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 = 14580m2

Cost of fence =₹ 3.25 m

Let length , breadth = 9x , 5x

Area = 14580

×b=145809x×5x=1458045x2=14580x2=1458045x2=324x=324

x=18 m

length =9x=9×18= 162m

breadth =5 x=5×18=90 m

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

Length of fence = 504m

Cost of fence = 504×3.25
=₹ 1638

Question 2

Sol :
Given 

Dimensions of rectangle = 16m ×9m

let side of square =x m

Perimeter of rectangle = area of square

l×b=x216×9=x2x=16×9x=4×3=12mx=12m

Side of square =12 m
Perimeter of square = 4x

=4×12
=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 × ten distance between the opposite sides

24×12=18×x

x=24×1218

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

Question 4

Sol :
Given













Plot dimension = 24 m×24 m

House dimensions =18 m×12 m.


∴ Garden Area = plot area - house area 

24×2418×12

Garden Area =360 m2

Given 
Cost of developing garden = Rs 50/m2

∴ Total cos of developing garden around house = 360×50

=Rs 18000

Question 5

Sol :
Dimension of tiles ( Parallelogram) =18 cm×6 cm Height

Floor area = 540m2

Area of one tile = 18 m×6cm(b×h)

108 cm2

Area of one tile = 108×104m2(1m=102m)

No.of tiles required =  A Total Area  Area of one tile 

=540108×104

No.of tiles required = 50000


Question 6

Sol :

(a) Diameter semi circle = 2.8cm





perimeter of semi circle =πd2

=π×2.82

=3.14×2.82

=3.14×1.4

Perimeter of semi circle = 4.398 cm


(b) 










Perimeter of given shape 

¯AB+¯BC+¯CD+ Semi circle perimeter
=1.5+2.8+1.5+4.398
= 110.198 cm

(c)  










Perimeter of given shape 

¯OA+ Semi circle AB+¯OB
=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 = 770cm2

Outer circle radius = 21 cm

Let inner circle radius = r cm


Outer circle area - Inner circle area = 770 cm2

π(21)2πr2=770π(212r2)=770212r2=245.098441r2=245.098r2=441245.098r2=195.90r=195.9r=13.99614 cm.

Radius of inner circle =14cm.

Question 8

Sol :
Given 

Area of square = 121 cm2

s2=121

S=121

S=11 cm

∴ Side of square = 11 cm

∴ Length of wire perimeter of square = 4×11 cm
=44 cm

Now wire is bent into a form of Circle 

∴Length of  wire= perimeter of circle 

44 =2πr      r = radius of circle 

πr=442

πr=22

r=22π

r=22314

r=7 cm

radius of circle = 7cm

Area of circle = πr2

=3.14×72

Area of circle =153.938cm2

Question 9

Sol :
(i) 











Area of Δle ABC

=12×b×h

=12×3×4 (∵right angle triangle )

=12×3×4

=12×12

=6cm2


(ii) BC2=AB2+AC2( Pythagorus theorem )

BC2=3v+42

BC2=9+16

BC2=25

BC=25

BC=5 cm


(iii) Area of triangle ABC=6cm2

By taking ¯BC As base

Area of triangle = 12×b×h

=12×CB×AD

6=12×6×AD

AD=6×26

AD=2 cm


Question 10

Sol :

Dimension of rectangular garden = 80m×40m

width  of path (w)=2.5 m

(i) Area of cross path = l×w+b×w(w×w)

=80×2.5+40×2.5(2.5×2.5)

293.75 m2


(ii) 




Area of unshaded portion 

=Area of garden - area of cross path 

80×40(293.75)

=2906.25 m2



Question 11

Sol :








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

=18×12[12×1×18+12×12×10+12×5×8]

=216[7×9+6×10+5×4]
=216-[63+60+20]
=216-143
=73 cm2

∴ Area of shaded portion 73cm2


Question 12

Sol :

Given 







Area of square EFGH = 729 m2

∴ side of lawn = 729

=27 m

Area of square ABCD = 295 m2

Side of ABCD = 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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