ID : pk-8-Mensuration [1]
Grade 8
Mensuration
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Answer t he quest ions
(1)
Find area of f ollowing trapezoid (All measurements are in cm).
(2)
T he diagonal of f ollowing quadrilateral ABCD is 50 cm long, and length of perpendiculars
dropped on diagonal are 20 cm and 20 cm. Find the area of quadrilateral.
(3)
If area of f ollowing trapezoid is 69 cm2, f ind the height of trapezoid.
(All measurements are in cm).
(4) An equilateral triangle has side length 11 cm. What is the area of a region of all the points
outside the equilateral triangle that are within 5 cm of the rectangle?
(5)
A rhombus ABCD has the side BC = X + 14 and AB = 6X - 16, then what is the value of X?
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ID : pk-8-Mensuration [2]
Choose correct answer(s) f rom given choice
(6) A mouse walks f rom one corner of a square shape plot to its opposite corner. If it walks along
the sides of the square, it takes 80 minutes to reach the opposite corner. How long would it
take, had it walked along the diagonal?
a. 40√2 minutes
c.
40
minutes
b. 40 minutes
d. 80√2 minutes
√2
(7) A rectangular screen has a diagonal of 10.96 m. If one side of the screen is 7.04 m, then what is
the area of the screen?
(8)
a. 58.836 sq. m.
b. 59.056 sq. m.
c. 59.416 sq. m.
d. 59.136 sq. m.
T he perimeter of a trapezoid of 8 cm height is 36 cm. If the sum of non-parallel sides is 24 cm,
f ind the area of trapezoid.
a. 50 cm2
b. 58 cm2
c. 46 cm2
d. 48 cm2
(9) Find the volume of a cylinder of radius 6 cm and height 7 cm (Assume π = 22/7)
a. 792 cm2
b. 786 cm3
c. 1584 cm3
d. 396 cm3
(10) T he length of a rectangle is increased by 25 %, and the breadth is reduced by 35 %. What is the
percentage change in the area of the rectangle?
a. 19.55
b. 18.75
c. 20.55
d. 17.25
Fill in t he blanks
(11)
If the sides of a square are increased by 50%, then the area of the square increases by
%.
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ID : pk-8-Mensuration [3]
(12)
If the side of the small squares that make up the shape below is 3.9 cm, then the total area of
the shape below is
sq. cm.
(13) If the diagonal of a square is decreased by 25 %, then the area of the square decreases by
%.
(14) A wire f rame is bent into a circle of radius 42 is reshaped as a rhombus. T he length of the side
of the resulting rhombus is
(assume π = 22/7)
(15) Laila f olds a rectangular paper of size 88 cm × 5 cm in cylindrical shape such that height of the
cylinder is 5 cm. T he volume of cylinder =
cm3. (Assume π = 22/7)
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ID : pk-8-Mensuration [4]
Answers
(1)
544 cm2
Step 1
It is given that,
Length of parallel sides are 28 cm and 40 cm.
Distance between parallel sides is 16 cm.
Step 2
Area of trapezoid =
1
(Sum of parallel sides) × (Distance between parallel sides)
2
=
1
(28 + 40) × 16
2
=
1
× 68 × 16
2
= 544
Step 3
T heref ore, the area of the trapezoid is 544 cm2.
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ID : pk-8-Mensuration [5]
(2)
1000 cm2
Step 1
Given,
Length of the diagonal, AC = 50 cm
Length of the perpendiculars, BE and DF are 20 cm and 20 cm
Step 2
We know that the area of a triangle =
Base × Height
2
Area of the triangle ABC =
AC × BE
2
=
50 × 20
2
= 500 cm2
Step 3
Area of the triangle ACD =
AC × DF
2
=
50 × 20
2
= 500 cm2
Step 4
T hus, the area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle
ACD
= 500 + 500
= 1000 cm2
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ID : pk-8-Mensuration [6]
(3)
6 cm
Step 1
If we look at the f igure, we notice that, the parallel sides of the trapezoid are 13 cm and 10
cm respectively.
Step 2
It is given that, the area of the trapezoid is 69 cm2.
We know that, the area of the trapezoid =
T he sum of the parallel sides of the trapezoid
2
× T he height of the trapezoid
T heref ore, 69 =
13 + 10
× T he height of the trapezoid
2
⇒ 69 ×
2
= T he height of the trapezoid
23
⇒ 6 = T he height of the trapezoid
⇒ T he height of the trapezoid = 6 cm
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ID : pk-8-Mensuration [7]
(4) 165 + 25π cm2
Step 1
If we look at the f igure, we notice that the area of the region of all the points outside the
equilateral triangle = 3(T he area of the rectangle + T he area of the sector of the circle)
Step 2
T he area of the rectangle = 11 × 5 = 55 cm2
Step 3
Let's assume the angle and the radius of the sector of the circle be 'θ' and 'r' respectively.
Step 4
We know that, each angle of an equilateral triangle and rectangle are 60° and 90°
respectively.
T heref ore, θ = 360° - (90° + 90° + 60°) = 120°,
r = 5 cm
T he area of the sector of the circle =
θ
× π r2
360
=
120
× π (5)2
360
= 25π /3 cm2
Step 5
Now, the area of the region of all the points outside the equilateral triangle within 5 cm =
3(55 + 25π /3) = 165 + 25π cm2
(5)
6
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ID : pk-8-Mensuration [8]
(6) a. 40√2 minutes
Step 1
Let's assume the length of the side of the square is 'a'.
Using Pythagorean T heorem, the length of the diagonal of the square = √(a2 + a2)
= √2(a)2
= a√2
Step 2
Since the mouse walks along the sides of the square, the distance covered by it f rom one
corner of the square to the opposite corner = 2 × √2a = 2a.
Step 3
It is given that the mouse takes 80 minutes to reach the opposite corner.
T heref ore, the time taken by the mouse to cover the distance 2a = 80 minutes.
T his means the time taken by the mouse to cover the distance 'a' = 80/2 minutes
T his means the time taken by the mouse to cover the distance a√2, the length of the
diagonal of the square = 80/2 × √2 = 40√2 minutes.
Step 4
T heref ore, the mouse would take 40√2 minutes, if it had walked along the diagonal.
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(7) d. 59.136 sq. m.
Step 1
Lets assume ABCD is a rectangular screen as shown below.
According to question one side of the screen is 7.04 m. Lets assume the second side of
the screen is x m.
In ΔABC
AB2 + BC2 = AC2
⇒ x2 + 7.04 2 = 10.962
⇒ x2 = 10.962 - 7.04 2
⇒ x = 8.4
Step 2
Now the area of the screen = 7.04 × 8.4
= 59.136 sq. m.
Step 3
T heref ore the area of the screen is 59.136 sq. m.
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(8)
d. 48 cm2
Step 1
T he f ollowing f igure shows the trapezoid ABCD.
It is given that the height of the trapezoid ABCD = 8 cm,
T he perimeter of the trapezoid ABCD = 36 cm.
T he sum of the non-parallel sides of the trapezoid ABCD = BC + DA = 24 cm.
Step 2
T he perimeter of the trapezoid ABCD = AB + BC + CD + DA
⇒ 36 = AB + CD + 24
⇒ 36 - 24 = AB + CD
⇒ 12 = AB + CD
⇒ AB + CD = 12 cm
Step 3
T he area of the trapezoid ABCD =
AB + CD
×h
2
=
12
×8
2
= 48 cm2
Step 4
T hus, the area of the trapezoid is 48 cm2.
(9) a. 792 cm2
Step 1
We know that the volume of a cylinder with height h and radius r is given by: V = π r2h
Here, radius (r) = 6 cm
and height (h) = 7 cm.
Step 2
T heref ore, the volume V =
22
× (6)2 × 7
7
=
22
× 36 × 7
7
= 792 cm3
Step 3
T hus, the volume of the cylinder is 7923.
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ID : pk-8-Mensuration [11]
(10) b. 18.75
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 35 %.
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 35 %,
theref ore the new breadth of the rectangle = b - (b ×
35
)
100
35
= b(1 -
)
100
= b(
100 - 35
)
100
= b(
65
)
100
=
65b
100
Step 5
Now new area of the rectangle =
125l
100
=
×
65b
100
8125lb
10000
Step 6
Change in area of the rectangle = lb -
8125 lb
10000
= lb(1 -
8125
)
10000
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= lb(
10000 - 8125
)
10000
=
1875 lb
10000
Percentage change in area of the rectangle =
1875 lb
10000
=
×
100
lb
1875
100
= 18.75 %
Step 7
T heref ore the percentage change in the area of the rectangle is 18.75 %.
(11)
125
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 50%
theref ore the side of the square = s + s ×
50
100
=
100s + 50s
100
= 1.5s
Now the area of the square = (1.5s)2
= 2.25s2
= 2.25A [Since A = s2]
Step 3
T he area of the square increase = 2.25A - A
= 1.25A
Step 4
T heref ore you can say that the area of the square increase =
1.25
%
100
= 125%
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ID : pk-8-Mensuration [13]
(12)
730.08
Step 1
Since the side of the small squares that make up the given shape is 3.9 cm.
T heref ore the area of the each small square = 3.9 × 3.9 = 15.21 cm2
Step 2
If you count the number of small squares that make up the given shape caref ully, you will
notice that the number of small squares are 48.
Step 3
Now the area of the shape = 48 × 15.21 = 730.08 sq. cm.
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(13)
43.75
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 25%, the new length of the diagonal will be:
25
=d-
d
100
= 0.75d
Step 3
Hence the new area will be 0.5(0.75d)2 = 0.5 × 0.5625d2.
Step 4
T he decrease in area = Old Area - New Area
= 0.5 d2 - 0.5 × 0.5625d2
= 0.5 × (1 - 0.5625) d2
= 0.5 × 0.4375 d2
Step 5
Percentage decrease in area =
Decrease in area
× 100 %
Old Area
=
0.5 × 0.4375 d2
× 100 %
0.5 d2
= 0.4375 × 100 %
= 43.75%
Step 6
Hence, when the diagonal of a square is decreased by 25%, then the area of the square
decreases by 43.75%.
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ID : pk-8-Mensuration [15]
(14)
66
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
× 42 [It is given that the radius of the circle is 42 and π = 22/7]
7
= 264
Step 3
Now, we know that the same wire f rame with length 264 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 264 divided by 4. T hat is:
264/4
= 66
Step 4
T hus the length of the side of the resulting rhombus is 66.
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(15)
3080
Step 1
T he paper has to be f olded along the length in order to get the cylinder of height 5 cm and
the perimeter of the base of the cylinder is equal to the length of the paper.
Step 2
Let the radius of cylinder be r cm.
Since, the base of the cylinder is circular, its perimeter is 2πr.
T heref ore, 2π r = 88
⇒ 2 × 22/7 × r = 88
⇒ 44/7 × r = 88
⇒ r = 88 × 7/44
⇒ r = 14 cm
Step 3
T he radius of the cylinder = 14 cm
Step 4
Now, the volume of the cylinder = π r2h
=
22
× 14 × 14 × 5
7
= 3080 cm3
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