WBBSE Solutions For Class 10 Maths Chapter 4 Rectangular Parallelopiped or Cuboid Exercise 4.1
Question 1. I am observing that the length, breadth, and height of the rectangular parallelopiped box, brought by my brother, are 40 cm, 25 cm, and 15 cm respectively.
Solution:
I am observing that the length, breadth, and height of the rectangular parallelopiped box, brought by my brother, are 40 cm, 25 cm, and 15 cm respectively.
Total surface area of a rectangular box
=2 (L x B+ L x H + B x H)
=2 (40 x 25+ 40 x 15+ 25 x 15) sqcm.
=2 (1000+ 600 + 375) sqcm.
= 2 x 1975 sqcm.
= 3950 sqcm. Ans.
Total surface area of a rectangular box = 3950 sqcm.
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Question 2. The length, breadth, and height of a rectangular parallelopiped box are 15 cm, 12 cm, and 20 cm respectively. Let me write its total surface area.
Solution:
The length, breadth, and height of a rectangular parallelopiped box are 15 cm, 12 cm, and 20 cm respectively.
Total Surface area of a rectangular parallelopiped
=2 (15 x 12 + 15 x 20 + 12 x 20) sqcm.
=2 (180 + 300 +240) sqcm = 2 x (720) sqcm
= 1440 sqcm. Ans.
Total Surface area of a rectangular parallelepiped = 1440 sqcm.
Question 3. By measuring, I see that the length of one side of the cube, brought by Rajia is 27 cm.
Solution: Length of one side of a cube = 27 cm.
∴ Total surface area of the cube = 6 x (27)2sqcm.
= 4374 sqcm.
Question 4. Let us write by calculating the number of colored papers required to cover the whole surface of a cube whose length of one side is 12 cm.
Solution: Total area of the colour paper required for the cube
= 6 x (12)2 sqcm.
= 6 x 144 sqcm.
= 884 sqcm.
Question 5. The length, breadth, and height of our living room are 7 m, 5 m, and 4 m respectively. The shape of the room is
[cube/rectangular parallelopiped] Let us calculate the total area to color four walls of the room.
Solution: The length, breadth & height of the room arc 7m, 5m, and 4m respectively.
∴ The room is a rectangular parallelopiped.
∴ The area of the 4 walls of the room
2 x (L + B) x H
= 2(7 + 5) x 4 sqcm.
= 96 sqcm.
Question 6. Let us write the length of one side of a cube whose total surface area is 150 sq. cm.
Solution: Let one side of a cube = a cm.
∴ Total surface area of the cube = 6a2
∴ 6a2 =
∴ \(a^2=\frac{150}{6}=25\)
∴ \(a= \pm \sqrt{25}= \pm 5\)
But a ≠ -5 as length of the cube is positive.
∴ Length of the cube = 5 cm.
Question 7. Let us calculate the length of one side of a cube whose total surface area is 486 sq.m.
Solution: Let one side of a cube = am.
∴ 6a2 = 486. or,
a2 = 81 or,
a = √81 = 9
One side of the cube = 9m.
Question 8. A cuboidal room has its length, breadth, and height as a, b, and c units respectively and if a + b + c = 25 units, ab + bc + ca= 240.5 units, then let us write the length of the longest rod that can be kept in this room.
Solution: \((a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)\)
or, \((25)^2=a^2+b^2+\dot{c}^2+2.240 .5\)
or, \(625=a^2+b^2+c^2+481\)
∴ \(a^2+b^2+c^2=625-481=144\)
∴ The diagonal of the room = \(\sqrt{a^2+b^2+c^2}\)
= √144 = 12 m.
Question 9. I have made a rectangular parallelopiped by joining two cubes side by side made by Mita, whose edge is 8 cm in length. Let us calculate the total surface area and the length of the diagonal of the rectangular parallelopiped which is made in this way.
Hints: The length of the rectangular parallelopiped which is made = (8 + 8)cm = 16 cm, breadth = 8 cm, height = 8 cm.
Solution: If two cubes are placed side by side. New length of the parallelopiped
= (8 + 8) = 16 cm.
Breadth = 8 cm.
Height = 8 cm.
∴ Total area of the rectangular parallelopiped
= 2(16 x 8 + 16 x 8 + 8 x 8)sqcm.
= 2(128 + 128 + 64)sqcm.
= 2 x 320
= 640 sqcm.
∴ Diagonal of the rectangular parallelopiped
= \(\sqrt{(16)^2+(8)^2+(8)^2}\)
= \(\sqrt{256+64+64}\)
= √384
= 8√6 cm.
If the height of a rectangular parallelopiped is increased, its volume will increased.
Application 10. By measuring the cuboidal box, I am observing that the length is 32 cm, breadth is 21 cm, and height is 15 cm.
Solution: Volume of the sand = 32 x 21 x 15 cu. cm = 10080 cu. cm.
Application 11. The length of one side of a cube is 5 cm and its volume is c.c = c.c
Solution: Length of one side of a cube = 5cm.
∴ Volume of the cube = (5)3cu.cm
= 125 cu.cm.
Question12. If the length, breadth, and volume of a cuboidal room are 8 m, 6 m, and 192 cubic m respectively, then let us calculate the height of the room and the area of the four walls of the room.
Solution: According to the problem,
8 x 6 x 4 = 192 h
= \frac{192}{8 \times 6}
= 4
∴ Area of the 4 walls of the room
= 2 x (8 + 6) x 4
= 2 x 14 x 4 sqm.
= 112 sqm.
Question 13. The length and breadth of cuboidal water land in the neighboring villages are 18 m and 11 m respectively. In this water land, water is being irrigated from a nearby pond with a pump. If the pump can irrigate 39,600 lit. water per hour then for how much time is required to raise the height of the water level by 3.5 cm of the water land [1 lit. – 1 cubic DCM.]
Solution: The pump will run = \(\frac{180 \times 110 \times 3.5}{39600} \mathrm{hr}\)
= \(\frac{180 \times 110 \times 35}{39600 \times 10} \times 60 \mathrm{~min}\)
= 35 x 3 = 105 min
= 1 hr 45 min.
Question 14. If the pump can fill 37,400 lit. water in 1 hour, then what time will be required to raise the height of water 17 dcm of a cuboidal water land whose length is 18 m and breadth is 11 m. Let us write by calculating it.
Solution: Length, Breadth & Height of the tanks are 18m, 11m & 17 dc,
Respectively volume of the tank = (180 x 110 x 17) cu.dcm.
= 180 x 110 x 17 litre.
∴ The pump will run = \(\frac{180 \times 110 \times 17}{37400}\)
= 9 hours.
Question 15. From a wooden log with a length of 5 cm., a breadth of 5 dcm. and the thickness of 3 cm, 40 planks are 2 cm in length and 2 cm in breadth are cloven. For cleaving, wood has been destroyed. But still now 108 cubic dcm of wood remains in the log. What is the thickness of each plank that is cloven.
Solution: Total volume of the log = 40 dcm x 5 dcm x 3 dcm.
= (40 x 5 x 3)cu.dcm.
= 6000 dcm.
Total volume of wood wasted = \(600 \times \frac{2}{10}\)
= 12 cu.dcm.
Let the thickness of each plate = x dcm.
∴ Volume of each plate = 20 x 2 x x cu.dcm.
Volume of 40 plate = 40 x (20 x 2 x x)
= 1600 x cu.dcm.
According to the problem,
1600x + 108 + 12 = 600
1600x = 600 – 120 = 480
∴ \(x=\frac{480}{1600}=0.3 \mathrm{dcm}\)
∴ Thickness of each plate = 0.3 dcm.