RS Aggarwal Solutions Class 8 Chapter-20 Volume and Surface Area of Solids (Ex 20A) Exercise 20.1 - Free PDF
FAQs on RS Aggarwal Solutions Class 8 Chapter-20 Volume and Surface Area of Solids (Ex 20A) Exercise 20.1
1. The length of a geometry box is 20 cm, its breadth is 5 cm, and has height of 1 cm. Find the volume of the geometry box.
Given,
The length of geometry box = 20 cm
The length of geometry box = 5 cm
The length of geometry box = 1 cm
The geometry box is in the shape of a cuboid.
So, the Volume of the geometry box is the same as the Volume of the cuboid.
The Volume of cuboid = lbh
Volume of geometry box = 20 cm * 5 cm * 1 cm = 100 cubic cm.
Hence, the Volume of the cuboid is 100 cubic cm.
2. A cheese cube has a length of 5 cm and is further cut into smaller cubes which have a volume of 1 cubic cm each. Find the number of such smaller cubes obtained.
Given,
Length of larger cube = 5cm
The Volume of cube = side3
The Volume of the larger cube = 53 = 125 cubic cm.
Number of cubes obtained = (Volume of the larger cube)/(Volume of smaller cubes)
Number of cubes = 125/1 = 125
Hence, the number of cubes obtained after cutting them into smaller cubes is 125.
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4. How can I do well in Class 8 Math?
The only way to do well in Class 9 Maths is to practise. Complete all of the Chapter exercises. This will enhance your problem-solving abilities as well as your speed and efficiency. Important formulas, definitions, and equations can be written down in a notebook and reviewed on a regular basis. Solve previous year's sample papers and question papers within a time limit. This will familiarise you with the paper pattern and question type, as well as help you improve your time management skills. Revise on a regular basis to ensure that you retain everything you've learned for a longer period of time.
5. A rectangular water reservoir contains 105 m3 of water. Find the depth of the water in the reservoir if its base measures 12 m by 3.5 m.
Given details are,
The capacity of water reservoir = 105 m3
Length of base of reservoir = 12 m
Width of base = 3.5 m
Let the depth of reservoir be ‘hm
l × b × h = 105
h = 105 / (l × b)
= 105 / (12×3.5)
= 105/42
= 2.5m
∴ The depth of the reservoir is 2.5 m