ACS Primary 2009 PSLE Math Prelim Paper 2 Q10

Given that the length of DB is 14 cm and the shaded region is 115 square cm, find the lenght of AC, correct to 2 decimal places.

Solution

Area of Triangle = Half base x height.
Area of Triangle ABC --> (1/2)(AC)(DX + 14)
Area of Triangle ACD --> (1/2)(AC)(DX)

Area of shaded area
--> Area of Triangle ABC - Area of Triangle ACD = 115

(1/2)(AC)(DX + 14) - (1/2)(AC)(DX) = 115

(1/2)(AC)(DX) + (1/2)(AC)(14) - (1/2)(AC)(DX) = 115

(1/2)(AC)(14) = 115
7(AC) =115
AC = 115 divided by 7
= 16.428
~ 16.43 (nearest to 2 decimal places)

Answer: 16.43 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 Q43

The area of the figure ABCD is 48.5 square cm . Find the length of BE.
(Give your answer to the nearest tenth)


Solution

Area of ACD
1/2 x 18 cm x 10.4 cm = 93.6 square cm

Area of ABC
93.6 square cm - 48.5 square cm = 45.1 square cm

Area of Triangle ABC ----- 1/2 x base x height
45.1 square cm = 1/2 x 18 cm x height

height = (45.1 sq cm x 2 ) divided by18 cm
= 5.01 cm ~ 5.0 cm (to the nearest tenth)

Answer: 5.0 cm

Rosyth Sch 2007 PSLE Math Prelim Q39

Find the sum of the six marked angles in the diagram.

The sum of the all the angles in the triangle in the centre is 180 degrees. The sum of all the 3 angles outside the centre triangle, in the figure above is therefore also 180 degrees (opposite angles).

The sum of all the six marked angles is hence,

Sum of marked angles in 1 triangle ----- 180 degrees
Sum of marked angles in 3 triangles ----- 180 degrees x 3 = 540 degrees
Sum of 6 marked angles ----- (540 – 180) degrees = 360 degrees

Answer: 360 degrees

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

Calculating the Area of a Triangle

Area of a triangle = ½ x base x height

A very common mistake students make is extending the base outside the triangle.
Example -
Question – Find the area of the shaded triangle.


Common Mistake (Wrong solution)

A common mistake students make in the above question is:-

Area of triangle = ½ x base x height
= ½ x 6cm x 3cm = 9 square cm (wrong)

The base should be 4 cm and not 6 cm. The question asks you to calculate the area of the SHADED triangle. The base of the shaded triangle is 4 cm.

If you put the base as 6 cm, you will calculate the area of the WHOLE figure as shown below.



The correct answer is as follows:

Area of triangle = ½ x base x height
= ½ x 4cm x 3cm = 6 square cm (correct answer)

In order to avoid making the mistake above, always remember that the base of the triangle must never extend outside the triangle.

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)

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)

Henry Park Pri Sch 2007 PSLE Math Prelim Q41

In the following figures, the area of the biggest equilateral triangle is 64 square cm as shown in Figure 1. A new triangle is formed by connecting the midpoints of the sides of the previous triangle. If the pattern continues, find the area of the smallest triangle in Figure 4.



Solution

The shaded area of Figure 2 is ¼ of the Area of Figure 1 -----
¼ x 64 square cm = 16 square cm

The shaded area of Figure 3 is ¼ of the shaded area of Figure 2 -----
¼ x 16 square cm = 4 square cm

The shaded area of Figure 4 is ¼ of the shaded area of Figure 3 -----
¼ x 4 square cm = 1 square 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)

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)

Ai Tong School P5 SA1 2006 Math (Q39)

ABCD is a rectangle (not drawn to scale). AX = XD and AY = YB. Find the area of triangle CXY.


Solution

Area of rectangle ----- length x breadth
= 14 cm x 8 cm = 112 square cm

Area of Triangle CDX ----- ½ x base x height
= ½ x 14 cm x 4 cm = 28 square cm

Area of Triangle AXY ----- ½ x 7 cm x 4 cm = 14 square cm

Area of Triangle CBY ----- ½ x 7 cm x 8 cm = 28 square cm

Area of Triangle CXY -----
Area of rectangle – Area of Triangle CDX – Area of Triangle AXY – Area of Triangle BCY
= (112 – 28 – 14 – 28) square cm
= 42 square cm

Answer: Area of Triangle CXY is 42 square cm.

Math Question on Area

The diagram below is made up of 3 squares. Find the area of the shaded area.







Solution






Area of shaded area is

Area of triangle ADC – Area of triangle BGC + Area of triangle EFG

Area of triangle ADC
½ x 16cm x 14cm = 112 square cm

Area of triangle BGC
CG = 6 cm because EG = EF = 2 cm
Therefore area = ½ x 6cm x 6cm = 18 square cm

Area of triangle EFG
½ x 2cm x 2cm = 2 square cm

Therefore area of shaded area is
(112 – 18 + 2) square cm
= 96 square cm (Answer)