCD?
Sorry. Please check the correct answer below.
You are Right
The figure below is made up of 3 identical isosceles right-angled triangle. What is the area of the figure?
Sorry. Please check the correct answer below.
You are Right
The figure below is made up 2 identical rectangles, with one overlapping the other, Given that the length of the rectangle is 15cm and the breadth is 3cm, find the area of the figure shown below.
Sorry. Please check the correct answer below.
You are Right
The figure below shown rectangle ABCD and 3 shaded triangles. The total area of the shaded parts is 300cm$^2$. Given that AB // EF// DC and BC is 10cm, FindAB.
Sorry. Please check the correct answer below.
You are Right
WXYZ is a square with an area of 64cm$^2$. Give that M is the midpoint of XY and N is the midpoint of ZY, Find the area of the shaded triangle WMN.
Sorry. Please check the correct answer below.
You are Right
The figure below is not drawn to scale. ADEF is a square and BCDG is a rectangle. Given that BC = 16cm, FE = 28cm CE = 41cm, Find the area of the shaded parts AEBG.
Sorry. Please check the correct answer below.
$\frac{1}{2} \times 16 \times 13 = 104$
$\frac{1}{2} \times 12 \times 28 = 168$
$168 + 104 = 272cm^2$
You are Right
$\frac{1}{2} \times 16 \times 13 = 104$
$\frac{1}{2} \times 12 \times 28 = 168$
$168 + 104 = 272cm^2$
In the figure, ABCD is a rectangle of area 96cm $^2$, with AE =EB .What is the total area of the shaded parts?
Sorry. Please check the correct answer below.
$96 ÷ 4 = 24cm^2$
You are Right
$96 ÷ 4 = 24cm^2$
The figure below shows a quadrant and a semicircle enclosed with in a square. O is the center of the square. The square has a length of 12cm. Find the area of the shaded part. $(Take \pi = 3.14)$
Sorry. Please check the correct answer below.
43.74cm$^2$ $\rightarrow$ $12 \times 12 = 144$
$144 ÷ 2 = 72$
$3.14 \times (12 ÷ 2) \times (12 ÷ 2) = 113.04$
$113.04 ÷ 4 = 28.26$
$72 – 28.26 = 43.74$
You are Right
43.74cm$^2$ $\rightarrow$ $12 \times 12 = 144$
$144 ÷ 2 = 72$
$3.14 \times (12 ÷ 2) \times (12 ÷ 2) = 113.04$
$113.04 ÷ 4 = 28.26$
$72 – 28.26 = 43.74$
The figure shows a circle inside a square. Find the area of the unshaded part. $ (Take\pi= \frac{22}{7})$.
Sorry. Please check the correct answer below.
You are Right
The figure ABCD below is made up 3 identical rectangles. What is the ratio of the area of the shaded parts to the area of the Unshaded part?
Sorry. Please check the correct answer below.
You are Right
Find the area of the shaded parts.
Sorry. Please check the correct answer below.
18cm$^2$ $\rightarrow$ $\frac{1}{2} \times 3 \times 6 = 9$
$\frac{1}{2} \times 2 \times 6 = 6$
$\frac{1}{2} \times 16 = 3$
$9 + 6 + 3 = 18$
You are Right
18cm$^2$ $\rightarrow$ $\frac{1}{2} \times 3 \times 6 = 9$
$\frac{1}{2} \times 2 \times 6 = 6$
$\frac{1}{2} \times 16 = 3$
$9 + 6 + 3 = 18$
The figure shows a cuboid with a volume of 640m$^3$. The shaded face, labelled B, is a square of area 64m$^2$. What is the area of A?
Sorry. Please check the correct answer below.
You are Right
The perimeter of the figure is 40cm. Find the area of the figure.
Sorry. Please check the correct answer below.
You are Right
In the figure below, rectangle ABCD is made up of 7 identical small rectangles.
(a) Find the perimeter of the rectangle ABCD.
(b) Find the area of the shaded triangle.
(b) 60
(b) 65
(b) 70
(b) 75
Sorry. Please check the correct answer below.
(a) $20 ÷ 5 = 4$
$20 ÷ 2 = 10$
$10 + 4 = 40$
$20 + 20 + 14 + 14 = 68$
(b) $\frac{1}{2} \times 14 \times 10 = 70$
(b) 70
You are Right
(a) $20 ÷ 5 = 4$
$20 ÷ 2 = 10$
$10 + 4 = 40$
$20 + 20 + 14 + 14 = 68$
(b) $\frac{1}{2} \times 14 \times 10 = 70$
The figure below is made up of 4 identical quadrants and a square. Find The area of the figure. $(Take \pi = \frac{22}{7})$
Sorry. Please check the correct answer below.
872cm$^2$ $\rightarrow$ CIRCLE $\rightarrow$ $\frac{22}{7} \times 14 \times 14 = 616$
SQUARE $\rightarrow$ $16 \times 16 = 256$
Total $\rightarrow$ $616 + 256 = 872$
You are Right
872cm$^2$ $\rightarrow$ CIRCLE $\rightarrow$ $\frac{22}{7} \times 14 \times 14 = 616$
SQUARE $\rightarrow$ $16 \times 16 = 256$
Total $\rightarrow$ $616 + 256 = 872$
In the figure below, a square is cut into 3 parts X, Y and Z. The area of X $\frac{1}{3}$ of the whole square and the area of Y is $\frac{7}{4}$ of the area of Z. What is the ration of the area of X to the area of Y to the area of Z?
Sorry. Please check the correct answer below.
$7 + 4 = 11$
$11 ÷ 2 = 5.5$
$X : Y : Z$
5.5 : 7 : 4
11 : 14 : 8
You are Right
$7 + 4 = 11$
$11 ÷ 2 = 5.5$
$X : Y : Z$
5.5 : 7 : 4
11 : 14 : 8
The figure below is formed by of 4 identical quarter circles, 1 semicircle and 1 rectangle. Find the area of the shaded figure. Leave your answer in terms of $\pi$.
Sorry. Please check the correct answer below.
You are Right
Find the area of the parallelogram PQRS in terms of k.
Sorry. Please check the correct answer below.
You are Right
In the figure below, XYZ is a right-triangle. Find the shaded area.
Sorry. Please check the correct answer below.
32cm$^2$ $\rightarrow \frac{1}{2} \times 12 \times 16 = 96$
$ 96 ÷ 3 = 32$
You are Right
32cm$^2$ $\rightarrow \frac{1}{2} \times 12 \times 16 = 96$
$ 96 ÷ 3 = 32$
In the square grid below, a shaded figure is drawn. What is the area of the shaded figure?
Sorry. Please check the correct answer below.
You are Right
ABCD is a square. What is the ratio of the shaded area to the area of square ABCD?
Sorry. Please check the correct answer below.
You are Right
The figure below shows 2 identical squares. Lines AF, FB, EF and FG are of the same length. The shaded area is 50cm$^2$. Find the area of the unshaded parts.
Sorry. Please check the correct answer below.
$50 \times 2 = 100cm^2$
You are Right
$50 \times 2 = 100cm^2$
The figure below shows 3 overlapping identical squares, with two overlapped area of the same size. If the area of each big square is 120cm$^2$ and the area of the whole figure is 300cm $^2$ , What is the area of each overlapped part?
Sorry. Please check the correct answer below.
You are Right
The figure below is created with 4 quarter circles and a square.
(a) The unshaded part marked A is enclosed by the quarter and quarter circle. Find the area of A.
(b) Find the area of the shaded figure. $ (Take\pi = 3.14)$.
(b) 8.12 cm$^2$
(b) 8.14 cm$^2$
(b) 9.12 cm$^2$
(b) 9.14 cm$^2$
Sorry. Please check the correct answer below.
$6cm \times 6cm \times \frac{1}{4} \times 3.14 = 28.26cm^2$
$6cm \times 6cm = 36 cm^2$
(a) $36 cm^2 – 28.26 cm^2 = 7.74 cm^2$
$4cm \times 4cm \times 3.14 \times \frac{1}{4} = 12.56 cm^2$
$4cm \times 4cm = 16 cm^2$
$16 cm^2 – 12.56 cm^2 = 3.44 cm^2$
$2cm \times 4cm = 8cm^2$
$36cm^2 – 3.44 cm^2 - 8 cm^2 = 24.56cm^2$
$24.56 cm^2 – 7.74 cm^2 = 16.82 cm^2$
$2cm \times 2cm = 4 cm^2$
$12.56 cm^2- 4 cm^2 = 8.56 cm^2$
$2cm \times 2cm \times 3.14 \times \frac{1}{4} = 3.14 cm^2$
$12.82 cm^2 – 3.14 cm^2 – 8.56 cm^2 = 5.12 cm^2$
(b) $5.12 cm^2 + 4 cm^2 = 9.12 cm^2$
(b) 9.12 cm$^2$
You are Right
$6cm \times 6cm \times \frac{1}{4} \times 3.14 = 28.26cm^2$
$6cm \times 6cm = 36 cm^2$
(a) $36 cm^2 – 28.26 cm^2 = 7.74 cm^2$
$4cm \times 4cm \times 3.14 \times \frac{1}{4} = 12.56 cm^2$
$4cm \times 4cm = 16 cm^2$
$16 cm^2 – 12.56 cm^2 = 3.44 cm^2$
$2cm \times 4cm = 8cm^2$
$36cm^2 – 3.44 cm^2 - 8 cm^2 = 24.56cm^2$
$24.56 cm^2 – 7.74 cm^2 = 16.82 cm^2$
$2cm \times 2cm = 4 cm^2$
$12.56 cm^2- 4 cm^2 = 8.56 cm^2$
$2cm \times 2cm \times 3.14 \times \frac{1}{4} = 3.14 cm^2$
$12.82 cm^2 – 3.14 cm^2 – 8.56 cm^2 = 5.12 cm^2$
(b) $5.12 cm^2 + 4 cm^2 = 9.12 cm^2$
The figure below is made up of two identical circles with radius 14cm each and a right- angle triangle ABC. Line AB and line AC are the diameters of the circles. What is the total area of the shaded prats? $(Take \pi = \frac{22}{7})$.
Sorry. Please check the correct answer below.
$\frac{22}{7} \times 14 \times 14 = 616$
$616 \times 2 = 1232$
$\frac{1}{2} \times 28 \times 28 = 392$
$1232 – 392 = 840cm^2$
You are Right
$\frac{22}{7} \times 14 \times 14 = 616$
$616 \times 2 = 1232$
$\frac{1}{2} \times 28 \times 28 = 392$
$1232 – 392 = 840cm^2$
The figure below is made up of a square and a triangle. Square QPSR has an area of 36cm$^2$. Find the area of triangle QRT.
Sorry. Please check the correct answer below.
$\frac{1}{2} \times 6 \times 11 = 33$
You are Right
$\frac{1}{2} \times 6 \times 11 = 33$
In the diagram below, the length of DE is twice the length of EF, G is the mid- point of AB and AE = AG. EFG and DCH are isosceles triangles. The area of ABCD is 72cm$^2$. What is the area of the shaded region?
Sorry. Please check the correct answer below.
You are Right
The figure below is made up 1 circle, 3 identical rectangles and 12 identical quarter circle of radius 10cm. Find the total area of the shaded parts. $(Take \pi = 3.14)$.
Sorry. Please check the correct answer below.
1514cm$^2$ $\rightarrow$ Area = $12 \times 10 \times 10 + 3.14 \times 10 \times 10 = 1514$cm$^2$
You are Right
1514cm$^2$ $\rightarrow$ Area = $12 \times 10 \times 10 + 3.14 \times 10 \times 10 = 1514$cm$^2$
In the figure, ABCF is a rectangle. Find the area of the shaded parts.
Sorry. Please check the correct answer below.
You are Right
Quadrilateral ABCD is drawn on a 1-cm square grid. Find the area of ABCD.
Sorry. Please check the correct answer below.
You are Right
The figure is made up of four semi – circle and a rectangle ABCD. AB = 9cm, BC = 12cm and AC 15cm. Find the total area of the shaded parts. $ (Take \pi = 3.14)$.
Sorry. Please check the correct answer below.
$\frac{1}{2} \times 9 \times 12 = 54cm^2$
$\frac{1}{2} \times 3.14 \times 4.5 \times 4.5 = 31.79cm^2$
$\frac{1}{2} \times 3.14 \times 6 \times 6 = 56.52cm^2$
$54 + 31.79 + 56.52 = 142.31cm^2$
$\frac{1}{2} \times 3.14 \times 7.5 \times 7.5 = 88.31cm^2$
$142.31 – 88.31 = 54cm^2$
$54 \times 2 = 108cm^2$
You are Right
$\frac{1}{2} \times 9 \times 12 = 54cm^2$
$\frac{1}{2} \times 3.14 \times 4.5 \times 4.5 = 31.79cm^2$
$\frac{1}{2} \times 3.14 \times 6 \times 6 = 56.52cm^2$
$54 + 31.79 + 56.52 = 142.31cm^2$
$\frac{1}{2} \times 3.14 \times 7.5 \times 7.5 = 88.31cm^2$
$142.31 – 88.31 = 54cm^2$
$54 \times 2 = 108cm^2$
From the figure, PQ and RS are 2 diameters which are perpendicular to each other, ABPQ is a square and OP = 14cm. $(Take \pi = \frac{22}{7})$, Find the unshaded area of the figure.
Sorry. Please check the correct answer below.
Area of square $\rightarrow$ $28 \times 28 = 784$
Area of quadrent $\rightarrow$ $\frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154$
Area of the unshaded part $\rightarrow$ $784 – 154 = 630cm^2$
You are Right
Area of square $\rightarrow$ $28 \times 28 = 784$
Area of quadrent $\rightarrow$ $\frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = 154$
Area of the unshaded part $\rightarrow$ $784 – 154 = 630cm^2$
Find the area of the shaded triangle.
Sorry. Please check the correct answer below.
You are Right
The figure shows two quadrants of circles, centered at C and D respectively. Find the difference between the area of the two shaded regions. $(Take \pi = \frac{22}{7})$.
Sorry. Please check the correct answer below.
Area of rectangle $\rightarrow$ $(28\times14) cm^2 = 392cm^2$
Area of quadrent $\rightarrow$ $(28\times 28 \times \frac{22}{7})cm^2 ÷ 4 = 616cm^2$
Area of A,B and small quadrent $\rightarrow$ $(616 - 392)cm^2 ÷ 4 = 224cm^2$
Area of A and B $\rightarrow$ $(224 -154)cm^2 = 70cm^2$
Area of H and B $\rightarrow$ $(616 - 154)cm^2 = 462cm^2$
Difference in two shaded parts $\rightarrow$ $(462-392)cm^2 = 70cm^2$
You are Right
Area of rectangle $\rightarrow$ $(28\times14) cm^2 = 392cm^2$
Area of quadrent $\rightarrow$ $(28\times 28 \times \frac{22}{7})cm^2 ÷ 4 = 616cm^2$
Area of A,B and small quadrent $\rightarrow$ $(616 - 392)cm^2 ÷ 4 = 224cm^2$
Area of A and B $\rightarrow$ $(224 -154)cm^2 = 70cm^2$
Area of H and B $\rightarrow$ $(616 - 154)cm^2 = 462cm^2$
Difference in two shaded parts $\rightarrow$ $(462-392)cm^2 = 70cm^2$
The volume of the cube shown below is 216cm $^3$ . Find the area of the face of the cube.
Sorry. Please check the correct answer below.
You are Right
The figure below shows a box will a height of 11cm. What is the base area of the box?
Sorry. Please check the correct answer below.
You are Right
Rectangle ABCD is divided into 6 identical small rectangle as shown below. Give that the perimeter of rectangle ABCD is 80cm, find the area of one small rectangle.
Sorry. Please check the correct answer below.
64cm$^2$ IU $\rightarrow$ $80 ÷ 2 = 4$
4u $\rightarrow$ $4 \times 4 = 16$
area $\rightarrow$ $16 \times 4 = 64$
You are Right
64cm$^2$ IU $\rightarrow$ $80 ÷ 2 = 4$
4u $\rightarrow$ $4 \times 4 = 16$
area $\rightarrow$ $16 \times 4 = 64$
Tow identical triangles ABC and DEF overlapped other to form six identical equilateral triangle as shown below. The area of the shaded part is 60cm$^2$. Find the area of triangle ABC.
Sorry. Please check the correct answer below.
You are Right
The figure below shown the design of a floor rug. The edges of the rug is made up of 4 semicircles and 4 quarter-circle, each of radius 21cm. $(Take \pi = \frac{22}{7})$ . Find the area of the rug.
Sorry. Please check the correct answer below.
8568cm$^2$ $\rightarrow$ Total area $\rightarrow$ $2772= 1386 + 882 + 3528 = 8568$
You are Right
8568cm$^2$ $\rightarrow$ Total area $\rightarrow$ $2772= 1386 + 882 + 3528 = 8568$
The figure shows three identical squares ABCD, BEFG and HJKB. What is the value of $\angle$GBK?
Sorry. Please check the correct answer below.
You are Right
A square piece of paper, ABCD, is shaded on one side as shown in figure1. It is then folded at its comer B to form an isosceles triangle as shown in figure 2. The perimeter and area of the remaining shaded region in figure 2 is 72cm and 180cm$^2$ respectively. Find the area of the isosceles triangle.
Sorry. Please check the correct answer below.
A. of ABCD $\rightarrow$ $18 \times 18 = 324cm^2$
A. of BEFG $\rightarrow$ $324 – 180 = 144cm^2$
A. of Triangle $\rightarrow$ $144 ÷ 2 = 72cm^2$
You are Right
A. of ABCD $\rightarrow$ $18 \times 18 = 324cm^2$
A. of BEFG $\rightarrow$ $324 – 180 = 144cm^2$
A. of Triangle $\rightarrow$ $144 ÷ 2 = 72cm^2$
The figure below shows 3 overlapping identical triangle. If the area of each big triangle is 80cm$^2$ and the area of the figure is 210cm$^2$, Find the area of the shaded triangle.
Sorry. Please check the correct answer below.
You are Right
The figure below is made up of 2 identical quadrants and 2 identical semicircles. Find the area of the shaded part of the figure. $(Take \pi = \frac{22}{7})$.
Sorry. Please check the correct answer below.
You are Right
Mysha cut out three identical right-angled triangles. She joined them to form a figure PQRS as shown below. SR = 20cm and QR = 8cm. The perimeter of the figure PQRS is 44cm.Find the area of the figure PQRS.
Sorry. Please check the correct answer below.
$44 – 30 = 14$
$14 – 8 = 6$
$\frac{1}{2} \times 8 \times 6 = 24$
$ 24 \times 3 = 72cm^2$
You are Right
$44 – 30 = 14$
$14 – 8 = 6$
$\frac{1}{2} \times 8 \times 6 = 24$
$ 24 \times 3 = 72cm^2$
The figure below, not drawn to scale, is made up of a square ABCD and two triangle EFG and FKH, overlapping one another. The square has an area of 81cm$^2$. FK is 14cm and it is $\frac{2}{3}$ of the length of FG. Find the area of the figure.
Sorry. Please check the correct answer below.
$\sqrt(81)=9$
$14 ÷ 2 = 7$
$14 + 7 = 21$
$\frac{1}{2} \times 19\times 21 = 199.5 (EFG)$
$\frac{1}{2} \times 14 \times 9 = 63$
$133 – 63 = 70 (HDF)$
$70 + 199.5 = 269.5cm^2$
You are Right
$\sqrt(81)=9$
$14 ÷ 2 = 7$
$14 + 7 = 21$
$\frac{1}{2} \times 19\times 21 = 199.5 (EFG)$
$\frac{1}{2} \times 14 \times 9 = 63$
$133 – 63 = 70 (HDF)$
$70 + 199.5 = 269.5cm^2$
The figure below is made up of a quadrant a semicircle. The area of the square is 144cm$^2$ . Find the area of the shaded parts. $ (Take \pi = 3.14)$
Sorry. Please check the correct answer below.
$\sqrt(144)= 12$
$\frac{1}{4} \times 12 \times 12 = 113.04$
$113.4 ÷ 2 = 56.52$
$\frac{1}{2} \times 12 \times 6 = 36$
$ 56.52 – 36 = 20.52$cm$^2$
You are Right
$\sqrt(144)= 12$
$\frac{1}{4} \times 12 \times 12 = 113.04$
$113.4 ÷ 2 = 56.52$
$\frac{1}{2} \times 12 \times 6 = 36$
$ 56.52 – 36 = 20.52$cm$^2$
ACE is a triangle. Which one of the following gives the area of ADE?
Sorry. Please check the correct answer below.
You are Right
In the diagram below, ABCD is a rhombus. E is the mid–point BC and AC. Find the area the shaded part.
Sorry. Please check the correct answer below.
You are Right
The figure below shown a cube. The total length of all the edges of the cube is 156cm. Find the area of the shaded face.
Sorry. Please check the correct answer below.
$12 edges \rightarrow 156cm$
$ 1 edges \rightarrow 13cm$
$Shaded face \rightarrow 13cm \times 13cm = 169cm^2$
You are Right
$12 edges \rightarrow 156cm$
$ 1 edges \rightarrow 13cm$
$Shaded face \rightarrow 13cm \times 13cm = 169cm^2$
The solid below is made up of five 2-cm cubes. James sprayed the whole solid with paint. What area of the solid is covered with paint?
Sorry. Please check the correct answer below.
You are Right
Study the diagram below. Find the area of the shaded triangle.
Sorry. Please check the correct answer below.
You are Right
The picture below is made up of 2 similar squares and 2 similar quadrants. The area of one square is 64cm$^2$. Find the area of the shaded region.
Sorry. Please check the correct answer below.
$\sqrt(64)= 8$
$0.25 3.14\times 8\times 8 = 50.24$
$8 \times 8 = 64$
$64 – 50.24 = 13.76$
$13.76 \times 2 = 27.52$
The area is 27.52cm$^2$
You are Right
$\sqrt(64)= 8$
$0.25 3.14\times 8\times 8 = 50.24$
$8 \times 8 = 64$
$64 – 50.24 = 13.76$
$13.76 \times 2 = 27.52$
The area is 27.52cm$^2$
The figure below is made up of 3 semi-circles. Find the area of the figure. $(Take \pi= \frac{22}{7})$
Sorry. Please check the correct answer below.
464cm$^2$ $\rightarrow$ $14 ÷ 2 =7$
$\frac{22}{7} \times 7 \times 7 = 154$
$\frac{22}{7} \times 14 \times 14 ÷ 2 = 44 \times 14 ÷ 2 = 616 ÷ 2 = 308$
$308 + 154 = 462$
You are Right
464cm$^2$ $\rightarrow$ $14 ÷ 2 =7$
$\frac{22}{7} \times 7 \times 7 = 154$
$\frac{22}{7} \times 14 \times 14 ÷ 2 = 44 \times 14 ÷ 2 = 616 ÷ 2 = 308$
$308 + 154 = 462$
The figure below is not drawn to scale. It shows a shaded quadrant in a semicircle. The diameter of the semicircle is 28cm. Find the total area of the unshaded parts. $(Take \pi = \frac{22}{7})$.
Sorry. Please check the correct answer below.
You are Right
The figure below is not drawn to scale. ABCD is a square. XYZ is a right-angled isosceles triangle of area 180cm$^2$. Find the area of Square ABCD.
Sorry. Please check the correct answer below.
You are Right
The rectangle below is divided into four prats W, X, Y and Z. The ratio of area W to area X is 3 : 5. The ratio of area Y to area Z is 1: 2. What fraction of the total area is area W?
Sorry. Please check the correct answer below.
You are Right
The figure below is formed by a square ABCD and a triangle DGC. AD = 9cm, EF =4cm and FC is a straight line. Find the area of the shaded part.
Sorry. Please check the correct answer below.
You are Right
In the figure below. Not draw to scale, ABCD is a rectangle of sides 36cm by 19cm. The area of the quadrilateral EFGH is 38cm $^2$. Find the area of the unshaded part.
Sorry. Please check the correct answer below.
304cm$^2$ $\rightarrow$ $\frac{1}{2} \times 19 \times 36 = 342$
You are Right
304cm$^2$ $\rightarrow$ $\frac{1}{2} \times 19 \times 36 = 342$
Find the area of the shaded triangle.
Sorry. Please check the correct answer below.
You are Right
The figure below is made up a big quadrant OBD, a small quadrant OAE and a square ACEO. The radius of the big quadrant OBD is 12cm. The area of the big quadrant OBD is twice the area of the small quadrant OAE. Using the calculator value of $\pi$, Find the area of the shaded parts, correct 2 decimal places.
Sorry. Please check the correct answer below.
$ 12 \times 12 \times \frac{1}{4} \times \pi = 36\pi$
$ 2A \rightarrow 36\pi ÷ 2 = 18\pi$
$ B \rightarrow(18\pi – 36)$
$(36\pi – 72) + (18\pi – 36) = 61.65cm^2$
You are Right
$ 12 \times 12 \times \frac{1}{4} \times \pi = 36\pi$
$ 2A \rightarrow 36\pi ÷ 2 = 18\pi$
$ B \rightarrow(18\pi – 36)$
$(36\pi – 72) + (18\pi – 36) = 61.65cm^2$
The figure below shows a rectangle ABCD. EFG and DFB are straight line. The area of rectangle ABCD is 960cm$^2$ and the total area of triangles DEF and BEG is 336cm$^2$. The ratio of length DG to the length GC is 7:5. What is the area of the triangle DFG?
Sorry. Please check the correct answer below.
112cm$^2$ $\rightarrow$ $\bigtriangleup$ DEG + $\bigtriangleup$ DBG = $\frac{7}{12} \times 960 = 560$
$\bigtriangleup$ DFG = $\frac{560 - 336}{2} = 112$
You are Right
112cm$^2$ $\rightarrow$ $\bigtriangleup$ DEG + $\bigtriangleup$ DBG = $\frac{7}{12} \times 960 = 560$
$\bigtriangleup$ DFG = $\frac{560 - 336}{2} = 112$
In the figure below, ABCD is a square. Find the area of Unshaded part. Leave your answer in terms of $\pi$.
Sorry. Please check the correct answer below.
16$\pi$ $\rightarrow$ $\pi \times 8 \times 8 =64\pi$
$64\pi ÷ 4 = 16\pi$
You are Right
16$\pi$ $\rightarrow$ $\pi \times 8 \times 8 =64\pi$
$64\pi ÷ 4 = 16\pi$
In the figure below, AC = 15cm, CD = 9cm and AD = 12cm. What is the area of triangle ABC?
Sorry. Please check the correct answer below.
You are Right
The figure below, not drawn to scale, is made up of tow identical right-angled triangles overlapping each other. Find the area of the shaded part ABCD, where AD=3cm.
Sorry. Please check the correct answer below.
37.5cm$^2$ $\rightarrow \frac{2}{2} \times 5 \times 3 = 7.5$
$9 – 3 = 6$
$ 5 \times 6 = 30$
You are Right
37.5cm$^2$ $\rightarrow \frac{2}{2} \times 5 \times 3 = 7.5$
$9 – 3 = 6$
$ 5 \times 6 = 30$
Ramesh had a rectangular block of wood 9cm by cm 7cm. He painted all the faces of the block. What is the total painted area?
Sorry. Please check the correct answer below.
$2 \times (9\times7) + 2 \times (9\times7) – 2 \times (9\times4) = 126 + 56 + 72 = 254cm^2$
You are Right
$2 \times (9\times7) + 2 \times (9\times7) – 2 \times (9\times4) = 126 + 56 + 72 = 254cm^2$
The figure below is not down to scale. ABCD is a trapezium. Find the area of triangle ABC.
Sorry. Please check the correct answer below.
60cm$^2$ $\rightarrow$ $\frac{1}{2} \times 12 \times 10cm = 60cm$
You are Right
60cm$^2$ $\rightarrow$ $\frac{1}{2} \times 12 \times 10cm = 60cm$
The figure below is made up of a semicircle, a square and 3 quadrants. The side of the square is 20cm. For each of the following, use the calculator value of $\pi$ to find the area of the shaded part, correct to 2 decimal places.
Sorry. Please check the correct answer below.
$\frac{3}{4} \times \pi \times 20\times 20 = 300\pi$
$\frac{1}{4} \times \pi \times 40\times 40 = 400\pi$
$20 \times 20 = 400$
$\frac{1}{4} \times \pi \times 20\times 20 = 100\pi$
$400\pi – 100\pi – 400 = 300\pi – 400$
$300\pi – 400 + 300\pi = 1481.96cm^2$
You are Right
$\frac{3}{4} \times \pi \times 20\times 20 = 300\pi$
$\frac{1}{4} \times \pi \times 40\times 40 = 400\pi$
$20 \times 20 = 400$
$\frac{1}{4} \times \pi \times 20\times 20 = 100\pi$
$400\pi – 100\pi – 400 = 300\pi – 400$
$300\pi – 400 + 300\pi = 1481.96cm^2$
The figure below is made up of three semi-circle and a circles. X is the center of the large semi-circle and WZ is 36cm. Find the area of the shaded part. Express your answer in terms of $\pi$.
Sorry. Please check the correct answer below.
Area of fig $\rightarrow$ $\frac{1}{2} \times 18 \times 18 \times \pi = 162\pi$
$\frac{1}{2} \times 36 \times 18 = 324$
$(162\pi - 324) ÷ 2 = 81\pi - 162$
$(20.25\pi – 40.5) \times 2 = 40.5\pi -81$
$(81\pi - 162) – (40.5\pi - 81) = (81\pi – 162)cm^2$
You are Right
Area of fig $\rightarrow$ $\frac{1}{2} \times 18 \times 18 \times \pi = 162\pi$
$\frac{1}{2} \times 36 \times 18 = 324$
$(162\pi - 324) ÷ 2 = 81\pi - 162$
$(20.25\pi – 40.5) \times 2 = 40.5\pi -81$
$(81\pi - 162) – (40.5\pi - 81) = (81\pi – 162)cm^2$
In the figure below, ABCD is a rectangle and DE = EA. If the area of triangle BED is 7cm$^2$, Find the area of the unshaded part.
Sorry. Please check the correct answer below.
You are Right
The figure is made up of a quarter circle CDF and a square ABEG. The corner B of the square line on the circumference of the quarter circle. Circle GFED is a straight lines. GB = FD = 10cm. Find the area of the shaded part BEF. $(Take\pi = 3.14)$.
Sorry. Please check the correct answer below.
Area of ABGE $\rightarrow$ $(\frac{1}{2} \times 10 \times 5) \times 2 = 50cm^2$
Area of BEF $\rightarrow$ $\frac{1}{4} \times 3.14 \times 10 \times 10 = 78.5cm^2$
Area of big triangle $\rightarrow$ $\frac{1}{2} \times 20 \times 10 = 100cm^2$
a + b $\rightarrow$ $100cm^2 – 78.5cm^2 = 21.5cm^2$
b $\rightarrow$ 10.75cm$^2$
Area of GBE $\rightarrow$ $\frac{1}{2} \times 10 \times 5 = 25cm^2$
Shaded area $\rightarrow$ $25cm^2 – 10.75cm^2 = 14.25cm^2$
You are Right
Area of ABGE $\rightarrow$ $(\frac{1}{2} \times 10 \times 5) \times 2 = 50cm^2$
Area of BEF $\rightarrow$ $\frac{1}{4} \times 3.14 \times 10 \times 10 = 78.5cm^2$
Area of big triangle $\rightarrow$ $\frac{1}{2} \times 20 \times 10 = 100cm^2$
a + b $\rightarrow$ $100cm^2 – 78.5cm^2 = 21.5cm^2$
b $\rightarrow$ 10.75cm$^2$
Area of GBE $\rightarrow$ $\frac{1}{2} \times 10 \times 5 = 25cm^2$
Shaded area $\rightarrow$ $25cm^2 – 10.75cm^2 = 14.25cm^2$
The table mat below is made up of identical semi – circles and a 31-cm square.
(a) Find the perimeter of the entire table mat.
(b) Half of the table mat shaded. Find the area of the shaded part. $(Take \pi = \frac{22}{7})$
(b) 553.5cm
(b) 555.5cm
(b) 557.5cm
(b) 559.5cm
Sorry. Please check the correct answer below.
(a) 156cm $\rightarrow$ $31 – 17 = 14$
$14 ÷ 2 = 7$
$\frac{22}{7} \times 7 = 22$
$22 \times 4 = 88$
$88 + (17\times4) = 156$
(b) 557.5cm$^2$ $\rightarrow$ $31 \times 31 = 961$
$961 ÷ 2 = 480.5$
$\frac{22}{7} \times 3.5 \times 3.5 = 38.5$
$38.5 \times 2 = 77$
$77 + 480.5 = 557.5$
(b) 557.5cm
You are Right
(a) 156cm $\rightarrow$ $31 – 17 = 14$
$14 ÷ 2 = 7$
$\frac{22}{7} \times 7 = 22$
$22 \times 4 = 88$
$88 + (17\times4) = 156$
(b) 557.5cm$^2$ $\rightarrow$ $31 \times 31 = 961$
$961 ÷ 2 = 480.5$
$\frac{22}{7} \times 3.5 \times 3.5 = 38.5$
$38.5 \times 2 = 77$
$77 + 480.5 = 557.5$
Mrs Lee drew 3 squares to form a figure. The area of the squares were in the ratio 1: 4 : 13. She then shaded some parts of the figure as shown below. What is the ratio of the shaded parts to the unshaded parts of the figure?
Sorry. Please check the correct answer below.
Shaded parts $\rightarrow$ $ 1 + (13-4) = 10$
Unshaded parts $\rightarrow$ $4 – 1 = 3$
S:U = 10:3
You are Right
Shaded parts $\rightarrow$ $ 1 + (13-4) = 10$
Unshaded parts $\rightarrow$ $4 – 1 = 3$
S:U = 10:3
The diagram below shown a quadrant and a right- angle triangle. The ratio of the length of AB to the length of AC is 2:3 . What is the difference in area between the 2 shaded parts? $(Take \pi = 3.14)$
Sorry. Please check the correct answer below.
You are Right
A piece of wire, forming 2 identical semi-circle as shown in figure 1, was straightened and bent into a square as shown in Figure2. Find the area of the square, giving your answer to 2 decimal places. $(Take \pi = 3.14)$
Sorry. Please check the correct answer below.
554.60cm$^3$ $\rightarrow$ $3.14 \times 30 = 94.2$
$94.2 ÷ 4 = 23.55$
$23.55 \times 23.50 = 554.60$
You are Right
554.60cm$^3$ $\rightarrow$ $3.14 \times 30 = 94.2$
$94.2 ÷ 4 = 23.55$
$23.55 \times 23.50 = 554.60$
The figure below is made up of identical quadrants. $(Take \pi= 3.14)$ .
(a) Find the area of the shaded part.
(b) Find the perimeter of the shaded part.
(b) 268.4cm
(b) 269.6cm
(b) 270.4cm
(b) 271.6cm
Sorry. Please check the correct answer below.
$ a 3.14 \times 10 \times 10 ÷ 4 = 78.5$
$ 10 \times 10 = 100$
$ 100 – 78.5 = 21.5$
$ 21.5 \times 2 = 43$
$ 100 – 43 = 57$
$ 4 \times 57 = 228$
$ 314 + 228 = 542cm^2$
$ b 2 \times 3.14 \times 10 = 62.8$
$ 62.8 \times 2 = 125.6
$ 62.8 + 125.6 + (8\times10) = 268.4cm
(b) 268.4cm
You are Right
$ a 3.14 \times 10 \times 10 ÷ 4 = 78.5$
$ 10 \times 10 = 100$
$ 100 – 78.5 = 21.5$
$ 21.5 \times 2 = 43$
$ 100 – 43 = 57$
$ 4 \times 57 = 228$
$ 314 + 228 = 542cm^2$
$ b 2 \times 3.14 \times 10 = 62.8$
$ 62.8 \times 2 = 125.6
$ 62.8 + 125.6 + (8\times10) = 268.4cm
The shaded figure below is formed by semicircles, quarter circle and squares. ABEF is a square. What is the area of the shaded region?
Sorry. Please check the correct answer below.
Area of rectangle: $30 \times 60 = 1800cm^2$
Area of semicircle: $\frac{1}{2} \times 30 \times 30 \times 3.14 = 141cm^2$
$1800cm^2 - 1413cm^2 = 387cm^2$
Area of circle: $15 \times 15 \times 3.14 = 706.5cm^2$
$706.5cm^2 + 387cm^2 = 1093.5cm^2$
You are Right
Area of rectangle: $30 \times 60 = 1800cm^2$
Area of semicircle: $\frac{1}{2} \times 30 \times 30 \times 3.14 = 141cm^2$
$1800cm^2 - 1413cm^2 = 387cm^2$
Area of circle: $15 \times 15 \times 3.14 = 706.5cm^2$
$706.5cm^2 + 387cm^2 = 1093.5cm^2$
A tank with a square base has a volume of 108cm$^3$. The length of one side of the base is $\frac{1}{4}$ that of its height. Find the area of its base.
Sorry. Please check the correct answer below.
You are Right
The figure below is made up of 2 identical squares STYZ and UVWX and a rectangle TUXY. Area A is twice the area of B and area D is half of area F. What fraction of the figure is shaded?
Sorry. Please check the correct answer below.
You are Right
The figure below is not drawn to scale. ABCD is a square of area 100cm$^2$. A semicircle and a quadrant lie within square ABCD. AE = ED. Find the area of the shaded part. Leave your answer in terms of $\pi$.
Sorry. Please check the correct answer below.
You are Right
Given that the radius of the circle is 14cm. Find its area. $(Take \pi = \frac{22}{7})$
Sorry. Please check the correct answer below.
You are Right
The figure is mane up of 6 identical quadrants, a semicircle K and a triangle. The total area of the unshaded parts is 370cm$^2$. Find the total area of the shaded parts. $(Take\pi = 3.14)$.
Sorry. Please check the correct answer below.
$3.14 \times 16 \times \frac{6}{4} = 1205.76$
$3.14 \times 16 ÷ 2 = 401.92$
$3.14 \times 8 ÷ 2 = 100.48$
$16 \times 16 \times 0.5 = 128$
$370 – 128 – 100.48 = 141.52$
$1205.76 – 128 – 100.48 – (141.52 \times 2) = 694.24cm^2$
You are Right
$3.14 \times 16 \times \frac{6}{4} = 1205.76$
$3.14 \times 16 ÷ 2 = 401.92$
$3.14 \times 8 ÷ 2 = 100.48$
$16 \times 16 \times 0.5 = 128$
$370 – 128 – 100.48 = 141.52$
$1205.76 – 128 – 100.48 – (141.52 \times 2) = 694.24cm^2$
PSLE-Math.com
Start winning at math! Be part of Singapore PSLE Math community and Unleash the Your Maths Potential!
General
Contact Us
admin@psle-math.com
https://psle-math.com
2024 © All Rights Reserved. Privacy Policy | Terms of Service
What subject area do you need help in?
Select level
Select subject
