MATHS 急

Volume of cone, V = 1/3 pi x r^2 x h
where r is radius and h is the height and pi is 3.1416
The ratio of r/h is always a constant r1/h1 = r2/h2, so r2 = r1 x h2/h1
r1 = radius of cone when the cone is full
h1 = depth of water when the cone is full, h1 = 5 cm (given)
r2 = radius of cone when the cone is half full
h2 = depth of water when the cone is half full

When the cone is full
V1 =1/3 pi x ( r1)^2 x h1
V1 =1/3 pi x ( r1)^2 x 5 cm ------------ (1)

When the cone is half full
½ V1 =1/3 pi x ( r2)^2 x h2
V1 =2/3 pi x ( r2)^2 x h2 --------------- (2)

r2 = r1 x h2/h1
r2 = r1 x h2/5 ------------------- (3)

Let equation 1 = equation 2, eliminate V

1/3 pi x ( r1)^2 x 5 = 2/3 pi x ( r2)^2 x h2
( r1)^2 x 5 = 2 x ( r2)^2 x h2

Substitute equation (3),

( r1)^2 x 5 = 2 x (r1x h2/5)^2 x h2
( r1)^2 x 5 = 2 x [(r1)^2 x (h2)^2/5^2] x h2
5 = 2 x (h2)^2/5^2 x h2
5 x 25/2 = (h2)^2 x h2
(h2)^3 = 62.5 cm^3
h2 = cubic root of (62.5) cm = 62.5^(1/3) cm = 3.9685 cm


The depth of water is 3.9685 cm after drinking half of the water.

The size of the right conical paper cup is not important.

The two right circular cones formed before and after drinking are similar.

The height of the right circular cone before drinking is 5 cm.
Let the height of the right circular cone after drinking is x cm.

2011-01-06 10:41:21 補充：
For simplicity, let V1 cm^3 and V2 cm^3 be the volumes of the water before and after drinking.
Then V1 = 2 V2
　V1/V2 = 2

The ratio of the volumes of two similar solids is equal to the ratio of the cubes of any two corresponding linear measurements.

2011-01-06 10:41:37 補充：
V1/V2 = (5^3)/(x^3)
2 = (5^3)/(x^3)
x^3 = 125/2 = 62.5
x = 3.9685

Therefore, after drinking half of the water, the depth of water is 3.97 cm.

1) Where is the figure?
2) 5 points too few to attract answers.