NCERT Solutions for Class 8 Maths Chapter 9 Mensuration  Free PDF Download
Class 8 students have a lot on their plate in terms of academics. A subject like Maths requires them to spend enough time in solving assignments and preparing for exams. Vedantu’s team of scholars understand the needs and understanding level of Class 8 students and have designed NCERT Solution for Class 8 Maths Chapter 9 keeping all of this in mind. NCERT Solution Class 8 Maths Chapter 9 Mensuration Solution is entirely based on the latest CBSE curriculum. It will be able to clarify all the doubts that you might have in the Mensuration chapter. The problems in the Class 8 Maths Chapter 9 are well explained with diagrams and stepbystep explanations to make it simple for you to follow the solution.
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Access NCERT Solutions for Class 8 Maths Chapter 9 – Mensuration
Exercise 9.1
Q1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Ans: Let Side of a Square be ‘\[{\text{a}}\]’.
Length and Breadth of Rectangle are ‘\[{\text{l}}\]’ and ‘\[{\text{b}}\]’ respectively.
Perimeter of Square =$4$(side of square) = $4$(\[{\text{a}}\])
= $4$($60$${\text{m}}$) = $240{\text{m}}$
Perimeter of Rectangle = $2$(Length+Breadth) = $2$(\[{\text{l + b}}\])
= $2$($80{\text{m}}$+ \[{\text{b}}\])
Now, It is given that the perimeter of the square and the perimeter of the rectangle are equal.
So, $240{\text{m}}$= $2$($80{\text{m}}$+ \[{\text{b}}\])
$240{\text{m}}$= $160{\text{m}}$+$2$\[{\text{b}}\]
$80{\text{m}}$= 2 \[{\text{b}}\]
$40{\text{m}}$= \[{\text{b}}\]
Area of square = $\mathop {{\text{(side)}}}\nolimits^{\text{2}} $= $\mathop {{\text{(60m)}}}\nolimits^{\text{2}} $=$\mathop {{\text{3600m}}}\nolimits^{\text{2}} $
Area of Rectangle = Length x Breadth = $(80 x 40)$$\mathop {\text{m}}\nolimits^{\text{2}} $ = $3200$${{\text{m}}^{\text{2}}}$
Thus, the area of the square field is larger than the area of the rectangular field.
Q2. Mrs. Kaushik has a square plot with the measurement as shown in the following figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ${\text{Rs}}{\text{.}}\;{\text{55}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$?
Ans: Area of the Square Plot = ${{\text{(side)}}^{\text{2}}}$= \[{{\text{(25m)}}^{\text{2}}}\]= ${\text{625}}{{\text{m}}^{\text{2}}}$
Area of the House = Length x Breadth = ${\text{15m x 20m}}$= ${\text{300}}{{\text{m}}^{\text{2}}}$
Now, Area of the remaining portion = Area of the square plot – Area of the house
= ${\text{625}}{{\text{m}}^{\text{2}}}$ ${\text{300}}{{\text{m}}^{\text{2}}}$
= ${\text{325}}{{\text{m}}^{\text{2}}}$
The cost of developing the garden around the house is ${\text{Rs}}{\text{.}}\;{\text{55}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$.
Therefore, total cost of developing the area${\text{325}}{{\text{m}}^{\text{2}}}$ is ${\text{Rs}}{\text{.}}\;{\text{(55 x 325)}}$ = ${\text{Rs}}{\text{.}}\;{\text{17875}}$.
Q3. The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of the garden (Length of rectangle is ${\text{20  (3}}{\text{.5 + 3}}{\text{.5)m}}$)?
Ans: As we have given that length of rectangle = ${\text{[20  (3}}{\text{.5 + 3}}{\text{.5)]m}}$
= ${\text{[20  7]m}}$= ${\text{13m}}$
Breadth = \[{\text{7m}}\].
Now, we have to find the circumference of both semi circles.
As, Diameter = ${\text{7m}}$, so, Radius(r) = $\dfrac{7}{{\text{2}}}{\text{m}}$= ${\text{3}}{\text{.5m}}$
Circumference of one semicircle = ${\text{$\pi$ r}}$= $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(3}}{\text{.5)m}}$= ${\text{11m}}$
Circumference of both circles = ${\text{2 x 11m}}$= ${\text{22m}}$
Now, Perimeter of the garden = AB + CD + Length of both semicircular regions AD & BC
= ${\text{(13 + 13 + 22)m}}$
= ${\text{48m}}$
Area of the garden = Area of the rectangle + $2 x $Area of two semicircular regions
= (Length x Breadth) + ${\text{2 x }}\dfrac{{\text{1}}}{{\text{2}}}{\text{$\pi$ }}{{\text{r}}^{\text{2}}}$
= ${\text{[(13 x 7) + 2 x (}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 3}}{\text{.5 x 3}}{\text{.5)]}}{{\text{m}}^{\text{2}}}$
= ${\text{(91 + 38}}{\text{.5)}}{{\text{m}}^{\text{2}}}$
= \[{\text{129}}{\text{.5}}{{\text{m}}^{\text{2}}}\]
Q4. A flooring tile has the shape of a parallelogram whose base is ${\text{24cm}}$and the corresponding height is ${\text{10cm}}$.How many such tiles are required to cover a floor of area ${\text{1080}}{{\text{m}}^{\text{2}}}$? (If required you can split the tiles in whatever way you want to fill up the corners).
Ans: Given that the Base of a parallelogram is ${\text{24cm}}$ and Height of a parallelogram is ${\text{10cm}}$.
Therefore, Area of Parallelogram = Base x Height
= ${\text{(24 x 10)c}}{{\text{m}}^2}$
= ${\text{240c}}{{\text{m}}^2}$.
Therefore, Area of parallelogram = Area of one tile.
Now, we have to find the number of tiles.
Given that Area of floor = ${\text{1080}}{{\text{m}}^{\text{2}}}$
So, Number of tiles = $\dfrac{{{\text{Area}}\;{\text{of}}\;{\text{floor}}}}{{{\text{Area}}\;{\text{of}}\;{\text{one}}\;{\text{tile}}}}$
= $\dfrac{{{\text{1080}}{{\text{m}}^{\text{2}}}}}{{{\text{240c}}{{\text{m}}^{\text{2}}}}}$ = $\dfrac{{{\text{1080 x 10000}}}}{{{\text{240}}}}{\text{c}}{{\text{m}}^{\text{2}}}$ \[(\because \;1{\text{m}} = 100{\text{cm}})\]
= $45000$tiles.
Hence, $45000$tiles are required to cover a floor of area ${\text{1080}}{{\text{m}}^{\text{2}}}$
Q5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food − piece would the ant have to take a longer round? Remember,
circumference of a circle can be obtained by using the expression c =${\text{2$\pi$ r}}$, where r is
the radius of the circle.
Ans: (a) Diameter(\[{\text{d}}\]) = ${\text{2}}{\text{.8cm}}$, so radius(\[{\text{r}}\]) = $\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{cm}}$= ${\text{1}}{\text{.4cm}}$
Therefore, Perimeter = ${\text{d + $\pi$ r}}$
= ${\text{2}}{\text{.8cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$
= ${\text{[2}}{\text{.8 + (0}}{\text{.2 x 22)]cm}}$
= ${\text{(2}}{\text{.8 + 4}}{\text{.4)cm}}$
= ${\text{7}}{\text{.2cm}}$
(b) Diameter( \[{\text{d}}\]) = ${\text{2}}{\text{.8cm}}$, so radius(\[{\text{r}}\]) = $\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{cm}}$= ${\text{1}}{\text{.4cm}}$
Therefore, Perimeter = ${\text{1}}{\text{.5cm + 2}}{\text{.8cm + 1}}{\text{.5cm + $\pi$ (1}}{\text{.4cm)}}$
= ${\text{5}}{\text{.8cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$
= ${\text{5}}{\text{.8cm + 4}}{\text{.4cm}}$
= ${\text{10}}{\text{.2cm}}$.
(c) Radius(\[{\text{r}}\]) = $\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{cm}}$= ${\text{1}}{\text{.4cm}}$
Perimeter = ${\text{2cm + $\pi$ r + 2cm}}$
= ${\text{4cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$
= ${\text{4cm + 4}}{\text{.4cm}}$
= ${\text{8}}{\text{.4cm}}$.
Thus, the ant will have to take a longer round for the food piece (b) because its perimeter is ${\text{10}}{\text{.2cm}}$ which is the greatest among all.
Exercise 9.2
Q1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are ${\text{1m}}$and ${\text{1}}{\text{.2m}}$and perpendicular distance between them is${\text{0}}{\text{.8m}}$.
Ans: Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ ( Sum of parallel sides) x (Distances between parallel sides) }}$
= $\dfrac{{\text{1}}}{{\text{2}}}{\text{[(1m + 1}}{\text{.2m) x 0}}{\text{.8m]}}$
= $\dfrac{{\text{1}}}{{\text{2}}}{\text{(1}}{\text{.76}}{{\text{m}}^{\text{2}}}{\text{)}}$= ${\text{0}}{\text{.88}}{{\text{m}}^{\text{2}}}$
Q2. The area of a trapezium is ${\text{34c}}{{\text{m}}^{\text{2}}}$and the length of one of the parallel sides is ${\text{10cm}}$and its height is ${\text{4cm}}$. Find the length of the other parallel side.
Ans: It is given that Area of Trapezium = ${\text{34c}}{{\text{m}}^{\text{2}}}$.
Length of one parallel side = ${\text{10cm}}$
Height = ${\text{4cm}}$
Now, let length of other parallel side = ‘\[{\text{a}}\]’ cm
Therefore, Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ ( Sum of parallel sides) x (Distances between parallel sides) }}$
${\text{34c}}{{\text{m}}^{\text{2}}}$ = $\dfrac{{\text{1}}}{{\text{2}}}{\text{[(10 + a) x 4]cm}}$
${\text{34c}}{{\text{m}}^{\text{2}}}$ = ${\text{(20 + 2a)cm}}$
$\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm}}$ = \[{\text{a}}\] cm
Thus, Length of the other parallel side (\[{\text{a}}\]) = ${\text{7cm}}$.
Q3. Length of the fence of a trapezium shaped field ABCD is ${\text{120m}}$. If BC = ${\text{48m}}$, CD =${\text{17m}}$ and AD =${\text{40m}}$, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.
Ans: Length of the fence of Trapezium ABCD = AB+BC+CD+DA
${\text{120m}}$ =${\text{(AB + 48 + 17 + 40)m}}$
${\text{120m}}$ =${\text{AB + 105m}}$
Therefore, AB = ${\text{15m}}$
Area of the field ABCD = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(AD + BC) x AB}}$
=$\dfrac{{\text{1}}}{{\text{2}}}{\text{(40 + 48) x 15}}$
=$\dfrac{{\text{1}}}{{\text{2}}}{\text{(88) x 15}}$
= ${\text{660}}{{\text{m}}^{\text{2}}}$
Q4. The diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8m and 13m. Find the area of the field.
Ans: It is given that Length of diagonal = ${\text{24m}}$
Length of the perpendiculars ${{\text{h}}_{\text{1}}}$and ${{\text{h}}_{\text{2}}}$ from the opposite vertices to the diagonal are
${{\text{h}}_{\text{1}}}$= ${\text{8m}}$ and ${{\text{h}}_{\text{2}}}$= ${\text{13m}}$
Area of Quadrilateral = $\dfrac{{\text{1}}}{{\text{2}}}{\text{d}}\left( {{{\text{h}}_{\text{1}}}{\text{ + }}{{\text{h}}_{\text{2}}}} \right)$
= $\dfrac{{\text{1}}}{{\text{2}}}{\text{(24m) x (13m + 8cm)}}$
= $\dfrac{{\text{1}}}{{\text{2}}}{\text{(24m)(21m)}}$
= ${\text{252}}{{\text{m}}^{\text{2}}}$
Thus, Area of field = ${\text{252}}{{\text{m}}^{\text{2}}}$
Q5. The diagonals of a rhombus are ${\text{7}}{\text{.5cm}}$and ${\text{12cm}}$. Find its area.
Ans: Area of Rhombus = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Product of its diagonals) }}$
= $\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 7}}{\text{.5cm x 12cm}}$
= ${\text{45c}}{{\text{m}}^{\text{2}}}$
Q6. Find the area of a rhombus whose side is ${\text{6cm}}$and whose altitude is ${\text{4cm}}$. If one of its diagonals is ${\text{8cm}}$long, find the length of the other diagonal.
Ans: Let the length of other diagonal of Rhombus be ‘\[{\text{X}}\]’
A Rhombus is a special case of Parallelogram
The area of Parallelogram is given by its base and height
Thus, Area of Rhombus = Base x Height
= ${\text{6cm x 4cm}}$ = ${\text{24c}}{{\text{m}}^{\text{2}}}$
So,
Area of Rhombus = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Product of its diagonals) }}$
${\text{24c}}{{\text{m}}^{\text{2}}}$ = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(8cm x X)}}$
${\text{X = }}\;{\text{6cm}}$
Therefore, Length of other diagonal of Rhombus = ${\text{6cm}}$
Q7. The floor of a building consists of $3000$tiles which are rhombus shaped and each of its diagonals are ${\text{45cm}}$and ${\text{30cm}}$in length. Find the total cost of polishing the floor, if the cost per ${{\text{m}}^{\text{2}}}$is Rs$4$.
Ans: Given that each diagonals of Rhombus are ${\text{45cm}}$and ${\text{30cm}}$
Area of Rhombus = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Product of its diagonals) }}$
= $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 45 x 30}}} \right){\text{c}}{{\text{m}}^{\text{2}}}$
= ${\text{675c}}{{\text{m}}^{\text{2}}}$
So, Area of $3000$ tiles = $(675 x 3000){\text{c}}{{\text{m}}^2} = 2025000\;{\text{c}}{{\text{m}}^2} = 202.5\;{{\text{m}}^2}$
Now, it is given that cost of polishing is ${\text{Rs}}{\text{.}}\;{\text{4}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$
So, Cost of Polishing ${\text{202}}{\text{.5}}{{\text{m}}^{\text{2}}}$ area = ${\text{Rs(4 x 202}}{\text{.5) = Rs 810}}$
Hence, Cost of polishing the floor is ${\text{Rs}}{\text{.}}\;{\text{810}}$
Q8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. It the area of this field is ${\text{10500}}{{\text{m}}^{\text{2}}}$and the perpendicular distance between the two parallel sides is ${\text{100m}}$, find the length of the side along the river.
Ans: Let the length of the side along the road = ‘\[{\text{l}}\]’
And Let the length of the side along the river = ‘${\text{2l}}$’
It is given that distance between two parallel sides = ${\text{100m}}$
and Area of Trapezium = ${\text{10500}}{{\text{m}}^{\text{2}}}$
Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$
${\text{10500}}{{\text{m}}^{\text{2}}}$ = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(l + 2l) x (100m)}}$
${\text{3l = }}\left( {\dfrac{{{\text{2 x 10500}}}}{{{\text{100}}}}} \right){\text{m = 210m}}$
${\text{l = }}\;\dfrac{{{\text{210}}}}{{\text{3}}}{\text{m}}\;{\text{ = }}\;{\text{70m}}$
Therefore, Length of the side along the river ‘${\text{2l}}$’ =${\text{140m}}$
Q9. Top 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.
Ans:
It is given in the figure that side of octagon = ${\text{5cm}}$
Area of Trapezium ABCH = Area of Trapezium DEFG
Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$
= $\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{(4)(11 + 5)}}} \right]{{\text{m}}^{\text{2}}}$
= $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 4 x 16}}} \right){{\text{m}}^{\text{2}}}{\text{ = 32}}{{\text{m}}^{\text{2}}}$
In rectangle HCDG, Length(\[{\text{l}}\]) = ${\text{11m}}$and Breadth(\[{\text{b}}\]) = ${\text{5m}}$
So, Area of rectangle = ${\text{(11 x 5)}}{{\text{m}}^{\text{2}}}$= ${\text{55}}{{\text{m}}^{\text{2}}}$
Therefore, Area of octagon = Area of Trapezium ABCH + Area of Trapezium DEFG +
Area of Rectangle
= ${\text{(32 + 32 + 55)}}{{\text{m}}^{\text{2}}}$
= ${\text{119}}{{\text{m}}^{\text{2}}}$
Q10. There 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?
Ans: From Jyoti’s Way of finding area ,
Area of Pentagon = $2$ (Area of Trapezium ABCF)
= $2$[ $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$]
=$\left[ {{\text{2 x }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(15 + 30)}}\left( {\dfrac{{{\text{15}}}}{{\text{2}}}} \right)} \right]{{\text{m}}^{\text{2}}}$
= ${\text{337}}{\text{.5}}{{\text{m}}^{\text{2}}}$
From Kavita’s Way of finding area ,
Area of Pentagon = Area of Triangle ABE + Area of Square BCDE
= [ $\dfrac{1}{2}$(basexheight)] + (sidexside)
= $\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 15 x (30  15) + (15}}{{\text{)}}^{\text{2}}}} \right]{{\text{m}}^{\text{2}}}$
= $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 15 x 15 + 225}}} \right){{\text{m}}^{\text{2}}}$
= ${\text{(112}}{\text{.5 + 225)}}{{\text{m}}^{\text{2}}}$
= ${\text{337}}{\text{.5}}{{\text{m}}^{\text{2}}}$
Q11. Diagram of the adjacent picture frame has outer dimensions = ${\text{24cm}}$× ${\text{28cm}}$and inner dimensions ${\text{16cm}}$× ${\text{20cm}}$ Find the area of each section of the frame, if the width of each section is same.
Ans:
Given that, the width of each section is the same.
Therefore,
IB = BJ = CK = CL = DM = DN = AO = AP
IL = IB + BC + CL
\[28\ =\ \text{IB}\ +\ 20\ +\ \text{CL}\]
$\begin{align} & 2820\ \text{=}\ \text{IB+}\ \text{CL} \\ & 8\ \text{=}\ \text{IB+}\ \text{CL} \\ \end{align}$
IB = CL = \[{\text{4cm}}\]
Hence, IB = BJ = CK = CL = DM = DN = AO = AP = $\text{4}\,\text{cm}$
Area of section BEFC = Area of section DGHA = Area of Trapezium
$\left[ \frac{\text{1}}{\text{2}}\text{(bas}{{\text{e}}_{1}}\text{ + bas}{{\text{e}}_{2}}\text{)(h)} \right]=\left[ \frac{1}{2}(20\text{ }+\text{ }28)(4) \right]\text{c}{{\text{m}}^{\text{2}}}\text{=}96\ \text{c}{{\text{m}}^{\text{2}}}$
Therefore, Area of section ABEH = Area of section CDGF.
Hence area of each section of frame is $96\ \text{c}{{\text{m}}^{\text{2}}}$
Exercise 9.3
Q1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Ans: From the given figure; Length(\[{\text{l}}\]), Breadth(\[{\text{b}}\]) and Height(\[{\text{h}}\]) of the Cuboid is
\[{\text{60cm}}\],${\text{40cm}}$, ${\text{50cm}}$ respectively.
And, Side of Cube is ${\text{50cm}}$.
Now, Total Surface area of Cuboid(a) = $2$(\[{\text{l}}\] \[{\text{h}}\] + \[{\text{b}}\] \[{\text{h}}\] +\[{\text{l}}\] \[{\text{b}}\])
= \[{\text{[2\{ (60)(40) + (40)(50) + (50)(60)\} ]c}}{{\text{m}}^{\text{2}}}\]
= ${\text{[2(2400 + 2000 + 3000)]c}}{{\text{m}}^{\text{2}}}$
= ${\text{(2 x 7400)c}}{{\text{m}}^{\text{2}}}$
= ${\text{14800c}}{{\text{m}}^{\text{2}}}$
Total Surface area of Cube(b) = \[{\text{6}}{{\text{l}}^{\text{2}}}\]
= $6{(50\;{\text{cm}})^2} = 15000\;{\text{c}}{{\text{m}}^2}$
Therefore, Cuboidal box(a) requires a lesser amount of material for making.
Q2. A suitcase with measure ${\text{80cm x 48cm x 24cm}}$is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width ${\text{96cm}}$is required to cover $100$ such suitcases?
Ans: Given that length(\[{\text{l}}\]),breadth(\[{\text{b}}\]),height(\[{\text{h}}\]) of suitcase is \[{\text{(80,48,24)cm}}\]respectively.
Total surface area of suitcase = $2$(\[{\text{l}}\] \[{\text{h}}\] + \[{\text{b}}\] \[{\text{h}}\] + \[{\text{l}}\] \[{\text{b}}\])
= ${\text{2[(80)(48) + (48)(24) + (24)(80)]}}$
= \[{\text{2[3840 + 1152 + 1920]}}\]
= ${\text{13824}}{{\text{m}}^{\text{2}}}$
Total surface area of $100$ suitcases = ${\text{100 x 13824c}}{{\text{m}}^{\text{2}}}$
= ${\text{1382400c}}{{\text{m}}^{\text{2}}}$
= required Tarpaulin
We have given that breadth of Tarpaulin is ${\text{96cm}}$ and we have to find Length
of tarpaulin.
Required Tarpaulin = (Length x Breadth) of Tarpaulin
${\text{1382400c}}{{\text{m}}^{\text{2}}}$ = Length x ${\text{96cm}}$
Length = $\left( {\dfrac{{{\text{1382400}}}}{{{\text{96}}}}} \right){\text{cm = 14400cm}}$
Therefore, Length = \[{\text{144m}}\;\;\;\;\;(\because \;{\text{1m = 100cm)}}\]
Thus, ${\text{144m}}$ of tarpaulin is required to cover $100$ suitcases.
Q3. Find the side of a cube whose surface area is ${\text{600c}}{{\text{m}}^{\text{2}}}$ .
Ans: It is given that the surface area of the cube is ${\text{600c}}{{\text{m}}^{\text{2}}}$ .
We have to find the side of the cube (\[{\text{a}}\]).
Surface area of cube = ${\text{6}}{{\text{a}}^{\text{2}}}$
${\text{600c}}{{\text{m}}^{\text{2}}}$ = ${\text{6}}{{\text{a}}^{\text{2}}}$
$\therefore \;{{\text{a}}^{\text{2}}}{\text{ = 100c}}{{\text{m}}^{\text{2}}}$
\[{\text{a}}\] = ${\text{10cm}}$
Thus, the side of the cube is ${\text{10cm}}$ .
Q4. Rukhsar painted the outside of the cabinet of measure${\text{1m x 2m x 1}}{\text{.5m}}$. How much surface area did she cover if she painted all except the bottom of the cabinet?
Ans: It is given that length (\[{\text{l}}\]), breadth(\[{\text{b}}\]), height(\[{\text{h}}\]) of the cabinet is
${\text{2m,1m,1}}{\text{.5m}}$respectively.
Area of the surface = ${\text{2h(1 + b) + lb}}$
= ${\text{[2 x 1}}{\text{.5 x (2 + 1) + (2)(1)]}}{{\text{m}}^{\text{2}}}$
= ${\text{[3(3) + 2]}}{{\text{m}}^{\text{2}}}$
= ${\text{11}}{{\text{m}}^{\text{2}}}$
Q5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of ${\text{15m,10m and 7m}}$respectively. From each can of paint ${\text{100}}{{\text{m}}^{\text{2}}}$ of area is painted. How many cans of paint will she need to paint the room?
Ans: Given that length (\[{\text{l}}\]), breadth(\[{\text{b}}\]), height(\[{\text{h}}\]) of Cuboid is
${\text{15m,10m and 7m}}$ respectively.
Area of the hall to be painted = Area of the wall + Area of ceiling
= $2$ \[{\text{h}}\] (\[{\text{l}}\] + \[{\text{b}}\]) + \[{\text{l}}\]\[{\text{b}}\]
= ${\text{[2 x 7(15 + 10) + (15 x 10)]}}{{\text{m}}^{\text{2}}}$
= ${\text{[(2 x 175) + 150]}}{{\text{m}}^{\text{2}}}$
= ${\text{500}}{{\text{m}}^{\text{2}}}$
It is given that ${\text{100}}{{\text{m}}^{\text{2}}}$ area is to be painted from each can.
Therefore, Number of cans required to paint an area of ${\text{500}}{{\text{m}}^{\text{2}}}$
= $\dfrac{{500}}{{100}}{\kern 1pt} \; = \;5$
Hence, $5$ cans of paint are required to paint the room.
Q6. Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?
Ans: The above given two figures alike for same height(\[{\text{h}}\])= ${\text{7cm}}$
And the difference between in these two figures is that one is cylinder and the other
One is a cube.
Now, we have to find the lateral surface area for both of the given figures.
Given that Side of cube(\[{\text{l}}\])= ${\text{7cm}}$
Height and Diameter of cylinder = ${\text{7cm}}$each
Radius of cylinder = $\dfrac{{{\text{Diameter}}}}{{\text{2}}}\;{\text{ = }}\;\dfrac{{\text{7}}}{{\text{2}}}{\text{cm = }}\;{\text{3}}{\text{.5cm}}$
First, Lateral surface area of Cube = ${\text{4}}{{\text{l}}^{\text{2}}}$=${\text{4(}}{{\text{7}}^{\text{2}}}{\text{)c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;{\text{4 x 49c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;{\text{196c}}{{\text{m}}^{\text{2}}}$
Second, Lateral surface area of cylinder = ${\text{2$\pi$ rh}}$
= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 3}}{\text{.5cm x 7cm}}\;{\text{ = }}\;{\text{154c}}{{\text{m}}^{\text{2}}}$
Therefore, the Cube has a larger lateral surface area.
Q7. A closed cylindrical tank of radius ${\text{7m}}$ and height ${\text{3m}}$ is made from a sheet of metal. How much sheet of metal is required?
Ans: Given that the radius and height of Cylinder is ${\text{7m}}$ and ${\text{3m}}$ respectively.
Therefore, Total surface area of Cylinder = ${\text{2$\pi$ }}$\[{\text{r}}\] (\[{\text{r}}\] + \[{\text{h}}\])
= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7m(7m + 3m)}}$
= ${\text{440}}{{\text{m}}^2}$
Thus, ${\text{440}}{{\text{m}}^2}$ of metal sheet is required.
Q8. The lateral surface area of a hollow cylinder is ${\text{4224c}}{{\text{m}}^{\text{2}}}$. It is cut along its height and formed a rectangular sheet of width ${\text{33cm}}$. Find the perimeter of the rectangular?
Ans: It is given that Hollow cylinder is cut along its height and formed a
Rectangular sheet.
Area of cylinder = ${\text{4224c}}{{\text{m}}^{\text{2}}}$
And, Breadth of rectangular sheet = ${\text{33cm}}$
So, Area of Cylinder = Area of Rectangular Sheet
${\text{4224c}}{{\text{m}}^{\text{2}}}$ = Length x Breadth
${\text{4224c}}{{\text{m}}^{\text{2}}}$ = Length x ${\text{33cm}}$
Length = \[\dfrac{{{\text{4224}}}}{{{\text{33}}}}{\text{cm}}\;{\text{ = }}\;{\text{128cm}}\]
Now, Perimeter of Rectangle = $2$(length + breadth)
= ${\text{2(128 + 33)cm}}\;{\text{ = }}\;{\text{2(161cm)}}\;{\text{ = }}\;{\text{322cm}}$
Q9. A road roller takes ${\text{750}}$complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is ${\text{84cm}}$ and length is ${\text{1m}}$.
Ans: In one revolution, the roller will cover an area equal to its lateral surface area.
Here, Radius = $\dfrac{{{\text{diameter}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{84}}}}{{\text{2}}}{\text{cm = 42cm = }}\dfrac{{{\text{42}}}}{{{\text{100}}}}{\text{m}}\;\;\;\;\;\;(\because {\text{1m = 100cm}})$
Height = ${\text{1m}}$
Thus, In One Revolution,
Area of the road covered = ${\text{2$\pi$ }}$rh
= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{{\text{42}}}}{{{\text{100}}}}{\text{ x 1}}{{\text{m}}^{\text{2}}}$
= $\dfrac{{{\text{264}}}}{{{\text{100}}}}{{\text{m}}^{\text{2}}}$
In ${\text{750}}$revolutions, area of road covered = ${\text{750 x }}\dfrac{{{\text{264}}}}{{{\text{100}}}}{{\text{m}}^{\text{2}}}$
= \[{\text{1980}}{{\text{m}}^{\text{2}}}\]
Q10. A company packages its milk powder in a cylindrical container whose base has a diameter of ${\text{14cm}}$and height ${\text{20cm}}$. Company places a label around the surface of the container (as shown in the figure). If the label is placed ${\text{2cm}}$from top and bottom, what is the area of the label?
Ans: It is given that Radius = $\dfrac{{{\text{diameter}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm = 7cm}}$
Height of label = ${\text{(20  2  2)cm}}\;\;\;\;\;\;(\because {\text{2cm}}\;{\text{deducted}}\;{\text{from}}\,{\text{top}}\;{\text{bottom}}\;{\text{each)}}$
= ${\text{16cm}}$
As shown in the figure, the label is in the shape of a cylinder.
So, Area of label(cylinder)= ${\text{2$\pi$ }}$rh
= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7cm x 16cm}}\;{\text{ = }}\;{\text{44 x 16c}}{{\text{m}}^{\text{2}}}$
= ${\text{704c}}{{\text{m}}^{\text{2}}}$
Exercise 9.4
Q1. 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.
Ans: (a) In this situation, we will find the volume.
(b) Number of cement bags required to plaster cylindrical bags so for that
situation, we will find the surface area.
(c) Number of smaller tanks that can be filled with so for that situation, we will
find the volume.
Q2. Diameter of cylinder A is ${\text{7cm}}$, and the height is ${\text{14cm}}$. Diameter of cylinder B is ${\text{14cm}}$ and height is ${\text{7cm}}$. 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?
Ans: The heights and diameters of these cylinders A and B are interchanged.
We know that,
Volume of cylinder = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$
If measures of radius(r) and height(h) are same, then the cylinder with greater
radius will have greater area.
Here, Radius of cylinder A = $\dfrac{{\text{7}}}{{\text{2}}}{\text{cm}}$
Radius of cylinder B = $\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm = 7cm}}$
As the radius of cylinder B is greater, therefore, the volume of cylinder B will
be greater.
Let us verify it by calculating the volume of both the cylinders.
Volume of Cylinder A = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$
= $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x 14c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{11 x 49c}}{{\text{m}}^{\text{3}}}\;{\text{ = 539c}}{{\text{m}}^{\text{3}}}$
Volume of Cylinder B = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$
= $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7 x 7 x 7c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{22 x 49c}}{{\text{m}}^{\text{3}}}\;{\text{ = 1078c}}{{\text{m}}^{\text{3}}}$
Therefore, the volume of cylinder B is greater.
Now, Surface area of cylinder A = ${\text{2$\pi$ r(r + h)}}$
= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x (}}\dfrac{{\text{7}}}{{\text{2}}}{\text{ + 14)c}}{{\text{m}}^{\text{2}}}$
= \[{\text{22 x }}\dfrac{{{\text{35}}}}{{\text{2}}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{ = }}\;{\text{385c}}{{\text{m}}^{\text{2}}}\]
Surface area of cylinder B = ${\text{2$\pi$ r(r + h)}}$
= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7 x (7 + 7)c}}{{\text{m}}^{\text{2}}}$
= \[{\text{44 x 14c}}{{\text{m}}^{\text{2}}}{\text{ = }}\;{\text{616c}}{{\text{m}}^{\text{2}}}\]
Q3. Find the height of a cuboid whose base area is ${\text{180c}}{{\text{m}}^{\text{2}}}$ and volume is ${\text{900c}}{{\text{m}}^{\text{3}}}$?
Ans. Here, we have given that Base Area of Cuboid = Length x Breadth
= ${\text{180c}}{{\text{m}}^{\text{2}}}$
Volume of Cuboid = Length x Breadth x Height
${\text{900c}}{{\text{m}}^{\text{3}}}$ = ${\text{180c}}{{\text{m}}^{\text{2}}}$ x Height
Height = $\dfrac{{{\text{900}}}}{{{\text{180}}}}{\text{cm = 5cm}}$
Q4. A cuboid is of dimensions ${\text{60cm x 54cm x 30cm}}$. How many small cubes with side${\text{6cm}}$can be placed in the given cuboid?
Ans. From given condition,
Volume of Cuboid = ${\text{60cm x 54cm x 30cm}}$
= ${\text{97200c}}{{\text{m}}^{\text{3}}}$
Given that side of cube = ${\text{6cm}}$
So, Volume of cube = ${\text{(6 x 6 x 6)c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{216c}}{{\text{m}}^{\text{3}}}$
Required number of cubes = \[\dfrac{{{\text{volume}}\;{\text{of}}\;{\text{cuboid}}}}{{{\text{volume}}\;{\text{of}}\;{\text{cube}}}}{\text{ = }}\dfrac{{{\text{97200}}}}{{{\text{216}}}}{\text{ = 450}}\]
Therefore, ${\text{450}}$ cubes can be placed in the given Cuboid.
Q5. Find the height of the cylinder whose volume is ${\text{1}}{\text{.54}}{{\text{m}}^{\text{3}}}$and diameter of the base is ${\text{140cm}}$?
Ans. It is given that Radius of Cylinder = $\dfrac{{{\text{140}}}}{{\text{2}}}{\text{cm = 70cm}}$
Volume of Cylinder = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$
${\text{1}}{\text{.54}}{{\text{m}}^{\text{3}}}$ =$\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{{\text{70}}}}{{{\text{100}}}}{\text{m x }}\dfrac{{{\text{70}}}}{{{\text{100}}}}{\text{m x h}}$
Height =$\dfrac{{{\text{1}}{\text{.54 x 100}}}}{{{\text{22 x 7}}}}{\text{m = 1m}}$
Hence, the height of cylinder = ${\text{1m}}$
Q6. A milk tank is in the form of cylinder whose radius is ${\text{1}}{\text{.5m}}$and length is ${\text{7m}}$. Find the quantity of milk in litres that can be stored in the tank?
Ans. It is given that Radius and height of the cylinder is ${\text{1}}{\text{.5m}}$ and ${\text{7m}}$ respectively.
Therefore, Volume of Cylinder = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$
= $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 1}}{\text{.5m x 1}}{\text{.5m x 7m}}$
= ${\text{22 x 2}}{\text{.25}}{{\text{m}}^{\text{3}}}$
= ${\text{49}}{\text{.5}}{{\text{m}}^{\text{3}}}$
As, ${\text{1}}{{\text{m}}^{\text{3}}}{\text{ = 1000}}\;{\text{litre}}$
So, Required Quantity = ${\text{49}}{\text{.5 x 1000}}\;{\text{litre}}\;{\text{ = }}\;{\text{49500}}\;{\text{litre}}$
Therefore, ${\text{49500}}\;{\text{litre}}$ can be stored in the tank.
Q7. If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Ans. (i) Let the edge of the cube be ‘\[{\text{a}}\]’.
Surface area of cube = ${\text{6}}{{\text{a}}^{\text{2}}}$
If each edge of the cube is doubled, then it becomes ${\text{2a}}$
Therefore, New surface area = ${\text{6(2a}}{{\text{)}}^{\text{2}}}{\text{ = 24}}{{\text{a}}^{\text{2}}}{\text{ = 4(6}}{{\text{a}}^{\text{2}}}{\text{)}}$
Clearly, the surface area will be increased by ${\text{4}}$ times.
(ii) Let Volume of the cube = ${{\text{a}}^{\text{3}}}$
When each edge of the cube is doubled, it becomes${\text{2a}}$.
New volume = \[{{\text{(2a)}}^{\text{3}}}{\text{ = 8}}{{\text{a}}^{\text{3}}}{\text{ = 8 x }}{{\text{a}}^{\text{3}}}\]
Clearly, the volume of the cube will be increased by ${\text{8}}$ times.
Q8. Water is pouring into a cubiodal reservoir at the rate of $60$litres per minute. If the volume of reservoir is ${\text{108}}{{\text{m}}^{\text{3}}}$, find the number of hours it will take to fill the reservoir.
Ans. Volume of cuboidal reservoir = ${\text{108}}{{\text{m}}^{\text{3}}}$ = $(108 x 1000)$ \[{\text{L}}\] = $108000$ \[{\text{L}}\]
It is given that water is being poured at the rate of $60$ L per minute.
That is, $(60 x 60)$ \[{\text{L}}\] = $3600$ \[{\text{L}}\] per hour
Required number of hours = $30$ hours
Thus, it will take $30$ hours to fill the reservoir.
NCERT Solutions for Class 8 Maths Chapter 9 Mensuration  PDF Download
It is not compulsory to be connected to the internet to access our solutions as Vedantu has made NCERT Solutions for Class 8 Maths Mensuration available in the PDF format on its official website. Now download the NCERT Solutions Class 8 Maths Chapter 9 PDF on your devices or get a printout and access it from anywhere, anytime. This mode of revision is quick and easy and can be done at your pace during a crucial exam period.
All Topics of NCERT Class 8 Maths Chapter 9  Mensuration
The topics covered under chapter 9 Mensuration are given below.
S.No.  Topic Name 
9.1  Introduction 
9.2  Let Us Recall 
9.3  Area of Trapezium 
9.4  Area of a General Quadrilateral 
9.5  Area of Polygon 
9.6  Solid Shapes 
9.7  Surface Area of Cube, Cuboid and Cylinder 
9.8  Volume of Cube, Cuboid and Cylinder 
9.9  Volume and Capacity 
Table of Important Formulas
Below given are the list of important formulas that you must remember to solve the exercise of chapter 9 NCERT Maths Class 8th.
S.No.  Shape of the Object  Formula 
1  Triangle  Area of triangle = 12 x Base x Height Perimeter of Triangle = Sum of all three sides 
2  Rectangle  Area of rectangle = Length x Breadth Perimeter of rectangle = 2( length + breadth) 
3  Square  Area of square = side x side Perimeter of square = 4 x side 
4  Circle  Area of circle = π x (radius)2 Perimeter of circle = 2π x radius 
5  Parallelogram  Area of parallelogram = Base x Height Perimeter of parallelogram = 2 (sum of two consecutive sides) 
6  Trapezium  Area of trapezium = height (sum of parallel sides2) 
7  Rhombus  Area of rhombus = 12 x diagonal1 x diagonal2 
8  Cuboid  Total surface area = 2 (length x breadth + breadth x height + height x length) Volume of Cuboid = Area of base x height = length x breadth x height 
9  Cube  Total surface area = 6 x (side)2 Volume of cube = (side)3 
10  Cylinder  Curved (lateral) surface area = 2πrh Total surface area = 2πr (r + h) Volume of cylinder = area of base x height = πr2 x h where r is the radius of base and h is the height of cylinder. 
You should also remember some of the basic conversion parameters. For your ease, the most important parameters are given below.
1 m3 = 1000000 cm3 = 1000 L
1 cm3 = 1 mL
1 L = 1000 cm3
Chapter 9 – Mensuration
Introduction
Mensuration is the process applied to different 2D and 3D solids of various shapes and figures to measure their lengths, volumes, area, heights, perimeters and several other dimensions. In this section of Ch 9 Maths Class 8, you will recall the areas of plane figures like triangles, circles, rectangles, etc., that you learned in the previous chapter. In NCERT Solutions for Class 8 Maths Ch 9, you would learn how to calculate perimeter and areas of other closed figures like Quadrilaterals.
Let us Recall
In this section of Mensuration Class 8 NCERT, students would recall the formula for calculating the perimeter and area of a park, flower bed, and amount of cement required to cover a given area. All the problems are based on areas of the following shape:
Rectangle  The area of a rectangle is x*y where x is the length and y is the breadth of the rectangle.
Square  The area of a square is x2 where x is the length of one side of a square.
Triangle  The area of a triangle is ½ *b*h where b is the length of the base and h is the height of the triable
Circle  The area of a circle is πr2 where r is the radius of the circle.
Area of a Trapezium
A trapezium is a quadrilateral where two of the sides are parallel to each other. In this portion of Mensuration Class 8 NCERT Solutions, students will learn how to derive the area of a trapezium which is given by:
Area of a trapezium = h * (x + y)/2, where h is the height of the trapezium, x and y are the lengths of its two sides.
Area of a general Quadrilateral
A simple definition of a general quadrilateral is a closed 2D shape having 4 straight sides. If we break the word Quadrilateral, Quad means 4, and lateral means sides. The area of a quadrilateral in Class 8th Maths Chapter 9 is calculated by splitting it into two triangles and then calculating and adding the areas of the two triangles.
(Image to be added soon)
So if PQRS is a quadrilateral, then its area = (area of triangle PQR) + (area of triangle PRS) = ½ * d * (h1 + h2), where d is the length of the diagonal from P to R and h1 and h2 are heights of perpendiculars dropped from Q and Pr and S or PR respectively.
9.8 Volume of Cube, Cuboid, and Cylinder
The amount of space that a 3D object occupies gives the volume of that object. To take examples from real life, the volume of a cupboard in a room is less than the volume of the room where it is placed.
9.8.1  Volume of a Cuboid  A 3D structure with 6 rectangular faces is a cuboid. Its volume is given by “l * h* b”, here l = length, b = breadth, and h= height.
9.8.2  Volume of a Cube  A cube is a special type of cuboid where its length, breadth, and height are the same. Hence the volume of a cube = length 3.
9.8.3  Volume of a Cylinder  A cylinder has two circular bases that are parallel to each other and separated by a distance. To measure the volume of a cylinder we use the formula πr2 * h. Here r is the radius of the circular base and h is the distance between the bases.
Key Features of NCERT Solutions for Class 8 Maths Chapter 9
You will find the NCERT Solutions of Class 8 Maths Chapter 9 by Vedantu extremely beneficial for your exams. The key features are:
Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.
Clear and structured presentation for easy comprehension.
Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.
Visual aids like diagrams and illustrations to simplify complex concepts.
Additional tips and insights to enhance students' performance.
Chapter summaries for quick revision.
Online accessibility and downloadable resources for flexible study and revision.
Other Study Material for CBSE Class 8 Maths Chapter 9
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Conclusion
NCERT Solutions play a crucial role in Class 8 exam prep. Start by thoroughly reading the textbook chapter. After that, solve the NCERT questions for Class 8 Chapter 9. You can find detailed solutions on Vedantu, aligning with CBSE guidelines. Download the free NCERT Solutions for Class 8 Chapter 9 to guide your exam preparation with expertreviewed answers.
Chapterwise NCERT Solutions for Class 8 Maths
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FAQs on NCERT Solutions for Class 8 Maths Chapter 9  Mensuration
1. What are some of the applications of mensuration in our everyday lives?
In real life, mensuration is applied in many fields like:
Measuring floor and site areas required for purchasing or selling land.
Measuring agricultural fields.
Measuring surface areas of a house required for estimating painting cost.
To know the volume of the level of water in rivers and tanks.
To find out the amount of carpet required for covering a specific room.
The volume of soil that is needed to fill a ditch.
2. What are solid shapes and what is the method of measuring the surface area of a solid shape?
Any 3D shapes occupying some space are called solid shapes like cube, sphere, cylinder, etc. To measure the surface area of a solid shape, we need to draw the net of that solid shape. From the net, we can see all the faces of the solid clearly. Then, we calculate the areas of each of the faces and add them up to get the total surface area of the solid shape. Surface area is measured in a square unit.
3. What are the concepts covered under Chapter 9 of Class 8 Maths?
The ideas or topics that are included in Chapter 9 “Mensuration” of Class 8 Maths are given below:
Introduction
Let Us Recall
Area of Trapezium
Area of a General Quadrilateral
Area of Special Quadrilaterals
Area of a Polygon
Solid Shapes
Surface Area of Cylinder, Cube and Cuboid
Cuboid
Cube
Cylinders
Volume of Cuboid, Cube and Cylinder
Cuboid
Cube
Cylinder
Volume and Capacity
What Have We Discussed?
4. The base area of the cuboid is 25cm sq. Its volume is 275 cubic cm. What will be the height of the cuboid?
In the question, we are given the base area of the cuboid which is equal to 25cm sq.
The volume of the cuboid is 275 cubic cm.
We know that according to the formula,
The volume of a cuboid = Height × Base Area
Therefore, the height of the cuboid will be = Volume of cuboid/ Base Area
Height = 275/25 = 11cm
Thus, 11cm is the height of the cuboid.
5. The distance between two parallel sides is 15m and the length of one parallel side is 20m. 480m sq. Is the area of the trapeziumshaped field. What is the length of the other parallel side?
Let one parallel side be a = 20m and the other parallel side is ‘b’.
The height of the field is 15m.
Given, the area of the trapezium is 480m sq.
The formula of trapezium is,
Area of trapezium = ½ (a + b) * h
480 = ½ (20 + b) * 15
20 + b = 480 × 2/ 15
64 = 20 + b
b = 44m
Thus, the length of the other parallel side is 44m.
6. What are the perks of NCERT Solutions of Chapter 9 of Class 8 Maths?
The perks of the NCERT Solutions of Chapter 9 of Class 8 Maths are given below:
The NCERT Solutions of Chapter 9 of Class 8 Maths offers comprehensive learning.
It enables students to develop their reasoning and logical skills.
These solutions assist students in understanding the difficult concepts.
By practising these, students will have a strong grip over the chapter.
You will get a hint of how to answer the questions in the proper format.
It helps in scoring good marks in Chapter 9 of Class 8 Maths.
The NCERT Solutions for Chapter 9 of Class 8 Maths are available free of cost on the Vedantu website and on the Vedantu app.
7. How can I make the best study plan for Chapter 9 of Class 8 Maths?
Keep the following points in mind while making an effective study plan for Chapter 9 of Class 8 Maths:
Have a timetable or schedule to manage your time.
Centralize the NCERT book to read Chapter 9 of Class 8 Maths.
Practice the NCERT Solutions to comprehend the Chapter.
Give yourself a break.
Do meditation and exercise to keep your body and mind fit.
Attend all your school lectures.