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ans:
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C
DEL
ANS
±
(
)
÷
×
7
8
9
–
4
5
6
+
1
2
3
=
0
.
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Question 1:
1 pts
Which expression can be used to find the area of a circle shown on the picture
A
=
16
π
c
m
2
A=16\pi cm^{2}
A
=
16
π
c
m
2
A
=
2
⋅
8
π
c
m
2
A=2\cdot8\pi cm^{2}
A
=
2
⋅
8
π
c
m
2
A
=
2
⋅
8
2
π
c
m
2
A=2\cdot 8^{2}\pi cm^{2}
A
=
2
⋅
8
2
π
c
m
2
A
=
8
2
π
c
m
2
A=8^{2}\pi cm^{2}
A
=
8
2
π
c
m
2
Question 2:
1 pts
Which expression can be used to find the perimeter of a circle shown on the picture.
P
=
4
π
c
m
P=4\pi cm
P
=
4
π
c
m
P
=
2
⋅
4
π
c
m
P=2\cdot4\pi cm
P
=
2
⋅
4
π
c
m
P
=
2
⋅
4
c
m
P=2\cdot 4 cm
P
=
2
⋅
4
c
m
P
=
4
2
π
c
m
P=4^{2}\pi cm
P
=
4
2
π
c
m
Question 3:
1 pts
Find the arc length of an arc formed by
3
6
∘
36^{\circ}
3
6
∘
of a circle with a radius of
5
5
5
inches.
l
=
5
2
π
3
6
∘
36
0
∘
l=\dfrac{5^{2}\pi 36^{\circ}}{360^{\circ}}
l
=
36
0
∘
5
2
π
3
6
∘
l
=
10
π
3
6
∘
36
0
∘
l=\dfrac{10\pi 36^{\circ}}{360^{\circ}}
l
=
36
0
∘
10
π
3
6
∘
l
=
5
π
3
6
∘
36
0
∘
l=\dfrac{5\pi 36^{\circ}}{360^{\circ}}
l
=
36
0
∘
5
π
3
6
∘
l
=
10
π
36
0
∘
3
6
∘
l=\dfrac{10\pi 360^{\circ}}{36^{\circ}}
l
=
3
6
∘
10
π
36
0
∘
Question 4:
1 pts
Which expression can be used to find the area of a sector,
A
s
A_{s}
A
s
, shown on the picture.
A
s
=
1.5
⋅
π
⋅
7
2
∘
36
0
∘
A_{s}=\dfrac{1.5\cdot \pi\cdot 72^{\circ}}{360^{\circ}}
A
s
=
36
0
∘
1.5
⋅
π
⋅
7
2
∘
A
s
=
(
1.5
)
2
⋅
π
⋅
7
2
∘
18
0
∘
A_{s}=\dfrac{(1.5)^{2}\cdot \pi\cdot 72^{\circ}}{180^{\circ}}
A
s
=
18
0
∘
(
1.5
)
2
⋅
π
⋅
7
2
∘
A
s
=
(
1.5
)
2
⋅
π
⋅
7
2
∘
36
0
∘
A_{s}=\dfrac{(1.5)^{2}\cdot \pi\cdot 72^{\circ}}{360^{\circ}}
A
s
=
36
0
∘
(
1.5
)
2
⋅
π
⋅
7
2
∘
A
s
=
1.5
⋅
π
⋅
7
2
∘
18
0
∘
A_{s}=\dfrac{1.5\cdot \pi\cdot 72^{\circ}}{180^{\circ}}
A
s
=
18
0
∘
1.5
⋅
π
⋅
7
2
∘
Question 5:
2 pts
The area of the sector is
18
π
c
m
2
18\pi cm^{2}
18
π
c
m
2
. Find the length of
x
.
x.
x
.
x
=
x=
x
=
Question 6:
2 pts
A path
8
m
8 m
8
m
wide surrounds a circular lawn whose diameter is
110
m
.
110 m.
110
m
.
Which expression can be used to find the area of the path.
A
=
π
⋅
(
6
3
2
−
5
5
2
)
c
m
2
A=\pi \cdot\left(63^{2}-55^{2}\right)cm^{2}
A
=
π
⋅
(
6
3
2
−
5
5
2
)
c
m
2
A
=
π
⋅
(
5
5
2
−
8
2
)
c
m
2
A=\pi \cdot\left(55^{2}-8^{2}\right)cm^{2}
A
=
π
⋅
(
5
5
2
−
8
2
)
c
m
2
A
=
π
⋅
(
5
5
2
−
4
7
2
)
c
m
2
A=\pi \cdot\left(55^{2}-47^{2}\right)cm^{2}
A
=
π
⋅
(
5
5
2
−
4
7
2
)
c
m
2
A
=
π
⋅
(
6
0
2
−
5
5
2
)
c
m
2
A=\pi \cdot\left(60^{2}-55^{2}\right)cm^{2}
A
=
π
⋅
(
6
0
2
−
5
5
2
)
c
m
2
Question 7:
2 pts
Find the distance between the centres of the circles shown on the picture.
Question 8:
2 pts
Find the perimeter of the figure shown on the picture.
P
=
2
⋅
(
2
+
2
π
)
c
m
P=2\cdot(2+2\pi)cm
P
=
2
⋅
(
2
+
2
π
)
c
m
P
=
π
⋅
(
2
+
π
)
c
m
P=\pi \cdot(2+\pi)cm
P
=
π
⋅
(
2
+
π
)
c
m
P
=
2
⋅
(
1
+
π
)
c
m
P=2\cdot(1+\pi)cm
P
=
2
⋅
(
1
+
π
)
c
m
P
=
2
⋅
(
2
+
π
)
c
m
P=2\cdot(2+\pi)cm
P
=
2
⋅
(
2
+
π
)
c
m
Question 9:
3 pts
Find the area of the shaded figure shown on the picture, if the area of shown rectangle is
32
c
m
2
32cm^{2}
32
c
m
2
.
A
=
A=
A
=
Question 10:
3 pts
Find the perimeter of the shaded area shown on the picture.
P
=
24
+
49
π
6
c
m
P=\dfrac{24+49\pi}{6}cm
P
=
6
24
+
49
π
c
m
P
=
25
π
6
c
m
P=\dfrac{25\pi}{6}cm
P
=
6
25
π
c
m
P
=
25
+
48
π
6
c
m
P=\dfrac{25+48\pi}{6}cm
P
=
6
25
+
48
π
c
m
P
=
49
π
6
c
m
P=\dfrac{49\pi}{6}cm
P
=
6
49
π
c
m
Question 11:
3 pts
Find the perimeter of the shaded area shown on the picture, if the radius of the biggest circle is
4
c
m
.
4cm.
4
c
m
.
P
=
P=
P
=
Question 12:
3 pts
Find the perimeter of the shaded area shown on the picture.
P
=
P=
P
=
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