Mensuration  Class 8  Mathematics
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Examples  Mensuration  NCERT  Mathematics  Class 8
The area of a trapezium shaped field is $480 \mathrm{~m}^{2}$, the distance between two parallel sides is $15 \mathrm{~m}$ and one of the parallel side is $20 \mathrm{~m}$Find the other parallel side.
The area of a trapezium can be calculated using the formula: $$ \text{Area} = \frac{1}{2} (\text{a} + \text{b}) \times h $$ where $\text{a}$ and $\text{b}$ are the lengths of the parallel sides, and $h$ is the distance between these sides.
Given the area of the trapezium is $480 , \text{m}^{2}$, the distance $h$ between the parallel sides is $15 , \text{m}$, and one of the parallel sides $\text{a}$ is $20 , \text{m}$, we can rearrange the formula to solve for the other parallel side $\text{b}$:
$$ 480 = \frac{1}{2} (20 + b) \times 15 $$
Solving for $b$, we get:
$$ \begin{align*} 480 & = \frac{1}{2} (20 + b) \times 15 \ \frac{480}{15} & = \frac{1}{2} (20 + b) \ \frac{960}{15} & = 20 + b \ 64 & = 20 + b \ b & = 44 \ \end{align*} $$
So, the length of the other parallel side is $44 , \text{m}$
The area of a rhombus is $240 \mathrm{~cm}^{2}$ and one of the diagonals is $16 \mathrm{~cm}$. Find the other diagonal.
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Sign up nowThere is a hexagon MNOPQR of side $5 \mathrm{~cm}$ (Fig 9.6). Aman and Ridhima divided it in two different ways (Fig 9.7). Find the area of this hexagon using both ways.
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Sign up nowAn aquarium is in the form of a cuboid whose external measures are $80 \mathrm{~cm} \times 30 \mathrm{~cm} \times 40 \mathrm{~cm}$. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed?
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Sign up nowThe internal measures of a cuboidal room are $12 \mathrm{~m} \times 8 \mathrm{~m} \times 4 \mathrm{~m}$. Find the total cost of whitewashing all four walls of a room, if the cost of white washing is ₹ 5 per $\mathrm{m}^{2}$. What will be the cost of white washing if the ceiling of the room is also whitewashed.
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Sign up nowIn a building there are 24 cylindrical pillars. The radius of each pillar is $28 \mathrm{~cm}$ and height is $4 \mathrm{~m}$. Find the total cost of painting the curved surface area of all pillars at the rate of $₹ 8$ per $\mathrm{m}^{2}$.
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Sign up nowFind the height of a cylinder whose radius is $7 \mathrm{~cm}$ and the total surface area is $968 \mathrm{~cm}^{2}$.
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Sign up nowFind the height of a cuboid whose volume is $275 \mathrm{~cm}^{3}$ and base area is $25 \mathrm{~cm}^{2}$.
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Sign up nowA godown is in the form of a cuboid of measures $60 \mathrm{~m} \times 40 \mathrm{~m} \times 30 \mathrm{~m}$. How many cuboidal boxes can be stored in it if the volume of one box is $0.8 \mathrm{~m}^{3}$ ?
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Sign up nowA rectangular paper of width $14 \mathrm{~cm}$ is rolled along its width and a cylinder of radius $20 \mathrm{~cm}$ is formed. Find the volume of the cylinder (Fig 9.31). (Take $\frac{22}{7}$ for $\pi$ )
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Sign up nowA rectangular piece of paper $11 \mathrm{~cm} \times 4 \mathrm{~cm}$ is folded without overlapping to make a cylinder of height $4 \mathrm{~cm}$. Find the volume of the cylinder.
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Sign up nowExercise 9.1  Mensuration  NCERT  Mathematics  Class 8
The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are $1 \mathrm{~m}$ and $1.2 \mathrm{~m}$ and perpendicular distance between them is $0.8 \mathrm{~m}$.
The area of a trapezium can be found using the formula:
$$ \text{Area} = \frac{1}{2} \times (\text{sum of the lengths of the parallel sides}) \times (\text{perpendicular distance between the parallel sides}) $$
Given the lengths of the parallel sides are $1 \mathrm{~m}$ and $1.2 \mathrm{~m}$, and the perpendicular distance between them is $0.8 \mathrm{~m}$, we can substitute these values into the formula:
$$ \text{Area} = \frac{1}{2} \times (1 + 1.2) \times 0.8 $$
$$ = \frac{1}{2} \times 2.2 \times 0.8 $$
$$ = 1.1 \times 0.8 $$
$$ = 0.88 \mathrm{~m}^2 $$
Therefore, the area of the top surface of the table, which is a trapezium, is $0.88 \mathrm{~m}^2$.
The area of a trapezium is $34 \mathrm{~cm}^{2}$ and the length of one of the parallel sides is$10 \mathrm{~cm}$ and its height is $4 \mathrm{~cm}$. Find the length of the other parallel side.
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Sign up nowLength of the fence of a trapezium shaped field $A B C D$ is $120 \mathrm{~m}$. If $B C=48 \mathrm{~m}, C D=17 \mathrm{~m}$ and $A D=40 \mathrm{~m}$, find the area of this field. Side $\mathrm{AB}$ is perpendicular to the parallel sides $\mathrm{AD}$ and $\mathrm{BC}$.
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Sign up nowThe diagonal of a quadrilateral shaped field is $24 \mathrm{~m}$ and the perpendiculars dropped on it from the remaining opposite vertices are $8 \mathrm{~m}$ and $13 \mathrm{~m}$. Find the area of the field.
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Sign up nowThe diagonals of a rhombus are $7.5 \mathrm{~cm}$ and $12 \mathrm{~cm}$. Find its area.
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Sign up nowFind the area of a rhombus whose side is $5 \mathrm{~cm}$ and whose altitude is $4.8 \mathrm{~cm}$. If one of its diagonals is $8 \mathrm{~cm}$ long, find the length of the other diagonal.
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Sign up nowThe floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are $45 \mathrm{~cm}$ and $30 \mathrm{~cm}$ in length. Find the total cost of polishing the floor, if the cost per $\mathrm{m}^{2}$ is $₹4$ .
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Sign up nowMohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is $10500 \mathrm{~m}^{2}$ and the perpendicular distance between the two parallel sides is $100 \mathrm{~m}$, find the length of the side along the river.
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Sign up nowTop surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
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Sign up nowThere is a pentagonal shaped park as shown in the figure.
For finding its area Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some other way of finding its area?
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Sign up nowDiagram of the adjacent picture frame has outer dimensions $=24 \mathrm{~cm} \times 28 \mathrm{~cm}$ and inner dimensions $16 \mathrm{~cm} \times 20 \mathrm{~cm}$. Find the area of each section of the frame, if the width of each section is same.
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Sign up nowExercise 9.2  Mensuration  NCERT  Mathematics  Class 8
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make? Image (a) shows a rectangular cuboid with dimensions 60 cm by 40 cm by 50 cm. Image (b) illustrates a cube with each edge measuring 50 cm
To find the amount of material required to make each box, we need to calculate the surface area of both and compare them. The surface area of a cuboid is given by the formula $$2(lw + lh + wh)$$, where $l$ is the length, $w$ is the width, and $h$ is the height of the cuboid. The surface area of a cube, on the other hand, is given by $$6a^2$$, where $a$ is the length of an edge of the cube.
Box (a)  Cuboidal Box
Given dimensions: $l = 60, \text{cm}$, $w = 40, \text{cm}$, and $h = 50, \text{cm}$.
Substituting these into the formula, we get: $$ \text{Surface area} = 2(60\cdot40 + 60\cdot50 + 40\cdot50) = 2(2400 + 3000 + 2000) = 2(7400) = 14800, \text{cm}^2 $$
Box (b)  Cube
Given edge length: $a = 50, \text{cm}$.
Substituting into the formula, we get: $$ \text{Surface area} = 6\cdot50^2 = 6\cdot2500 = 15000, \text{cm}^2 $$
Conclusion:
The cuboidal box (a) requires $14800, \text{cm}^2$ of material to make, while the cube (b) requires $15000, \text{cm}^2$ of material. Therefore, box (a), the cuboidal box, requires the lesser amount of material to make.
A suitcase with measures $80 \mathrm{~cm} \times$ $48 \mathrm{~cm} \times 24 \mathrm{~cm}$ is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width $96 \mathrm{~cm}$ is required to cover 100 such suitcases?
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Sign up nowFind the side of a cube whose surface area is $600 \mathrm{~cm}^{2}$.
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Sign up nowRukhsar painted the outside of the cabinet of measure $1 \mathrm{~m} \times 2 \mathrm{~m} \times 1.5 \mathrm{~m}$. How much surface area did she cover if she painted all except the bottom of the cabinet.
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Sign up nowDaniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of $15 \mathrm{~m}, 10 \mathrm{~m}$ and $7 \mathrm{~m}$ respectively. From each can of paint $100 \mathrm{~m}^{2}$ of area is painted. How many cans of paint will she need to paint the room?
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Sign up nowDescribe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area? The illustration depicts two 3D shapes: a cylinder with a height and diameter of 7 cm, and a cube with each edge measuring 7 cm
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Sign up nowA closed cylindrical tank of radius $7 \mathrm{~m}$ and height $3 \mathrm{~m}$ is made from a sheet of metal. How much sheet of metal is required?
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Sign up nowThe lateral surface area of a hollow cylinder is $4224 \mathrm{~cm}^{2}$. It is cut along its height and formed a rectangular sheet of width $33 \mathrm{~cm}$. Find the perimeter of rectangular sheet?
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Sign up nowA road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is $84 \mathrm{~cm}$ and length is $1 \mathrm{~m}$.
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Sign up nowA company packages its milk powder in cylindrical container whose base has a diameter of $14 \mathrm{~cm}$ and height $20 \mathrm{~cm}$. Company places a label around the surface of the container (as shown in the figure). If the label is placed $2 \mathrm{~cm}$ from top and bottom, what is the area of the label.
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Sign up nowExercise 9.3  Mensuration  NCERT  Mathematics  Class 8
Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.

To find how much it can hold: This situation requires finding the volume of the cylindrical tank. The volume indicates the capacity of the tank, showing how much liquid it can store.

Number of cement bags required to plaster it: For this, you will need to find the surface area of the cylindrical tank. The surface area is necessary to estimate the amount of plaster (or any other surface covering material) required to cover the entire exterior of the tank.

To find the number of smaller tanks that can be filled with water from it: Again, this situation calls for finding the volume of the cylindrical tank. Knowing the volume of the larger tank and the volume of each smaller tank, you can calculate how many smaller tanks can be filled.
Diameter of cylinder A is $7 \mathrm{~cm}$, and the height is $14 \mathrm{~cm}$. Diameter of cylinder $\mathrm{B}$ is $14 \mathrm{~cm}$ and height is $7 \mathrm{~cm}$. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
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Sign up nowFind the height of a cuboid whose base area is $180 \mathrm{~cm}^{2}$ and volume is $900 \mathrm{~cm}^{3}$ ?
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Sign up nowA cuboid is of dimensions $60 \mathrm{~cm} \times 54 \mathrm{~cm} \times 30 \mathrm{~cm}$. How many small cubes with side $6 \mathrm{~cm}$ can be placed in the given cuboid?
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Sign up nowFind the height of the cylinder whose volume is $1.54 \mathrm{~m}^{3}$ and diameter of the base is $140 \mathrm{~cm}$ ?
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Sign up nowA milk tank is in the form of cylinder whose radius is $1.5 \mathrm{~m}$ and length is $7 \mathrm{~m}$. Find the quantity of milk in litres that can be stored in the tank?
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Sign up nowIf each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
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Sign up nowWater is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is $108 \mathrm{~m}^{3}$, find the number of hours it will take to fill the reservoir.
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Sign up nowExtra Questions  Mensuration  NCERT  Mathematics  Class 8
The average weight of a group of 25 men was calculated to be $78.4 , \mathrm{kg}$. It was discovered later that one weight was wrongly entered as $69 , \mathrm{kg}$ instead of $96 , \mathrm{kg}$. What is the correct average?
(A) 80.30
B) 79.48
C) 77.56
(D) 75.76
To find the correct average weight of the group after adjusting for the error:

Calculate the initial total weight using the average given for 25 men: $$ \text{Initial total weight} = 25 \times 78.4 = 1960 , \text{kg} $$

Identify the error where the weight was wrongly recorded as $69 , \text{kg}$ instead of the correct $96 , \text{kg}$. The difference due to error is: $$ 96 , \text{kg}  69 , \text{kg} = 27 , \text{kg} $$

Add this difference to the initial total weight to correct the error: $$ \text{Correct total weight} = 1960 , \text{kg} + 27 , \text{kg} = 1987 , \text{kg} $$

Calculate the correct average weight by dividing the correct total weight by the number of men: $$ \text{Correct average weight} = \frac{1987}{25} = 79.48 , \text{kg} $$
Thus, option (B) 79.48 is the correct average weight of the group after correcting the error.
Two full turns of the circular scale of a screw gauge cover a distance of $1 \mathrm{~mm}$ on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of $0.03 \mathrm{~mm}$. While measuring the diameter of a thin wire, a student notes the main scale reading of $3 \mathrm{~mm}$ and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is:
A $\quad 3.32 \mathrm{~mm}$
B $\quad 3.73 \mathrm{~mm}$
C $\quad 3.67 \mathrm{~mm}$
D $\quad 3.38 \mathrm{~mm}$
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