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# Mensuration

A solid occupies fixed amount of space. Solids occur in different shapes. Observe the following shapes. These shapes (cuboid, cube, triangular prism, cylinder, cone etc) are known as three dimensional objects.

4.1.2 Surface area of cuboid. Lateral Surface area of cuboid = 2h(l + b)

Total surface area of cuboid = 2(lb +bh + lh)

Here h – height ; l – length; b – breadth.

Exercise 4.1.2

1. Find the total surface area of the cuboid with l = 4m, b = 3m and h = 1.5m

Solution:

Total surface area of cuboid = 2(lb +bh + lh)

We have, l = 4m; b = 3m; h = 1.5m

Therefore,

Total surface area of cuboid = 2[(4×3) +(3×1.5) + (4×1.5)]

= 2[ 12 + 4.5 + 6.0]

= 45 m²

2. Find the area of four walls of a room whose length 3.5m, breadth 2.5m and height 3m

Solution:

Area = 2h(l + b)

We have, l = 3.5m; b = 2.5m; h = 3m

Therefore,

Area = 2 x 3 [3.5 + 2.5]

= 6[ 6]

= 36 m²

3. The dimensions of a room are l = 8m, b = 5m, h = 4m. Find the cost of distempering its four walls at the rate of Rs. 40/m²

Solution:

We know that, l = 8m, b = 5m, h = 4m

Area = 2h(l + b)

We have, l = 8m; b = 5m; h = 4m

Therefore,

Area = 2 x 4 [ 8 + 5]

= 8[ 13]

= 104 m²

Thus, the cost of distempering its four walls at the rate of Rs. 40/m² = 104 x 40 = Rs. 4160.

4.A room is 4.8m long, 3.6m broad and 2m high. Find the cost of laying tiles on its floor and its four walls at the rate of Rs. 100/m²

Solution:

Area = 2h(l + b)

We have, l = 4.8m; b = 3.6m; h = 2m

Therefore,

Area = 2 x 2 [ 4.8 + 3.6]

= 4[ 8.4]

=  33.6 m²

Therefore, the area of four walls is 33.6 m²

Now, we have find the area of the floor = (l x b) = ( 4.8 x 3.6) = 17.28 m²

Thus, area of floor + area of 4 walls =  17.28m² + 33.6 m²  = 50.88m²

Thus, the cost of laying tiles on its floor and its four walls at the rate of Rs. 100/m²

= 50.88 x 100

= 50.88 Rs.

5. A closed box is 40cm long, 50cm wide and 60cm deep. Find the area of the foil needed for covering it.

Solution:

Total surface area of cuboid = 2(lb +bh + lh)

We have, l = 40 cm; b = 50 cm; h = 60 cm

Therefore,

Total surface area of cuboid = 2[(40 x 50) +(50 x 60) + (60 x 40)]

= 2[ 2000 + 3000 + 2400]

= 2[ 7400]

= 14800 cm²

6. The total surface area of a cube  is 384cm². Calculate the side of the cube.

Solution:

Total surface area of cube = 6(l²)

384 = 6(l²)

l² = 384/6 = 64 cm²

l = √64 = 8m

Thus, The side of the cube is 8m.

7. The L.S.A of a cube is 64m². Calculate the side of the cube.

Solution:

Lateral surface area of cube = 4(l²)

64 = 4(l²)

l² = 64/4 = 16

Therefore, l = 4m

Thus, the side of the cube is 4l.

8.Find the cost of whitewashing the four walls of a cuboidal room of side 4m at the rate of Rs. 20/m²

Solution:

Lateral surface area of the cuboidal room = 4(l²)

= 4 ( 4²)

= 4 ( 16)

= 64 m²

The cost of whitewashing the cuboidal room at the rate of Rs. 20/m² = 64 x 20 = Rs. 1280

9. A cubical box has edge 10 cm and another cuboidal box is 12.5cm long, 10cm wide and 8cm high.

(i) Which box has smaller total surface total surface area?

(ii) If each edge of the cube is doubled, how many times will its T.S.A increase?

Solution:

(i)

Cube:

Total surface area = 6l² = 6 x 10² = 6 x 100 = 600m²

Cuboid:

Total surface area = 2(lb + bh + lh)

= 2[(12.5 x 10) + (10 x 8) + (8 x 12.5)]

= 2 [ 125 + 80 + 100]

= 610 m²

Thus, Cube has a smaller total surface area.

(ii)

If each edge of the cube is doubled,

Then,

Cube:

Total surface area = 6l² = 6 x 20² = 6 x 400 = 2400m²

Cuboid:

Total surface area = 2(lb + bh + lh)

= 2[(25 x 20) + (20 x 16) + (16 x 25)]

= 2 [ 500 + 320 + 400]

= 2

= 2440 m²

Thus,

If each edge of the cube is doubled, then its T.S.A will increase 4 times.

4.1.3 Volume of cubes and cuboids:

Volume of cube = l³ , where l = length

Volume of cuboid = area of base x height;

where area of base = l x b, here l = length, b = breadth

Exercise 4.1.3

1. Three metal cubes whose edges measure 3cm, 4cm and 5cm respectively are melted to form a single cube. Find (i) Side length (ii) total surface area of the new cube. What is the difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes?

Solution:

Three metal cubes whose edges measure 3cm, 4cm and 5cm respectively are melted to form a single cube.

(i) Side length of the new cube = 3 + 4 + 5 = 12cm

(ii) total surface area of the new cube = 6 (l²) = 6(12²) = 6 x 144 = 864 cm²

Total surface area of the cube with edge 3cm = 6 (l²) = 6 x 9 = 54cm²

Total surface area of the cube with edge 4cm = 6 (l²) = 6 x 16 = 96cm²

Total surface area of the cube with edge 5cm = 6 (l²) = 6 x 25 = 150cm²

The sum of total surface area of the cubes = 54 + 96 + 150 = 300cm²

The difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes = 864 cm² – 300 cm² = 564 cm²

2.Two cubes, each of volume 512cm³ are joined end to end. Find lateral and total surface areas of the resulting cuboid.

Solution:

since its a cube , Volume of cube = l³
512 = l³ ; l = 8
side is 8 cm
when you join two cubes , we will get a cuboid whose length is 16 cm , h= 8 , b=8
total s.area of cuboid = 2(lh+bh+lb)
= 2( 128+128+64)
= 640 cm sq
latral .s area = 2h(l+b)
= 2*8 (16+8)
= 384cm²

3. The length, breadth and height of a cuboid are in the ratio 6:5:3. If the total surface area is 504cm², find its dimension. Also find the volume of the cuboid.

Solution:

total surface area is 504cm²

The length, breadth and height of a cuboid are in the ratio 6:5:3

total surface area = 2(lb + bh + lh)

504 = 2(30x² + 15x² + 18x²)

504 = 2 x 63x²

504 = 126x²

x² = 504/126 = 4

x = 2cm

Therefore,

length= 6x = 12cm

height = 3x = 6cm

Volume of cuboid = area of the base x height = l x b x h = 12 x 10 x 6 = 720cm³

4. How many m³ of soil has to be excavated from a rectangular well 28m deep and whose base dimensions are 10m and 8m. Also find the cost of the plastering its vertical walls at the rate of Rs. 15/m²

Solution:

Volume of cuboid = area of the base x height = l x b x h = 28 x 10 x 8 = 2240cm³

Lateral surface area = 2h(l + b) = 2 x 28 (10 + 8) = 2 x 28 (18) = 1008

the cost of the plastering its vertical walls at the rate of Rs. 15/m² = 1008 x 15 = Rs. 15120

5. A solid cubical box of fine wood costs Rs. 256 at the rate Rs. 500/m³. Find its volume and length of each side.

Solution:

A solid cubical box of fine wood costs Rs. 256 at the rate Rs. 500/m³

Cost for one meter cube = Rs. 500

Original Cost = Rs. 256

Volume * cost for one meter cube = 256

volume = 256/500 = 0.512 m³

Volume of Cube = l³(l is the side)
l³=0.512m³

Now to convert it into cm³ we multiply 1000000 to it.
Thus, 0.512m³ = 0.512 x 1000000 = 512000cm³
So we have the volume as 512000cm³. Thus the length of the side of the cube = ∛512000
= 80cm