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Helppppppppp

Guest Feb 16, 2015

#1
+93691
+5

This is the cross section of the cone of solid X.

The top bit is INSIDE the hemisphere.  So I am only interested in the bottom smaller cone.

Volume of little cone:

$$\\=\frac{1}{3}\pi r^2 h\\\\ =\frac{1}{3}\pi* 25* 15\\\\ =\frac{1}{3}\pi* 25* 15\\\\ =75\pi\\\\$$

Volume of hemisphere

$$\\=\frac{1}{2}*\frac{4}{3} \pi r^3\\\\ =\frac{4}{6}\pi* 9^3\\\\ =\frac{2}{3}\pi* 9^3\\\\ =486 \pi$$

total volume of  solid X = $$486\pi + 75\pi = 561 \pi\;\;cm^3$$

Now the ratio of surface areas of X:Y   =     25:36

so the ratio of lengths of   X:Y              =    5:6

and the ratio of volumes      X:Y           =   125: 216

$$\\\frac{216}{125}=\frac{Y}{561\pi}\\\\ \frac{216}{125}\times 561\pi=Y\\\\ Y=\frac{216}{125}\times 561\pi\\\\ Volume\;of\;Y=\frac{121176\pi}{125}\;\;cm^3\\\\$$

Melody  Feb 16, 2015
#1
+93691
+5

This is the cross section of the cone of solid X.

The top bit is INSIDE the hemisphere.  So I am only interested in the bottom smaller cone.

Volume of little cone:

$$\\=\frac{1}{3}\pi r^2 h\\\\ =\frac{1}{3}\pi* 25* 15\\\\ =\frac{1}{3}\pi* 25* 15\\\\ =75\pi\\\\$$

Volume of hemisphere

$$\\=\frac{1}{2}*\frac{4}{3} \pi r^3\\\\ =\frac{4}{6}\pi* 9^3\\\\ =\frac{2}{3}\pi* 9^3\\\\ =486 \pi$$

total volume of  solid X = $$486\pi + 75\pi = 561 \pi\;\;cm^3$$

Now the ratio of surface areas of X:Y   =     25:36

so the ratio of lengths of   X:Y              =    5:6

and the ratio of volumes      X:Y           =   125: 216

$$\\\frac{216}{125}=\frac{Y}{561\pi}\\\\ \frac{216}{125}\times 561\pi=Y\\\\ Y=\frac{216}{125}\times 561\pi\\\\ Volume\;of\;Y=\frac{121176\pi}{125}\;\;cm^3\\\\$$

Melody  Feb 16, 2015
#2
0

the radius of the hemisphere must be 15cm

so all you have to do is change the R of the equation of what Melody had done from 9 to 15

Guest Feb 16, 2015