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Question 1 of 84
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?
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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 = $\frac{560 - 336}{2} = 112$
The perimeter of the figure is 40cm. Find the area of the figure.
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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. $(Take \pi = \frac{22}{7})$.
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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?
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$7 + 4 = 11$
$11 ÷ 2 = 5.5$
$X : Y : Z$
5.5 : 7 : 4
11 : 14 : 8
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$7 + 4 = 11$
$11 ÷ 2 = 5.5$
$X : Y : Z$
5.5 : 7 : 4
11 : 14 : 8
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)$
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$\sqrt(144)= 12$
$\frac{1}{4} \times 12 \times 12 = 113.04$
$113.4 ÷ 2 = 56.52$
$\frac{1}{2} \times 12 \times 6 = 36$
$ 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$
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.
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$30 + 30 = 45 + WTU$
$WTU = 15$
$5u --- 30$
$3u ---- 18$
$18 + 15 = 33$
You are Right
$30 + 30 = 45 + WTU$
$WTU = 15$
$5u --- 30$
$3u ---- 18$
$18 + 15 = 33$
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.
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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?
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Shaded parts $\rightarrow$ $ 1 + (13-4) = 10$
Unshaded parts $\rightarrow$ $4 – 1 = 3$
S:U = 10:3
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Shaded parts $\rightarrow$ $ 1 + (13-4) = 10$
Unshaded parts $\rightarrow$ $4 – 1 = 3$
S:U = 10:3
The figure shows two quadrants of circles, centered at C and D respectively. Find the difference between the area of the two shaded regions. $(Take \pi = \frac{22}{7})$.
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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 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. $(Take\pi = 3.14)$.
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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 figure below shows a box will a height of 11cm. What is the base area of the box?
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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?
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The shaded figure below is formed by semicircles, quarter circle and squares. ABEF is a square. What is the area of the shaded region?
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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: $15 \times 15 \times 3.14 = 706.5cm^2$
$706.5cm^2 + 387cm^2 = 1093.5cm^2$
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.
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$\frac{1}{2} \times 16 \times 13 = 104$
$\frac{1}{2} \times 12 \times 28 = 168$
$168 + 104 = 272cm^2$
You are Right
$\frac{1}{2} \times 16 \times 13 = 104$
$\frac{1}{2} \times 12 \times 28 = 168$
$168 + 104 = 272cm^2$
The figure shows a circle inside a square. Find the area of the unshaded part. $ (Take\pi= \frac{22}{7})$.
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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)$.
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$\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
$\frac{1}{2} \times 9 \times 12 = 54cm^2$
$\frac{1}{2} \times 3.14 \times 4.5 \times 4.5 = 31.79cm^2$
$\frac{1}{2} \times 3.14 \times 6 \times 6 = 56.52cm^2$
$54 + 31.79 + 56.52 = 142.31cm^2$
$\frac{1}{2} \times 3.14 \times 7.5 \times 7.5 = 88.31cm^2$
$142.31 – 88.31 = 54cm^2$
$54 \times 2 = 108cm^2$
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?
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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 $\pi$.
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Find the area of the shaded triangle.
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In the figure below, ABCD is a square. Find the area of Unshaded part. Leave your answer in terms of $\pi$.
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16$\pi$ $\rightarrow$ $\pi \times 8 \times 8 =64\pi$
$64\pi ÷ 4 = 16\pi$
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16$\pi$ $\rightarrow$ $\pi \times 8 \times 8 =64\pi$
$64\pi ÷ 4 = 16\pi$
In the figure, ABCF is a rectangle. Find the area of the shaded parts.
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Given that the radius of the circle is 14cm. Find its area. $(Take \pi = \frac{22}{7})$
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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? $(Take \pi = 3.14)$
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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
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$ 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
In the figure, ABCD is a rectangle of area 96cm $^2$, with AE =EB .What is the total area of the shaded parts?
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$96 ÷ 4 = 24cm^2$
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$96 ÷ 4 = 24cm^2$
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.
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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.
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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
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(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$^2$ $\rightarrow$ $31 \times 31 = 961$
$961 ÷ 2 = 480.5$
$\frac{22}{7} \times 3.5 \times 3.5 = 38.5$
$38.5 \times 2 = 77$
$77 + 480.5 = 557.5$
The figure shows three identical squares ABCD, BEFG and HJKB. What is the value of $\angle$GBK?
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Find the area of the shaded parts.
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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$^2$ $\rightarrow$ $\frac{1}{2} \times 3 \times 6 = 9$
$\frac{1}{2} \times 2 \times 6 = 6$
$\frac{1}{2} \times 16 = 3$
$9 + 6 + 3 = 18$
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. $(Take \pi = 3.14)$
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554.60cm$^3$ $\rightarrow$ $3.14 \times 30 = 94.2$
$94.2 ÷ 4 = 23.55$
$23.55 \times 23.50 = 554.60$
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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 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.
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$50 \times 2 = 100cm^2$
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$50 \times 2 = 100cm^2$
The figure below is made up of 4 identical quadrants and a square. Find The area of the figure. $(Take \pi = \frac{22}{7})$
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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 diagram below, ABCD is a rhombus. E is the mid–point BC and AC. Find the area the shaded part.
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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.
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$ 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)$
$(36\pi – 72) + (18\pi – 36) = 61.65cm^2$
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?
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$2 \times (9\times7) + 2 \times (9\times7) – 2 \times (9\times4) = 126 + 56 + 72 = 254cm^2$
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$2 \times (9\times7) + 2 \times (9\times7) – 2 \times (9\times4) = 126 + 56 + 72 = 254cm^2$
In the figure below, AC = 15cm, CD = 9cm and AD = 12cm. What is the area of triangle ABC?
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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 $\pi$.
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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.
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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 $\rightarrow$ $784 – 154 = 630cm^2$
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.
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304cm$^2$ $\rightarrow$ $\frac{1}{2} \times 19 \times 36 = 342$
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304cm$^2$ $\rightarrow$ $\frac{1}{2} \times 19 \times 36 = 342$
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.
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$12 edges \rightarrow 156cm$
$ 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 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.
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In the square grid below, a shaded figure is drawn. What is the area of the shaded figure?
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Quadrilateral ABCD is drawn on a 1-cm square grid. Find the area of ABCD.
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ABCD is a square. What is the ratio of the shaded area to the area of square ABCD?
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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.
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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.
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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 is made up of 3 semi-circles. Find the area of the figure. $(Take \pi= \frac{22}{7})$
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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$^2$ $\rightarrow$ $14 ÷ 2 =7$
$\frac{22}{7} \times 7 \times 7 = 154$
$\frac{22}{7} \times 14 \times 14 ÷ 2 = 44 \times 14 ÷ 2 = 616 ÷ 2 = 308$
$308 + 154 = 462$
Find the area of the shaded triangle.
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The figure below is made up of 3 identical isosceles right-angled triangle. What is the area of the figure?
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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)$.
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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?
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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 down to scale. ABCD is a trapezium. Find the area of triangle ABC.
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60cm$^2$ $\rightarrow$ $\frac{1}{2} \times 12 \times 10cm = 60cm$
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60cm$^2$ $\rightarrow$ $\frac{1}{2} \times 12 \times 10cm = 60cm$
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.
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64cm$^2$ IU $\rightarrow$ $80 ÷ 2 = 4$
4u $\rightarrow$ $4 \times 4 = 16$
area $\rightarrow$ $16 \times 4 = 64$
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64cm$^2$ IU $\rightarrow$ $80 ÷ 2 = 4$
4u $\rightarrow$ $4 \times 4 = 16$
area $\rightarrow$ $16 \times 4 = 64$
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.
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$\frac{3}{4} \times \pi \times 20\times 20 = 300\pi$
$\frac{1}{4} \times \pi \times 40\times 40 = 400\pi$
$20 \times 20 = 400$
$\frac{1}{4} \times \pi \times 20\times 20 = 100\pi$
$400\pi – 100\pi – 400 = 300\pi – 400$
$300\pi – 400 + 300\pi = 1481.96cm^2$
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 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?
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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? $(Take \pi = \frac{22}{7})$.
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$\frac{22}{7} \times 14 \times 14 = 616$
$616 \times 2 = 1232$
$\frac{1}{2} \times 28 \times 28 = 392$
$1232 – 392 = 840cm^2$
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 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?
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In the figure below, XYZ is a right-triangle. Find the shaded area.
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32cm$^2$ $\rightarrow \frac{1}{2} \times 12 \times 16 = 96$
$ 96 ÷ 3 = 32$
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32cm$^2$ $\rightarrow \frac{1}{2} \times 12 \times 16 = 96$
$ 96 ÷ 3 = 32$
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. $(Take\pi = 3.14)$.
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$3.14 \times 16 \times \frac{6}{4} = 1205.76$
$3.14 \times 16 ÷ 2 = 401.92$
$3.14 \times 8 ÷ 2 = 100.48$
$16 \times 16 \times 0.5 = 128$
$370 – 128 – 100.48 = 141.52$
$1205.76 – 128 – 100.48 – (141.52 \times 2) = 694.24cm^2$
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$3.14 \times 16 \times \frac{6}{4} = 1205.76$
$3.14 \times 16 ÷ 2 = 401.92$
$3.14 \times 8 ÷ 2 = 100.48$
$16 \times 16 \times 0.5 = 128$
$370 – 128 – 100.48 = 141.52$
$1205.76 – 128 – 100.48 – (141.52 \times 2) = 694.24cm^2$
The volume of the cube shown below is 216cm $^3$ . Find the area of the face of the cube.
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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. $(Take \pi = 3.14)$.
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1514cm$^2$ $\rightarrow$ Area = $12 \times 10 \times 10 + 3.14 \times 10 \times 10 = 1514$cm$^2$
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1514cm$^2$ $\rightarrow$ Area = $12 \times 10 \times 10 + 3.14 \times 10 \times 10 = 1514$cm$^2$
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?
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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.
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$44 – 30 = 14$
$14 – 8 = 6$
$\frac{1}{2} \times 8 \times 6 = 24$
$ 24 \times 3 = 72cm^2$
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$44 – 30 = 14$
$14 – 8 = 6$
$\frac{1}{2} \times 8 \times 6 = 24$
$ 24 \times 3 = 72cm^2$
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
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(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
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(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 a square and a triangle. Square QPSR has an area of 36cm$^2$. Find the area of triangle QRT.
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$\frac{1}{2} \times 6 \times 11 = 33$
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$\frac{1}{2} \times 6 \times 11 = 33$
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.
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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$^2$
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$6cm \times 6cm \times \frac{1}{4} \times 3.14 = 28.26cm^2$
$6cm \times 6cm = 36 cm^2$
(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$^2$
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$6cm \times 6cm \times \frac{1}{4} \times 3.14 = 28.26cm^2$
$6cm \times 6cm = 36 cm^2$
(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 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.
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8568cm$^2$ $\rightarrow$ Total area $\rightarrow$ $2772= 1386 + 882 + 3528 = 8568$
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8568cm$^2$ $\rightarrow$ Total area $\rightarrow$ $2772= 1386 + 882 + 3528 = 8568$
The figure below is made up of 2 identical quadrants and 2 identical semicircles. Find the area of the shaded part of the figure. $(Take \pi = \frac{22}{7})$.
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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.
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37.5cm$^2$ $\rightarrow \frac{2}{2} \times 5 \times 3 = 7.5$
$9 – 3 = 6$
$ 5 \times 6 = 30$
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37.5cm$^2$ $\rightarrow \frac{2}{2} \times 5 \times 3 = 7.5$
$9 – 3 = 6$
$ 5 \times 6 = 30$
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.
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$\sqrt(64)= 8$
$0.25 3.14\times 8\times 8 = 50.24$
$8 \times 8 = 64$
$64 – 50.24 = 13.76$
$13.76 \times 2 = 27.52$
The area is 27.52cm$^2$
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$\sqrt(64)= 8$
$0.25 3.14\times 8\times 8 = 50.24$
$8 \times 8 = 64$
$64 – 50.24 = 13.76$
$13.76 \times 2 = 27.52$
The area is 27.52cm$^2$
The figure below 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.
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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 $\pi$.
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Area of fig $\rightarrow$ $\frac{1}{2} \times 18 \times 18 \times \pi = 162\pi$
$\frac{1}{2} \times 36 \times 18 = 324$
$(162\pi - 324) ÷ 2 = 81\pi - 162$
$(20.25\pi – 40.5) \times 2 = 40.5\pi -81$
$(81\pi - 162) – (40.5\pi - 81) = (81\pi – 162)cm^2$
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Area of fig $\rightarrow$ $\frac{1}{2} \times 18 \times 18 \times \pi = 162\pi$
$\frac{1}{2} \times 36 \times 18 = 324$
$(162\pi - 324) ÷ 2 = 81\pi - 162$
$(20.25\pi – 40.5) \times 2 = 40.5\pi -81$
$(81\pi - 162) – (40.5\pi - 81) = (81\pi – 162)cm^2$
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.
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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. $(Take \pi = 3.14)$
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43.74cm$^2$ $\rightarrow$ $12 \times 12 = 144$
$144 ÷ 2 = 72$
$3.14 \times (12 ÷ 2) \times (12 ÷ 2) = 113.04$
$113.04 ÷ 4 = 28.26$
$72 – 28.26 = 43.74$
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43.74cm$^2$ $\rightarrow$ $12 \times 12 = 144$
$144 ÷ 2 = 72$
$3.14 \times (12 ÷ 2) \times (12 ÷ 2) = 113.04$
$113.04 ÷ 4 = 28.26$
$72 – 28.26 = 43.74$
Study the diagram below. Find the area of the shaded triangle.
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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?
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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?
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ACE is a triangle. Which one of the following gives the area of ADE?
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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.
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$\sqrt(81)=9$
$14 ÷ 2 = 7$
$14 + 7 = 21$
$\frac{1}{2} \times 19\times 21 = 199.5 (EFG)$
$\frac{1}{2} \times 14 \times 9 = 63$
$133 – 63 = 70 (HDF)$
$70 + 199.5 = 269.5cm^2$
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$\sqrt(81)=9$
$14 ÷ 2 = 7$
$14 + 7 = 21$
$\frac{1}{2} \times 19\times 21 = 199.5 (EFG)$
$\frac{1}{2} \times 14 \times 9 = 63$
$133 – 63 = 70 (HDF)$
$70 + 199.5 = 269.5cm^2$
Find the area of the parallelogram PQRS in terms of k.
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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?
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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.
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