ID : ca-7-Mensuration-Perimeter-Area-Volume [1]
Grade 7
Mensuration - Perimeter, Area, Volume
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Answer t he quest ions
(1)
If the sides of the small blocks below are 1 cm each, then what is the perimeter of the f igure
below?
(2)
What is the radius of a circle if the circumf erence of the circle is 263.76 (assume that the value
of π is 3.14)?
(3)
A square and an equilateral triangle have the same perimeter. T he diagonal of the square is 9
cm. What is the area of the triangle?
(4)
If each small square is of size 1cm x 1cm, f ind the perimeter of next shape in the series.
Choose correct answer(s) f rom given choice
(5)
T he length of a rectangle is increased by 55%. By what percentage should the width be reduced
to keep the area of the rectangle the same?
a. 35
c. 40
15
31
17
34
b. 40
d. 40
12
29
17
30
(6) A square park has an area of 3025 sq. m. It has to be f enced. T he f encing will require use of a
wire that must be able to enclose the park 4 times, and each circuit of the wire would be 5 %
greater than the perimeter of the park. What is the length of wire (in metres) needed f or this?
a. 930.88
b. 914.26
c. 924
d. 922.55
(7) If the side of cube is halved, its volume will reduce by ______ times.
a.
1
8
c.
1
b.
1
2
d. none of these
4
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ID : ca-7-Mensuration-Perimeter-Area-Volume [2]
(8)
T he length of a rectangle is increased by 15 %, and the breadth is reduced by 23 %. What is the
percentage change in the area of the rectangle?
a. 10.05
b. 11.45
c. 12.25
d. 13.15
(9) If each side of a cube is doubled, f ind the f actor by which volume of the cube will change.
a. 4 times
b. 6 times
c. 8 times
d. 2 times
Fill in t he blanks
(10)
Radius of two concentric circles are 7 and 3 meters, area of space between them =
m2.
(assume π = 3 f or this question)
(11) If the sides of a square are increased by 25%, then the area of the square increases by
%.
(12)
T he area of f ollowing shape is
m2 (all measures are in meters).
(13) A f armer wants to tie his cow with rope so that it can graze the grass in an area of 192 m2. T he
length of the rope should be
meters. (assume π = 3 f or this question).
(14) What is the area of a rectangle with length 2 m 9 dm 3 cm and width as 10 m 1 dm 9 cm is
sq. m.
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ID : ca-7-Mensuration-Perimeter-Area-Volume [3]
(15)
Area of f ollowing f igure is
cm2(Area of each square on graph paper is 1 cm2).
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ID : ca-7-Mensuration-Perimeter-Area-Volume [4]
Answers
(1)
28 cm
Step 1
If you read the question caref ully, you will notice that the length of each side of the small
block is 1 cm.
Step 2
Now if you count the sides of the small blocks at the perimeter of f igure, you will notice
that the number of sides of small blocks are 28.
T heref ore the perimeter of the f igure is 28 cm.
(2)
42
Step 1
According to question the circumf erence of the circle is 263.76
Step 2
T he circumf erence of a circle = 2π r
⇒ 263.76 = 2 × 3.14 × r
⇒ 2 × 3.14 × r = 263.76
⇒r=
263.76
6.28
⇒ r = 42
Step 3
Now the radius of a circle is 42.
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ID : ca-7-Mensuration-Perimeter-Area-Volume [5]
(3)
18√3
Step 1
Lets assume s and t are the sides of square and triangle respectively as shown in the
f ollowing f igures.
Square
Equilateral triangle
According to question the diagonal AC of the square is 9 cm.
Now in right angle triangle ABC
AB2 + BC2 = AC2
⇒ s2 + s2 = (9)2
⇒ 2(s)2 = (9)2
(9)2
⇒ s2 =
2
⇒ s2 = 40.5 cm
Step 2
Perimeter of the given square = 4s
Perimeter of the given equilateral triangle = 3t
According to question the perimeter of a square is equal to the perimeter of an equilateral
triangle.
T heref ore 3t = 4s
Squaring both sides
(3t)2 = (4s)2
⇒ 9t 2 = 16s2
16 × 40.5
⇒ t2 =
9
[Since s2 = 40.5]
⇒ t 2 = 72 cm
Step 3
Now the area of an equilateral traingle =
t 2√3
4
=
72√3
[Since t 2 = 72 ]
4
= 18√3 cm2
Step 4
T heref ore the area of the triangle is 18√3 cm2.
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ID : ca-7-Mensuration-Perimeter-Area-Volume [6]
(4) 24 cm
Step 1
Perimeter is the distance around a two dimensional shape.
Perimeter of a square = 4 × length of the side of a square
Step 2
If you look at the given shapes caref ully, you will notice that the perimeter of the f irst
shape in the series = 1 × 4 = 4 cm
Perimeter of the second shape in the series = 2 × 4 = 8 cm
T heref ore the perimeter of next shape (i.e 6th shape) in the series = 6 × 4 = 24 cm
Step 3
T he perimeter of next shape in the series is 24 cm.
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ID : ca-7-Mensuration-Perimeter-Area-Volume [7]
(5)
a. 35
15
31
Step 1
Let L and W be the original length and width of the rectangle.
Step 2
Length of the rectangle is increased by 55%. Let us assume that the width of the rectangle
needs to be reduced by x% to keep the area of the rectangle same. T hus:
55
New Length = L + 55% of L = L +
L=
100
x
New Width = W - x% of W = W -
100
155L
100
W=
(100 - x)W
100
Step 3
We know that the area of a rectangle is the product of its length and width. Since the area
of the rectangle af ter changing the length and width of the rectangle remains the same, we
can write:
New Area = Original Area
⇒ New Length × New Width = Original Length × Original Width
⇒
155L
(100 - x)W
×
100
⇒
155
= LW
100
×
100 - x
100
= 1 (Dividing both sides by LW)
100
10000
⇒ 100 - x =
155
10000
⇒ x = 100 -
155
⇒x=
15500 - 10000
155
⇒x=
1100
31
⇒ x = 35
15
%
31
Step 4
Hence, when the length of the rectangle is increased by 55%, then the width of the
rectangle should be reduced to 35
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15
% to keep the area of the rectangle the same.
31
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ID : ca-7-Mensuration-Perimeter-Area-Volume [8]
(6) c. 924
Step 1
Let's assume that a be the length of the park.
T heref ore the area of a square park = a2 = 3025
⇒ a2 = 552
⇒ a = 55 m
Step 2
Now the perimeter of a square = 4a = 4 X 55 = 220 m
Step 3
Since each circuit of the wire be 5% greater than the perimeter of the park.
T heref ore the length of the each circuit of the wire = perimeter of the square park + 5% of
the perimeter of the square park
= 220 + 220 X
5
100
= 220 +
1100
100
= 220 + 11
= 231 m
Step 4
T he f encing will require use of a wire that must be able to enclose the park 4 times.
T heref ore the length of wire needed f or this = 4 X 231 = 924 m.
(7)
a.
1
8
Step 1
Lets assume a is the side of cube,
the volume of cube = a3
Step 2
Since the side of cube is halved,
the new side of cube =
a
2
Step 3
Now new volume of cube = (
a
2
)3 =
a3
23
=
a3
8
Step 4
T heref ore we can say that the volume of cube is reduced by
1
times.
8
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ID : ca-7-Mensuration-Perimeter-Area-Volume [9]
(8)
b. 11.45
Step 1
If you look at the question caref ully, you will notice that the length of a rectangle is
increased by 15 %, and the breadth is reduced by 23 %.
Step 2
Let assume the original length and breadth of a rectangle are l and b respectively and
hence initial area is lb.
Step 3
Since the length of a rectangle is increased by 15 %,
theref ore new length of the rectangle = l + (l ×
15
)
100
15
= l(1 +
)
100
100 + 15
= l(
)
100
115
= l(
)
100
=
115l
100
Step 4
Since the breadth of a rectangle is reduced by 23 %,
theref ore the new breadth of the rectangle = b - (b ×
23
)
100
23
= b(1 -
)
100
= b(
100 - 23
)
100
= b(
77
)
100
=
77b
100
Step 5
Now new area of the rectangle =
115l
100
=
×
77b
100
8855lb
10000
Step 6
Change in area of the rectangle = lb -
8855 lb
10000
= lb(1 -
8855
)
10000
= lb(
10000 - 8855
)
10000
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ID : ca-7-Mensuration-Perimeter-Area-Volume [10]
=
1145 lb
10000
Percentage change in area of the rectangle =
1145 lb
10000
=
×
100
lb
1145
100
= 11.45 %
Step 7
T heref ore the percentage change in the area of the rectangle is 11.45 %.
(9) c. 8 times
Step 1
Lets assume a is the side of the cube.
Volume of the cube = a3
Step 2
According to question each side of a cube is doubled.
Now Volume of the cube = (2a)3
= 8a3
Step 3
Now you can say that the f actor by which volume of the cube will change be 8 times.
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ID : ca-7-Mensuration-Perimeter-Area-Volume [11]
(10)
120
Step 1
Lets assume circles C1 and C2 are two concentric circles. r1 and r2 are the radius of the
circles C1 and C2 respectively.
Concentric Circles
According to question the radius of two concentric circles are 3 and 7 meters respectively.
T heref ore r1 = 3 meters
r2 = 7 meters
Step 2
Area of circle C1 = π(r1)2
=3×3×3
= 27 m2
Step 3
Area of circle C2 = π(r2)2
=3×7 ×7
= 147 m2
Step 4
Area of space between the concentric circles C1 and C2 = Area of circle C2 - Area of circle
C1
= 147 - 27
= 120 m2
Step 5
Now area of space between the concentric circles is 120 m2.
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ID : ca-7-Mensuration-Perimeter-Area-Volume [12]
(11)
56.25
Step 1
Lets assume s and A are the side of the square and area of the square respectively.
Step 2
According to question the sides of the square are increased by 25%
theref ore the side of the square = s + s ×
25
100
=
100s + 25s
100
= 1.25s
Now the area of the square = (1.25s)2
= 1.5625s2
= 1.5625A [Since A = s2]
Step 3
T he area of the square increase = 1.5625A - A
= 0.5625A
Step 4
T heref ore you can say that the area of the square increase =
0.5625
%
100
= 56.25%
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ID : ca-7-Mensuration-Perimeter-Area-Volume [13]
(12)
302
Step 1
We need to f ind the area of the f ollowing shape:
Step 2
First let us divide the shape into three rectangles. T he area of the shape in question will be sum of areas of these
three rectangles.
rectangle 1
rectangle 2
rectangle 3
Step 3
Let us now f ind the area of each rectangle:
Area of rectangle 1 = Length × Width
= 6 m × 17 m
= 102 m2
Area of rectangle 2 = Length × Width
= 20 m × 6 m
= 120 m2
Area of rectangle 3 = Length × Width
= 4 m × 20 m
= 80 m2
Step 4
Area of the shape given = Area of rectangle 1 + Area of rectangle 2 + Area of rectangle 3
= 102 m2 + 120 m2 + 80 m2
= 302 m2
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ID : ca-7-Mensuration-Perimeter-Area-Volume [14]
(13)
8
Step 1
Lets assume the length of the rope is r meters by which the f armer wants to tie his cow.
Since cow is tie with rope, the cow can graze the grass in a circular area, which radius is
equals to the length of the rope.
Step 2
According to question cow can graze the grass in an area of 192 m2.
or πr 2 = 192 [Since area of a circle is = πr 2]
192
⇒ r2 =
π
192
⇒ r2 =
3
⇒ r 2 = 64
⇒ r = 8 meters
Step 3
T heref ore the length of the rope should be 8 meters.
(14)
29.8567
Step 1
T o calculate the area of a rectangle f irst of all we have to change the length and width of a
rectangle into same unit.
Step 2
1
Since 1cm =
m
100
and 1dm =
1
m
10
theref ore the length of a rectangle = 2 m 9 dm 3 cm = 2 +
9
+
10
and the width of a rectangle = 10 m 1 dm 9 cm = 10 +
1
10
+
3
= 2.93 m
100
9
= 10.19 m
100
Step 3
Now the area of a rectangle = length × width
= 2.93 × 10.19
= 29.8567 sq. m.
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