Question

kuta software - infinite geometry
similar right triangles
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

The altitude to hypotenuse: $\text{altitude} = \sqrt{\text{segment}_1 \times \text{segment}_2}$
First find the other hypotenuse segment: $100 - 36 = 64$
$x = \sqrt{36 \times 64}$

Step2: Simplify the radical

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

2)

Step1: Use geometric mean for leg

Leg length: $\text{leg} = \sqrt{\text{hypotenuse} \times \text{adjacent segment}}$
$x = \sqrt{25 \times 9}$

Step2: Simplify the radical

$\sqrt{25 \times 9} = \sqrt{25} \times \sqrt{9} = 5 \times 3$

3)

Step1: Use geometric mean for altitude

Altitude to hypotenuse: $x = \sqrt{9 \times (25 - 9)}$
Calculate the other segment: $25 - 9 = 16$
$x = \sqrt{9 \times 16}$

Step2: Simplify the radical

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

4)

Step1: Use geometric mean for leg

First find the other hypotenuse segment: $81 - 45 = 36$
Leg length: $x = \sqrt{81 \times 36}$

Step2: Simplify the radical

$\sqrt{81 \times 36} = \sqrt{81} \times \sqrt{36} = 9 \times 6$

5)

Step1: Use geometric mean for altitude

First find the hypotenuse segments: $7$ and $9$, hypotenuse total $7+9=16$
First find the legs: $\text{leg}_1 = \sqrt{16 \times 7}$, $\text{leg}_2 = \sqrt{16 \times 9}$
Area of triangle: $\frac{1}{2} \times \text{leg}_1 \times \text{leg}_2 = \frac{1}{2} \times \text{hypotenuse} \times x$
$\frac{1}{2} \times \sqrt{112} \times \sqrt{144} = \frac{1}{2} \times 16 \times x$

Step2: Solve for x

Simplify: $\sqrt{112} \times 12 = 16x$
$4\sqrt{7} \times 12 = 16x$
$48\sqrt{7} = 16x$
$x = \frac{48\sqrt{7}}{16}$

6)

Step1: Find hypotenuse of large triangle

Hypotenuse: $\sqrt{84^2 + 16^2} = \sqrt{7056 + 256} = \sqrt{7312} = 4\sqrt{457}$
Area of triangle: $\frac{1}{2} \times 84 \times 16 = \frac{1}{2} \times 4\sqrt{457} \times x$

Step2: Solve for x

Simplify: $84 \times 16 = 4\sqrt{457}x$
$1344 = 4\sqrt{457}x$
$x = \frac{1344}{4\sqrt{457}} = \frac{336}{\sqrt{457}}$
Rationalize: $x = \frac{336\sqrt{457}}{457}$

7)

Step1: Use geometric mean for segment

Segment length: $\text{segment} = \frac{\text{altitude}^2}{\text{other segment}}$
$x = \frac{12^2}{16}$

Step2: Calculate the value

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

8)

Step1: Find the other hypotenuse segment

First find the hypotenuse: let total hypotenuse be $c$, use area: $\frac{1}{2} \times 64 \times x = \frac{1}{2} \times c \times 48$
Also use Pythagoras: $x^2 + (\text{leg})^2 = 64^2$, and $\text{leg} = \sqrt{64 \times (c - x)}$
Alternatively, use geometric mean: $48^2 = x \times (64 - x)$
$2304 = 64x - x^2$
$x^2 -64x +2304=0$
Use quadratic formula: $x = \frac{64 \pm \sqrt{64^2 -4\times1\times2304}}{2}$

Step2: Simplify the quadratic

$\sqrt{4096 - 9216} = \sqrt{-5120}$ (invalid, correct approach: altitude formula: $48 = \sqrt{x \times (64 - x)}$ is wrong. Correct: leg adjacent to x: $\sqrt{64x}$, other leg $\sqrt{64(64-x)}$, area: $\frac{1}{2}\sqrt{64x}\sqrt{64(64-x)} = \frac{1}{2} \times 64 \times 48$
Simplify: $8\sqrt{x} \times 8\sqrt{64-x} = 64 \times 48$
$64\sqrt{x(64-x)} = 3072$
$\sqrt{x(64-x)} = 48$
Square both sides: $64x -x^2 = 2304$
$x^2 -64x +2304=0$
Discriminant: $64^2 -4\times1\times2304=4096-9216=-5120$, so use similar triangles: $\frac{x}{64} = \frac{48}{\text{other leg}}$, and $\text{other leg} = \sqrt{64^2 -x^2}$
$\frac{x}{64} = \frac{48}{\sqrt{4096 -x^2}}$
Cross multiply: $x\sqrt{4096 -x^2} = 64 \times 48$
Square: $x^2(4096 -x^2) = 3072^2$
Let $y=x^2$: $4096y - y^2 = 9437184$
$y^2 -4096y +9437184=0$
Discriminant: $4096^2 -4\times9437184=16777216-37748736=-20971520$ (error, correct: altitude to hypotenuse: $48 = \frac{x \times \text{other leg}}{64}$, and $\text{other leg} = \sqrt{64^2 -x^2}$
$48 \times 64 = x\sqrt{4096 -x^2}$
$3072 = x\sqrt{4096 -x^2}$
Square: $9437184 = x^2(4096 -x^2)$
$x^4 -4096x^2 +9437184=0$
Let $z=x^2$: $z^2 -4096z +9437184=0$
$z = \frac{4096 \pm \sqrt{4096^2 -4\times9437184}}{2} = \frac{4096 \pm \sqrt{16777216-37748736}}{2} = \frac{4096 \pm \sqrt{-20971520}}{2}$ (invalid, correct approach: the triangle has altitude 48, hypotenuse part x, total hypotenuse 64: use $(\text{leg})^2 = 64x$, and $(\text{leg})^2 = 48^2 + (64-x)^2$
$64x = 2304 + 4096 -128x +x^2$
$64x = 6400 -128x +x^2$
$x^2 -192x +6400=0$
Discriminant: $192^2 -4\times6400=36864-25600=11264$
$\sqrt{11264} = 106.13$
$x = \frac{192 \pm 106.13}{2}$
Take smaller root: $x = \frac{192-106.13}{2}=42.93$ (exact: $\sqrt{11264}=16\sqrt{44}$? No, $11264=64\times176=64\times16\times11=1024\times11$, so $\sqrt{11264}=32\sqrt{11}$
$x = \frac{192 \pm 32\sqrt{11}}{2} = 96 \pm 16\sqrt{11}$
Take the value less than 64: $x = 96 - 16\sqrt{11}$

Answer:

1. $48$
2. $15$
3. $12$
4. $54$
5. $3\sqrt{7}$
6. $\frac{336\sqrt{457}}{457}$
7. $9$
8. $96 - 16\sqrt{11}$