Topic: Circles
Name: ______________________
1
Class: ____
Date: ________
The figure below is not drawn to scale.
A piece of wire is bent to form the shape as shown below.
The shape is made up of a large semicircle and 2 smaller identical semicircles.
(a) Find the length of the wire used.
(b) Find the area of the shaded part.
[Leave both part (a) and part (b) in 1decimal place.]
10 m
Ans: ____________________________ (2)
Ans: ____________________________ (3)
2.
The figure below is made up of squares and quadrants. Find the total area of the
22
shaded parts. (Take  =
)
7
3.
In the figure below, find its
a)
area,
b)
perimeter.
Give the answers to the nearest 2 decimal places.
Ans: ______________________ (1)
Ans: ______________________ (2)
4.
Find the area of the shaded part of the figure below. Correct your answer to the
nearest 2 decimal places.
18 cm
5.
The figure below shows a large square WXYZ and a small square. There are 4 quarter
circles with centres W, X, Y, Z, respectively. There are also 4 semicircles with centres at
the midpoint of each side of the large square. Each of the quarter circles has a radius of
7 cm. Find the area of the shaded part.
(Take  = Ошибка!)
W
X
Z
Y
6.
The diagram below shows a rectangle AOCD with a quadrant BOC
enclosed in it. CX = XW and CW = WO. UXO is
1
of another circle
8
which has XO as its radius. Find the total area of the unshaded parts.
(Take  = 3.14)
D
79 cm
C
X
C
U
W
V
A
B
O
40 cm
7.
The figure below is made up of a circle, a triangle and two identical semi-circles. The
diameter of the circle is 8 cm and the diameter of the semi-circle is 4 cm. Find the area
of the shaded regions. ( Take  =3.14)
4 cm
8 cm
4 cm
8.
The figure below is made up of semicircles and quadrants. Find
a) the area of A (2m)
b) the perimeter B (3m)
Leave your answers correct to 1 decimal place.
B
B
A
12 cm
Topic: Circles
Answer Key
1(a)
Method 1
Diameter of smaller circle = 5 m
Circumference of smaller circle =   5 = 15.7 m
Half circumference of large circle = (  10)  2 = 15.7 m
Length of wire = 15.7  2 = 31.4 m
[1st M1] – finding circumference of any circle (correct use of formula)
Method 2
=   10
Length of wire
= 31.4 m
1(b)
Area of bigger semi-circle =  x 5 x 5 x ½
≈ 39.270
Area of 2 smaller semi-circles
=  x 2.5 x 2.5
≈ 19.635
Area of shaded part
= 39.270 – 19.635
= 19.635
≈ 19.64
Ans (a) : 31.4 m 2
Ans (b): 19.6 m 2
2.
14 cm  28 cm = 392 cm²
OR ¼ 
(M1, A1)
22
14  14 cm² = 154 cm² (area of quadrant)
7
14  14 cm² = 196 cm² (area of 1 square)
196 cm² − 154 cm² = 42 cm²
(154 +154 +42 + 42) cm² = 392 cm² (M1,A1)
OR
14  14  2 cm² = 392 cm² (M1,A1)
OR
14  14 cm² = 196 cm²
2 × 196cm² = 392 cm²
(M1,A1)
3(a)
x 14 x14 ≈615.75cm2
(A1)
3(b)
x 28 + 14 + 14 ≈115.96 cm
(M1 A1)
4.
Area of half the rectangle = (18  2) x (18  3)
=9x6
= 54 cm2
Area of circle
=x3x3
= 28.274 cm2
Area of shaded part
(M1)
(M1)
= 54 - 28.274
= 25.726
25.73cm2
(M1 A1)
5.
Length of side of WXYZ = 28 cm
Area of WXYZ = 28 x 28 = 784 cm2
Area of small square = 14 x14 =196 cm2
4 quarter circles and 4 semi-circles form 3 circles
Area of the circles = 3 x
22
x 7 x 7 = 462 cm2
7
Area of the shaded part
= 784 – 196 – 462
= 126 cm2
6.
(M1)
(M1)
(A1)
Mthd 1
Radius of UXO ---
Area of UXO ---
1
x 3.14 x 30 x 30 = 353.25 cm² (M1)
8
Radius of VWO ---
Area of VWO ---
3
x 40 = 30 cm
4
1
x 40 = 20 cm
2
1
x 3.14 x 20 x 20 = 157 cm²
8
Area of shaded part UXWV --- 353.25 – 157 = 196.25 cm² (A1)
Area of triangle ABD ---
1
x (79 - 40) x 40 = 780 cm²
2
Area of rect. --- 79 x 40 = 3160 cm² (M1)
Area of unshaded parts --- 3160 - 780 - 196.25 = 2183.75 cm² (M1/A1)
Mthd 2
Radius of UXO ---
Area of UXO ---
1
x 3.14 x 30 x 30 = 353.25 cm² (M1)
8
Radius of VWO ---
Area of VWO ---
3
x 40 = 30 cm
4
1
x 40 = 20 cm
2
1
x 3.14 x 20 x 20 = 157 cm²
8
Area of BOCX --- 40 x 40 = 1600 cm²
Area of shaded part UXWV --- 353.25 – 157 = 196.25 cm² (A1)
Area of unshaded part in BOCX --- 1600 – 196.25 = 1403.75 cm² (M1)
Area of unshaded triangle DXB ---
1
x (79 - 40) x 40 = 780 cm²
2
Total area of unshaded parts --- 1403.75 + 780 = 2183.75 (M1/A1)
7.
Area of the circle = 3.14 x 4 cm x 4 cm = 50.24 cm2
[M1]
Area of the 2 semi-circles = 3.14 x 2 cm x 2 cm = 12.56 cm2
Area of the triangle =
1
x 8 x 8 cm2 = 32 cm2
2
[M1]
Area of the shaded regions = ( 50.24 + 12.56 ‒ 32) cm2 [M1]
= 30.8 cm2
[A1]
[M1]
8.
Area of A = Small Semi-circle + square
=
1
X3X3X+6X6
2
(M2)
= 14.137 + 36

50.1 cm2
(A1)
Perimeter of B =
=
3
big circle + small semi-circle + 6 cm
4
3
1
X 12 X  + X 6 X  + 6
4
2
(M1)
= 12 + 6

43.7 cm
(A1)
Note:
¼ x  x 6 x 6 or ½ x 3 x 3 or 6 x 6----------------M1
b) ( x 6)/2 + ( x 6) /2 + ( x 12)/2 + 6 ---------M1,A1
or
9.42 x 4 + 6 = 43.7cm---------------------------------M1,A1
Topical Test: Circles
Answer Key
1.
28 cm x 28 cm x 3.14 = 2461.76 cm2 (M1)
2461.76  4 = 615.44cm2  615.4 cm2 (M1, A1))
The area of the shaded region is 615.4 cm2
2.
√36 = 6 cm
Area of 1 triangle
(M1)
=(
1
x 6 x 3)
2
= 9 cm2
Area of shaded square
(M1)
= area of 2 triangles
= 9 cm2 x 2
= 18 cm2
3.
Area of big quadrant =
(A1)
1
x  x 24 x 24
4
≈ 452.389 cm2
M1
1
x 24 x 24
M1
2
= 288 cm2
Shaded area = 452.389 cm2 - 288 cm2 = 164.389 cm2 ≈ 164.39 cm2 (M1, A1)
Area of triangle
=
OR
1
x 12 x 12
2
= 72 cm2
1
Area of small quadrant
=
x  x 12 x 12
4
≈ 113.097 cm2
(113.097 – 72) x 2 = 82.194 cm2
M1
Area of triangle
=
Area of circle
=  x 12 x 12
≈ 452.389 cm2
= 370.195 cm2
452.389 – 82.194
M1
Area of big quadrant =
1
x  x 24 x 24
4
≈ 452.389 cm2
(452.389 – 370.195) = 82.194 cm2
= 164.388 cm2
≈ 164.39 cm2
Shaded area is 164.39 cm2
M1
82.194 + 82.194
4.
A1
Area of square  16 x 16
= 256 cm2
Area of quadrant 
1
x  x 16 x 16
4
[M1]
 201.061 cm2
Difference  256 – 201.061
[M1]
= 54.939 cm2
Area of X 54.939  2
 27.469 cm2
1
x  x 16 x 16
2
Area of semicircle 
[M1]
 402.122 cm2
Area of shaded parts  402.122 – 27.469
[M1]
= 374.653 cm²

374.65 cm²
[A1]
St Andrew’s Junior School
Primary 6
Topical Test: Circles
Name: _________________________
Class: _____
Date: _____________
15
Show your working clearly in the space provided for each question and write your
answers in the spaces provided. Give your answers in the units stated.
1.
The figure below shows four identical circles, each with radius 14 cm, in a bigger
circle. Find the area of the shaded region. (Take  = 3.14). (Round off your
answer to one decimal place.)
2.
In the figure below, the circle is touching each of the two squares at exactly four
points. If the area of the bigger square is 36 cm2, find the area of the shaded
square.
3.
The figure below shows a square, a quarter circle and two semicircles. Find the
total shaded area in the figure. (Round off your answer to 2 decimal places) (4m)
24 cm
4.
In the figure below, O is the centre of the circle.
ABCO is a square and OC is 16 cm.
Quadrant OAC is equal to Quadrant BAC.
Find the area of the shaded parts.
Round off your answer to the nearest 1 decimal place.
O
C
A
B