A garden roller whose length is 3 m long and whose diameter is 2.8 m is rolled to level a garden. How much area will it cover in 8 revolutions?
Answers
Area covered = Curved surface × Number of revolutions
r=21.4=0.7 m
h=2 m
Curved surface = 2πrh=2×722×0.7×2=8.8 m2
Hence,
Area covered = 8.8×5=44 m2
Step-by-step explanation:
correct answer
211.2 of (a)
528 of (b)
25 of (c)
4158 of (d)
7 of (e)
which one is easy you.
The correct answer
1.
circumference of circle = 2pir
= 2 x pi x 1.4
= 8.81 m
Area = length x breadth
= 8.81 x 3
= 26 .43 m^{2}
Area cover in 8 revolutions = 26.43 x 8
= 211.44 m^{2}------------------------------------------------------------------------2.
Area of the canvas = Curved surface area of the conical tent
Since the canvas is rectangular in shape, its area is = length × width
Curved surface area of a cone =πrl, where r is the radius of the cone and l is the slant height.
For \: a \: cone \: , l= \sqrt{h { }^{2} } + \sqrt{r {}^{2} }
where l is the slant height.
Hence, l= \sqrt{24 {}^{2} } + \sqrt{7 {}^{2} }
⇒l= \: \: \sqrt{625}
⇒l=25 cm
Hence, length ×5= 22/7 ×7×25
∴ length =110 m------------------------------------------------------------------------3.Given :-
The total surface area of a cone of radius 7 cm is 704 cm².
To find :-
The slant height.
Solution :-
Let the slant height of the cone is l cm.
Radius = 7 cm
According to the question,
πr(r+l) = 704
→ (22/7) ×7 (7+l) = 704
→ 22(7+l) = 704
→ 7+l = 704/22
→ 7+l = 32
→ l = 32-7
→ l = 25
Therefore, the slant height of the cone is 25 cm.------------------------------------------------------------------------4.
Let r be the radius of the hemisphere.
Given that, base area = πr2 = 1386 sq. m
T.S.A. = 3πr2 sq.m
= 3 ×1386 = 4158
Therefore, T.S.A. of the hemispherical solid is 4158 m2------------------------------------------------------------------------5.
d≈7
ASurface area
154
Unit Conversion:
Using the formulas
A=4πr2
d=2r
Solving for d
d=A
π=154
π≈7.00141------------------------------------------------------------------------