S Chand Class 10 CHAPTER 12 Similar triangles Exercise 12D

Exercise 12D

Question 1

Ans: In $\triangle P Q R$.

XY 11 QR

$P X=1 \mathrm{~cm}, X Q=3 \mathrm{~cm}, Y R=4.5 \mathrm{~cm}$ and

$Q R=9 \mathrm{~cm}$

(IMAGE TO BE ADDED)

$P Y=x \mathrm{~cm}$ and $XY=y \mathrm{~cm}$

if $X Y \| Q R$

So $P X Y \sim \triangle P Q R$

so $\frac{P X}{P Q}=\frac{P Y}{P R}=\frac{X Y}{Q R}$

$\begin{aligned}&\Rightarrow \frac{1}{1+3}=\frac{x}{x+4.5}=\frac{y}{9} \\&\Rightarrow \frac{1}{4}=\frac{x}{x+4.5}=\frac{y}{9} \\&\frac{x}{x+4.5}=\frac{1}{4} \Rightarrow 4 x=x+4.5 \\&\Rightarrow 4 x-x=4.5 \\&\Rightarrow 3 x=4.5\end{aligned}$

(i) $\Rightarrow x=\frac{4.5}{3}=1.5 \mathrm{~cm}$

and $\frac{y}{9}=\frac{1}{4}=4 y=9$

$\Rightarrow y=\frac{9}{4}=2.25 \mathrm{~cm}$

(ii) so $\mathrm{Py}=1.5 \mathrm{~cm}$ and $\mathrm{Xy}=2.25 \mathrm{~cm}$

(iii) if $\triangle P XY \sim \triangle P Q R$

So area of $\triangle P X Y$

area of $\triangle P Q R=\frac{P X^{2}}{P Q^{2}}$

$\Rightarrow \frac{A}{\text { area } \triangle P Q R}=\frac{1^{2}}{4^{2}}=\frac{1}{16}$

So Area $\triangle P Q R=16 A \mathrm{~cm}^{2}$

(iv) Now area of figure X.4RQ = Area of $\triangle P Q R$ - area of $\triangle P \times 4$

$=16 \mathrm{~A}-\mathrm{A}=15 \mathrm{~A} \mathrm{~cm}^{2}$

Question 2

Ans: In the higure two triangles are similar 

i-e. $\triangle A B C \sim \triangle A P Q, P Q$ is not parallel

To $B C, P C=4 \mathrm{~cm}, A Q=3 \mathrm{~cm}, Q B=12 \mathrm{~cm}$,

$\begin{gathered}B C=15 \mathrm{~cm} \\A P=x, \text { and } P Q=y\end{gathered}$

(image to be added)

Then $\frac{A Q}{A C}=\frac{A P}{A B}=\frac{P Q}{B C}$

$\begin{aligned}&\Rightarrow \frac{3}{x+4}=\frac{x}{3+12}=\frac{y}{15} \\&\Rightarrow \frac{3}{x+4}=\frac{x}{15} \Rightarrow x^{2}+4 x=45 \\&\Rightarrow x^{2}+4 x-45=0 \\&\Rightarrow x^{2}+9 x-5 x-45=0\end{aligned}$

$\Rightarrow x(x+9)-5(x+9)=0$

$\Rightarrow(x+9)(x-5)=0$

Either $x+9=0$ then $x=-9$ which is not possible being negative

or $x-5=0$ then $x=5$

So $\quad A P=5$

and $\frac{3}{x+4}=\frac{y}{15}$

$\Rightarrow \frac{3}{5+4}=\frac{y}{15}$

$\Rightarrow y=\frac{3 \times 15}{9}=5$

is $\triangle A P Q \sim \triangle A B C$

$\text {If } \frac{\text { area } \triangle A P Q}{\text { area } \triangle A D C}=\frac{P Q^{2}}{BC^{2}}$

$=\frac{(5)^{2}}{(15)^{2}}=\frac{25}{225}=\frac{1}{9}$

So

Ratio between the areas of $\triangle A P Q$ and $\triangle A B C$ $=1: 9$

Question 3

Ans: $\triangle A B C \approx \triangle D E F$

$A B=2, D E=4$ 

(image to be added)

(a) if $\Delta s$ are similar

So $\frac{\text { area } \triangle A B C}{\text { arer } \triangle D E F}=\frac{A B^{2}}{D E^{2}}=\frac{(2)^{2}}{(4)^{2}}=\frac{4}{16}=\frac{1}{4}$

So Ratio between their areas $=1: 4$

(b)$\triangle A B C: \triangle O E F=16: 25$

$\Rightarrow \frac{\text { ara } D A B C}{\text { area } \triangle D E F}=\frac{16}{25}$

$\Rightarrow \frac{\text { area } \triangle A B C}{\text { area } \triangle D E F}=\frac{A B^{2}}{D E^{2}}$

So $\frac{A B^{2}}{D E^{2}}=\frac{16}{25}=\left(\left.\frac{4}{5}\right)^{2}\right.$

So $\frac{A B}{D E}=\frac{4}{5}$

Question 4

Ans: In $\triangle A B C$

$P$ is a point on $A B$ such that

$A P: P B=1: 2$

and $Q$ is a point on $A C$ soch that $P Q \| B C$

(image to be added)

if $P Q \| B C$

so $\triangle A P Q \backsim \triangle A B^{\prime} C$

if $\frac{\text { area of } \triangle A P Q}{\text { area of } D A B C}=\frac{A P^{2}}{A B^{2}}$

$=\frac{(1)^{2}}{(1+2)^{2}}=\frac{(1)^{2}}{(3)^{2}}=\frac{1}{9}$

$\Rightarrow 9$ areas of $\triangle A P Q=$ area of $\triangle A B C$

subtracting area of $\triangle A P Q$ from both sides area of $\triangle A B C$ - area of $\triangle A P Q=9$ are of $\triangle A P Q$-area of $\triangle A P Q$

$\Rightarrow$ area of trap. $B P Q C=8$ area of $\triangle A P Q$

$\Rightarrow \frac{\text { area of } \triangle A P Q}{\text { area of trap. } B P Q C}=\frac{1}{8}$

So area of $\triangle A P Q$ : area as trap. $B P Q C=1: 8$

Question 5

Ans: Area of $\triangle A B C=36 \mathrm{~cm}^{2}$

and Area of $\triangle D E F=81 \mathrm{~cm}^{2}$

$E F=6.9 \mathrm{~cm}$

is $\triangle A B C 〜\triangle D E F$

so $\frac{\text { area of } \triangle A B C}{\text { area of } \triangle D E F}=\frac{B C^{2}}{E F^{2}}$

$\Rightarrow \frac{36}{81}=\frac{(B C)^{2}}{(6.9)^{2}} \Rightarrow \frac{(6)^{2}}{(9)^{2}}=\frac{(B C)^{2}}{(6.9)^{2}}$

$\Rightarrow \frac{B C}{C .9}=\frac{6}{9}$

$\Rightarrow B C=\frac{6.9 \times 6}{9}$

$\Rightarrow B C=4.6 \mathrm{~cm}$

Question 6

Ans: In $\triangle A D E, B C \| D E$

$A D=3 \mathrm{~cm} D B=2 \mathrm{~cm}$

(image to be added)

So $A B=A D-D B=3-2=1 \mathrm{~cm}$ Area of $\triangle A B C=10 \mathrm{~cm}$

 if $\triangle A D E: B C || D E$

So $D A B C \backsim \triangle A D E$

So $\frac{\text { area } \triangle A B C}{\text { area } △ A D E}=\frac{A D^{2}}{A D^{2}}$

$\Rightarrow \frac{10}{\text { area } \triangle A D E}=\frac{(1)^{2}}{(3)^{2}}=\frac{1}{9}$

$\Rightarrow$ area $D A D E=10 \times 9=90 \mathrm{~cm}^{2}$

Question 7

Ans:  In $\triangle P R S$ and $\triangle P Q R$

$\angle P R S=\angle P Q R$

$\angle R P S=\angle R P Q$

So $\triangle P R S \sim \triangle P Q R$

So $\frac{\text { area of }  △P R S}{\text { area of } △ P Q R}=\frac{P S^{2}}{P R^{2}}$

$\Rightarrow \frac{2}{\text { area of } \triangle P Q R}=\frac{(2)^{2}}{(3)^{2}}=\frac{1}{9}$

$\Rightarrow$ anea of $\triangle P Q R=\frac{2 \times 9}{4}=\frac{9}{2}=4.5 \mathrm{~cm}^{2}$

Question 8

Ans: In $\triangle A B C: A D: D B=5: 4 \Rightarrow \frac{A D}{D B}=\frac{5}{4}$

$\begin{aligned}&\Rightarrow \frac{A D}{A D+D B}=\frac{5}{5+4} \\&\Rightarrow \frac{A D}{A B}=\frac{5}{9}\end{aligned}$

$\Rightarrow \frac{A D}{A D+D B}=$ $\Rightarrow \frac{A D}{A B}=$ and $D E \| B C$

(image to be added)

So $\triangle A D E \approx \triangle A B C$

So $\frac{\text { Area }(\triangle A D E)}{\text { Area }(\triangle A B C)}=\frac{A D^{2}}{A B^{2}}$

{Area of two similar triangles are proportional to the squares of their corresponding sides }

$=\frac{(5)^{2}}{(9)^{2}}=\frac{25}{81}$

So $\frac{\text { Area }(\triangle A D E)}{\text { Area }(\triangle A B C)}=\frac{25}{81}$

Question 9

Ans: In $\triangle P A B$ and $\triangle P Q R$.

$\begin{aligned}\angle A B P &=\angle Q \\\angle P &=\angle P\end{aligned}$

So $\triangle P A B \sim \triangle P Q R$

In right $\triangle P Q R, P Q=8 \mathrm{~cm}$ and $P R=10 \mathrm{~cm}$

But $P R^{2}=P Q^{2}+Q R^{2}$

$\begin{aligned}&\Rightarrow(10)^{2}=(8)^{2}+(Q R)^{2} \\&\Rightarrow 100=64+(Q R)^{2} \\&\Rightarrow Q R^{2}=100-64=36=(6)^{2}\end{aligned}$

So $Q R=6 \mathrm{~cm}$

if $\triangle P A B \sim \triangle P Q R$

So $\frac{A B}{Q R}=\frac{P B}{P Q} \Rightarrow \frac{2}{6}=\frac{P B}{8}$

$\Rightarrow P B=\frac{2}{6} \times 8=\frac{8}{3} \mathrm{~cm}$

Again $\triangle P A B \sim \triangle P Q R$

So

$\frac{\operatorname{arca}(\Delta P A B)}{\operatorname{arcr}(\triangle P Q R)}=\frac{A B^{2}}{QR^{2}}$

{Area of similar triangles are proportional to the squares of their corresponding sides}

$=\frac{(2)^{2}}{(6)^{2}}=\frac{4}{36}=\frac{1}{9}$

So area $(\triangle P Q R)=9$ area $(\triangle P A B)--(i)$ subtracting area ( $\triangle P A B$ ) from both sides,

Area $(\triangle P Q R)$ - are $(\triangle P A B)=9$ area $(\triangle P A B)$-area ( $ \triangle P A B$ )

$\begin{aligned}\Rightarrow & \text { area quad. } A Q R B=8 \text { area }(\triangle P A B) \\& \frac{\text { area }(\text { quad } \cdot A Q R B)}{\text { area }(\triangle P Q R)}=\frac{8 \text {area}(\triangle P A B)}{9 \text { area }(\triangle P A B)}=\frac{8}{9}\end{aligned}$

So Area quad. $A Q R B$ : area $(\triangle P Q R)=8: 9$

Question 10

Ans: $\triangle A B C \sim \triangle P Q R$

Area $(\triangle A B C)=64 \mathrm{~cm}$ and arer $(\triangle P Q R)=121 \mathrm{~cm}$

$Q R=15.4 \mathrm{~cm}$

if triangles are similar.

$\text { So } \frac{\text { area }(\triangle A B C)}{\text { area }(\triangle P Q R)}=\frac{B C^{2}}{QR^{2}} \text {. }$

$\Rightarrow \frac{64}{121}=\frac{(B C)^{2}}{(15.4)^{2}} \Rightarrow \frac{(8)^{2}}{(11)^{2}}=\frac{(B C)^{2}}{(15.4)^{2}}$

$\Rightarrow \frac{B C}{15.4}=\frac{8}{11}$

$\Rightarrow B C=\frac{8}{11} \times 15.4$

$\Rightarrow B C=11.2 \mathrm{~cm}$

Question 11

Ans: DE and F are the mid- point of the sides BC, CA and AB of △ABC

DE,EF and FD are joined 

If E and F are mid points of AC and AB 

So EF ||BC and = $\frac{1}{2} B C$

$\Rightarrow \frac{E F}{B C}=\frac{1}{2}$............(i)

similanly $D$ and $E$ are the mid-point of $B C$ and $A C$ 

if $D E \| A B$ and $=\frac{1}{2} A B$

$\Rightarrow \frac{D E}{A B}=\frac{1}{2}$.......(ii)

and $D$ and $F$ are mid-point of $B C$ and $A B$

So $D F \| A C$ and $=\frac{1}{2} A C$

$\Rightarrow \frac{DF}{A C}=\frac{1}{2}$.......(iii)

From (i) (ii) and (iii)

$\frac{E F}{D C}=\frac{D E}{A B}=\frac{D F}{A C} \quad \text { (each }=\frac{1}{2} \text { ) }$

So $\triangle D E F \backsim \triangle A B C$

$\text { So } \frac{\text { area } \triangle D E F}{\text { area } \triangle A B C}=\frac{E F^{2}}{BC^{2}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$

So area $\triangle D E F:$ area $\triangle A B C=1: 4$

Question 12

Ans:  $A B C D$ is a trapezium in which $A B \| D C$ and $A B=2 D C$ $A C$ and $B D$ intersect each other at $O$

In $\triangle A O B$ and $\triangle C O D$.

$\angle A O B=\angle C O D$

$\angle A B O=\angle O D C$

So $\triangle A O B \sim \triangle C O D$

if $\frac{\text { area }(D A O B)}{\text { area }(\triangle C O D)}=\frac{A B^{2}}{D C^{2}}=\frac{(2 D C)^{2}}{(D C)^{2}}$

$\frac{4 D C^{2}}{D C^{2}}=\frac{4}{1}$

So area $(\Delta A O Q)=4$ area $(\Delta C OD)$

Hence proved.

Question 13

Ans:  $\triangle A B C$ and $\triangle D E F$ are similar 

and area $(\triangle A B C)=$ area $(\triangle D E F)$

(image to be added)

$\frac{\text { area }(△ A B C)}{\text { area }(△D E F)}=1$.........(i)

$\frac{\text { area }(\triangle A B C)}{\text { area }(\triangle D E F)}=\frac{B C^{2}}{E F^{2}}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}$.........(ii)

From (i.) and (ii)

$\begin{aligned}& \frac{B C^{2}}{E F^{2}}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}=1 \\\Rightarrow & \frac{B C^{2}}{E F^{2}}=1 \Rightarrow B C^{2}=E F^{2} \\\Rightarrow & B C=E F\end{aligned}$

similarly $A B=D E$ and $A C=D F$ 

Hence $\triangle A B C \cong \triangle D E F$

Question 14

Ans: In $\triangle A B C$ ard $\triangle D E F$,

$A P \perp B C$ and $D Q \perp E F$

 if $\triangle A B C \backsim \triangle D E F$

So $\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}$ and $\angle B=\angle E$

Now in $\triangle A B P$ and $\triangle P E Q$

$\begin{aligned}&\angle B=\angle E \quad(\text ) \\&\angle P=\angle Q\end{aligned}$

So $\triangle A B P \sim \triangle D E Q$

$\begin{aligned}&\text { So } \frac{A B}{D E}=\frac{A P}{D C} \\&\text { But } \frac{A B}{D E}=\frac{B C}{E F} \\&\text { So } \frac{A P}{D C}=\frac{B C}{E F}\end{aligned}$

Hence proved 

Question 15

Ans: In $\triangle A B C, P$ and $Q$ are the points on $A B$ and $A C$ such that

$P Q \quad || B C$

and area $(\triangle A P Q)=$ area (quad. $P B C Q$ )

$\begin{aligned} \Rightarrow \text { area }(\Delta A P Q)+\text { area }(\Delta A P Q) &=\text { area }(\text { quad }) P B C O) \\ &+\text { area }(\Delta A P Q) \end{aligned}$

$\Rightarrow 2$ area $(A P Q)=$ area $(\triangle A B C)$

$\Rightarrow$ area $(\triangle A P Q)=\frac{1}{2}$ area $(\triangle A B C)$

$\Rightarrow \frac{\text { area }(\triangle A P Q)}{\text { area }(\triangle A B C)}=\frac{1}{2}$

is. $P Q \| B C$

So $\triangle A P Q \backsim \triangle A B C$

So $\frac{\text { area }(\triangle A P Q)}{\text { area }(\triangle A B C)}=\frac{A P^{2}}{A B^{2}}=\frac{1}{2}$

So $\frac{A P^{2}}{A B^{2}}=\frac{1}{2} \Rightarrow \frac{A P}{A B}=\frac{1}{\sqrt{2}}$

$\begin{aligned}& \frac{A B-B P}{A B}=\frac{1}{\sqrt{2}} \\\Rightarrow & \frac{A B}{A B}=\frac{B P}{A B}=\frac{1}{\sqrt{2}} \\& 1-\frac{B P}{A B}=\frac{1}{\sqrt{2}}\end{aligned}$

$\Rightarrow \frac{B P}{A B}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$

Hence $\frac{B P}{A B}=\frac{\sqrt{2}-1}{\sqrt{2}}$

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

 Exercise 12D

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Q1 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 1

Find the curved surface area and whole surface area for the cylinders shown below. Measures are in centimetres.

(a)

h=5 ,r=7, $\pi =\frac{22}{7}$

∴C.S.A=2πrh

$=2\times \frac{22}{7}\times 7\times 5 $

=220cm²

T.S.A=2πrh+(πr²×2)

$=220+\frac{22}{7}\times 7\times 7\times 2$

=220+(154×2)

=220+308=528cm²

(b)

h=8 ,r=5 ,π=3.14

∴C.S.A=2πrh

=2×3.14×5×8

=251.20cm²

T.S.A=2πrh+(πr²×2)

=251.20+{(3.14×5²)×2}

=251.20+157

=408.20cm²

(c)

h=10 ,r=2, π=3.14

C.S.A=2πrh

=2×3.14×2×10

=125.60cm²

T.S.A=2πrh+2πr²

=125.6+(2×3.14×2²)

=125.6+(2×3.14×4)

=125.6+25.12=150.72cm²

(d)

h=3 ,r=7, $\pi=\frac{22}{7}$

C.S.A=2πrh

$=2\times \frac{22}{7}\times 7\times 3$

=132cm²

T.S.A=2πrh+2πr²

$=132+\left(2\times \frac{22}{7}\times 7\times 7\right)$

=132+308=440cm²

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Q2 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 2

The table given below contains information about a few cylinders. Complete it.

	Radius	Height	Curved Surface Area	Area of the base	Whole Surface
(a)	7 cm	12 cm
(b)	2.8 m	5 m
(c)	35 mm	50 mm

Sol :

(a)

r=7cm , h=12cm

T.S.A=2πrh+2πr²

$=\left(2\times \frac{22}{7}\times 7\times 12\right)+\left(2\times \frac{22}{7}\times 7 \times 7\right)$

=528+308=836cm²

Area of the base=πr²

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

=154cm²

C.S.A=2πrh

$=2\times \frac{22}{7} \times 7\times 12$

=528cm²

(b)

r=2.8m ,h=5m

C.S.A=2πrh

=2×2.8×5×3.14

=87.92m²

T.S.A=2πrh+2πr²

=87.92+2×3.14×2.8×2.8

=87.92+49.22

=137.15m²

Area of the base=πr²

=3.14×2.8×2.8

=24.61m²

(c)

r=35mm ,h=50mm

C.S.A=2πrh

$=2\times \frac{22}{7}\times 35 \times 50$

=11000m²

Area of the base=πr²

$=\frac{22}{7}\times 35\times 35$

=3850m²

T.S.A=2πrh+2πr²

=11000+(2×3850)

=18700m²

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Q3 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 3

Find the missing dimension for a cylinder.

(a) Radius: Curved surface area $=220 \mathrm{~cm}^{2}$. Height =10 cm

Sol :

(a) 

C.S.A=220cm² ,h=10cm , r=?

⇒C.S.A=2πrh

$r=\frac{\text{C.S.A}}{2 \pi h}=\frac{220}{2\times \frac{22}{7}\times 10}$

$r=\frac{220\times 7}{2\times 22\times 10}=\frac{7}{2}=3.5$cm

(b) Height: Curved surface area = $3168 \mathrm{~cm}^{2}$, Radius = 21 cm

Sol :

(b)

C.S.A=2πrh

$h=\frac{\text{C.S.A}}{2\pi r}=\frac{3168}{2\times \frac{22}{7}}\times 21$

$=\frac{3168}{2\times 22\times 3}$

h=24cm

(c) Height: Radius=5 cm. Total surface area $=120 \pi\mathrm{cm}^{2}$

Sol :

(c)

T.S.A=120π , r=5 , h=?

T.S.A=2πrh+2πr²

120π=2π(rh+r²)

120π=2π(5h+5²)

60=5h+25

5h=60-25=35

h=7cm

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Question 4

The circumference of a right circular cylinder is 440 cm and its height is 5 cm. Find the lateral surface area of the cylinder.

Sol :

c=440cm , h=5cm

ATQ ,

2πr=440

$r=\frac{440}{2\times \pi}$

$r=\frac{440}{2\times \frac{22}{7}}$

$=\frac{440\times 7}{2\times 22}$

=70cm

∴L.S.A=2πrh

$=2\times \frac{22}{7}\times 70\times 5$

=2200cm²

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Q5 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

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Question 5

A cylindrical soup can is 15.4 cm in diameter and 20 cm in length. What is the area of the label that covers the side of the can?

Sol :

d=15.4cm

∴$r=\frac{15.4}{2}$cm ,h=20cm

⇒Lateral Surface Area=2πrh

$=2\times \frac{22}{7}\times \frac{15.4}{2}\times 20$

$=2\times \frac{22}{7}\times \frac{154}{2\times 10}\times 20$

=968cm²

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Question 6

In a building, there are 30 cylindrical pillars. The radius of each pillar is 42 m and height is 7 m. Find the total cost of painting the curved surface area of all pillars at the rate of ₹ 10 per m²?

Sol :

r=42m , h=7m

⇒C.S.A of a pillar=2πrh

$=2\times \frac{22}{7}\times 42\times 7$

=1848m²

C.S.A of 30 pillar=1848×30

=55440m²

∴Paint cost for 30 pillar's=55440×10

=554400

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Question 7

A cylinder has a diameter 20 cm and height 7 cm. Explain whether tripling the height would have the same effect on the surface area as tripling the radius or not.

Sol :

d=20cm , h=7cm ,$r=\frac{20}{2}=10$cm

For 1st case:-

h₁=3×7=21cm

∴T.S.A for 1st case=2πr(r+h₁)

=2×3.14×10×(10+21)

=1946.8 cm²

For 2nd case:-

r₁=10×3=30cm

∴T.S.A to 2nd case=2πr₁(r₁+h)h

=2×3.14×30×(30+7)

=6970.8 cm²

∴Both case surfaces are will be different.

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Question 8

The diameter of a garden roller is 1.8 m and it is 2.8 m long. How much area will it cover is 25 revolutions in levelling a field?

Sol :

d=1.8m , h=2.8m , $r=\frac{1.8}{2}=0.9$m

⇒C.S.A=2πrh

$=2\times \frac{22}{7}\times 0.9\times 2.8$

$=2\times \frac{22}{7}\times \frac{9}{10}\times \frac{28}{10}$

$=\frac{2\times 22\times 9\times 4}{100}=\frac{396}{25}$m²

∴1 revolution need$=\frac{396}{25}$m² area 

∴25 revolution need$=\frac{396}{25}\times 25$=396m²  

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Question 9

The diameter of a road roller is 90 cm and is 2 m 10 cm in length. If it takes 400 revolutions to level a playground , find the cost of levelling the ground at 2.25 per square meter.

Sol :

d=90cm ,$r=\frac{90}{2}$=45cm

Length=2m10cm=(2×100+10)=210cm

⇒1 revolution=C.S.A of roller=2πrh

$=2\times \frac{22}{7}\times 45 \times 210$

=2×22×45×30 cm²

=59400 cm² or 5.94 m² 

If takes 400 revolution to level 5.94 m² area

It takes 400 revolutions , so total area of ground = 5.94×400=2376 m²

∴Cost of levelling at 2.25 for 1m²

∴Cost of levelling 

(2376×2.25)₹=(2376×1) m²

5346 ₹=2376 m²

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Question 10

The inner diameter of a 15 m deep circular well is 4.2 m. Find the cost of plastering its inner covered surface at the rate of 40 per m²

Sol :

Inner diagram=4.2m or radius$=\frac{4.2}{2}$=2.1m

Height(h)=15m

⇒Inner C.S.A=2πrh

$=2\times \frac{22}{7}\times 2.1\times 15$

$=2\times \frac{22}{7}\times \frac{21}{10}\times 15$

=198m²

Cost of plastering of 1m² area=40

∴Cost of plastering of 198 m² area=198×40=7920

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Question 11

The inner circumference of a hollow cylindrical pipe is 4.5 m and outer circumference is 4.6 m. Find the cost of painting both its inner and outer sides at the rate of 20 per m², if it is 8 m tall

Sol :

Inner Circumference=4.5m

Outer Circumference=4.6m

Height(h)=8m

⇒Inner Circumference=2πr=4.5

∴$r=\frac{4.5}{2\pi}$

$=\frac{4.5}{2\times 3.14}=\frac{4.5}{6.28}$

=0.715m

Outer circumference=2πR=4.6

$R=\frac{4.6}{2\pi}=\frac{4.6}{2\times 3.14}$

=0.731m

Inner C.S.A=2πrh

$=2\times \frac{22}{7}\times 0.731\times 8$

=35.95m²

Outer C.S.A=2πRh

$=2\times \frac{22}{7}\times 0.731\times 8$

=36.75m²

∴Total C.S.A=35.95+36.75=72.7m²

∵Cost of painting 1m²=20

∴Cost of painting hollow cylinder=72.7×20=1454

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Q12 | Ex-12D |Class 8 |Mensuration | S.Chand | New Learning | Composite mathematics | myhelper

OPEN IN YOUTUBE

Question 12

The external diameter of a 20 cm long and 1 cm thick hollow iron pipe is 25 cm. Determine the whole surface of the iron pipe.

Sol :

D=25cm ,t=1cm , $R=\frac{25}{2}=12.5$cm

r=12.5-1=11.5cm

h=20cm

C.S.A of Iron pipe=2πh(R+r)

$=2\times \frac{22}{7}\times 20(12.5+11.5)$

$=2\times \frac{22}{7}\times 20 \times 24$

$=\frac{44\times 480}{7}$ cm²

∴Area of one circular end=2(πR²-πr²)

$=2\times \frac{22}{7} \left\{(12.5+11.5)(12.5-11.5)\right\}$

$=\frac{44}{7}\times 24$ cm²

∴T.S.A of Iron pipe$=\left(\frac{44\times 480}{7}\right)+\left(\frac{44}{7}\times 24\right)$

$=\frac{44}{7}\left(480+24\right)$

$=\frac{44}{7}\times 504$

=44×72=3168 cm²