Mensuration Made Great: With Word Problems For Class 8th,9th [2]

Q.1) A rectangular lawn is 30m x 20m. It has two flower beds 1.5m wide running in the middle of it, one parallel to the length and the other parallel to the breadth. There is a lamp post in the center on a cubical cemented platform of edge 1.5m. There is a walking path on the outside of the rectangular park. Find the :

(a) Area of the flower beds
(b)Area of walking path
(c)Area of grassy portion

SOLUTION:

Given:
ABCD = Rectangle lawn
EFGH = Rectangle park
TUVW = Flower bed
PQRS = Flower bed
XYZJ = Cubical cemented platform

SOLUTION:

(a) To find area of the flower beds = ?

Area of the flower beds = Area of rectangle TUVW + Area of rectangle PQRS – Area of rectangle XYZJ

1] Length of rectangle TUVW = 20m
Breadth of rectangle TUVW = 1.5 m

Thus , Area of rectangle TUVW is = L * B = 20 * 1.5 = 30m²


2] Length of rectangle PQRS = 30 m
Breadth of rectangle PQRS = 1.5m

Thus , Area of rectangle PQRS = L *B = 30 * 1.5 = 45m²


3] Area of Square XYZJ = S * S = 1.5 * 1.5 = 2.25m²


4] Area of flower beds = 30 + 45 – 2.25 = 72 .75m²
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(b)
To find Area of walking paths = ?

Area of walking paths = Area of rectangle EFGH – Area of rectangle ABCD

1] Length of rectangle EFGH = 30m + (2 +2) = 30 + 4 = 34m
Breadth of rectangle EFGH = 20 + ( 2 + 2 ) = 20 + 4 = 24m

Area of Rectangle EFGH = L * B = 34 * 24 = 816 m²


2] Area of Rectangle ABCD = L * B = 30 * 20 = 600m²


3] Area of walking paths = 816 – 600 = 216m²
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(c)
Area of grassy portion = ?
Area of grassy portion = Area of rectangle lawn – Area of the flower beds
= 600 – 72.75
= 527.25m²
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Q 2 ]. A circular plot has a radius of 42m . A 3.5 m wide road runs on the inside of the plot around it . Find the area of the road . Also find the cost of paving the road at rupees 45 per m².

SOLUTION :

Given : Radius of outer circle = 42m
Width of path = 3.5m
Rate of paving road = Rs . 45 per m²

FIND : Area of road = ?
Cost of paving = ?

Radius of inner circle = 42 – 3.5 = 38.5 m

(a)
Area of road = Area of outer circle – Area of inner circle
= pie(R)² – pie(r)²
= pie (R² – r²)
= 22/7 ( 42² – 38.5²)
= 22/7 ( 1764 – 1482.25)
= 22/7 ( 281.75 )
= 885.5m²


(b)

Cost of paving = Area * rate
= 885.5 * 45
= Rs. 39,847.5


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Q 3 ] . Find the area of the shaded region in given figure if PQRS is a square .

SOLUTION :

Area of shaded region = Area of square – (Area of triangle SOM + Area of triangle QUV)


1] Side of square = SM + MR = 7 + 5 = 12 cm

Therefore, Area of a square PQRS = S * S = 12 * 12 = 144cm²


2] Area of triangle SMO = 1/2 * B * H = 1/2 * 7 * 7 = 49/2 = 24.5cm²

Area of triangle QUV = 1/2 * B * H = 1/2 * 7 * 7 = 24.5cm²


3] Area of shaded = 144 – ( 24.5 + 24.5 )
= 144 – 49
= 95 cm²