Question

Given right triangle ABCABC with altitude \overline{BD}

Answer 1

Answer:

$x = 2\sqrt{11}$

Step-by-step explanation:

AD = 4

AC = 11

DC = 11 - 4 = 7

First, find BD using the right triangle altitude theorem:

$BD = \sqrt{AD*DC}$

Plug in the values

$BD = \sqrt{4*7}$

$BD = 2\sqrt{7}$

Use pythagorean theorem to find x:

x² = AD² + BD²

Plug in the values

$x^2 = 4^2 + (2\sqrt{7})^2$

$x^2 = 4^2 + (2\sqrt{7})^2$

$x^2 = 16 + (4*7)$

$x^2 = 16 + 28$

$x^2 = 44$

Take the square root of both sides

$\sqrt{x^2} = \sqrt{44}$

$x = \sqrt{44}$

$x = \sqrt{4 * 11}$

$x = 2\sqrt{11}$

Answer 2

Answer:

6

Step-by-step explanation:

\frac{\color{darkorange}{x}}{\color{darkgreen}{4}}=

4

x

​

=

\,\,\frac{\color{darkorange}{9}}{\color{darkgreen}{x}}

x

9

​

Use the longer leg over the shorter leg in the smaller triangle and the middle triangle

x^2=

x

2

=

\,\,36

36

Multiply both sides by 44 and xx

\sqrt{x^2}=

x

2

​

=

\,\,\sqrt{36}

36

​

Square root both sides

x=

x=

\,\,6

6