Mensuration – Maharashtra Board Class 9 Solutions for Geometry
AlgebraGeometryScience and TechnologyHindi
Exercise – 9.1
Solution 1:
Solution 2:
Solution 3:
Solution 4:
Solution 5:
Exercise – 9.2
Solution 1:
Let PQRS be the parallelogram having SQ as one diagonal of length 6.8 cm and RT of length 7.5 cm perpendicular to SQ.
The diagonal of the parallelogram divides it into two congruent triangles.
Solution 2:
Solution 3:
The area of the square-shaped field
= (side)2
= (300)2
= 90000 m2
Cost of levelling the field per square metre = Rs. 1.25
∴The cost of levelling = rate × area
= Rs. 1.25 × 90000
= Rs. 1,12,500
The cost of levelling the field is Rs. 1,12,500.
Solution 4:
Let the length and breadth of the hall be l metres and b metres.
The length of the rectangular hall is 5 m more than its breadth.
∴l = b + 5 ………(1)
The area of the rectangular hall = l × b
∴l × b = 750 ………(2)
Substitute (1) in (2) we get
∴(b + 5) × b = 750
b2 + 5b -750 = 0
b2 – 25b + 30b – 750 = 0
b(b – 25) + 30(b – 25) = 0
(b + 30)(b – 25)=0
b + 30 = 0 or b – 25 = 0
b = -30 or b = 25
But breadth of the hall cannot be negative.
∴ b = 25 m
Substitute b = 25 in equation (1) we get
l = 25 + 5 = 30m
The perimeter of the rectangular hall
= 2 (l + b)
= 2 (30 + 25)
= 110
The perimeter of the rectangular hall is 110 m.
Solution 5:
Solution 6:
Solution 7:
Let the length of the lawn = 75 m and its breadth = 60m.
Road ABCD is parallel to the length and road EFGH is parallel to the breadth.
AB = FG = 4 m
AD = 75 m
EF = 60 m
The area of the road ABCD
= AD × AB
= 75 × 4
= 300 m2 …(1)
The area of the road EFGH
= EF× FG
= 60 × 4
= 240 m2 …(2)
□UVWX which is 4 × 4, is common to both the roads.
Area of □UVWX = 4 × 4 = 16 m2 …(3)
∴The area of the road
= A(□ABCD) + A(□EFGH) – A(□UVWX)
= (300 + 240 -16) m2…..[From (1),(2) and (3)]
= 524 m2
The total cost of gravelling the road
= rate × area
= Rs. 4.50 × 524
= Rs. 2358
The total cost of gravelling the road is Rs.2358.
Exercise – 9.3
Solution 1:
A horse is tethered at a point by a 10 m long rope.
The horse can graze around that point in a circular path.
The horse can graze in a circular path with the point where horse is tethered as the centre and radius equal to the length of the rope, i.e. 10 m.
So the area of the region where the horse can graze is area of the circle.
The area of circle = πr2
= 3.14 × (10)2
= 3.14 × 100
= 314 m2
The area of the region where the horse can graze is 314 m2.
Solution 2:
The diameter of the circle = 2r = 20 cm
The diameter of the circle is equal to the side of the square.
∴ The side of the square is 20 cm.
Area of a square= (side)2 = 400 cm2
Area of a circle = πr2
= 3.14 × (10) 2
= 314 cm 2
Area of the shaded portion
= Area of the square – Area of the circle
=(400 – 314)cm2
= 86 cm2
The area of the shaded portion is 86 cm2.
Solution 3:
Area of the circle with radius 3 m
= πr2
= π (3)2
= 9π m2 …(1)
Area of the circle with radius 4 m
= πr2
= π (4)2
= 16 π m2 …(2)
∴Area of the required circle
= 9 π + 16 π … [From (1) and (2)]
= 25π m2 …(3)
Let the radius of the required circle be R.
Then its area = πR2 = 25 π …[From (3)]
∴ R2 = 25 ∴ R= 5 m
The radius of the required circle is 5 m.
Solution 4:
Solution 5:
