RGS Primary 2009 PSLE Math Prelim Paper 2 Q13

The figure below is made up of a big circle, square and a small circle. The area of the square is 400 square cm. Find the area of the shaded region. (Correct your answer to 2 decimal places)

Solution

Area of square --> 400 square cm
Length of square --> 20 cm (square root of 400 sq cm)

Area of 1/4 square

--> 400 square cm divided by 4

= 100 square cm

Area of one right-angle triangle -->(1/2)(b)(h)

100 square cm = (1/2)(r)(r)

(r)(r) = (100 square cm) x 2 = 200 square cm

Area of circle = (3.14)(r)(r)

=(3.14)(200 square cm)

= 628 square cm

Shaded area -->(628 - 400) square cm

= 228 square cm
= 228.00 square cm (correct to 2 decimal places)

Answer: 228.00 square cm

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CHIJ St Nicholas Girls' Sch 2009 P6 SA1 Paper 2 Q7

The figure below is made up of a square and a rectangle. The perimeter of the figure is 178 cm and the area of the square, PQRS, is 784 square cm. What is the length of ST?

Solution

Square root of 784 square cm = 28 cm (length of 1 side of square)
2 lengths of rectangle --> (178 - 28 - 28 - 28 - 28) cm = 66 cm
1 length --> 66 cm divided by 2 = 33 cm
ST --> 33 cm + 28 cm = 61 cm

Answer: 61 cm

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Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q13

The length of a rectangle is thrice as long as its width. Its perimeter is 16p cm.

a) Find the length of the rectangle in terms of p.

b) Find the area of the rectangle if p = 4.

Solution

Perimeter

--> 3 units + 1 unit + 3 units + 1 unit

= 8 units

8 units --> 16p cm

1 unit --> 16p cm divided 8 = 2p cm

a)

Length

3 units --> 3 x 2p cm = 6p cm

Answer: 6p cm

Answer: 192 square cm

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Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q8

The rectangle ACEG is divided into 4 parts. BCEF is a square. Each part has a different area. Find the area X.

Area of X

--> 4 cm x 3 cm = 12 square cm

Answer: 12 square cm

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Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q6

Zen had a piece of rectangular paper. He cut away 1/9 of the breadth and 2/5 of the length. Then he measured the remaining piece of rectangular paper. He found that the breadth to be 16 cm and the length to be 15 cm. Find the area of the original piece of paper.

Solution

Breadth

8/9 --> 16 cm

1/9 --> 16 cm divided by 8 = 2 cm

9/9 --> 9 x 2 cm = 18 cm

Length

3/5 --> 15 cm

1/5 --> 15 cm divided by 3 = 5 cm

5/5 --> 5 x 5 cm = 25 cm

Original area --> 25 cm x 18 cm = 450 square cm

Answer: 450 square cm

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Ai Tong Sch 2009 P6 CA1 Math Paper 2 Q4

In a right-angled triangle, the two sides which form the right angle are 16 cm by 12 cm respectively. How many such triangles are needed to form the smallest square?

Horizontal --> 4 x 12 cm = 48 cm

Vertical --> 3 x 16 cm = 48 cm

The above figure (not drawn to scale) is therefore a square.

The smallest 1 unit rectangle has 2 triangles.

There are 3 rows and 4 columns of the smallest 1 unit rectangle.

3 x 4 = 12 smallest unit of rectangles

2 triangles x 12 = 24 triangles

Answer: 24 triangles

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Rosyth Sch 2007 PSLE Math Prelim Q47

The perimeter of a rectangle to that of a square is in the ratio of 11:6. If the square has an area of 144 square m and the length and breadth of the rectangle are in the ratio of 6:5, find the length and breadth of the rectangle.

Solution



Perimeter
(6 + 6 + 5 + 5) units ----- 88 m
22 units ----- 88 m
1 unit ----- 88m divided by 22 = 4 m

(Length) 6 units ----- 6 x 4 m = 24 m
(Breadth) 5 units ----- 5 x 4 m = 20 m

Answer: Length = 24 m; Breadth = 20 m

Rosyth Sch 2007 PSLE Math Prelim Q44

The figure is not drawn to scale. What fraction of the figure is the total area of the shaded regions A, B, C, D, E, F, G and H?


Solution

The base of the whole figure is (4 + 3 + 6 + 3 + 2) cm = 18 cm

1st row from the top
Fig A has a base of 6 cm.

2nd row
The sum of the bases of Figs B and C is 6 cm.

3rd row
The sum of the bases of Figs D and E is 6 cm.

4th row
The sum of the bases of Figs F and G is 6 cm.

5th row
Fig H has a base of 6 cm.

The average length of the bases of all shaded figures is 6 cm.

The length of the base of the whole figure is 18 cm.

Therefore, the fraction of whole figure that is shaded is
6 cm divided by 18 cm = 1/3

Answer: 1/3

Note that the above method can be used because the vertical columns also correspond to the 6 cm shaded out of the 18 cm total vertical length of the figure.

Raffles Girls Pri Sch 2007 PSLE Math Prelim Q45

The figure below shows a rectangle PQRS. The lines are extended from point P, Q, R and S and they meet at point Y. The length of QR is 20 cm and the length of XY is 4 cm. Given that QS is a straight line, the area of Triangle PQY is 72 square cm and the area of Triangle SRY is 84 square cm, find the shaded area of Triangle QSY.



Solution

Area of Triangle PQY + Area of Triangle SRY
72 sq cm + 84 sq cm = 156 square cm

Area covered by Triangles PQY and SRY is also 1/2 of area of Rectangle PQRS. This also means that Area of Triangle PQS is also 156 sq cm.

Area of Triangle PSY = 1/2 x 20 cm x 4 cm = 40 sq cm

Area of Triangle PQY = 72 sq cm

Shaded area
Area of PQS – Area of PQY – Area of PSY
=156 sq cm - 72 sq cm - 40 sq cm= 44 square cm

Answer: 44 square cm

Nanyang Pri Sch P6 CA1 2008 Math 46

The figure below is made up of a rectangle, square and a shaded triangle. Find the area of the shaded triangle.



Solution



Area of Triangle A ----- ½ x 1cm x 6 cm = 3 square cm
Area of Rectangle B ----- 4cm x 1cm = 4 square cm
Area of Triangle C ----- ½ x 7cm x 4cm = 14 square cm

Area of A + B + C + shaded triangle -----
½ x 8cm x 10cm = 40 square cm

Area of shaded triangle -----
(40 – 14 – 4 – 3) square cm = 19 square cm (Answer)

Tao Nan School P5 SA2 2007 Math Q40

WXYZ is a rectangle. Find the area of the shaded part.



Solution

Area of rectangle ---- 12 cm x 6 cm = 72 square cm

Area of unshaded triangle ----
½ x base x height
= ½ x 7 cm x 6 cm = 21 square cm

Area of shaded part -----
(72 – 21) square cm= 51 square cm (Answer)

Catholic High Sch 2006 PSLE Math Prelim Q41

In the figure below, not drawn to scale, the square, ABCD is made up of four rectangles. Given that the area of the square ABCD = 144 square cm, area of rectangle DFHG = 20 square cm and the area of rectangle AEHG = 28 square cm, find the area of rectangle EBIH.




Solution



Area of EBIH ----- 8 cm x 7 cm = 56 square cm (Answer)

Pei Chun Public Sch 2007 PSLE Math Prelim Q43

Taufik arranged a rectangle and a square and painted them in three colours as shown in the figure below. The ratio of the area of the rectangle to that of the square is 3:1. The ratio of the area of the red part to that of the blue part is 4:1. The length of the square is 9 cm.
a) What is the area of the purple part?
b) What is the ratio of the area of the purple part to that of the figure?






Area of Rectangle : Area of Square
3 : 1
9 units : 3 units

Red Area : Blue Area
4 : 1
8 units : 2 units

a)
(Square) 3 units ----- 81 square cm
(Purple) 1 unit ----- 81 square cm divided by 3 = 27 square cm (Answer)

b)
Purple ----- 1 unit
Whole figure 11 units

Area of purple part : Area of whole figure
1 : 11 (Answer)

Pei Chun Public Sch 2007 PSLE Math Prelim Q42

The figure below shows a park which is made up of a triangular fitness area, a rectangular pond and a field in the shape of a trapezium. The length of the pond is twice its breadth.



a) The cost of fencing material is $3 per meter. How much will it cost to fence up the pond?
b) What is the area of the park?

Solution

a)
Length of pond ----- 2 x 4 m = 8 m
Perimeter of pond ----- 2 x (8 + 4) m = 24 m

1 m ----- $3
24 m ----- $3 x 24 = $72

Answer: It will cost $72 to fence up the pond.

b)
Area of rectangle -----
21 m x 4 m = 84 square m

Area of triangle -----
½ x 14 m x 21 m = 147 square m

Total area of the park ----- (84 + 147) square m = 231 square meters (Answer)

Pei Chun Public Sch 2007 PSLE Math Prelim Q38

A rectangular piece of cardboard measures 17 cm by 12 cm. Sushila cuts the greatest number of rectangular pieces, each measuring 3 cm by 2 cm, from the cardboard. What is the total area of all the pieces cut?

Solution
The diagram below is not drawn to scale



34 pieces of small rectangles can be cut without any cardboard left.

Area used ----- 17 cm x 12 cm = 204 square cm (Answer)

Maha Bodhi Sch 2007 PSLE Math Prelim Q46

Each corner of the floor mat shown above is made up of a quadrant of radius 4 cm. Find the perimeter of the floor mat.

Solution

4 quadrants make 1 full circle
Perimeter of arcs of 4 quadrants above -----
2 x 3.14 x 4 cm = 25.12 square cm

Consider length of mat -----
40 cm – 4 cm – 4 cm = 32 cm

Consider breadth of mat -----
30 cm – 4 cm – 4 cm = 22 cm

Perimeter of mat -----
(25.12 + 32 cm + 32 cm + 22 cm + 22 cm)

= 133.12 cm (Answer)

Henry Park Pri Sch 2007 PSLE Math Prelim Q37

In the figure, ABCD is a square of side 8 cm. BE is 2 cm and FC is 3 cm. Find the area of Triangle AEF.


Solution

Area of Square ABCD ----- 8 cm x 8 cm = 64 square cm
Area of Triangle ABE ----- ½ x 8 cm x 2 cm = 8 square cm
Area of Triangle CEF ----- ½ x 3 cm x 6 cm= 9 square cm
Area of Triangle ADF ----- ½ x 5 cm x 8 cm = 20 square cm

Area of Triangle AEF -----
(64 – 8 – 9 – 20) square cm = 27 square cm (Answer)

S’pore Hokkien Huay Kuan 2007 PSLE Math Prelim Q37

The rectangle is divided into 4 parts. Each part has a different area. Find the area of X.



Area of X ----- 4 cm x 3 cm = 12 square cm (Answer)

Note – “The Singapore Hokkien Huay Kuan 5-School Combined Prelim Maths” is the common Maths Prelim Exam for Tao Nan, Ai Tong, Chongfu, Nan Chiau and Kong Hwa schools.

Ai Tong School P6 SA1 2006 Math (Q47)

The figure shown below is made up of a square, a rectangle and a right-angled triangle. The area of the square MNQR is 49 square cm and the area of the rectangle NOPS is 60 square cm. OP is 4 cm.
(a) Find the length of MR
(b) Find the length of SP
(c) Find the area of the triangle PSQ


Solution

(a)
Area of square MNQR = 49 square cm.
1 side is therefore 7 cm (7 x 7 =49)
Answer: 7 cm

(b)
Area of rectangle NOPS is 60 square cm.
60 square cm = SP x 4 cm
SP = 60 square cm divided by 4 cm = 15 cm
Answer: 15 cm

(c)


NQ = 7 cm, therefore
SQ = 7 cm – 4 cm = 3 cm

Area of triangle PSQ = ½ x base x height
= ½ x 15 cm x 3 cm
= 22.5 square cm

Answer: 22.5 square cm

Ai Tong School P5 SA1 2006 Math (Q44)

ABCD is a square. AD = 48 cm. Given that DN is twice as long as NC and BM = MC, find the area of the unshaded part.

Solution

Area of Triangle ADN ----- ½ x base x height
= ½ x 32 cm x 48 cm = 768 square cm

Area of OMCN ----- 24 cm x 16 cm = 384 square cm

Total unshaded area ----- (768 + 384) square cm = 1152 square cm (Answer)