Table of contents |
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Match the column |
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Fill in the blanks |
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True & False |
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Answer the following questions |
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Sol :
Q2:
Ans:
(p) → (ii)
(q) → (iii)
(r) → (i)
Sol:
(i) 8πab
Sol: S = 2πrH
= 2π × 2b × 2a
= 8πab
(ii) πp2q
Sol: V=π r2H
Substituting value of radius = p and Height = q
We get ,
V=π p2q
(iii) 65,500cm2
Sol: 1 m2 = 10,000 cm2
6.55m2 = 6.55×10,000cm2 = 65,500cm2
(iv)double
Sol:V1= πr2H,
V2=π(2r)2(1/2)H
=2πr2H,
So it becomes double
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Important Questions: Mensuration
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Q1: The parallel sides of a trapezium are 20 cm and 10 cm. Its nonparallel sides are both equal, each being 13 cm. Find the area of the trapezium.
Sol:Let ABCD is the trapezium with AB and CD are the parallel sides.
Now
AB=10 cm, CD=20 cm, BC=13 cm
Now draw line BE || AD and draw a perpendicular from B on ECNow ABED is a parallelogram, then BE=13 cm
In triangle BEC, BE =BC, So Isoceles triangle,So perpendicular will bisect the EC
Hence EF = FC
Now EC= DC -DE = 20 -10 =10 cm
Therefore EF = FC = EC/2 = 5 cm
Now in Triangle BEF, it is right angle at F,So by pythagorous theorem
BE2 = BF2 + EF2
169 = BF2 + 25
BF= 12cm
This is also the perpendicular distance between the parallel sides. So now coming back to Area of trapezium
A = 1/2[a + b] × h
Here a = 10 cm , b = 20 cm, h = 12 cm., A =?
Therefore
A = 1/2[10 + 20]×12
A = 180cm2
Q2: The area of a rhombus is 300 cm², and one of the diagonals is 20 cm. Find the other diagonal.
Sol:Q3: A circular garden has a radius of 14 m. Find the cost of fencing the garden along its circumference at the rate of ₹12 per meter.
Sol:Q4: Find the area of a trapezium whose parallel sides are 38.7 cm and 22.3 cm, and the distance between them is 18 cm.
Sol: Area of trapezium = Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
A = 1/2[a + b] × h
Here a = 38.7 cm , b = 22.3 cm h = 18 cm
Therefore
A = 1/2[38.7 + 22.3] × 18
= 549cm2
Q5: Find the area of a trapezium whose parallel sides are 12 cm and 20 cm and the distance between them is 15 cm.
Sol:
Area of trapezium = Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
A = 12[a + b] × h
Here a = 12 cm , b = 20 cm h =15 cm
Therefore
A = 12[12 + 20] × 15
= 240cm2
Q6: A box in the form of a cuboid has external dimensions of 60 cm × 40 cm × 30 cm. The top face, two side faces, and the front face are to be painted with a glossy finish. Calculate the area that needs to be painted.
Sol:Q7: The area of a trapezium is 384cm2. Its parallel sides are in the ratio 2: 6 and the perpendicular distance between them is 12 cm. Find the length of each of the parallel sides.
Sol:
Area of trapezium=Half the product of the sum of the lengths of parallel sides and the perpendicular distance between them gives the
A = 12[a + b] × h
Here a = 2x cm , b = 6x, h = 12 cm., A = 384 cm2
Therefore
384 = 12[2x + 6x] × 12
x = 8cm
So parallel sides are 16 cm and 48 cm
Q8: A cylindrical road roller makes 750 complete revolutions to level a road. The diameter of the roller is 84 cm, and its length is 1 m. Calculate the total area leveled by the road roller.
Sol:
Q9: A rectangular piece of cardboard measuring 18 cm × 7 cm is folded without overlapping to make a cylinder of height 7 cm. Find the volume of the cylinder.
Sol:
Q10: A storage room is in the shape of a cuboid with dimensions 50 m × 30 m × 20 m. How many smaller cuboidal containers of volume 1.2 m³ can be stored in it?
Sol:
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