# Question #6b302

Mar 8, 2016

The length of the altitude $C D$ is $4$.

#### Explanation:

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] $\text{ } A {C}^{2} + B {C}^{2} = A {B}^{2}$

[2] $\text{ } A {D}^{2} + C {D}^{2} = A {C}^{2}$

[3] $\text{ } B {D}^{2} + C {D}^{2} = B {C}^{2}$

Furthermore, you know that:

• $A D$ is $12$ more than the altitude, so $A D = C D + 12$

• $B D$ is $3$ less than the altitude, so $B D = C D - 3$

Thus, you have

[4] $\text{ } A D = C D + 12$

[5] $\text{ } B D = C D - 3$

Last piece of information is that

[6] $\text{ } A B = A D + B D$

Now, let's try to find $C D$ with the help of those 6 equations.

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

$\textcolor{b l u e}{A {C}^{2}} \text{ " + color(green)(BC^2) " } = A {B}^{2}$

$\implies \text{ } \left(\textcolor{b l u e}{A {D}^{2} + C {D}^{2}}\right) + \left(\textcolor{g r e e n}{B {D}^{2} + C {D}^{2}}\right) = A {B}^{2}$

Now, let's use [6] and plug $A D + B D$ for $\textcolor{b r o w n}{A B}$:

$\implies \text{ } \left(A {D}^{2} + C {D}^{2}\right) + \left(B {D}^{2} + C {D}^{2}\right) = {\left(\textcolor{b r o w n}{A D + B D}\right)}^{2}$

Let's simplify this equation:

$\implies \text{ } A {D}^{2} + B {D}^{2} + 2 C {D}^{2} = {\left(A D + B D\right)}^{2}$

Use the formula ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$ to expand the right side:

$\implies \text{ } A {D}^{2} + B {D}^{2} + 2 C {D}^{2} = A {D}^{2} + 2 \cdot A D \cdot B D + B {D}^{2}$

$\implies \text{ } \cancel{A {D}^{2}} + \cancel{B {D}^{2}} + 2 C {D}^{2} = \cancel{A {D}^{2}} + 2 \cdot A D \cdot B D + \cancel{B {D}^{2}}$

$\implies \text{ } 2 \cdot C {D}^{2} = 2 \cdot A D \cdot B D$

Divide both sides by $2$...

$\implies \text{ } C {D}^{2} = A D \cdot B D$

Now, we can use [4] and [5]: plug $C D + 12$ for $A D$ and $C D - 3$ for $B D$:

$\implies \text{ } C {D}^{2} = \textcolor{\mathmr{and} a n \ge}{A D} \cdot \textcolor{p u r p \le}{B D}$

$\implies \text{ } C {D}^{2} = \left(\textcolor{\mathmr{and} a n \ge}{C D + 12}\right) \cdot \left(\textcolor{p u r p \le}{C D - 3}\right)$

Expand the right side:

$\implies \text{ } C {D}^{2} = C {D}^{2} + 9 C D - 36$

Subtract $C {D}^{2}$ from both sides...

$\implies \text{ } 0 = 9 C D - 36$

Solve for $C D$:

$\implies \text{ } C D = 4$

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

• $A D = 4 + 12 = 16$
• $B D = 4 - 3 = 1$
• $A C = \sqrt{A {D}^{2} + C {D}^{2}} = \sqrt{{16}^{2} + {4}^{2}} = \sqrt{272}$
• $B C = \sqrt{B {D}^{2} + C {D}^{2}} = \sqrt{{1}^{2} + {4}^{2}} = \sqrt{17}$
• $A B = A D + B D = 16 + 1 = 17$

or

• $A B = \sqrt{A {C}^{2} + B {C}^{2}} = \sqrt{272 + 17} = 17$