ID : us-8-Mensuration [1]
Grade 8
Mensuration
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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)
Brandon is painting the f ollowing box. If he painted all surf aces except f ront side of the box,
f ind the area of surf ace painted by him.
(All measurements are in cm).
(3)
If the diagonal of a square is decreased by 5%, then by what percent does the area of the
square decrease?
(4) A triangle ABC with ∠ABC = 90 has length of the side AB = 72 cm and length of BC = 54 cm.
What is the length of the perpendicular line f rom side AC to point B?
(5)
Area of the f our walls of a hall is 1980 m2. If the height and length of the room are in the ratio
of 5:2 and the height and breadth in the ratio 3:1, then f ind the area of the f loor.
(6) A square park has an area of 5625 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 2 %
greater than the perimeter of the park. What is the length of wire needed f or this?
(7) T he area of a rhombus is 39 cm2 and one of its is diagonal is 6 cm. Find the length of the other
diagonal.
(8)
If angle ∠B is a right angle, f ind area of the trapezium ABCD.(All dimensions are in cm)
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ID : us-8-Mensuration [2]
Choose correct answer(s) f rom given choice
(9) A square and an equilateral triangle have the same perimeter. T he diagonal of the square is
12√2 cm. What is the area of the triangle?
a. 64√3
b. p - √3
c. p + √2
d.
64
√3
(10) T he length of a rectangle is increased by 25 %, and the breadth is reduced by 33 %. What is the
percentage change in the area of the rectangle?
a. 17.45
b. 14.35
c. 16.55
d. 16.25
(11) Find the amount of water that can be f illed in this cuboid.
(All measurements are in cm).
a. 246 cm3
b. 258 cm3
c. 264 cm3
d. 252 cm3
(12) Area of parallelogram ABCD is x cm2. If E,F,G and H are mid-points of the sides, f ind the area of
EFGH.
a. x/4 cm2
b. x/2 cm2
c. x/√2 cm2
d. x/3 cm2
(13) If the diagonals of a rhombus are 16 and 12 cm, f ind its perimeter.
a. 28 cm
b. 44 cm
c. 40 cm
d. 36 cm
Fill in t he blanks
(14) A wire f rame is bent into a circle of radius 49 is reshaped as a rhombus. T he length of the side
of the resulting rhombus is
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(assume π = 22/7)
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ID : us-8-Mensuration [3]
(15) In a trapezoid of area 120 cm2, the length of two parallel sides are in ratio 2:3. If the height of
trapezoid is 8 cm, the length of larger of the two parallel sides is
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cm.
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generated at www.edugain.com
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ID : us-8-Mensuration [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)
378 cm2
Step 1
If we look at the f igure, we notice that the length, width and height of the box are 9 cm, 9
cm and 8 cm respectively.
Step 2
T he surf ace area painted by him = T he total surf ace area of the box - T he area of the
f ront side of the box
= 2{(9 × 9) + (9 × 8) + (8 × 9)} - (8 × 9)
= 450 - 72
= 378 cm2
Step 3
T heref ore, the area of surf ace painted by him is 378 cm2.
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ID : us-8-Mensuration [5]
(3)
9.75%
Step 1
Let the length of the diagonal of the square be d. T he length of the side of the square will
then be d / √2, and the area of the square will be (d / √2) × (d / √2) = 0.5d2
Step 2
Af ter reducing the length of the diagonal by 5%, the new length of the diagonal will be:
5
=d-
d
100
= 0.95d
Step 3
Hence the new area will be 0.5(0.95d)2 = 0.5 × 0.9025d2.
Step 4
T he decrease in area = Old Area - New Area
= 0.5 d2 - 0.5 × 0.9025d2
= 0.5 × (1 - 0.9025) d2
= 0.5 × 0.0975 d2
Step 5
Percentage decrease in area =
Decrease in area
× 100 %
Old Area
=
0.5 × 0.0975 d2
× 100 %
0.5 d2
= 0.0975 × 100 %
= 9.75%
Step 6
Hence, when the diagonal of a square is decreased by 5%, then the area of the square
decreases by 9.75%.
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ID : us-8-Mensuration [6]
(4) 43.2 cm
Step 1
In the right angled triangle ABC, AB = 72 cm, BC = 54 cm
Now, AC =
...[As per Pythagoras T heorem]
=
=
=
= 90 cm
Step 2
T he area of the right angled triangle ABC when base is BC =
AB × BC
2
=
72 × 54
2
= 1944 cm2 -----(1)
Step 3
T he area of the right angled triangle ABC when base is AC =
AC × BD
2
=
90 × BD
-----(2)
2
Step 4
Since, area in equation (1) and (2) should be same, equating the area in equation (2), we
get:
⇒
90 × BD
= 1944
2
⇒ 90 × BD = 3888
⇒ BD =
3888
90
⇒ BD = 43.2 cm
Step 5
T heref ore, the length of the perpendicular line f rom side AC to point B is 43.2 cm.
(5)
180 m2
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ID : us-8-Mensuration [7]
(6) 1224 m
Step 1
Let's assume that a be the length of the park.
T heref ore the area of a square park = a2 = 5625
⇒ a2 = 752
⇒ a = 75 m
Step 2
Now the perimeter of a square = 4a = 4 X 75 = 300 m
Step 3
Since each circuit of the wire be 2% greater than the perimeter of the park.
T heref ore the length of the each circuit of the wire = perimeter of the square park + 2% of
the perimeter of the square park
2
= 300 + 300 X
100
600
= 300 +
100
= 300 + 6
= 306 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 306 = 1224 m.
(7) 13 cm
Step 1
T he area of the rhombus = 39 cm2.
One diagonal of the rhombus = 6 cm.
Let us assume that the other diagonal of the rhombus is b cm.
Step 2
We know that the area of a rhombus can be calculated by multiplying the length of its
diagonals and then dividing by 2.
Step 3
T hus, the area of the rhombus =
6×b
2
⇒ 39 =
6×b
2
⇒ 39 × 2 = 6 × b
⇒b=
78
6
⇒ b = 13
Step 4
T heref ore, the length of the other diagonal of the rhombus is 13.
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ID : us-8-Mensuration [8]
(8)
592 cm2
Step 1
It is given that, ∠B is a right angle,
the height of the trapezium ABCD = BC = 16 cm,
AB = 52 cm,
CD = 22 cm.
Step 2
T he area of the trapezium =
Sum of parallel sides
× Height
2
⇒ Area (ABCD) =
AB + CD
×h
2
⇒ Area (ABCD) =
52 + 22
× 16
2
⇒ Area (ABCD) = 592 cm2
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ID : us-8-Mensuration [9]
(9) a. 64√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 12√2 cm.
Now in right angle triangle ABC
AB2 + BC2 = AC2
⇒ s2 + s2 = (12√2)2
⇒ 2(s)2 = (12√2)2
(12√2)2
⇒ s2 =
2
⇒ s2 = 144 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 × 144
⇒ t2 =
9
[Since s2 = 144]
⇒ t 2 = 256 cm
Step 3
Now the area of an equilateral traingle =
t 2√3
4
=
256√3
[Since t 2 = 256 ]
4
= 64√3 cm2
Step 4
T heref ore the area of the triangle is 64√3 cm2.
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ID : us-8-Mensuration [10]
(10) d. 16.25
Step 1
If you look at the question caref ully, you will notice that the length of a rectangle is
increased by 25 %, and the breadth is reduced by 33 %.
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 25 %,
theref ore new length of the rectangle = l + (l ×
25
)
100
25
= l(1 +
)
100
100 + 25
= l(
)
100
125
= l(
)
100
=
125l
100
Step 4
Since the breadth of a rectangle is reduced by 33 %,
theref ore the new breadth of the rectangle = b - (b ×
33
)
100
33
= b(1 -
)
100
= b(
100 - 33
)
100
= b(
67
)
100
=
67b
100
Step 5
Now new area of the rectangle =
125l
100
=
×
67b
100
8375lb
10000
Step 6
Change in area of the rectangle = lb -
8375 lb
10000
= lb(1 -
8375
)
10000
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ID : us-8-Mensuration [11]
= lb(
10000 - 8375
)
10000
=
1625 lb
10000
Percentage change in area of the rectangle =
1625 lb
×
10000
=
100
lb
1625
100
= 16.25 %
Step 7
T heref ore the percentage change in the area of the rectangle is 16.25 %.
(11) d. 252 cm3
Step 1
,
If we look at the f igure, we notice that the length, width and height of the cuboid are 6 cm, 7
cm and 6 cm respectively.
Step 2
T he amount of water that can be f illed in the cuboid will be equal to the volume of the
cuboid. T he volume of the cuboid = Length × Width × Height
= 6 cm × 7 cm × 6 cm
= 252 cm3
Step 3
T hus, 252 cm3 water can be f illed in this cuboid.
(12) b. x/2 cm2
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ID : us-8-Mensuration [12]
(13) c. 40 cm
Step 1
We have been given the (length of ) diagonals of a rhombus and we are required to f ind its
perimeter. If we can f ind the side of the rhombus, f inding the perimeter will be easy. Let us
f irst f ind the side of the rhombus.
Step 2
We know that the diagonals of a rhombus are perpendicular bisectors of each other. Let us
use this f act in combination with the Pythagorean T heorem to f ind the side of the rhombus.
Step 3
Let us ref er to the f igure below showing a rhombus ABCD with diagonals of 16 and 12 cm
intersecting at point O:
Step 4
Since the diagonals of a rhombus are perpendicular bisectors of each other, f ollowing
f acts hold true f or the triangle COB:
T riangle COB is a right angled triangle with CB being the hypotenuse.
T he length of the side CO = Half of diagonal CA = 16/2 = 8 cm.
T he length of the side BO = Half of diagonal BD = 12/2 = 6 cm.
Step 5
Using the Pythagorean T heorem f or triangle ΔCOB, we can write:
CB2 = (8)2 + (6)2
⇒ CB2 = 64 + 36
⇒ CB2 = 100
⇒ CB = √100
⇒ CB = 10
Step 6
We know that the perimeter of a rhombus is 4 times its side.
T hus, the perimeter of the rhombus = 4 × CB
= 4 × 10
= 40 cm
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ID : us-8-Mensuration [13]
(14)
77
Step 1
A wire f rame of some length was f irst bent into a circle and then reshaped as a rhombus:
Wire
Circle
Rhombus
Step 2
Let us f irst f ind the length of the wire f rame. We know that the total length of the boundary of a circle is called
its circumf erence and is given by:
Circumf erence = 2π r, where r is the radius of the circle.
Since the circle is f ormed by the wire f rame, the length of the wire f rame = 2π r
=2×
22
× 49 [It is given that the radius of the circle is 49 and π = 22/7]
7
= 308
Step 3
Now, we know that the same wire f rame with length 308 is reshaped as a rhombus. A rhombus has 4 sides
and all sides are equal. T his means, the length of a side of the rhombus will be 308 divided by 4. T hat is:
308/4
= 77
Step 4
T hus the length of the side of the resulting rhombus is 77.
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ID : us-8-Mensuration [14]
(15)
18
Step 1
Let's assume that 'a' and 'b' are the lengths of the parallel sides of the trapezoid and 'h' is
the height of the trapezoid.
T hen, the area of the trapezoid =
a+b
× h.
2
Step 2
It is given that the length of the two parallel sides of the trapezoid are in ratio 2:3.
Let's assume 'x' is the ratio f actor of the parallel sides.
T hen, the parallel sides 'a' and 'b' of the trapezoid can be considered to be 2x and 3x
respectively.
Step 3
It is also given that the area of the trapezoid = 120 cm2,
and the height of the trapezoid(h) = 8 cm.
Step 4
T hus, the area of the trapezoid =
2x + 3x
×8
2
⇒ 120 =
5x
×8
2
⇒
120
=
8
⇒
5x
2
120 × 2
= 5x
8
⇒
240
=x
8×5
⇒x=6
Step 5
Now that we know the value of 'x',
a = 2x = 2 × 6 = 12 cm,
b = 3x = 3 × 6 = 18 cm.
Step 6
T hus, the length of the larger of the two parallel sides is 18 cm.
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