Area P6 2011 PSLE Q

Area P6 2011 Nanyang P2 Q11

Area P6 2011 Nanyang P2 Q8

Area P6 Ang Mo Kio 2011 SA2 P2 Q44

Area P6 Rosyth 2011 SA2 P1 Q11

Perimeter PSLE 2010 P1 Q28

The shaded figure below is formed using 3 squares and 3 equilateral triangles. The length of the straight line AB is 15 cm. Find the perimeter of the shaded figure.

15 x 3  =  45        Squares


15 x 2  =  30          Triangles


 45 + 30  =  75 cm


Alternative method
----------------------

15  x  5  =  75  cm

Area & Perimeter PSLE 2010 P2 Q4

In the figure below, AB=7 cm, BC=9 cm, CD=3 cm and DA=11 cm. Angle ABC and Angle CDA are right angles.

Find the area of the figure ABCD.


1/2 x  9 x 7  =  31.5


1/2 x 11 x 3  =  16.5


31.5  +  16.5  =  48  cm2

It is  48  cm2.

Area & Perimeter PSLE 2010 P2 Q10

10. ADEF is a rectangular cardboard with AF = 7 cm. Two quarter circles have been cut from it as shown below. The remaining cardboard, which is the shaded part, has an area of 56 cm².
Using π=22/7, find the length of BC.

1/2  x  22/7  x  7  x  7  =  77

77  +  56  =  133

133/7  =  19

19  -  7  -  7 =  5 cm

It is 5 cm.

Area & Perimeter P6 2010 SA2 NY P2 Q5

Area P6 2010 SA2 Maha Bodhi P2 Q7

10 / 2 = 5    Right-angled isosceles triangle(s)

1/2 x 10 x 5 = 25

25 + 133 = 158 cm2

It is 158 cm2.

Area P6 2010 SA2 Nanyang P1 Q30

The figure below is formed by stacking 4 pieces of square paper one on top of another. The papers have different prints and sizes of sides 3cm, 4cm, 5cm and 6cm. The 6-cm piece is placed at the bottom of the stack, followed by the 5-cm piece, then the 4-cm piece and the 3-cm piece is placed right on top.

Find the sum of the area A and B. (ans : 8 cm2)

6 - 1 = 5

5 - 3 = 2

6 - 4 = 2

2 x 2 = 4

4 x 2 = 8 cm2

Area P6 2010 SA2 Rosyth P2 Q15

Given ABCD is a square so Angle DAC = BAC = 45 degree. Thus AGFH is also a square and hence GF = FH.

Area of triangle ABF = Area of triangle AFI = Area of triangle BFI (since base and height are the same)

1/2 x 6 x 12 = 36

36/3 = 12 cm2

1/2 x 12 x 12 = 72 cm2

72 - 12 = 60 cm2

It is 60 cm2.

Area P6 2010 SA2 Pei Chun P2 Q14

The figure below shows three square plots of land E, F and G. The total area of the land is 246 square meters. The length of each square is a whole number. The area of E is 1/4 the area of F. (Sorry can't get the figure uploaded. It comprises of 3 adjoining squares, E,F & G. E being the smallest & G being the biggest).

(a) It costs $29 to cover 1 m2 of land with grass. How much does it cost to cover F and G with grass altogether?

(b) It costs $41 to build 1m of wooden fencing. How much does it cost to build wooden fencing along the perimeter of the figure?


L --> 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11

L x L --> 1 ,  4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100 , 121 , ....

25 + 100 + 121 = 246, so L = 5, 10, 11 for E, F and G respectively

100 + 121 = 221

221 x 29 = $6409

a) It was $6409.

5 + 10 + 11 + 11 = 37

37 x 2 = 74

74 x 41 = $3034

b) It was $3034.

Fraction P6 2009 SA1 P2 RGS Q13

13a. The figure below is made up of unit squares. What fraction of the figure is shaded? [1]

1/2 x 3 x 4 = 6

½ x 2 x 1 = 1

1 + 6 = 7

5 x 4 = 20

7/20

13b. The figure below is not drawn to scale. Given that the area of WXYZ is 40 cm², find the area of the shaded part. [3]

The two shaded triangles may be adjusted to form a big triangle (base 12 cm and height 18 cm) such that they do not overlap.

1/2  x 12 x 18 = 108

108 – 40 = 68 cm²

The area of the shaded part is 68 cm².

Area P6 2009 SA1 P2 RGS Q14

14. The figure below is made up of two identical right-angled triangles, a small square and a big square. The perimeter of the shaded region is 50 cm, and the total area of the two unshaded squares is 254.5 cm. Find the total area of the two shaded right-angled triangles. [5]

Firstly, we may adjust the figure and add a red rectangle to form another square as follows:

2 L + 2 B

50 – 3 – 3 = 44

L + B

44 ÷ 2 = 22

Area of biggest square (L + B) x (L + B)

22 x 22 = 484

484 – 254.5 = 229.5

229.5 ÷ 2 = 114.75 cm²

The total area of the two shaded right-angled triangles is 114.75 cm².

Area P6 2009 SA2 P2 MGS Q5

5. The diagram below, not drawn to scale, shows a shaded triangle drawn within a rectangle. The ratio of the length AB to the length of the rectangle is 1 : 3. Calculate the area of the shaded triangle.

36 ÷ 3 = 12

1/2 x 12 x 28 = 168 cm²

Area P6 2009 SA2 P2 Rulang Q2

2. In the diagram below, ABCD is a rectangle. DE, CE, DF, FC and GH are straight lines. The ratio of GO to OH is 1 : 3. Find the ratio of the area of ABCD to the area of DFC to the area of DOC.

Area of DFC = 1/2 Area of ABCD

Area of DOC = 1/2 x 3/4 Area of ABCD

                   = 3/8 Area of ABCD

1 : 1/2 : 3/8 = 8 : 4 : 3

Area P6 2009 SA2 P2 Rulang Q13

13. In Figure ABCD, there are 6 identical rectangles. Study the figure carefully and find the total shaded area.

Area of ABCD.

30 x 18 = 540

Find the length and breadth of each small rectangle.

18 ÷ 3 = 6

6 x 2 = 12

Find the area of each rectangle and the 6 rectangles.

12 x 6 = 72

72 x 6 = 432

540 – 432 = 108 cm².

The total shaded area is 108 cm².