Question #6b302

1 Answer
Mar 8, 2016

The length of the altitude #CD# is #4#.

Explanation:

Your situation looks like this:

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There is a right angle between AC and BC and a right angle between AD and CD (and BD and CD, of course).

As we have three right triangles, we can apply the Pythagorean Theorem to all three of them:

[1] #" "AC^2 + BC^2 = AB^2#

[2] #" "AD^2 + CD^2 = AC^2#

[3] #" "BD^2 + CD^2 = BC^2#

Furthermore, you know that:

  • #AD# is #12# more than the altitude, so #AD = CD + 12#

  • #BD# is #3# less than the altitude, so #BD = CD - 3#

Thus, you have

[4] #" "AD = CD + 12#

[5] #" "BD = CD -3#

Last piece of information is that

[6] #" "AB = AD + BD#

Now, let's try to find #CD# with the help of those 6 equations.

First of all, let's plug [2] and [3] into [1]:

#color(blue)(AC^2)" " + color(green)(BC^2) " "= AB^2#

#=>" "(color(blue)(AD^2 + CD^2)) + (color(green)(BD^2 + CD^2)) = AB^2#

Now, let's use [6] and plug #AD + BD# for #color(brown)(AB)#:

#=>" "(AD^2 + CD^2) + (BD^2 + CD^2) = (color(brown)(AD + BD))^2#

Let's simplify this equation:

#=>" "AD^2 + BD^2 + 2CD^2 = (AD + BD)^2#

Use the formula #(a+b)^2 = a^2 + 2ab + b^2# to expand the right side:

#=>" "AD^2 + BD^2 + 2CD^2 = AD^2 + 2*AD * BD + BD^2#

#=>" "cancel(AD^2) + cancel(BD^2) + 2CD^2 = cancel(AD^2) + 2*AD * BD + cancel(BD^2)#

#=> " " 2 * CD^2 = 2 * AD * BD#

Divide both sides by #2#...

#=> " " CD^2 = AD * BD#

Now, we can use [4] and [5]: plug #CD + 12# for #AD# and #CD - 3# for #BD#:

#=> " " CD^2 = color(orange)(AD) * color(purple)(BD)#

#=> " " CD^2 = (color(orange)(CD + 12)) * (color(purple)(CD - 3))#

Expand the right side:

#=> " " CD^2 = CD^2 + 9 CD - 36#

Subtract #CD^2# from both sides...

#=> " " 0 = 9CD - 36#

Solve for #CD#:

#=> " " CD = 4#

Thus, the altitude is #4#, and we can also compute all the other sides from the triangle:

  • #AD = 4 + 12 = 16#
  • #BD = 4 - 3 = 1#
  • #AC = sqrt(AD^2 + CD^2) = sqrt(16^2 + 4^2) = sqrt(272)#
  • #BC = sqrt(BD^2 + CD^2) = sqrt(1^2 + 4^2) = sqrt(17)#
  • #AB = AD + BD = 16 + 1 = 17#

or

  • #AB = sqrt(AC^2 + BC^2) = sqrt(272+17) = 17#