MENSURATION (CH.11) EXTRA QUESTIONS
1. If each side of a cube is doubled, how many times will its surface area increase?
2. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3
.3. A cuboid is of dimensions (60*50*30)cm.How many small cubes with side 6 cm can be placed in the given cuboid?
4. Find the height of the cylinder whose volume is 1.54 m3 and diameter of base is 140 cm.
5. Find the area of trapezium where length of parallel sides are 15 cm and 25 cm and the third side measures 12 cm.
6. Find the area of rhombus whose diagonals are 8cm and 10cm.
7. If each side of a cube is doubled, how many times will its volume increase?
8. A rectangular sheet of paper is having measures 11 cm* 4 cm. it is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.
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Answers
Answer:
. it will increase double
98000
your answer are up
1 . four times
So the new surface area of the cube is four times the old surface area.
2 . Given
the base area of the cuboid is 180 cm2
Th volume of cuboid is 900 cm3
Find out
We have to find out the height of a cuboid
Solution
Volume of Cuboid = length x breadth x height = 900 cm3
So, Area x height = Volume
180 cm2 x height = 900 cm3
900 = 180×h
On substituting the known values, we get
h= 900/180 = 5
Answer
Hence the height of the cuboid is 5 cm
3 . 450 small cubes
Summary: A cuboid is of dimensions 60 cm × 54 cm × 30 cm. 450 small cubes with side 6 cm can be placed in the given cuboid.
4 . Let's construct a cylinder with the given dimensions as shown below.
Class 8 Maths Chapter 11 Exercise 11.4 Question 5
If the height of the cylinder = h
Radius of the base (r) = 140 / 2 cm = 70 cm = 0.7 m
Given that the volume of the cylinder is 1.54 m³
Volume of the cylinder = πr²h
1.54 = (22/7) × 0.7 × 0.7 × h
h = (1.54 × 7) / (0.7 × 0.7 × 22)
h = 1 m
Thus the height of the cylinder is 1 m.
5 . You have not mentioned anything about the fourth side or the shape of the figure whether it is isosceles trapezium or two adjacent angles are 90 deg. Hence there are two possibilities:
1. The figure is an isosceles trapezium: ABCD, having AB = 15 cm, CD = 25 cm and BC = AD = 12 cm.
The distance between AB and CD = [12^2–5^2]^0.5 = [144–25]^0.5 = 119^0.5 = 10.90871211 cm. The area of the isosceles trapezium = (15+25)*10.90871211/2 = 218.1742423 sq cm.
2. The figure is a trapezium: ABCD, having AB = 15 cm, CD = 25 cm, <A = <D = 90 deg and BC = 12 cm. The distance between AB and CD = [12^2–10^2]^0.5 = [144–100]^0.5 = 44^0.5 = 6.633249581 cm. The area of the trapezium = (15+25)*6.633249581/2 = 132.6649916 sq cm.
You have two choices for the area of the trapezium as 218.1742423 sq cm. or 132.6649916 sq cm.
6 . Hello,
Lets make it easy.
Basic thing we need is area of triangle formula i.e 1/2*(base)*(height).
Here is a rough diagram, take a look
Now we can see there are 4 triangles of similar area.
Area of one triangle = 1/2 *(4)*(5) = 10 cm sq.
Total area of rhombus = 4* area of one triangle
= 4*10 = 40 cm sq.
Thank you
Hope u find it useful