# NCERT Solutions for Class 8 Maths Chapter 11 Mensuration  Written by Team Trustudies
Updated at 2021-02-11

## NCERT solutions for class 8 Maths Chapter 11 Mensuration Exercise 11.1

1.A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area? NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

We know that Perimeter of square=$4×side=4×60=240m$
Also, we have perimeter of rectangle=2(l+b)

Given that these both have same parimeter.

Therefore, we can say that:

perimeter of rectangle=Perimeter of square

$2\left(l+b\right)=240$

$?2\left(80+b\right)=240$

$?160+2b=240$

$?2b=240?160=80$

$?b=40m$

Hence we have, Area of figure (a) i.e, square =$\left(side{\right)}^{2}$ = $60×60=3600{m}^{2}$

Area of figure (b) i.e, rectangle =$l×b=80×40=3200{m}^{2}$

So, the area of figure (a) i.e, square is more than the area of figure (b) i.e, rectangle.

2.Mrs Kaushik has a square plot with the measurement as shown in the 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 ? 55 per ${m}^{2}$. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

We have the area of the square plot =$side×side=25m×25m=625{m}^{2}$

Area of the rectangular house =$l×b=20m×15m=300{m}^{2}$

Therefore, Area of the garden to be developed = Area of the plot – Area of the house = $625{m}^{2}–300{m}^{2}=325{m}^{2}$

Cost of developing the garden = $?325×55=?17875$

3.The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden. [Length of rectangle is 20 – (3.5 + 3.5) metres] NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

We have,Total area of the garden = Area of the rectangular portion + The sum of the areas of the pair of semi-circles
We have, Length of the rectangle = $20–\left(3.5+3.5\right)=20–7=13m$
Therefore, area of the rectangle =$l×b=13×7=91{m}^{2}$

Area of two circular ends = $2×\frac{1}{2}?{r}^{2}$

$=?{r}^{2}$

$=\frac{22}{7}×\frac{7}{2}×\frac{7}{2}$

$=\frac{77}{2}{m}^{2}$

$=38.5{m}^{2}$

Total area = Area of the rectangle + Area of two ends = $91{m}^{2}+38.5{m}^{2}=129.5{m}^{2}$

Total perimeter = Perimeter of the rectangular portion + Perimeter of two circular ends

$=2\left(l+b\right)+2×\left(?r\right)–2\left(2r\right)$

$=2\left(13+7\right)+2\left(\frac{22}{7}×\frac{7}{2}\right)–4×\frac{7}{2}$

$=2×20+22–14$

$=40+22–14$

$=48m$

So the total perimeter of the figure =48m

4.A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given that area of floor=$1080{m}^{2}=1080×10000c{m}^{2}=10800000c{m}^{2}$[? $1{m}^{2}=100c{m}^{2}$]

Area of one tile= $1×base×height=1×24×10=240c{m}^{2}$

So, the number of tiles required=$\frac{Area\phantom{\rule{thickmathspace}{0ex}}of\phantom{\rule{thickmathspace}{0ex}}the\phantom{\rule{thickmathspace}{0ex}}floor}{Area\phantom{\rule{thickmathspace}{0ex}}of\phantom{\rule{thickmathspace}{0ex}}1\phantom{\rule{thickmathspace}{0ex}}tile}$

$=\frac{10800000}{240}$

$=45000tiles$

Hence, the reuired number of tiles=45000 tiles

5.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, the circumference of a circle can be obtained by using the expression C = $2?r$, where r is the radius of the circle. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

(a)Distance covered by the ant to take a round:
$=\frac{1}{2}×2?r+2r$

$=?r+2r$

$=\frac{22}{7}×1.4+2×1.4$

$=22×0.2+2.8$

$=4.4+2.8=7.2cm$

(b)Distance covered by the ant to take a round:
$=\frac{1}{2}×2?r+1.5+1.5+2.8$

$=?r+5.8$

$=\frac{22}{7}×0.2+5.8$

$=4.4+5.8$

$=10.2cm$

(c)Distance covered by the ant to take a round:
$=\frac{1}{2}×2?r+2+2$

$=?r+4$

$=\frac{22}{7}×1.4+4$

$=4.4+4$

$=8.4cm$

So, among all the given paths, in path (b) ant has to travel a larger distance for food piece than others.

## NCERT solutions for class 8 Maths Chapter 11 Mensuration Exercise 11.2

1.The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Area of top trapezium surface of the table=(\frac{1}2\times (sum\; of \;parallel \;sides)\times (distances \;between\; them)\)

$=\frac{1}{2}×\left(1+1.2\right)×0.8{m}^{2}$

$=\frac{1}{2}×2.2×0.8{m}^{2}$

$=\left(1.1×0.8\right)=0.88{m}^{2}$

So, the area of top trapezium surface of the table=$0.88{m}^{2}$

2.The area of a trapezium is $34c{m}^{2}$ and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, area of trapezium is $34c{m}^{2}$.
Let the required side be x cm.

We know that, area of trapezium=$\frac{1}{2}×\left(sum\phantom{\rule{thickmathspace}{0ex}}of\phantom{\rule{thickmathspace}{0ex}}parallel\phantom{\rule{thickmathspace}{0ex}}sides\right)×\left(distances\phantom{\rule{thickmathspace}{0ex}}between\phantom{\rule{thickmathspace}{0ex}}them\right)$

$34=\frac{1}{2}×\left(10+x\right)×4$

$34=2\left(10+x\right)$

$10+x=17$

$x=7$

Hence, the required side is 7cm.

3.Length of the fence of a trapezium A shaped field ABCD is 120 m. IF BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given:
BC = 48 m, CD = 17 m, AD = 40 m

AB + BC + CD + DA = 120 m .

AB = 120 m – (48 m + 17 m + 40 m) = 120 – 105 m = 15 m

So the area of trapezium fence ABCD=$\frac{1}{2}×\left(sum\phantom{\rule{thickmathspace}{0ex}}of\phantom{\rule{thickmathspace}{0ex}}parallel\phantom{\rule{thickmathspace}{0ex}}sides\right)×\left(distances\phantom{\rule{thickmathspace}{0ex}}between\phantom{\rule{thickmathspace}{0ex}}them\right)$

$=\frac{1}{2}×\left(BC+AD\right)×AB$

$=\frac{1}{2}×\left(48+40\right)×15$

$=\frac{1}{2}×88×15$

$=44×15=660{m}^{2}$

So, the required area=$660{m}^{2}$

4.The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration As per the figure we have area of field=area of ?ABD+area of ?BCD

$=\frac{1}{2}×base×height+\frac{1}{2}×base×height$

$=\frac{1}{2}×24×13+\frac{1}{2}×24×8$

$=12×13+12×8$

$=12×\left(13+8\right)$

$=12×21$

$=252{m}^{2}$

Hence the required area of the given field=$252{m}^{2}$

5.The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given,d1 = 7.5 cm, d2 = 12 cm Area of a rhombus =$\frac{1}{2}x\left(productofdiagonals\right)$

$=\frac{1}{2}×\left(d1×d2\right)$

$=\frac{1}{2}×\left(7.5×12\right)$

$=7.5×6$

$=45c{m}^{2}$

So the area of rhombus=45 $c{m}^{2}$

6.Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, Side = 5 cm, Altitude = 4.8 cm and length of one diagonal(d1) = 8 cm
We know, area of the rhombus = $Side×Altitude=5×4.8=24c{m}^{2}$

Area of the rhombus =$\frac{1}{2}×d1×d2$

$?24=\frac{1}{2}×8×d2$

$?24=4×d2$

$?d2=6cm$

Hence, the length of other diagonal = 6 cm.

7.The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per ${m}^{2}$ is ? 4.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, number of tiles = 3000 and the length of the two diagonals of a tile = 45 cm and 30 cm
We have area of one rhombus shaped tile =$\frac{1}{2}×d1×d2$

$=\frac{1}{2}×45×30$

$=45×15$

$=675c{m}^{2}$

So, area of one tile=675$c{m}^{2}$

Therefore, Area covered by 3000 tiles = $3000×675c{m}^{2}=2025000c{m}^{2}=202.5{m}^{2}$

So, the cost of polishing the floor = $202.5×4=?810\phantom{\rule{1em}{0ex}}$[? cost per ${m}^{2}$ is ? 4. ]

Hence, the required cost = ? 810.

8.Mohan 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 ${m}^{2}$ and the perpendicular distance between the two parallel sides is 100m, find the length of the side along the river. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, area of trapezium field area=10500${m}^{2}$, Altitude of the field(h)=100m.
Let, the side of the trapezium near road be x cm and the opposite parallel side = 2x m

We know, area of trapezium = $\frac{1}{2}×\left(a+b\right)×h$

$?10500=\frac{1}{2}×\left(2x+x\right)×100$

$?2×10500=3x×100$

$?21000=300x$

$?x=70m$

So, the other side = $2x=2×70=140m$

Hence, the required length of the side along the river= 140 m.

9.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. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

We can observe, Area of the octagonal surface= area of trapezium ABCH + area of rectangle HCDG + area of trapezium GDEF
Finding area of different parts:

Area of trapezium ABCH = Area of trapezium GDEF

$=\frac{1}{2}×\left(a+b\right)×h$

$=\frac{1}{2}×\left(11+5\right)×4$

$=\frac{1}{2}×16×4$

$=32{m}^{2}$

Area of rectangle HCDG =$l×b=11m×5m=55{m}^{2}$

Area of the octagonal surface = $32{m}^{2}+55{m}^{2}+32{m}^{2}=119{m}^{2}$

Hence, the required area = $119{m}^{2}$.

10.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?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

(i) From Jyoti’s diagram: We can observe that, Area of the pentagonal shape = Area of trapezium ABCD + Area of trapezium ADEF

$=2×Area\phantom{\rule{thickmathspace}{0ex}}of\phantom{\rule{thickmathspace}{0ex}}trapezium\phantom{\rule{thickmathspace}{0ex}}ABCD$

$=2×\frac{1}{2}×\left(a+b\right)×h$

$=\left(15+30\right)×7.5$

$=45×7.5$

$=337.5{m}^{2}$

So, the area of the pentagonal shape=337.5${m}^{2}$

(ii) From Kavita’s diagram: We can see that, area of the pentagonal shape = Area of triangle ABE + Area of square BCDE

$=\frac{1}{2}×b×h+15×15$

$=\frac{1}{2}×15×15+225$

$=112.5+225=337.5{m}^{2}$

So, the area of the pentagonal shape =$337.5{m}^{2}$

Yes, there is other way to find the area which is as follows: Required area ABCDEA=Area of rect. ABPQ-2$×$area of triangle CPD

$=\left(15×30\right)?\left(2×\frac{1}{2}×\frac{15}{2}\right)$

$=\left(450?\frac{225}{2}\right)=\frac{\left(900?225\right)}{2}{m}^{2}$

$=\frac{675}{2}{m}^{2}=337.5{m}^{2}$

11.Diagram of the picture frame has outer dimensions = $24cm×28cm$ and inner dimensions $16cm×20cm$. Find the area of each section of the frame, if the width of each section is the same. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

We have:
${h}_{1}=\frac{1}{2}\left(28?20\right)=\frac{1}{2}×8=4cm$

${h}_{2}=\frac{1}{2}\left(24?16\right)=\frac{1}{2}×8=4cm$

We also know, area of trapezium A=$\frac{1}{2}×\left(a+b\right)×{h}_{1}$

$=\frac{1}{2}×\left(24+16\right)×4$

$=\frac{1}{2}×48×4=80c{m}^{2}$

so, the area of trapezium A=area of trapezium C=80$c{m}^{2}$

Now, area of trapezium B=Area of trapezium D

$=\frac{1}{2}×\left(28+20\right)×4$

$=\frac{1}{2}×48×4=96c{m}^{2}$

Hence, we have areas of all the parts as:

Part A=$80c{m}^{2}$

Part B=$96c{m}^{2}$

Part C=$80c{m}^{2}$

Part D=$96c{m}^{2}$

## NCERT solutions for class 8 Maths Chapter 11 Mensuration Exercise 11.3

1.There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make? NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

We have, the total surface area of first box=$2\left(lb+bh+lh\right)$
$=2\left(60×40+40×50+60×50\right)c{m}^{2}$

$=200\left(24+20+30\right)c{m}^{2}$

$=200×74c{m}^{2}=14800c{m}^{2}$

Also, the total surface area of second box= $6\left(Edge{\right)}^{2}$

$=6×50×50c{m}^{2}=15000c{m}^{2}$

So, the first box i.e., (a) requires the least amount of material to make because the total surface area of first box is less than that of the second.

2.A suitcase of measures $80cm×48cm×24cm$ is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Measurement of the suitcase =$80cm×48cm×24cm$

We can say that l = 80 cm, b = 48 cm and h = 24 cm So the total surface area of the suitcase = $2\left[lb+bh+hl\right]$

$=2\left(80×48+48×24+80×24\right)c{m}^{2}$

$=2\left(3840+1152+1920\right)c{m}^{2}$

$=2×6912c{m}^{2}=13824c{m}^{2}$

We have, Area of tarpaulin = Area of 100 suitcase

Area of Area of 100 suitcase=$100×13824$

Also, we have the Area of tarpaulin = $length×breadth=l×96=96lc{m}^{2}$

Equating both we have:

$96l=100×13824$

$?l=100×144=14400cm=144m$

Hence, the required length of the cloth = 144 m.

3.Find the side of a cube whose surface area is $600c{m}^{2}$?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, the surface area of cube is $600c{m}^{2}$
Total surface area of a cube =$6{l}^{2}$

$?6{l}^{2}=600$

$?{l}^{2}=100$

$?l=\sqrt{100}=10cm$

Hence, the required length of side = 10 cm.

4.Rukhsar painted the outside of the cabinet of measure $1m×2m×1.5m$. How much surface area did she cover if she painted all except the bottom of the cabinet. NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, l = 2 m, b = 1.5 m, h = 1 m
Area of the surface to be painted = Total surface area of box – Area of base of box

$=2\left(lb+bh+hl\right)–lb$

$=2\left(2×1.5+1.5×1+1×2\right)–2×1$

$=2\left(3+1.5+2\right)–2$

$=2\left(6.5\right)–2$

$=13–2=11{m}^{2}$

Hence, the required area = 11 ${m}^{2}$.

5.Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 ${m}^{2}$ of area is painted.How many cans of paint will she need to paint the room?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, l = 15 m, b = 10 m and h = 7 m
Area of the paint in one can = $100{m}^{2}$

So,Surface area of a cuboidal hall without bottom = Total surface area – Area of base

$=2\left(lb+bh+hl\right)–lb$

$=2\left(15×10+10×7+7×15\right)–15×10$

$=2\left(150+70+105\right)–150$

$=2\left(325\right)–150$

$=650–150=500{m}^{2}$

Therefore, the number of cans required =$\frac{500}{100}=5cans$.

6.Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area? NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

The two figures given differ in their shapes.One of them is cylinder and the other is cube.

Similarity: Both the given figures have the same height.

Given, Cylinder: d = 1 cm, h = 7 cm

Cube: Length of each side a = 7 cm

So, Lateral surface of cylinder = 2$?$rh

$=2×\frac{22}{7}×\frac{7}{2}×7=154c{m}^{2}$

Also, the Lateral surface of the cube =$4{l}^{2}=4×\left(7{\right)}^{2}=4×49=196c{m}^{2}$

So, the cube has the larger lateral surface =$196c{m}^{2}$.

7.A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, r = 7 m and h = 3 m.
Sheet required to make a closed cylinder= Total surface area of the cylinder

$=2?r\left(r+h\right)$ sq. units

$=2×\frac{22}{7}×7×\left(3+7\right)$

$=2×\frac{22}{7}×7×10$

$=440{m}^{2}$

So, the required metal sheet=$440{m}^{2}$

8.The lateral surface area of a hollow cylinder is 4224 $c{m}^{2}$. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Given, The lateral surface area of a hollow cylinder is 4224 $c{m}^{2}$ and the width of rectangular sheet =33cm
$?l×33=4224$