NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

1.A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Answer :

We know that Perimeter of square=$4\times side=4\times 60=240m$
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

$2(l+b)=240$

$?2(80+b)=240$

$?160+2b=240$

$?2b=240?160=80$

$?b=40m$

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

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

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 $m^2$.

Answer :

We have the area of the square plot =$side\times side=25m\times 25m=625m^2$

Area of the rectangular house =$l\times b=20m\times 15m=300m^2$

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

Cost of developing the garden = $?325\times 55=?17875$

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]

Answer :

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 = $20–(3.5+3.5)=20–7=13m$
Therefore, area of the rectangle =$l\times b=13\times 7=91m^2$

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

$=?r^2$

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

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

$=38.5m^2$

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

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

$=2(l+b)+2\times (?r)–2(2r)$

$=2(13+7)+2(\frac{22}{7}\times \frac{7}{2})–4\times \frac{7}{2}$

$=2\times 20+22–14$

$=40+22–14$

$=48m$

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).

Answer :

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

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

So, the number of tiles required=$\frac{Areaofthefloor}{Areaof1tile}$

$=\frac{10800000}{240}$

$=45000tiles$

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.

Answer :

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

$=?r+2r$

$=\frac{22}{7}\times 1.4+2\times 1.4$

$=22\times 0.2+2.8$

$=4.4+2.8=7.2cm$

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

$=?r+5.8$

$=\frac{22}{7}\times 0.2+5.8$

$=4.4+5.8$

$=10.2cm$

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

$=?r+4$

$=\frac{22}{7}\times 1.4+4$

$=4.4+4$

$=8.4cm$

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

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.

Answer :

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

$=\frac{1}{2}\times (1+1.2)\times 0.8m^2$

$=\frac{1}{2}\times 2.2\times 0.8m^2$

$=(1.1\times 0.8)=0.88m^2$

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

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.

Answer :

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

We know that, area of trapezium=$\frac{1}{2}\times (sumofparallelsides)\times (distancesbetweenthem)$

$34=\frac{1}{2}\times (10+x)\times 4$

$34=2(10+x)$

$10+x=17$

$x=7$

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.

Answer :

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=$\frac{1}{2}\times (sumofparallelsides)\times (distancesbetweenthem)$

$=\frac{1}{2}\times (BC+AD)\times AB$

$=\frac{1}{2}\times (48+40)\times 15$

$=\frac{1}{2}\times 88\times 15$

$=44\times 15=660m^2$

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

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.

Answer :

As per the figure we have area of field=area of ?ABD+area of ?BCD

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

$=\frac{1}{2}\times 24\times 13+\frac{1}{2}\times 24\times 8$

$=12\times 13+12\times 8$

$=12\times (13+8)$

$=12\times 21$

$=252m^2$

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

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

Answer :

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

$=\frac{1}{2}\times (d1\times d2)$

$=\frac{1}{2}\times (7.5\times 12)$

$=7.5\times 6$

$=45c{m}^2$

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.

Answer :

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

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

$?24=\frac{1}{2}\times 8\times d2$

$?24=4\times d2$

$?d2=6cm$

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.

Answer :

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 =$\frac{1}{2}\times d1\times d2$

$=\frac{1}{2}\times 45\times 30$

$=45\times 15$

$=675c{m}^2$

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

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

So, the cost of polishing the floor = $202.5\times 4=?810$[? 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.

Answer :

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 = $\frac{1}{2}\times (a+b)\times h$

$?10500=\frac{1}{2}\times (2x+x)\times 100$

$?2\times 10500=3x\times 100$

$?21000=300x$

$?x=70m$

So, the other side = $2x=2\times 70=140m$

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.

Answer :

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

$=\frac{1}{2}\times (a+b)\times h$

$=\frac{1}{2}\times (11+5)\times 4$

$=\frac{1}{2}\times 16\times 4$

$=32m^2$

Area of rectangle HCDG =$l\times b=11m\times 5m=55m^2$

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

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

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?

Answer :

(i) From Jyoti’s diagram:


We can observe that, Area of the pentagonal shape = Area of trapezium ABCD + Area of trapezium ADEF

$=2\times AreaoftrapeziumABCD$

$=2\times \frac{1}{2}\times (a+b)\times h$

$=(15+30)\times 7.5$

$=45\times 7.5$

$=337.5m^2$

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

$=\frac{1}{2}\times b\times h+15\times 15$

$=\frac{1}{2}\times 15\times 15+225$

$=112.5+225=337.5m^2$

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

Yes, there is other way to find the area which is as follows:



Required area ABCDEA=Area of rect. ABPQ-2$\times$area of triangle CPD

$=(15\times 30)?(2\times \frac{1}{2}\times \frac{15}{2})$

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

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

11.Diagram of the picture frame has outer dimensions = $24cm\times 28cm$ and inner dimensions $16cm\times 20cm$. Find the area of each section of the frame, if the width of each section is the same.

Answer :

We have:
$h_1=\frac{1}{2}(28?20)=\frac{1}{2}\times 8=4cm$

$h_2=\frac{1}{2}(24?16)=\frac{1}{2}\times 8=4cm$

We also know, area of trapezium A=$\frac{1}{2}\times (a+b)\times h_1$

$=\frac{1}{2}\times (24+16)\times 4$

$=\frac{1}{2}\times 48\times 4=80c{m}^2$

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

Now, area of trapezium B=Area of trapezium D

$=\frac{1}{2}\times (28+20)\times 4$

$=\frac{1}{2}\times 48\times 4=96c{m}^2$

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=$96c{m}^2$

1.There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Answer :

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

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

$=200\times 74c{m}^2=14800c{m}^2$

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

$=6\times 50\times 50c{m}^2=15000c{m}^2$

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 $80cm\times 48cm\times 24cm$ is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Answer :

Measurement of the suitcase =$80cm\times 48cm\times 24cm$

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

$=2(80\times 48+48\times 24+80\times 24)c{m}^2$

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

$=2\times 6912c{m}^2=13824c{m}^2$

We have, Area of tarpaulin = Area of 100 suitcase

Area of Area of 100 suitcase=$100\times 13824$

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

Equating both we have:

$96l=100\times 13824$

$?l=100\times 144=14400cm=144m$

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

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

Answer :

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

$?6l^2=600$

$?l^2=100$

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

Hence, the required length of side = 10 cm.

4.Rukhsar painted the outside of the cabinet of measure $1m\times 2m\times 1.5m$. How much surface area did she cover if she painted all except the bottom of the cabinet.

Answer :

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

$=2(lb+bh+hl)–lb$

$=2(2\times 1.5+1.5\times 1+1\times 2)–2\times 1$

$=2(3+1.5+2)–2$

$=2(6.5)–2$

$=13–2=11m^2$

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?

Answer :

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

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

$=2(lb+bh+hl)–lb$

$=2(15\times 10+10\times 7+7\times 15)–15\times 10$

$=2(150+70+105)–150$

$=2(325)–150$

$=650–150=500m^2$

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?

Answer :

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

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

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

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?

Answer :

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

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

$=2\times \frac{22}{7}\times 7\times (3+7)$

$=2\times \frac{22}{7}\times 7\times 10$

$=440m^2$

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

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.

Answer :

Given, The lateral surface area of a hollow cylinder is 4224 $c{m}^2$ and the width of rectangular sheet =33cm
We know that the circumference of the base of the cylinder is equal to the length of the sheet and the height of the cylinder is equal to the breadth of the sheet when a hollow cylinder is cut and a rectangular sheet is formed. Therefore, the lateral surface area of the cylinder is equal to the area of the rectangular sheet.

Let the length of the sheet be l.

? we have ,area of the sheet = Lateral surface area of cylinder

$?l\times 33=4224$