Hokkien Huay Kuan Combined Primary 2010 PSLE Math Prelim Paper 2 Q13

The figure below shows 2 quarter circles and a rectangle. The radius of the big quarter circle is 8 cm. The radius of the small quarter circle is 4 cm. Find the difference in area between the two shaded parts of X and Y. Use the calculator value of pi and give your answer correct to 1 decimal place.

Solution

Area of rectangle ----- 8 cm x 4 cm = 32 square cm

Area of large quadrant -----
(1/4) x pi x 8cm x 8 cm
= 16(pi) square cm

Area of small quadrant ------
(1/4) x (pi) x 4cm x 4cm
= 4(pi) square cm

Area of Large quadrant - Area of small quadrant -----
16(pi) square cm - 4(pi) square cm
= 12(pi) square cm

Difference between the two shaded parts X and Y -----
(12 x pi) square cm - 32 square cm
approximately ----- 5.7 square cm

Answer: 5.7 square cm

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ACS Primary 2010 SA1 Math Paper 2 Q17

The shaded figure below is formed by semicircles, quarter circles and straight lines of 15 cm each. For each of the following, use the calculator value of to find
a) the perimeter of the shaded figure, correct to 2 decimal places.
b) the area of the shaded figure, correct to 2 decimal places.

Solution

(a)
Circumference of 2 semicircles (1 circle)
--> 2r
= 2 x 7.5cm
= 47.12 cm

Circumference of 2 quarter circles (1 semi circle)
-->()2r
= ()(2)() x 15cm
= 47.12 cm

Length of vertical line on the left side of figure
--> 2 x 15cm = 30cm


Perimeter of shaded figure
--> 47.12cm + 47.12cm + 30cm
= 124.25cm

Answer: 124.25 cm


(b)
Area of 2 quarter circles (1 semi circle)
-->()
= () x 15cm x 15cm
= 353.43


Area of 2 semi circles (1 circle)
-->
= x 7.5cm x 7.5 cm
= 176.71

Area of whole figure
--> 30cm x 30cm
= 900

Shaded area
--> 900 - 353.43 - 176.71
= 369.86

Answer: 369.86

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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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ACS Primary 2009 PSLE Math Prelim Paper 2 Q16

The figure below is made up of semicircles and quadrants. Find
a) the area of A
b) the perimeter of B
Leave your answer correct to 1 decimal place.

Solution

a)
Area of unshaded semicircle (on the left of the figure above)
--> (1/2)(3.14)(3)(3)sq cm = 14.13 sq cm

Area of unshaded square (at the bottom left of the figure)
--> 6cm x 6cm = 36 sq cm
(The unshaded portion outside the shaded area B on the bottom right square, can be shifted left to form a square at the bottom left of the figure.)

Total Area of A
--> (14.13 + 36) sq cm
= 50.13 sq cm
~ 50.1 sq cm

Answer: 50.1 sq cm


b)
Perimeter of B
--> There are 3 quadrants of radius 6cm, 1 semicircle of radius 3 cm and one straight line measuring 6 cm
--> (3/4)(2)(3.14)(6)cm + (1/2)(2)(3.14)(3)cm + 6 cm
= (28.26 + 9.42 + 6) cm
= 43.68 cm
~ 43.7 cm

Answer: 43.7 cm

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ACS Primary 2009 PSLE Math Prelim Paper 2 Q3

The circumference of a circular disc is 154 cm What is the radius of the circular disc?

Solution

154 cm = (22/7) d

d = (154cm)(7/22) = 49 cm

d = 49 cm divided by 2 = 24.5 cm

Answer: 24.5 cm

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Request for help from a reader - 10 Jul 2010

Posted by a reader which can be found in this link.

Aden and John started jogging along a circular track. Aden started at Point X while John started at Point Y where the line XY formed the diameter of the circle. Aden and John jogged toward each other along the circular track from their respective starting point and first met at Point W which was 80 m from Point X. After they met for the first time, they continued jogging along the track and finally met again for the second time at Point Z which was 60 m from Point Y. Find the distance of the circular track. (Answer : 360 m)

Any kind soul would like to help? Please post solution here. Registration is not required.

Rosyth Sch 2007 PSLE Math Prelim Q42

Find the area of the figure.



Answer: 196 square cm

Anglo Chinese School 2007 PSLE Math Prelim Q45

The figure is made up of three circles and two semi-circles. O is the center of the circle. AB is 42 cm. Find the total area of the shaded part.





Area of 1 square ----- (10.5 x 10.5)square cm = 110.25 square cm

Area of 2 squares ----- 2 x 110.5 square cm = 220.5 square cm

Area of shaded area is ---- Area of whole figure – area of 1 small circle – area of 4 small quadrants – area of 2 squares

(1386 - 346.5 - 346.5 - 220.5) square cm = 472.5 square cm

Answer: 472.5 square cm

Catholic High Sch 2006 PSLE Math Prelim Q45

Q45 can be found in the link below.

http://road-to-psle.blogspot.com/2007/11/catholic-high-school-2006-psle-math.html

Catholic High Sch 2006 PSLE Math Prelim Q26

The figure below is made up of a square and three identical semi-circles.
Calculate the perimeter of the shaded region.




= (2 x 3.14 x 5 cm) + 10 cm + 10 cm
= 31.4 cm + 20 cm= 51.4 cm (Answer)

Pei Chun Public Sch 2007 PSLE Math Prelim Q44

The figure below is made up of a rectangle, a semi-circle and 4 identical quadrants.
a) What is the total area of the shaded parts?
b) What is the perimeter of the whole figure?


Solution

a) Redrawing


Area of shaded area
= Area of rectangle + Area of quadrant
= (7 cm x 3.5 cm) + (¼ x 22/7 x 3.5 cm 3.5 cm)
= (24.5 + 9.625) square cm
= 34.125 square cm (Answer)

b)
Perimeter -----
Lengths of 4 quarter arcs + 6r
= 4 x (¼ x 2 x 22/7 x 3.5 cm) + (6 x 3.5) cm
= 22 cm + 21 cm
= 43 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 Q46

In the figure, ABCD is a square with a perimeter of 84 cm. It is made up of identical squares and quarter-circles.
a) Find the perimeter of the shaded region.
b) Find the area of the shaded region.



Solution

a)
The perimeter of square ABCD is 84 cm.
1 side of square ABCD ----- 84 cm divided by 4 = 21 cm
Radius of the 6 quadrants ----- 21 cm divided by 3 = 7 cm

The shaded area is made up of 6 quadrants -----
6 x ¼ x 2 x 22/7 x 7 cm = 66 cm (Answer)


b) Redrawing the figure….


Area of 1 quadrant -----
¼ x 22/7 x 7 cm x 7 cm = 38.5 square cm

Area of 1 small partially shaded square ----
Area of 1 small square – Area of 1 quadrant ----- (7 cm x 7 cm) – 38.5 square cm = 10.5 square cm

Area of 2 such small partially shaded squares -----
10.5 square cm x 2 = 21 square cm

Area of 3 small shaded squares -----
(3 x 7 x 7) square cm = 147 square cm

Total shaded area ------
(21 + 147) square cm = 168 square cm (Answer)

Common Mistake - Perimeter of Partial Circles

S’pore Hokkien Huay Kuan 2007 PSLE Math Prelim Q43

The figure below is made up of a semicircle and a rectangle. Find the area of the unshaded region.






Solution

The base of the figure is 28 cm. The diameter of the unshaded circle is half the base, hence, it is 14 cm. This also means that the radius of the circle is 7 cm.

The radius of the 2 quadrants is 14 cm.

Area of circle ----- 22/7 x 7 cm x 7 cm = 154 square cm

Area of 1 quadrant ----- 22/7 x 14 cm x 14 cm x ¼ = 154 square cm

Area of 2 quadrants ----- 154 square cm x 2 = 308 square cm

Total unshaded area ----- (154 + 308) square cm

= 462 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 (Q48)

The figure below shows a circle and two quadrants enclosed within a square. O is the centre of the circle. Find the total area of the shaded parts in the figure shown. (Take =3.14)

Solution

Redrawing…..



Area of A ----- Area of small square – Area of small quadrant

= (10 cm x 10 cm) – (3.14 x 10 cm x 10 cm x ¼)
= 100 square cm – 78.5 square cm
= 21.5 square cm


Area of 1 shaded portion ----
Area of large square – Area of large quadrant – Area of A

= (20 cm x 20cm) – (3.14 x 20 cm x 20 cm x ¼) – 21. 5 square cm
= (400 – 314 – 21.5) square cm
= 64.5 square cm

Area of total shaded portion therefore is 64.5 square cm x 2 = 129 square cm (Answer)