Radius of cone (r) = 6/2 = 3 cm
and height of cone (h) = 4cm
Slant height (l) = ?h2 + r2
= ?(4)2 + (3)2
= ?16 + 9
= 5 cm
Now, curved surface area = ?rl
= (22/7) x 3 x 5
= 330/7 = 47 cm2
Volume = (1/3)?r2h
According to the question,
(1/3)?r2h = 100?
? (1/3)?r2 x 12 = 100?
? r2 = 25
? r = ?25 = 5 cm
? Slant height (l) = ?h2 + r2
= ?122 + 52
= ?169 = 13 cm
volume = 48 ? cm3 and h = 9 cm
? (1/3)?r2h = 48?
? (1/3) x ? x r2 x 9 = 48?
? r2 = (48? x 3) / (? x 9) = 16
? r = ?16 = 4 cm
? Diameter = 2r = 2 x 4 = 8 cm
Slant height (l) = ?r2 + h2
=?72 + 242
= ?49 + 576
= ?625
= 25 cm
Curved surface area = ?rl = (22/7) x 7 x 25 = 550 sq cm
? Area of 5 caps = 550 x 5 = 2750 sq cm
In first situation.
Radius = r1, height = h1 and volume = v1
In second situation,
Radius =2r1, height = h2 and volume = v2
In the volume is fixed, then
v1 = v2
? (1/3)?r21h1 = (1/3)?(2r1)2h2
? h1 = 4h2
? h2 = h1/4
Therefore, height of the the cone will be one-fourth of the previous height.
Given, l = 9 m
and diameter = 14 m ? r = 14/2 = 7 m
Now, volume = (1/3)?r2h
= (1/3)? x 49 x ?l2 - r2
= (1/3)? x 49 x ?81 - 49
= (1/3) x 49? x ?32 = 49??32 /3 m3
[(1/3)?r12h1]/[(1/3)?r22h2] = 2/3
? (r1/r2)2 x (h1/h2) = 2/3
? (1/2)2 x (h1/h2) = 2/3 [ &bacaus; r1/r2 = 1/2]
? h1/h2 = 8/3
? h1 : h2 = 8 : 3
Let radius = 5k, height = 12k
According to the question
= 1/3 x 22/7 x (5k)2 x 12k = 2200/7
? k = 1
? r = 5, h = 12
? Slant height (l) = ?r2 + h2
= ?25 + 144
= ?169
=13 cm
Here, x = -50%, y = 200%
According to the formula Net effect
= [2x + y + x2 + 2xy/100 + x2y/1002] %
= [-100 + 200 + 2500 - 20000/100 + 500000/10000]%
= [100 - 175 + 50] % = -25%
Let diameter, radius and height of first cone are d1, r1 and h1, respectively and that of second cone are d2, r2 and h2, respectively.
r1/r2 = d1/d2 = 3/5,
h1/h2 = ?
Given,
[(1/3)?r21h1] / [(1/3)?r22h2] = 1/3
? (3/5)2 x h1/h2 = 1/3
? h1/h2 = (1/3) x (25/9) = 25/27
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