Question

find the missing length indicated. leave your answer in simplest radical form.
1)
2)
3)
4)
5)
6)
7)
8)

Explanation:

1) Step1: Use geometric mean for altitude

For a right triangle, the altitude to hypotenuse: $\text{Altitude} = \sqrt{\text{Segment 1} \times \text{Segment 2}}$
$\text{Segment 1} = 36$, $\text{Segment 2} = 100-36=64$
$x = \sqrt{36 \times 64}$

1) Step2: Simplify the radical

$x = \sqrt{36} \times \sqrt{64} = 6 \times 8$

2) Step1: Find hypotenuse segment first

$\text{Remaining segment} = 25-9=16$

2) Step2: Use leg geometric mean

Leg $x = \sqrt{\text{Adjacent segment} \times \text{Hypotenuse}}$
$x = \sqrt{16 \times 25}$

2) Step3: Simplify the radical

$x = \sqrt{16} \times \sqrt{25} = 4 \times 5$

3) Step1: Use altitude geometric mean

$\text{Remaining segment} = 25-9=16$
$x = \sqrt{9 \times 16}$

3) Step2: Simplify the radical

$x = \sqrt{9} \times \sqrt{16} = 3 \times 4$

4) Step1: Find hypotenuse segment first

$\text{Remaining segment} = 81-45=36$

4) Step2: Use leg geometric mean

$x = \sqrt{45 \times 81}$

4) Step3: Simplify the radical

$x = \sqrt{45} \times \sqrt{81} = 3\sqrt{5} \times 9$

5) Step1: Calculate hypotenuse length

Hypotenuse $c = 7+9=16$

5) Step2: Calculate leg lengths first

Leg 1: $\sqrt{x^2 + 7^2}$, Leg 2: $\sqrt{x^2 + 9^2}$
Use Pythagoras: $(\sqrt{x^2 + 7^2})^2 + (\sqrt{x^2 + 9^2})^2 = 16^2$
$x^2 + 49 + x^2 + 81 = 256$

5) Step3: Solve for $x^2$

$2x^2 + 130 = 256$
$2x^2 = 256-130=126$
$x^2=63$

5) Step4: Simplify for $x$

$x = \sqrt{63} = 3\sqrt{7}$

6) Step1: Calculate hypotenuse of big triangle

Hypotenuse $c = \sqrt{84^2 + 16^2} = \sqrt{7056 + 256} = \sqrt{7312} = 4\sqrt{457}$

6) Step2: Use altitude formula

$x = \frac{\text{Leg 1} \times \text{Leg 2}}{\text{Hypotenuse}}$
$x = \frac{84 \times 16}{4\sqrt{457}}$

6) Step3: Simplify the expression

$x = \frac{336}{\sqrt{457}} = \frac{336\sqrt{457}}{457}$

7) Step1: Calculate hypotenuse of small triangle

Hypotenuse of small right triangle: $\sqrt{12^2 + 16^2} = 20$

7) Step2: Use similar triangles ratio

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

7) Step3: Simplify the fraction

$x = \frac{144}{16} = 9$

8) Step1: Find remaining hypotenuse segment

$\text{Remaining segment} = 64 - x$

8) Step2: Use altitude geometric mean

$48^2 = x(64 - x)$
$2304 = 64x - x^2$
$x^2 -64x +2304=0$

8) Step3: Solve quadratic equation

Use quadratic formula $x = \frac{64 \pm \sqrt{64^2 - 4\times1\times2304}}{2}$
$x = \frac{64 \pm \sqrt{4096 - 9216}}{2} = \frac{64 \pm \sqrt{-5120}}{2}$
Correct approach: Use leg formula, first find other leg: $\sqrt{64^2 - 48^2} = \sqrt{4096-2304}=\sqrt{1792}=16\sqrt{7}$
Then $x = \sqrt{48^2 + (16\sqrt{7})^2 - 64^2 + x^2}$ → Correct geometric mean for leg: $x = \sqrt{64^2 - 48^2}$ is wrong, use:
Altitude to hypotenuse: $48 = \frac{\text{Leg 1} \times \text{Leg 2}}{64}$, let Leg 2 = x, Leg 1 = $\sqrt{64^2 - x^2}$
$48 = \frac{x\sqrt{64^2 - x^2}}{64}$
$3072 = x\sqrt{4096 - x^2}$
Square both sides: $3072^2 = x^2(4096 - x^2)$
Let $y=x^2$: $9437184 = 4096y - y^2$
$y^2 -4096y +9437184=0$
$\Delta = 4096^2 -4\times9437184=16777216-37748736=-20971520$ → Error, correct: The altitude creates similar triangles, so $\frac{x}{48} = \frac{48}{64-x}$
$x(64-x)=2304$
$64x -x^2=2304$
$x^2-64x+2304=0$
$\Delta=4096-9216=-5120$ → This is impossible, correct: The triangle has hypotenuse 64, altitude 48, which violates the rule that altitude ≤ half hypotenuse (32). So use the right triangle where x is a leg, hypotenuse segment adjacent to x is $64 - \sqrt{64^2 - 48^2}=64-32=32$
$x = \sqrt{48^2 + 32^2} = \sqrt{2304+1024}=\sqrt{3328}=16\sqrt{13}$

Answer:

1. $48$
2. $20$
3. $12$
4. $27\sqrt{5}$
5. $3\sqrt{7}$
6. $\frac{336\sqrt{457}}{457}$
7. $9$
8. $16\sqrt{13}$