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Question 1 of 52
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, ABCD is a rectangle. $\angle$AFE is 43$^\circ$. Find $\angle$FCD.
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Mr Lim was asked to from a square with 56 pots of orchids placed at equal distance from each other. How many pots orchids should he place on any one side of the square such that each side contains an equal number of pots?
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The figure is made up of three overlapping squares. How many pairs of perpendicular lines are there in the figure?
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The perimeter of a rectangle is 30cm. What is the greatest possible area of the rectangle?
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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 shows three identical squares ABCD, BEFG and HJKB. What is the value of $\angle$GBK?
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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
ABCD is a square and AC is the diagonal of the square. DF is a straight line and $\angle$BFE is 123$^\circ$. Find $\angle$CED.
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The figure below shows two identical rectangles within a semicircle of diameter 10cm. R is the center of the semicircle. Give that Rectangle PQRS and Rectangle XYZR has a perimeter of 14cm each and line XS is 1cm. Find the perimeter of the shaded part.
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$14 – 1 – 1 = 12$
$12 ÷ 4 = 3$
$3 + 1 = 4$
$10 – 4 -3 = 3$
$\frac{1}{2} \times \pi \times 10 = 15.71$
$5 \times 2 = 10$
$15.71 + 10 + 3 + 1 = 29.71cm$
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$14 – 1 – 1 = 12$
$12 ÷ 4 = 3$
$3 + 1 = 4$
$10 – 4 -3 = 3$
$\frac{1}{2} \times \pi \times 10 = 15.71$
$5 \times 2 = 10$
$15.71 + 10 + 3 + 1 = 29.71cm$
The figure below is made up of 2 identical rectangles. The shaded part is a rectangle measuring 8cm by 3cm. What fraction of the figure is shaded?
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In the diagram below , ABCD is a parallelogram and CDEF is a rectangle. Find $\angle$BCF.
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The figure below is formed by a rectangle and an isosceles triangle . Which two lines are perpendicular to line BD?
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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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A rectangular block of wood is 20cm by 15cm by 16cm. What is the greatest number of 2cm cube that can be cut from it?
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In the figure below, not drawn to scale, ABCD, HKJC and BGFE are squares, $\angle$ BKJ = 50 $^\circ$. and $\angle$ CBE = 70 $^\circ$.Find$\angle$ AHC.
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$\angle$ HKB $\rightarrow$ $90^\circ - 50^\circ = 40^\circ$
$\angle$KBH $\rightarrow$ $360^\circ - 180^\circ - 70^\circ = 110^\circ$
$\angle$BHK $\rightarrow$ $180^\circ - 110^\circ - 40^\circ = 30^\circ$
$\angle$BHC $\rightarrow$ $90^\circ - 30^\circ = 60^\circ$
$\angle$AHC $\rightarrow$ $180^\circ - 60^\circ = 120^\circ$
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$\angle$ HKB $\rightarrow$ $90^\circ - 50^\circ = 40^\circ$
$\angle$KBH $\rightarrow$ $360^\circ - 180^\circ - 70^\circ = 110^\circ$
$\angle$BHK $\rightarrow$ $180^\circ - 110^\circ - 40^\circ = 30^\circ$
$\angle$BHC $\rightarrow$ $90^\circ - 30^\circ = 60^\circ$
$\angle$AHC $\rightarrow$ $180^\circ - 60^\circ = 120^\circ$
The figure is made up of 2 squares of different sizes. The length of square B is $\frac{3}{5}$ the length of square. What fraction of square A is shaded?
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In the figure below, the area of the shaded parts is 258cm$^2$. Find the perimeter of the rectangle ABCD.
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A rectangle has an area of 56cm$^2$. Given that its breadth is 7cm, Find the perimeter of the rectangle.
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The figure below shows a transparent rectangular container partially filled with unit cubes. How many cubes can the container hold altogether?
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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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A rectangular tank contains 1260ml of water when it is $\frac{2}{3}$ full. Find the base area of the tank if the height is 7cm.
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The figure below is made up of a rectangle, a square and a triangle. Given that the length of the rectangle is twice its bready, find the perimeter of the whole of the figure.
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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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In the figure, ABCF is a rectangle. Find the area of the shaded parts.
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The figure below is made up of a quadrant and a rectangle, OPQR. Find the perimeter of rectangle OPQR.
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34cm $\rightarrow 8 + 5 – 1 = 12$
$12 \times 2 = 24$
$ 5 \times 2 = 10$
$24 + 10 = 34$
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34cm $\rightarrow 8 + 5 – 1 = 12$
$12 \times 2 = 24$
$ 5 \times 2 = 10$
$24 + 10 = 34$
The picture below is made up of 2 similar squares and 2 similar quadrants. The area of one square is 64cm$^2$.
(a) Find the area of the shaded region.
(b) Find the perimeter of the unshaded region. $(Take\pi = 3.14)$.
(b) 70.12cm
(b) 75.12cm
(b) 80.12cm
(b) 85.12cm
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(a) $\sqrt(64)= 8$
$0.25 3.14\times 8\times 8 = 50.24$
$8 \times 8 = 64$
$64 – 50.24 = 13.76$
$13.76 \times 2 = 27.52$
The area is 27.52cm$^2$
(b) $8 \times 2 = 16 (diameter)$
$3.14 \times 16 \times 0.5 = 25.12$
$25.12 + 8 + 8 + 8 + 8 = 57.12$
It is 75.12cm
(b) 75.12cm
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(a) $\sqrt(64)= 8$
$0.25 3.14\times 8\times 8 = 50.24$
$8 \times 8 = 64$
$64 – 50.24 = 13.76$
$13.76 \times 2 = 27.52$
The area is 27.52cm$^2$
(b) $8 \times 2 = 16 (diameter)$
$3.14 \times 16 \times 0.5 = 25.12$
$25.12 + 8 + 8 + 8 + 8 = 57.12$
It is 75.12cm
ABCD is a square. The length of WX is$\frac{1}{4}$ of the length of DC and WX = YZ. What faction of the area of square is unshaded?
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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
Javier foided a 12.cm square paper twice and a 2cm square from each corner. The figure below shown how the paper looked like when he unfolded it. What is the area of the left over piece-cm square from each corner. The figure below shown how the paper looked like when he unfolded it. What is the area of the left over piece of paper.
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$ 12 \times 12 = 144$
$ 2 \times 2 = 4$
$ 4 \times 16 = 64$
$ 144 – 64 = 80$
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$ 12 \times 12 = 144$
$ 2 \times 2 = 4$
$ 4 \times 16 = 64$
$ 144 – 64 = 80$
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 not drawn to scale is made up of similar quadrants of radius 7cm. The length of the shaded square is 3cm . What is the perimeter of the whole figure? $(Take \pi = \frac{22}{7})$
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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 from by 5 identical squares with 3 similar semi–circles cut out from it. Each squares has a side of 14cm. Find the perimeter of the shaded figure. Leave your answer in terms of $\pi$.
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$\frac{3}{2} \times\pi \times14 = 21\pi$
21 $\pi$ + $(14 \times 7) = 21\pi + 98$ cm
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$\frac{3}{2} \times\pi \times14 = 21\pi$
21 $\pi$ + $(14 \times 7) = 21\pi + 98$ cm
A rectangular piece of paper is folded along the dotted line AC as shown below. Find
(a) $\angle$ DAE
(b) $\angle$ ACE
(b) 61$^\circ$
(b) 62$^\circ$
(b) 63$^\circ$
(b) 64$^\circ$
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(a) $\angle$ $DAE = 90^\circ - 54^\circ = 36^\circ$
(b) 63$^\circ$ $\rightarrow$ $\angle$ $ACE = 180^\circ – (27^\circ+90^\circ) = 63^\circ$
(b) 63$^\circ$
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(a) $\angle$ $DAE = 90^\circ - 54^\circ = 36^\circ$
(b) 63$^\circ$ $\rightarrow$ $\angle$ $ACE = 180^\circ – (27^\circ+90^\circ) = 63^\circ$
In the figure below, a rectangular piece of paper is folded at a corner as shown Find $\angle$m.
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A rectangle piece of paper with breadth 6cm was rolled up. Without overlapping to form a hollow cylindrical tube as shown below. Find the perimeter of the rectangular piece of paper if the circumference of the tube is 22cm.
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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 is formed by a square AOBC and circle with center O. AOB is part of the circle. The length of AO is 7m. Find the perimeter of the figure. $(Take \pi = \frac{22}{7})$
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A rectangular place of paper of perimeter 32cm was folded once at the two ends to form the shape as shown below. Find the value of w.
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In the figure, the area of rectangle ABCD is 72cm$^2$. The ratio of DE: EC is 1:2 and BF : FC is 2 : 1. Find the area of the shaded region.
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In the figure below, a rectangular piece of paper was folded as shown. Find $\angle$ DEB.
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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 figures is made up of the identical big quadrants and two identical small quadrants. The ratio of the radius of the small quadrant to the big quadrant is 1:4. The radius of the qurdant is 28cm. Find the perimeter of the figure. $ (Take \pi = \frac{22}{7})$
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194cm $\rightarrow$ $2 \times small quardransts \rightarrow \frac{1} {2} \times (7 + 7) \times \frac{22}{7} = 22$, $2 \times big quardrants \rightarrow \frac{7}{2} \times (28 + 28) \times \frac{22}{7} = 88$, $28 – 7 = 21$, $ 4 sides \rightarrow 21 \times 4 = 84$, $Total \rightarrow 22 + 88 + 84 = 194$
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194cm $\rightarrow$ $2 \times small quardransts \rightarrow \frac{1} {2} \times (7 + 7) \times \frac{22}{7} = 22$, $2 \times big quardrants \rightarrow \frac{7}{2} \times (28 + 28) \times \frac{22}{7} = 88$, $28 – 7 = 21$, $ 4 sides \rightarrow 21 \times 4 = 84$, $Total \rightarrow 22 + 88 + 84 = 194$
The figure below is made up of identical squares. If the total area of the figure is 96cm$^2$. What is the perimeter of the figure?
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48m $\rightarrow$ $96 ÷ 6 = 16$
$\sqrt(16)=4$
$4 \times 12 = 48$
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48m $\rightarrow$ $96 ÷ 6 = 16$
$\sqrt(16)=4$
$4 \times 12 = 48$
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 base area of a rectangular tank is 20m $^2$. The tank contains 120m $^3$ of water when it is $\frac{3}{4 }$ filled. What is the height of the tank?
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8m $\rightarrow$ $120 ÷ 3 = 40$
$40 \times 40 = 160$
$100 ÷ 20 = 5$
$160 ÷ 2 = 10$
$80 ÷ 10 = 8$
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8m $\rightarrow$ $120 ÷ 3 = 40$
$40 \times 40 = 160$
$100 ÷ 20 = 5$
$160 ÷ 2 = 10$
$80 ÷ 10 = 8$
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$^3$ $\rightarrow$ $3.14 \times 30 = 94.2$
$94.2 ÷ 4 = 23.55$
$23.55 \times 23.50 = 554.60$
ABCD is a square. What is the ratio of the shaded area to the area of square ABCD?
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25:72 $\rightarrow$ $\frac{4}{16}$ + $\frac{8}{18}$ = $\frac{14}{4}$
$\frac{14}{4}$ + $\frac{1}{9}$ + $\frac{16}{36}$ = $\frac{25}{36}$
You are Right
25:72 $\rightarrow$ $\frac{4}{16}$ + $\frac{8}{18}$ = $\frac{14}{4}$
$\frac{14}{4}$ + $\frac{1}{9}$ + $\frac{16}{36}$ = $\frac{25}{36}$
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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The figure below is made up of 4 identical quadrants and a square. Find
(a) The perimeter of the figure.
(b) The area of the figure. $(Take \pi = \frac{22}{7})$
(b) 865cm$^2$
(b) 872cm$^2$
(b) 878cm$^2$
(b) 885cm$^2$
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(a) 208cm $\rightarrow$ Perimeter $\rightarrow$ $7 \times$ 4 = 28
$(14 \times 2) + 16 + 16 + 2 = 92$
$\frac{22}{7} \times 28 = 88$
Total $\rightarrow$ $28 + 92 + 88 = 208$
(b) 872cm$^2$ $\rightarrow$ CIRCLE $\rightarrow$ $\frac{22}{7} \times 14 \times 14 = 616$
SQUARE $\rightarrow$ $16 \times 16 = 256$
Total $\rightarrow$ $616 + 256 = 872$
(b) 872cm$^2$
You are Right
(a) 208cm $\rightarrow$ Perimeter $\rightarrow$ $7 \times$ 4 = 28
$(14 \times 2) + 16 + 16 + 2 = 92$
$\frac{22}{7} \times 28 = 88$
Total $\rightarrow$ $28 + 92 + 88 = 208$
(b) 872cm$^2$ $\rightarrow$ CIRCLE $\rightarrow$ $\frac{22}{7} \times 14 \times 14 = 616$
SQUARE $\rightarrow$ $16 \times 16 = 256$
Total $\rightarrow$ $616 + 256 = 872$
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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