ML AGGARWAL CLASS 8 Chapter 18 Mensuration Exercise 18.1

 Exercise 18.1

Question 1

Sol :
Given 

Ratio of length and breadth of rectangular field = 9.5 

Area of field = 14580$m^{2}$

Cost of fence =₹ 3.25 m

Let length , breadth = 9x , 5x

Area = 14580

$\begin{aligned} \ell \times b=& 14580 \\ 9 x \times 5 x &=14580 \\ 45 x^{2} &=14580 \\ x^{2} &=\frac{14580}{45} \\ x^{2} &=324 \\ x &=\sqrt{324} \end{aligned}$

x=18 m

$\therefore$ length $=9 x=9 \times 18=$ 162m

breadth =5 x$=5 \times 18$=90 m

Length of fence= perimeter of rectangle section 
=2(1+b)
=2(162+90)
=2(252)

Length of fence = 504m

Cost of fence = $504 \times 3.25$
=₹ 1638

Question 2

Sol :
Given 

Dimensions of rectangle = 16m $\times 9m$

let side of square =x m

Perimeter of rectangle = area of square

$\begin{aligned} l\times b &=x^{2} \\ 16 \times 9 &=x^{2} \\ x &=\sqrt{16 \times 9} \\ x &=4 \times 3=12 m \\ x &=12 m \end{aligned}$

$\therefore$ Side of square =12 m
Perimeter of square = 4x

$=4 \times 12$
=48 m

∴ Perimeter rectangle exceeds perimeter of square by 50 - 48 = 2m


Question 3

Sol :

Given 

Length of adjacent sides = 24 cm and 18 cm







Distance between longer sides = 12 cm
Let ten distance between shorter sides = x cm

Area of parallelogram = Side $\times$ ten distance between the opposite sides

$\therefore \quad 24 \times 12=18 \times x$

$x=\frac{24 \times 12}{18}$

x=16 cm
∴ Ten distance between shorter sides = 16 cm 

Question 4

Sol :
Given













Plot dimension = $24 \mathrm{~m} \times 24 \mathrm{~m}$

House dimensions $=18 \mathrm{~m} \times 12 \mathrm{~m}$.


∴ Garden Area = plot area - house area 

= $24 \times 24-18 \times 12$

Garden Area $=360 \mathrm{~m}^{2}$

Given 
Cost of developing garden = Rs 50/$m^{2}$

∴ Total cos of developing garden around house = 360$\times 50$

=Rs 18000

Question 5

Sol :
Dimension of tiles ( Parallelogram) $=18 \mathrm{~cm} \times 6 \mathrm{~cm}$ Height

Floor area = 540$m^{2}$

Area of one tile = $18 \mathrm{~m} \times \operatorname{6cm}(b \times h)$

= $108 \mathrm{~cm}^{2}$

Area of one tile = $108 \times 10^{-4} m^{2}\left(∴-1 m=10^{-2} m\right)$

No.of tiles required = $\frac{\text { A Total Area }}{\text { Area of one tile }}$

$=\frac{540}{108 \times 10^{-4}}$

No.of tiles required = 50000


Question 6

Sol :

(a) Diameter semi circle = 2.8cm





perimeter of semi circle $=\frac{\pi d}{2}$

$=\frac{\pi \times 2.8}{2}$

$=\frac{3.14 \times 2.8}{2}$

$=3.14 \times 1.4$

Perimeter of semi circle = 4.398 cm


(b) 










Perimeter of given shape 

= $\overline{A B}+\overline{B C}+\overline{C D}+$ Semi circle perimeter
=1.5+2.8+1.5+4.398
= 110.198 cm

(c)  










Perimeter of given shape 

= $\overline{O A}+$ Semi circle $A B+\overline{O B}$
=2+4.398+2
=8.398 cm

∴ Comparing three figure perimeter values , we can 
Say in case of figure ' B' and has covered more distance 

Question 7

Sol :

Given









Area between concentric

Circle = $770 \mathrm{cm}^{2}$

Outer circle radius = 21 cm

Let inner circle radius = r cm


Outer circle area - Inner circle area = $770 \mathrm{~cm}^{2}$

$\begin{aligned} \pi(21)^{2}-\pi r^{2}=& 770 \\ \pi\left(21^{2}-r^{2}\right) &=770 \\ 21^{2}-r^{2}=& 245.098 \\ 441-r^{2} &=245.098 \\ r^{2} &=441-245.098 \\ r^{2} &=195.90 \\ r &=\sqrt{195.9} \\ r &=13.996 \approx 14 \mathrm{~cm} . \end{aligned}$

Radius of inner circle =14cm.

Question 8

Sol :
Given 

Area of square = $121 \mathrm{~cm}^{2}$

$s^{2}=121$

$S=\sqrt{121}$

S=11 cm

∴ Side of square = 11 cm

∴ Length of wire perimeter of square = $4 \times 11 \mathrm{~cm}$
=44 cm

Now wire is bent into a form of Circle 

∴Length of  wire= perimeter of circle 

44 =2$\pi r$      r = radius of circle 

$\pi r=\frac{44}{2}$

$\pi r=22$

$r= \frac{22}{\pi}$

$r=\frac{22}{3 \cdot 14}$

$r=7 \mathrm{~cm}$

radius of circle = 7cm

Area of circle = $\pi r^{2}$

$=3.14 \times 7^{2}$

Area of circle $=153.938 \mathrm{cm}^{2}$

Question 9

Sol :
(i) 











Area of $\Delta^{\text {le}}$ ABC

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

$=\frac{1}{2} \times 3 \times 4$ (∵right angle triangle )

$=\frac{1}{2} \times 3 \times 4$

$=\frac{1}{2} \times 12$

$=6 \mathrm{cm}^{2}$


(ii) $B C^{2}=A B^{2}+A C^{2} \quad(\therefore$ Pythagorus theorem $)$

$B C^{2}=3^{v}+4^{2}$

$B C^{2}=9+16$

$B C^{2}=25$

$B C=\sqrt{25}$

BC=5 cm


(iii) Area of triangle ABC=6$c m^{2}$

By taking $\overline{B C}$ As base

Area of triangle = $\frac{1}{2} \times b \times h$

$=\frac{1}{2} \times CB\times A D$

$6=\frac{1}{2} \times 6 \times A D$

$A D=\frac{6 \times 2}{6}$

AD=2 cm


Question 10

Sol :

Dimension of rectangular garden = 80m$\times$40m

width  of path (w)=2.5 m

(i) Area of cross path = $l  \times w+b \times w-(w \times w)$

$=80 \times 2.5+40 \times 2.5-(2.5 \times 2.5)$

$293.75 \mathrm{~m}^{2}$


(ii) 




Area of unshaded portion 

=Area of garden - area of cross path 

= $80 \times 40-(293.75)$

=$2906.25 \mathrm{~m}^{2}$



Question 11

Sol :








Area of shaded portion = Area of square ABCD - [Area of triangle ABC + Area of triangle AFD + Triangle EFC]

=$18 \times 12-\left[\frac{1}{2} \times 1 \times 18+\frac{1}{2} \times 12 \times 10+\frac{1}{2} \times 5 \times 8\right]$

=$216-[7 \times 9+6 \times 10+5 \times 4]$
=216-[63+60+20]
=216-143
=$73 \mathrm{~cm}^{2}$

∴ Area of shaded portion 73$\mathrm{cm}^{2}$


Question 12

Sol :

Given 







Area of square EFGH = $729 \mathrm{~m}^{2}$

∴ side of lawn = $\sqrt{729}$

=27 m

Area of square ABCD = $295 \mathrm{~m}^{2}$

Side of ABCD = $\sqrt{295}$

 Side of ABCD = 17.175m

(i) ∴ Length of square filed including lawn and path =27 m

(ii) Width of the path = side of EFGH - side of ABCD

=27  - 17.175

width of the path = 9.825m

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