Question

kuta software - infinite geometry
name elijah gallegos
similar right triangles
date 1-27-26
find the missing length indicated. leave your answer in simplest radical form.
1)
2)
3)
4)
5)
6)
7)
8)

Explanation:

Problem 1

Step1: Use geometric mean for altitude

For a right triangle, the altitude to the hypotenuse satisfies $\frac{\text{hypotenuse segment 1}}{x} = \frac{x}{\text{hypotenuse segment 2}}$. Here, hypotenuse segments are $36$ and $100-36=64$.
$\frac{36}{x} = \frac{x}{64}$

Step2: Solve for $x$

Cross multiply: $x^2 = 36 \times 64$
$x = \sqrt{36 \times 64} = 6 \times 8 = 48$

Problem 2

Step1: Find full hypotenuse length

Total hypotenuse: $25 + 9 = 34$

Step2: Use leg geometric mean theorem

The leg $x$ satisfies $\frac{\text{full hypotenuse}}{x} = \frac{x}{\text{adjacent segment}}$. Adjacent segment is $25$.
$\frac{34}{x} = \frac{x}{25}$

Step3: Solve for $x$

$x^2 = 34 \times 25$
$x = \sqrt{34 \times 25} = 5\sqrt{34}$

Problem 3

Step1: Use altitude geometric mean theorem

Altitude $x$ satisfies $\frac{\text{segment 1}}{x} = \frac{x}{\text{segment 2}}$. Segments are $9$ and $25$.
$\frac{9}{x} = \frac{x}{25}$

Step2: Solve for $x$

$x^2 = 9 \times 25$
$x = \sqrt{9 \times 25} = 3 \times 5 = 15$

Problem 4

Step1: Find full hypotenuse length

Total hypotenuse: $81 + 45 = 126$

Step2: Use leg geometric mean theorem

Leg $x$ satisfies $\frac{\text{full hypotenuse}}{x} = \frac{x}{\text{adjacent segment}}$. Adjacent segment is $45$.
$\frac{126}{x} = \frac{x}{45}$

Step3: Solve for $x$

$x^2 = 126 \times 45 = 5670$
$x = \sqrt{5670} = \sqrt{81 \times 70} = 9\sqrt{70}$

Problem 5

Step1: Find hypotenuse of big triangle

First, find the legs of the big right triangle:
Leg1: $\sqrt{7^2 + x^2}$, Leg2: $\sqrt{9^2 + x^2}$
Big hypotenuse: $7 + 9 = 16$

Step2: Apply Pythagorean theorem

$(\sqrt{7^2 + x^2})^2 + (\sqrt{9^2 + x^2})^2 = 16^2$
$49 + x^2 + 81 + x^2 = 256$

Step3: Solve for $x$

$2x^2 = 256 - 130 = 126$
$x^2 = 63$
$x = \sqrt{63} = 3\sqrt{7}$

Problem 6

Step1: Find hypotenuse of big triangle

Big triangle hypotenuse: $\sqrt{84^2 + 16^2} = \sqrt{7056 + 256} = \sqrt{7312} = \sqrt{64 \times 114.25} = 8\sqrt{114.25} = 8 \times \frac{\sqrt{457}}{2} = 4\sqrt{457}$ (or use altitude formula directly)

Step2: Use altitude formula

Altitude $x = \frac{\text{leg1} \times \text{leg2}}{\text{hypotenuse}}$
$x = \frac{84 \times 16}{4\sqrt{457}} = \frac{336}{\sqrt{457}} = \frac{336\sqrt{457}}{457}$
Alternative simpler method: Recognize the small right triangles are similar. $\frac{x}{16} = \frac{84}{\sqrt{84^2+16^2}}$, same result.

Problem 7

Step1: Find one leg of big triangle

Leg adjacent to $16$: $\sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20$

Step2: Use similar triangles

$\frac{x}{12} = \frac{12}{16}$

Step3: Solve for $x$

$x = \frac{12 \times 12}{16} = 9$

Problem 8

Step1: Find hypotenuse segment length

Remaining hypotenuse segment: $64 - 48 = 16$

Step2: Use leg geometric mean theorem

Leg $x$ satisfies $\frac{64}{x} = \frac{x}{16}$

Step3: Solve for $x$

$x^2 = 64 \times 16$
$x = \sqrt{64 \times 16} = 8 \times 4 = 32$

Answer:

1. $48$
2. $5\sqrt{34}$
3. $15$
4. $9\sqrt{70}$
5. $3\sqrt{7}$
6. $\frac{336\sqrt{457}}{457}$
7. $9$
8. $32$