Question 900733
The radius of the base of each of two right circular cones is 8 inches.
 The altitude of the first cone is 9 greater than that of the second cone
 and its slant height is 7 inches greater than that of the second.
 Find the altitude of the taller cone.
:
This can be done by considering two right triangles, using pythag formulas
let x = the height of taller cone
then
(x-9) = the height of the smaller cone
The slant dimensions are the hypotenuses
{{{sqrt(x^2+8^2)}}} - 7 = {{{sqrt((x-9)^2 + 8^2)}}}
{{{sqrt(x^2+64)}}} - 7 = {{{sqrt((x^2-18x+81)+64)}}}
{{{sqrt(x^2+64)}}} - 7 = {{{sqrt((x^2-18x+145))}}}
square both sides
{{{x^2 + 64 - 14sqrt(x^2+64) + 49 = x^2 - 18x + 145}}}
subtract x^2 from both sides
{{{64 - 14sqrt(x^2+64) + 49 = - 18x + 145}}}
{{{-14sqrt(x^2+64) + 113 = - 18x + 145}}}
{{{-14sqrt(x^2+64) = - 18x + 145 - 113}}}
{{{-14sqrt(x^2+64) = -18x + 32}}}
simplify, change the signs, divide by -2
{{{7sqrt(x^2+64) = 9x - 16}}}
square both sides
49(x^2 + 64) = 81x^2 - 288x + 256
49x^2 + 3136 = 81x^2 - 288x + 256
combine like terms on the right
0 = 81x^2 - 49x^2 - 288x + 256 - 3136
0 = 32x^2 - 288x - 2880
simplify divide by 32
x^2 - 9x - 90 = 0
Factors to
(x-15)(x+6) = 
the positive solution here
x = 15 in is the altitude of the taller cone
:
:
See if that checks out the height of the smaller will be 6
{{{sqrt(15^2+8^2)}}} = 17
{{{sqrt(6^2+8^2)}}}  = 10
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slant differences 7 in