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Section A: For each question, four options are given. (10 marks)
(1) What is the area of the shaded triangle shown below?
(2) The squares below are identical. Which one the of shaded area is the biggest?
(3) The area of a rectangle is 180cm². The ratio of its length to its breadth is 5:4.
Find the length of the rectangle.
(A) 20cm (B) 45cm (C) 15cm (D) 80cm
(4) The ratio of the area a square to the area of a rectangle is 2:3. The side of the
square is 6cm and the length of the rectangle is 9cm. What is the perimeter of
the rectangle?
(A) 15cm (B) 24cm (C) 30cm (D) 54cm (C)
(5) The figure below is made up of two triangles. Find the difference between the
area of the shaded triangle and the area of the unshaded triangle.
(A) 55 cm²
(B) 20cm²
(C) 75cm²
(D) 85cm²
(A) 712⁄ cm²
(B) 12cm²
(C) 40cm²
(D) 11cm²
Area of Triangle = ½ x 11 x 10 = 55 cm ² (A)
Area of shaded triangle = ½ x 13 x 4 = 26 Area of unshaded triangle = ½ x 4 x 7 = 14 Difference = 26-14 = 12cm² (B)
(D) The other areas
are the same.
(3) Factors of 180 1x 180 , 2x 90, 3 x 60 4x 45, 6x30, 9x20, 10x18, 12 x 15 15:12 5:4 , Hence the length is 15cm (C)
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(6) In the figure below, point A is midpoint of the line WZ of rectangle WXYZ. Line
AB is 4 5⁄ of WX. Find the shaded area
(7) A rectangular piece of paper of perimeter 42cm was fold once twice at both ends
to form the shape in the figure below. Find the value of x
(8) WXYZ is a square. The length of AB is 1 4⁄ of YZ. AB=CD. What fraction of the
square is unshaded?
(9) The figure below is made up of 3 overlapping triangles. The area of the figure is
240cm² and the area of the each big triangle is 90cm². Find the area of the
shaded triangle
(A) 96cm²
(B) 48cm²
(C) 60cm ²
(D) 72cm²
(A) 10
(B) 8
(C) 6
(D) 4
(A) 1 2⁄
(B) 1 4⁄
(C) 3 4⁄
(D) 1 6⁄
(A) 10cm²
(B) 15cm²
(C) 30cm²
(D) 45cm²
Length of WX = Length of XY = 4 Length of AB = 1u Area of Triangle AB = ½ x 1 x 4 = 2u Area of 2 shaded triangle = 2 x 2 = 4u Area of square = 4 x 4 = 16u Area of unshaded part = 16-4 = 12u 12
16⁄ = 3 4⁄ (C )
Let the area of the shaded part be x Area of figure = 90 + 2(90-x) 240 = 270 -2x 2x= 270-240 = 3- X = 15cm² (B)
4cm
m 4
c
m
m
Perimeter of Rectangle = (4x2) + (6x2) + (7x2)+2x 8+12+14 +2x = 42 2x= 42-34 = 8 X= 4 (D)
Area of AXY = ½x15x8=60
Area of XBY=½x8x (1 5⁄ x 15)=12
Area of shaded area = 60-12=48 (B)
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(10) WXYZ is a square. The shaded triangles in the squares are isosceles triangles of
the same size. What percentage of the squares is unshaded?
Section B
Questions 11 to 15 carry 1 mark each. Questions 16 -20 carry 2 marks each. Show your workings clearly in the space provided for each question and write your answers in the space provided. For questions which require units, give your answers in the units stated. (15 marks)
(11) A square has an area of 81cm². What is its perimeter?
Ans: _____________________
(12) In the figure below, four identical squares have been place side by side to form
a rectangle of perimeter 80cm. Find the area of each square.
Ans: _____________________
(`13) Each side of a 6cm square is increased by 50%. Find the ratio of the original
area to the new area of the square.
Ans: _____________________
(14) A rectangle is divided into three area, A,B and C. The ratio of area B to area C is
4:7. Find the area of A to Area of rectangle.
Ans: _____________________
(A) 16 2 3⁄ %
(B) 3313⁄ %
(C) 66 1 3⁄ %
(D) 83 1 3⁄ %
Area of A = Area of B + C = 4 + 7 = 11u
Area of rectangle = Area of A + B + C = 11x2 = 22u
Area of A : Area of Rectangle = 11: 22 = 1: 2
Length of square = 81 = 9cm Perimeter = 9 x 4 = 36cm
Length of square = 80 10 = 8cm Area of 1 square = 8 x8 = 64cm ²
New length of square = 6 x1.5 = 9cm Area of square (original) = 6 x6 = 36cm² Area of square (new) = 9 x 9 = 81cm ² Original Area: New Area 36: 81 = 12: 27 = 4: 9
WXYZ can be cut equally into 9 squares = 18 triangles Number of unshaded triangle = 15 15
18 ⁄ x 100 = 8313⁄ % (D)
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(15) In the figure below, what is the perimeter of the figure?
Ans: _____________________
(16) The figure below is made up of two identical equilateral triangles. Find the
perimeter of the figure.
Ans: _____________________
(17) The figure below shows 6 identical squares and a shaded triangle. Find the area
of the shaded triangle.
Ans: _____________________ (18) Tammy has a rectangular plot of land that measures 22m by 15m. She uses 35
identical square tiles to cover the land leaving an area of 15m² uncovered. Find
the length of one tile.
Ans: _____________________
10.4 x 4 =41.6 5.6 x 2 = 11.2 41.6 + 11.2 = 52.8cm (Perimeter)
Base of Triangle = 11x2 = 22 Height of triangle = 22-7 = 15 Area of triangle = ½ x 22 x 15 = 165cm ²
Perimeter = 27 + 12+ 13+5 +
18+27+12= 114cm
Area of land = 22x15 = 330 Area of land covered by tiles = 330-15 = 315
Area of 1 tile = 31535= 9 Length of 1 tile = 3cm
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(19) The length of a rectangle is 8 times its breadth. Given the area of the rectangle
is 1352cm², find the perimeter of the rectangle.
Ans: _____________________ (20) The total surface area of a cube is 486cm². What is the length of the cube?
Ans: _____________________
Area of 1 face of a cube = 4866 = 81
Length of the cube = 81 = 9cm
Factors of 1352 Perimeter =(104 x 2) + (13x2) = 234cm 1x 1352 13 x 104
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Section C
Show your working clearly in the space provided for each question and write the answer in the space provided. The number of marks available is shown in the brackets ( ) at the end of each questions or part questions. (37marks)
(21) Figure A is a trapezium. Figure B is a rectangle. Figure A and B have the same area.
Find the perimeter of figure B. (3m)
Ans: ____________________
(22) The rectangle below is divided into four parts; A, B, C and D. Part D is 3 5⁄ of
Part C. The ratio of Part D to Part A is 2:1.
(a) Find the ratio of Part C to Part D to Part A. (1m)
(b) Given that the area of Part B is 39cm², find the area of the rectangle. (2m)
Ans(a) ___________________________
Ans(b) ____________________________
(23) The figure below is made up of a rectangle and a square. What is the ratio of the
area of unshaded part to area of shaded part? (3m)
Area of A + B = Area of D + C Area D :Area C =( 3:5) x 2 = 6:10 Area of D : Area of A
(2 : 1) x 3 = 6:3 (a) Area of C : Area of D : Area of A
10: 6:3 (b) Area of D + C = 10+6 = 16u
Area of rectangle = 16 x 2 = 32u Area of B = 16-3 = 13u 13u = 39 1u = 3 32u=96cm²
Area of A = (½x 4x7) + (17x7) = 133 Area of B = 133
Breadth of B = 13319=7 Perimeter of B = (19x2) +(7x2) = 52cm
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Ans: ____________________
(24) Shawn had 1.89m of wire. He cut some of the wire and folded it into 5 squares
of different sizes as shown in the figure 1 below. The length of AB is 27cm long.
With the remaining wire, Shawn bent it into an equilateral triangle as shown in
figure 2. Find the length of the sides of the triangle. Give your answer in metres.
(3m)
Ans: ____________________
Perimeter of Figure 1 = 0.27 x 4 =1.08 Perimeter of Figure 2 = 1.89 -1.08 =0.81
Length of triangle = 0.813 =0.27m
Area of rectangle = 30 x 12 = 360 Area of square = 16x16=256 Area of big triangle = ½ x (30+16) x 16 = 368 (unshaded part) Area of Figure = 360+256 = 616 Area of shaded part = 616=368=248 Area of unshaded part: Area of shaded part 368: 248 92:62 46:31
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(25) The figure below shows a rectangle WXYZ. K is the mid-point of WX. J is the mid-
point of WK and O is the mid-point of WZ. Find the area of the shaded part. (3m)
Ans: ____________________
(26) A rectangle piece of paper (not drawn to scale) is folded along the dotted lines
to form the figure below. JRST is a square, JMR is a triangle and MNOR is a
rectangle. MR = 15cm and R is the mid-point of MS. The ratio of the shaded
triangle JMR to the area of the shaded rectangle NOPQ to the rectangular piece
of paper is 3:4:24. Find the area of the rectangle MNQR. (4m)
Ans: _____________________
A
B C
Area of rectangle WXYZ = 24 x 74 = 1776
Height of A = 242= 12 Area of A = ½x 12x74 = 444
Length of WK = 742= 37 Area of B = ½x 37 x 12=222 Area of C = ½x 37x24=444 Area of unshaded part= 444x2 + 222=1110 Area of shaded part = 1776-1110= 666cm²
Area of Square JRST = 15x15= 225 Area of Shaded triangle JMR = ½ x 15x15= 112.5 JMR: Area if the shaded rectangle NOPQ: rectangular piece of paper 3: 4: 24 3u= 112.5 1u= 37.5 4u= 150 (NOPQ) 24= 900 (Area of paper) Area of MNQR = 900 – 225-(150x2) – (112.5x2) = 150cm²
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(27) The figure below is made up of 4 identical rectangles and a shaded square. The
perimeter of each rectangle is 36cm. The area of the shaded square is 136cm²
more than the total area of the 4 rectangles. Find the length of the shaded
square. (Give your answer correct to 2 decimals) (4m)
Ans: _____________________
(28) The figure below shows two overlapping rectangles X and Y. Area of rectangle
X is 2013⁄ of rectangle Y. 25% of rectangle X is shaded. The total unshaded area
of X and Y is 391cm². Find the area of the shaded part. (4m)
Ans: _____________________
1Length + 1 Breadth of the rectangle = 36 2 = 18cm Length of the big square = 18cm Area of the big square = 18x18 = 324cm² Area of 4 rectangle + area of shaded square = 324cm² Area of square = Area of four rectangle + 136cm² Hence Area of 4 rectangles + Area of 4 rectangles + 136 = 324 (Area of shaded square) Area of 8 rectangle = 324-136 =188
Area of 1 rectangle = 188 8 = 23.5cm² Area of 4 rectangles = 23.5 x 4=94cm² Area of shaded square = 94 + 136 =230
Length of square = 230 = 15.165= 15.17cm
25% of X = 20u x 0.25= 5u
Total area of X + Y = 20+13=33u
Area of unshaded part = 33u-5u-5u=23u
23u = 391
1u= 17
5u= 85cm²
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(29) In the figure below, ABCD is a square of sides 36cm. F is the midpoint of AD.
AB=BC. AF is 4 times of GF. The ratio of EH to HF is 3:8. Find the total shaded
area (5m)
Ans: _____________________
Area of square ACDE= 36x36=1296cm²
Area of triangle ABD= ½of triangle ACD= 1 4⁄ of ACDE
Area of triangle EGF= 1 4⁄ of triangle EAG
Area of triangle EGF = 1 4⁄ of 1 4⁄ of ACDE = 1 16⁄ of ACDE
Area of triangle EHD = 3 8⁄ of triangle EFD
= 3 8⁄ of 1 4⁄ of ACDE = 3 32⁄ of ACDE
Area of shaded area = 1 4⁄ + 1 16⁄ + 3 32⁄ = 1332⁄
Area of shaded part = 1332⁄ x 1296 =526.5cm²
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(30) The figure below shows 2 overlapping semicircles and a triangle. Given that WX =
12 cm, WY = 20 cm, and XY = 16 cm, letting π = 3.14, find
(a) The perimeter of the shaded region. (2m)
(b) The area of the shaded region (3m)
Ans(a) _________________ (2m)
Ans: (b) _________________ (3m)
a) 0.5 x 3.14 x 12 = 18.84
0.5 x 3.14 x 16 = 25.12
18.84 + 25.12 + 20 = 63.96 cm
b) Small semicircle – A + B + C
Triangle – B + C + D
Big Semicircle – C + D + E
Shaded Region = Small semicircle – triangle + big semicircle
0.5 x 3.14 x 6 x 6 – 0.5 x 12 x 16 + 0.5 x 3.14 x 8 x 8
= 61 cm2
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Answer:
Section A
(1) A (2) B (3) C (4) C (5) B
(6) B (7)D (8) C (9) B (10) D
Section B
(11) 36 (12) 64 (13) 4:9 (14) 1:2` (15) 114
(16) 52.8 (17) 165 (18) 3 (19) 234 (20) 9
Section C
(21) 52cm (22)(a) 10:6:3 (b) 96
(23)46:31 (24) 0.27m (25) 666
(26) 150 (27) 15.17 (28) 85
(29) 526.5 (30a) 63.96 (30b) 61
