Mensuration Made Interesting : With Word Problems For Class 7th,8th,9th,10th [7]

Q.1) Find the diagonal of a rectangle whose area is 60cm² and wide 5 cm.

SOLUTION : RECTANGLE

GIVEN : Area = 60 cm² width = B = 5 cm
Find : diagonal of the rectangle = hypotenuse side = ?

Step 1) Area = l * b
60 = l * 5
60/5 = l
12 cm = l

Step 2) Diagonal² = L² + b²
Diagonal² = 12² + 5²
D² = 144 + 25
D² = 169
D = 13 cm

Ans: The length of the diagonal is 13 cm.
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Q.2) The minute hand of a clock is 5.6 cm long. Find the distance covered by it in
2 days?

SOLUTION : CIRCULAR CLOCK

GIVEN : The minute hand will go round the clock in 2 days radius = 5.6 cm
No. of rounds it will take in 2 days = 48 rounds
Step 1) The distance covered in 1 round = circumference = 2 * Pie * r
= 2 * 22/7 * 5.6
= 35.2 cm

Step 2) Distance covered in 48 rounds = 35.2 * 48
= 1689.6 cm

Ans: The distance covered in 2 days is 1689.6 cm
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Q.3) A wooden door has a trapezium shaped glass in the center. One of the parallel
sides of the glass is thrice the other. If the area of the glass is 216 m² and the distance between the parallel sides is 12m, find the parallel sides.

SOLUTION: LET one of the parallel sides be x
the other side = 3x
Area of the glass = 216 m²
Distance between the parallel sides = 12 m = h
FIND: The parallel sides =?

Area of the trapezium glass =1/2 * ( sum of parallel sides ) * h
216 = 1/2 * (x + 3x) * 12
216 *2 *1/12 = 4x
36 / 4 = x
9 m = x
Thus, the parallel sides are x = 9m and 3x = 3 * 9 =27 m
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Q.4) Harish has dug out a cuboidal well with dimensions 2m * 1.5m * 10m in his field. Find the cost of cementing the walls of the wall at the rate of Rs. 52 per m².

SOLUTION: WELL ( CUBOIDAL )

Given :
Length = L = 2m
Breadth = B = 1.5m
Height = 10m
Rate of cementing the walls = Rs. 52 per m²
Cost of cementing the walls = ?

STEP – 1) Area of CSA = LSA = ?
LSA = 2 ( BH + HL )
= 2 H ( B + L )
= 2 * 10 * ( 2 + 1.5 )
= 20 * 3.5
= 70m²

STEP – 2) Cost of cementing = LSA * Rate
= 70 * 52
= Rs. 3640
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Q.5) Nisha is making a twin – color cuboidal box with dimensions 16cm * 6cm * 5cm. Walls of the box are red in color while the base and the lid are black in color. Find the areas of both the colored sheets required to make the box.

SOLUTION: CUBOIDAL BOX

Given:
Length = L = 16cm
Breadth = B = 6cm
Height = 5cm
Area of four walls = LSA = ? in red color
Area of base and lid = 2 * Area of rectangle in black color = ?

STEP – 1) LSA = 2 * ( BH + HL )
= 2 * H ( B + L )
= 2 * 5 *( 16 + 6 )
= 10 * 22
= 220cm ²

Thus, Area of red sheet required is 220cm ²

STEP – 2) Area of base + Area of lid = ?
= L * B + L * B
= 2 * ( L * B )
= 2 * ( 16 * 6 )
= 2 * 96
= 192 cm ²

Thus, the area of black sheet required is 192 cm ².
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Q.6) A room has a door of dimensions 2.5m * 1m and two windows of dimensions 1m * 0.5m. If the dimensions of the room are 6m * 5m * 10m, find the cost of white washing the four walls and the ceiling at the rate of Rs. 102 per m².

SOLUTION: DOOR ( RECTANGLE )

Given:
( STEP – 1 )
Length = L = 2.5m
Breadth = B = 1m
Area of the door = L * B
= 2.5 * 1
= 2.5m ²

STEP – 2) WINDOWS ( RECTANGLE )

Given:
Length = L = 1m
Breadth = B = 0.5m
Area of two windows = 2 * ( L * B )
= 2 * ( 1 * 0.5 )
= 1m²

STEP – 3) ROOM

Given:
Length = L = 6m Breadth = B = 5m Height = H = 10m
Area of the four walls = 2 * ( BH + HL )
= 2 * H ( B + L )
= 2 * 10 ( 6 + 5 )
= 20 * 11
= 220m²

STEP – 4) CEILING ( RECTANGLE )

Given:
Length = L = 6m Breadth = B = 5m
Area of the ceiling = L * B
= 6 * 5 = 30m²

STEP – 5 ) Area to be white washed = Area of the four walls + Area of the ceiling –
( Area of the door and two windows )
= 220 + 30 – ( 2.5 + 1 )
= 250 – 3.5
= 246.5m²

STEP – 6) Rate of white washing =Rs. 102 per m²
Cost of white washing = Area to be white washed * Rate
= 246.5 * 102 = Rs. 25, 143
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