Answer:
x = 16
AB = 25
BC = 15
AC = 20
Explanation:
I have marked the vertices to explain
There are three right triangles that can be seen in the picture
Δ ABC with legs BC and AC and hypotenuse AB
ΔACD with legs AD and CD and hypotenuse AC
ΔBCD with legs CD and BD and hypotenuse BC
We will using the Pythagorean formula in all three triangles:
hypotenuse² = sum of squares of legs
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(1) Let's first find missing side BC
In right triangle ΔBCD ,
BC² = CD² + BD²
We have CD = 12, BD = 9
BC² = 12² + 9²
= 144 + 81
= 225
BC = √225 = 15
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(2) Now let's focus on ΔABC
AB is the hypotenuse
AC and BC are the legs
AB² = AC² + BC²
or
AC² = AB² - BC²
Since AB = AD + BD = x + 9 and BC = 15
AC² = (x + 9)² - 15²
(x + 9)² = x² + 2 · 9 · x + 9² = x² + 18x + 81
15² = 225
AC² = x² + 18x + 81 -225
AC² = x² + 18x - 144 (1)
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(3) Now let's turn our attention to ΔACD with legs AD and CD and hypotenuse AC
AC² = AD² + CD²
With AD = x and CD = 12,
AC² = x² + 12²
AC² = x² + 144 (2)
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Equations (1) and (2) have the same left side, AC²
So the expressions on the right side must be equal
Therefore:
x² + 18x - 144 = x² + 144
x² terms cancel out from left and right sides giving
18x - 144 = 144
Add 144 on both sides:
18x - 144 + 144 = 144 + 144
18x = 288
x = 288/18 = 16
Plugging this into equation (2) AC² = x² + 144 gives
AC² = 16² + 144
AC² = 256 + 144
AC² = 400
AC = √400
or
AC = 20
Since AB is the third unknown side and AB = x + 9 we get
AB = 16 + 9 = 25