If the radius remains unchanged
Then, x = y = 0 and z = 17.5 %
? Net increase in volume = 17.5%
The volume of the original cone is V = (?R2h)/3
The height and the radius of the smaller cone are 2h/3 and 2R/3, respectively
? 1/3? (2R/3)2 x 2h/3 = 8V/27
? Volume of the frustum = (V - 8V/27) = 19V/27
So, the required ratio is 8 : 19 .
Volume of the Earth taken out = 30 x 20 x 12
= 7200 m3
Area of the remaining portion (leaving the area of dug out portion) = 470 x 30 + 30 x 10
= 14100 + 300 = 14400 m3
Let h be height to which the field is raised by spreading the Earth dug out.
Then, 14400 x h = 7200
? h = 7200/14400 = 0.5 m
? l + b + h = 19
and l2 + b2 + h2 = (5?5)2 = 125
? (l + b + h)2 = (19)2
? (l2 + b2+ h2) + 2(lb + bh + lh) = 361
? 2 (lb + bh + lh) = (361 - 125) = 236
? Surface area = 236 cm2
Let the edges of the two cubes be x and y metres
Then, x3 - y3 = 152
and x2 - y2 = 20
Also, x + y = 10
So, x - y = (x2 - y2) / (x + y) = 20/10 = 2
Now, (x3 - y3) / (x - y) = 152/2
? x2 + y2 + xy = 76
? (x + y)2 - xy = 76
? xy = (x + y)2 - 76 = (10)2 - 76 = 24
Let original length of cube = L
Then, its surface area = 6L2
New edge = (150 x L)/10 = 3L/2
New surface area = 6 x (3L/2)2 = (6 x 9 x L2)/4 = (27/2) x L2
Increase in surface area = (27/2 - 6) x L2 = 15 x L2/2
? Increase percent = (15 x L2/2 ) / (6L2) x 100 % = 125 %
Let x, y and z be the length breadth and deapth of a cuboid.
? x + y + z = 19
x2 + y2 + z2 = (5?5)2 = 125
Surface area of the cuboid = 2(xy + yz + zx)
= (x + y + z)2) - (x2 + y2 + z2)
= 361 - 125 = 236 cm3
L + B + H = 19 cm, where L is the length, B is the breadth and
H is the height of the cuboid.
? (?L2 + B2 + H2
= (5?5)
On squaring both sides, we get
L2 + B2 + H2 + = 125
(L + B + H)2 = L2 + B2 + H2 + 2LB + 2LH + 2BH
? (19)2 = 125 + 2(LB + LH + BH)
? Surface area = 2(LB + LH + BH)
= 192 = 125 + 2(LB + LH + BH)
= 192 - 125 = 361 - 125 = 192 - 125
= 361 - 125 = 236 cm2
Let the side of cube be x .
Then, Volume of cube = Surface area of cube
? x3 = 6x2
? x = 6
Volume of the tank = 2.6 cu m
Base area of the tank = 6500 cm = 0.65 sq m
? Depth of the tank = 26/0.65 = 4m
Total area of four walls = 2h(l + b)
Perimeter of the celling = 2l + 2b = 2(l + b)
Perimeter of the ceiling in percentage = 2(l +b ) x 100/2h(l + b) %
= 100/h %
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