A compact bowl (CD) is made such that the shortest distance between the edge of the centre hole and also the sheet of the disc is 53.0 mm. Find the radius of the centre-hole if 1.36% the the key is removed in making the hole.

You are watching: What is the radius of a cd  Suppose the facility hole"s radius is \$,r,\$ , then

\$\$pi r^2=(0.0136)pi(53+r)^2implies 0.9824 r^2-1.4416r-38.2024=0\$\$

Now just solve the above straightforward albeit pretty stroked nerves quadratic in \$,r,\$ ... Let the radius of the center hole it is in x. Climate from the conditions we have.

\$\$r = 53 + x, ext whereby r is the radius the the CD\$\$

For the area that the facility hole we have:

\$\$P_1 = x^2pi\$\$

and because that the totality disk we have:

\$\$P = r^2pi\$\$

From the problems we have

\$\$P_1 = frac1.36100 P\$\$\$\$x^2pi = frac13610000 r^2pi\$\$\$\$10000x^2 = 136(53+x)^2\$\$\$\$10000x^2 = 136(2809+106x+x^2)\$\$\$\$10000x^2 = 382024 + 14416x + 136x^2\$\$\$\$9864x^2 - 14416x - 382024 = 0\$\$

Solving this quadratic equation we obtain two solutions:

\$\$x = frac53 (17+25 sqrt34)1233\$\$\$\$x = frac53 (17-25 sqrt34)1233\$\$

Because the radius need to be confident number we have:

\$\$ x = frac53 (17+25 sqrt34)1233 approx 7\$\$

So the radius the the center hole is 7 mm

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reply Aug 8 "13 in ~ 12:26 Stefan4024Stefan4024
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Lets say that the radius that the disk is \$r\$.

So because of this \$r=53+x\$ where \$x\$ is radius that hole.

The area, \$A\$ of the disk is\$\$A_disk=pi r^2\$\$\$\$A_disk=pi (53+x)^2\$\$

We likewise know that

\$\$A_hole=pi x^2\$\$\$\$A_hole=0.0136A_disk\$\$

This is since we know that the area that the feet is what castle took out of the CD

So because \$A_hole=pi x^2\$, and also \$A_hole=0.0136A_disk\$, for this reason we deserve to make the statement

\$\$pi x^2=0.0136A_disk\$\$

And due to the fact that we recognize the area that the disk,

\$\$pi x^2=0.0136\$\$\$\$x^2=0.0136(53+x)^2\$\$\$\$10000x^2=136(x+53)^2\$\$\$\$10000x^2=136(x^2+106x+2809)\$\$\$\$10000x^2=136x^2+14416x+382024\$\$\$\$9864x^2-14416x-382024=0\$\$\$\$2466x^2-3604x-95506=0\$\$\$\$1233x^2-1802x-47753=0\$\$\$\$x=frac1802pm sqrt3247204+2355177962466\$\$\$\$x=frac1802pm sqrt2387650002466\$\$

Accounting because that the hopeful only,

\$\$x=frac1802+sqrt2387650002466\$\$

X is about 6.9967 i beg your pardon is about 7 devices so as such the radius that the hole is 7 units.

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edited Aug 9 "13 at 5:30
reply Aug 9 "13 in ~ 4:57 \$endgroup\$
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