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
ℓ×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
= 24×24−18×12
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 = 108×10−4m2(∴−1m=10−2m)
No.of tiles required = A Total Area Area of one tile
=540108×10−4
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
Perimeter of given shape
= ¯AB+¯BC+¯CD+ Semi circle perimeter
=1.5+2.8+1.5+4.398
= 110.198 cm
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)2−πr2=770π(212−r2)=770212−r2=245.098441−r2=245.098r2=441−245.098r2=195.90r=√195.9r=13.996≈14 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=223⋅14
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
Area of unshaded portion
=Area of garden - area of cross path
= 80×40−(293.75)
=2906.25 m2
Question 11
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 :
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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