myhelper: Exercise 12E

Exercise 12E

Question 1

Ans: side of a square $=3 \mathrm{~cm}$

when it is enlarged the scale factor $=2$ i.e. $k=2$

Area of the square $=$ side$^{2}=(3)^{2}=9 \mathrm{~cm}^{2}$

so area of the image of square $=k^{2}$

$=(2)^{2} \times 9=4 \times 9=36 \mathrm{~cm}^{2}$

Question 2

Ans: Area of the riangle $=12 \mathrm{~cm}^{2}$

Area of enlarged triangle $=108 \mathrm{~cm} 2$

scale factor $=k$ and then scale factor of area $=k^{2}$

Then $k^{2} \times 12=108$

$\begin{aligned}&k^{2}=\frac{108}{12}=9 \\&\text { So } k=\sqrt{9}=3\end{aligned}$

Question 3

Ans: Sides of given $\triangle A B C, A B=10 \mathrm{~cm}, B C=12 \mathrm{~cm}$ and $A C=16 \mathrm{~cm}$

So longest side AC = 16cm

But longest side of Enlarged $\triangle A^{\prime} B^{\prime} C^{\prime}=24 \mathrm{~cm}$

i.e. $A^{\prime} \mathrm{c'}=24 \mathrm{~cm}$

Let k be the scale factor

Now length of side A'B' = 10cm $times \frac{3}{2}=15 \mathrm{~cm}$

and length of side $B^{\prime} C^{\prime}=12 \mathrm{~cm} \times \frac{3}{2}=18 \mathrm{~cm}$

Question 4

Ans: In right angled $\triangle 1 Q R, \angle Q=90^{\circ}$

$P Q=8 \mathrm{~cm}$ and $Q R=15 \mathrm{~cm}$

So $\begin{aligned} P R &=\sqrt{P Q^{2}+Q R^{2}}=\sqrt{(8)^{2}+1151^{2}} \\=& \sqrt{64+225}=\sqrt{289}=17 \mathrm{~cm} \end{aligned}$

$\triangle P Q R$ is enlarged into $\triangle P'Q'R$ in which image of P is P' of Q is Q' and of R is R' length of P'R'= 42.5cm

(i) $K$ be the scale factor of enlargement

So $42.5=k \times P R=k \times 17$

So $k=\frac{42.5}{17}=2.5=\frac{25}{10}=\frac{5}{2}$.

(ii) Now length of $P^{\prime} Q^{\prime}=\frac{5}{2} \times P Q=\frac{5}{2} \times 8=20 \mathrm{~cm}$ and length of $Q' K^{\prime}=\frac{5}{2} Q K=\frac{5}{2} \times 15=\frac{75}{2}=37.5 \mathrm{~cm}$

(iii) Now area of $\triangle P Q R=\frac{1}{2} \times P Q \times Q R$

$=\frac{1}{2} \times 8 \times 15 \mathrm{~cm}^{2}=60 \mathrm{~cm}^{2}$

So area of $\triangle P^{\prime} Q^{\prime} R^{\prime}=K^{2}$ \times area of $\triangle P Q R$

$=\left(\frac{5}{2}\right)^{2} \times 60 \mathrm{~cm}^{2}=\frac{25}{4} \times 60=375 \mathrm{~cm}^{2}$

Question 5

Ans: scale of map $=1 \mathrm{~cm}: 1 \mathrm{~km}$ or $1: 100000$ Now area of island on the map $=3.5 \mathrm{~cm}^{2}$ so Actual area $=k^{2} \times$ area as the map $=(100000)^{2} \times 3.5 \mathrm{~cm}^{2}$ $=100000 \times 100000 \times 3.5 \mathrm{~cm}^{2}$

$=\frac{100000 \times 100000 \times 3.5}{100000 \times 100000} \mathrm{~km}^{2}$

$=3.5 \mathrm{~km}^{2}$

Question 6

Ans: Scale of map $=5 \mathrm{~m}$ to $1 \mathrm{~cm}$

$\text { or } 500: 1 \Rightarrow k=\frac{500}{1}$

side of a square on the map $=3 \mathrm{~cm}$

So its area $=(\text { side })^{2}=(3 \mathrm{~cm})^{2}=9 \mathrm{~cm}^{2}$

(i) So Actual area of the square = $k^{2} \times$ area of

square on the map $=(500)^{2} \times 9 \mathrm{~cm}^{2}$

$=500 \times 500 \times 9 \mathrm{~cm}^{2}$

$=\frac{500 \times 500 \times 9}{100 \times 100}=225 \mathrm{~m}^{2}$

(ii) Actual area of a square $=50625 \mathrm{~m}^{2}$

So Area on the map $=\frac{1}{k^{2}} \times$ actual area

$\begin{aligned}&=\frac{1}{(500)^{2}} \times 50625 \\&=\frac{50625 \times 100 \times 100}{500 \times 500} \mathrm{~cm}^{2} \\&=2025 \mathrm{~cm}^{2} \\&\text { So side }=\sqrt{2025}=45 \mathrm{~cm}\end{aligned}$

Question 7

Ans: scale af a ground plan of a nouse $=1 \mathrm{~cm}: 15 \mathrm{~m}$

So scale factor (k)= 1cm : 1500cm

$=1 :1500=\frac{1}{1500}$

Now length of a room = 18m and breadth = 12m

So length on plan = $18 \times \frac{1}{1500} m$

$=\frac{18 \times 100}{1500} \mathrm{~cm}=\frac{6}{5} \mathrm{~cm}=1.2 \mathrm{~cm}$

and bread th $=12 \times \frac{1}{1500} \mathrm{~m}=\frac{1200}{1500} \mathrm{~cm}$

$=\frac{4}{5} \mathrm{~cm}=0.8 \mathrm{~cm}$

Area on the plan $=1 \mathrm{~cm}^{2}$

Area on the ground $=1 \times(1500)^{2} \mathrm{~cm}^{2}$

$=\frac{1 \times 1500 \times 1500}{100 \times 100}=225 \mathrm{~m}^{2}$

Question 8

Ans: scale on the map $=1 \mathrm{~cm}$ to $4 \mathrm{~km}$

$\begin{aligned}&=1 \mathrm{~cm}: 400000 \mathrm{dm} \\&=1: 400000 \\&\Rightarrow 1<=\frac{1}{400000}\end{aligned}$

Area of an estate on the map $=9.37 \mathrm{~cm}^{2}$

So Actual area $=k^{2}$ (area on the map $=\left(4000001^{2} \times 9.37 \mathrm{~cm}^{2}\right.$

$=\frac{937 \times(400000)^{2}}{100 \times 100000 \times 100000}$

$=\frac{14992}{100}=149.92 \mathrm{~km}^{2}$

$=150 \mathrm{~km}^{2}$ (nearest $\left.\mathrm{km}^{2}\right)$

Question 9

Ans: scalc of a map $=1: 20000$

measure of hield on the map $=20 \mathrm{~cm} \times 15 \mathrm{~cm}=300 \mathrm{~cm}^{2}$

So Actual area $=K^{2}$ ( Pree of the field on the mapl

$=\left(200001^{2} \times 300 \mathrm{~cm}^{2}\right.$

$=20000 \times 20000 \times 300 \mathrm{~cm}^{2}$

$=\frac{20000 \times 20000 \times 300}{100000 \times 100000} \mathrm{~km}^{2}$

$=12 \mathrm{~km}^{2}$

Question 10

Ans: scale on the map $=1: 50000$

on the map in right angled $\triangle A B C, A B=2 \mathrm{~cm}$

$B C=3.5 \mathrm{~cm}$ and $\angle A B C=90^{\circ}$

So Area $=\frac{1}{2} \times A B \times B C=\frac{1}{2} \times 2 \times 3.5 \mathrm{~cm}^{2}=3.5 \mathrm{~cm} 2$

(i) Now Actual length of $B C=3.5 \mathrm{~cm} \times 50000$

$\begin{aligned}&=\frac{3.5 \times 50000}{100000} \mathrm{~km} \\&=\frac{35 \times 50000}{10 \times 100000}=\frac{35 \times 5}{100} \mathrm{~km}\end{aligned}$

$=\frac{175}{100}=1.75 \mathrm{~km}$

and actual area $=3.5 \mathrm{~cm}^{2} \times\left(500001^{2}\right)$

$=\frac{3.5 \times 50000 \times 50000}{100000 \times 100000} \mathrm{~km}^{2}$

$=\frac{3.5 \times 25}{100}=\frac{3.5}{4} \mathrm{~km}^{2}=0.875 \mathrm{~km}^{2}$

Question 11

Ans Scale factor of the model $=1: 50$

Dimensions of a model of a multistory

building are $1 \mathrm{~m} \times 60 \mathrm{~cm} \times 1.20 \mathrm{~m}$ or $100 \mathrm{~cm} \times 60 \mathrm{~cm} \times 120 \mathrm{~cm}$

So Actual length $=100 \times 50 \mathrm{~cm}=\frac{100 \times 50 \mathrm{~m}}{100}=50 \mathrm{~m}$

$\text { Breadth } m=60 \times 50 \mathrm{~cm}$

$=\frac{60 \times 50}{100}=30 \mathrm{~m}$

and height $=120 \times 50 \mathrm{~cm}$

$=\frac{120 \times 50}{100}=60 \mathrm{~m}$

So Actual dimensions $=50 \mathrm{~m} \times 30 \mathrm{~m} \times 60 \mathrm{~m}$

(i) floor area of a room on the model

$=50 \mathrm{sq} \cdot \mathrm{cm}$

So Actual area $=50 \times k^{2}=50 \times 50 \times 50 \mathrm{~cm}^{2}$

$=\frac{50 \times 2500}{100 \times 100} \mathrm{~m}^{2}=12.5 \mathrm{~m}^{2}$

(ii) Actual volume of the room space $=90 \mathrm{~m}^{3}$

So volume on the model $=\frac{90}{(50)^{3}} \mathrm{~m}^{3}$

$=\frac{90}{50 \times 50 \times 50} \mathrm{~m}^{3}$

$=\frac{90 \times 100 \times 100 \times 100}{50 \times 50 \times 50} \mathrm{~cm}^{3}$

$=\frac{90000}{125} \mathrm{~cm}^{2}$

$=720 \mathrm{~cm}^{3}$

Exercise 12E

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Question 1

Find the volume of each of the following cuboids

(a)

Volume=l×b×h

=6×4×5=120cm³

(b)

Volume=l×b×h

=7×7×10=490cm³

(c)

Volume=l×b×h

=11×8×2=176cm³

(d)

Volume=l×b×h

=3×2×9=54cm³

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Question 2

Find the volume of each of the following cuboids

(a) Length =12 cm, width =7 cm, height =8 cm

Sol :

(a)

Length=12cm, Width=7cm , Height=8cm

Volume=l×b×h=12×7×8=672cm³

(b) Length =16 cm, width =16 cm, height =10 cm

Sol :

(b)

Length=16cm, Width=16cm , Height=10cm

Volume=l×b×h=16×16×10=2560cm³

(c) Length =3.5 m, width =2.8 m, height =2 m

Sol :

(c)

Length=3.5m, Width=2.8m , Height=2m

Volume=l×b×h=3.5×2.8×2=19.6 m³

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Question 3

Find the volume of each cube of side length:

(a) 5 cm

(b) 8 cm

(c) 0.3 m

(d) 50 mm

(e) 1.2 m

Sol :

(a) volume=l³=5³=125 cm³

(b) volume=l³=8³=612 cm³

(d) volume=l³=(50)³=125000 mm³

(e) volume=l³=(1.2)³=1.728 m³

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Question 4

A cuboidal water tank is 7 m long, 5 m width and 4 m deep. How many litres of water can it hold?

Sol :

Tank can hold water= l×b×h

=7×5×4=140 m³

∵1m³=1000l

multiplying both side by 140

(140×1) m³=(140×1000) l

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Question 5

A cuboidal tank is 12 m long and 8 m wide. What should be its height so that it may hold $960 \mathrm{~m}^{3}$ of water?

Sol :

Water it can hold=960m³

Length of tank=l=12m

Width of tank=8m

ATQ,

l×w×h=960

$h=\frac{960}{l\times w}=\frac{960}{12\times 8}$

=10 m

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Question 6

A solid cube of side 12 cm is cut into eight cubes of equal volume. Calculate the size of the new cube.

Sol :

Large cube volume=l³=(12)³

=1728 cm³

∴Large cube is cut into 8 cube

∴Small cube volume$=\frac{1728}{8}$

=216 cm³

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Question 7

A cube of side length 8 cm is immersed completely in a rectangular vessel containing water. If the base is 16 cm long and 10 cm wide, then, find the rise in water level in the vessel.

Sol :

Volume of cube=8³

=512cm³

The water level rise due to immerse of cube.

∴Rise in water level= h cm

The volume of water rise is equal to volume of the cube.

∴Volume of water rise=16×10×h=512

$h=\frac{512}{16\times 10}$

=3.2cm

∴Water level rise=3.2cm

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Question 8

A godown is in the form of a cuboid measuring 40 m by 30 m by 20 m. How many cuboidal boxes of volume $1.2 \mathrm{~m}^{3}$ each can be stored in the godown?

Sol :

Volume of godown=40×30×30=24000m³

1.2m³of volume need=1 cuboidal box

∴24000m³of volume needed$=\frac{24000}{1.2}$

$=\frac{240000}{12}=20000$

∴Number of cuboidal box=20000

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Question 9

A machine for making ice bricks freezes 5.76 litres of water into ice cubes measuring 4 cm by 3 cm by 2 cm. How many ice bricks will be made?

Sol :

We know that:-

1litre=1000cm³

∴5.76litre=5.76×1000=5760cm³

∴Volume of Ice cubes=4×3×2=24cm³

∴Ice bricks will be made$=\frac{5760}{24}=240$

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Question 10

A storage tank at a petrol station is a cuboid 4 m long , 3 m wide and 2 m deep. In one day if the station sells 7200 litres of petrol from the tank, what is the fall in the level of the petrol in the tank?

Sol :

Volume of petrol storage tank=4×3×2=24m³

We know that:-

1Litre=0.001m³

∴7200litre=7200×0.001$=\frac{7200}{1000}$=7.2m³

Let , the fall level=x m

∴ATQ,

4×3×2=7.2

12x=7.2

$x=\frac{7.2}{12}$=0.6m

=60cm or 0.6 m

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Question 11

The platform at the base of a statue is to be built to a height of 1.5 m on a square base measuring 2 m by 2 m. It is to be constructed from bricks of dimensions 25 cm, 15 cm and 10 cm. How many bricks will be required to build the base of the statue?

Sol :

∵1m=100 cm

∴2m=200 cm and 1.5m=150 cm

Volume of required area=200×200×150

=6000000cm³

Single brick dimension=25×15×10cm

=3750cm³

∴Brick needed$=\frac{6000000}{3750}$

=1600

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Question 12

On a certain day 8 cm of rain fell. Calculate the volume of water that fell on 1.8 ha of land.

Sol :

∴Depth of water on the land=8cm or 0.08m

We know that:-

1h=10000m²

1.8h=10000×1.8=18000m²

∴Volume of water=0.08×18000=1440m³

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Question 13

A covered wooden box has the inner measure as 115 cm, 75 cm and 35 cm and the thickness of wood is 2.5 cm. Find the volume of the wood.

Sol :

Inner volume=115×75×35cm

=301875cm³

∴External length=115+(2.5×2)=120cm

External Breadth=75+(2.5×2)=80cm

External Height=35+(2.5×2)=40cm

∴External volume=120×80×40

=384000cm³

∴Only wooden volume=384000-301875

=82125cm³

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Question 14

Each side of a cube is decreased by 25% . Find the ratio of the volume of the original cube and the resulting cube.

Sol :

Let original cube's side=100unit

∴Original cube volume=(100)³unit³

Decreasing 25% of side

∴New cube's side$=100-\left(100\times \frac{25}{100}\right)$

=100-25=75unit

∴New cube volume=(75)³

∴Original cube's volume : New cube's volume

=(100)³: (75)

$=\frac{100\times 100 \times 100}{75 \times 75 \times 75}=\frac{4^3}{3^3}$

$=\frac{64}{27}$

=64 : 27

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Question 15

100 men took a dip in a pond 50 m long and 18 m wide. If the average displacement of water by a man was 3.6m³ , then calculate the rise in the water level.

Sol :

Each man displacement of water=3.6m³

∴100 man displacement of water=3.6×100=360m³

Rise in water level in pond is same as height of volume of cuboid

ATQ ,

50×18×h=360

$h=\frac{360}{50\times 8}$

=0.9m=90cm

S Chand Class 10 CHAPTER 12 Similar triangles Exercise 12E

Exercise 12E

S.chand publication New Learning Composite mathematics solution of class 8 Chapter 12 Mensuration Exercise 12E

Exercise 12E

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

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