Question

for exercises 15 and 16, find the unknown side length of each triangle. see example 1

1. rs

1. xy

1. given △abc with (a^2 + b^2 = c^2), write a paragraph proof of the converse of the pythagorean theorem. see example 2

1. write a two - column proof of the (45^{circ}-45^{circ}-90^{circ}) triangle theorem. see example 3

1. write a paragraph proof of the (30^{circ}-60^{circ}-90^{circ}) triangle theorem. see example 4

for exercise 20 and 21, find the side lengths of each triangle. see examples 3 and 4

1. what are gj and hj?

1. what are xy and yz?

1. what is qs? see example 5

Explanation:

Step1: Use Pythagorean theorem for RS

Let $RS = x$, hypotenuse $RT=5\sqrt{10}$, leg $ST=9$.
$$x^2 + 9^2 = (5\sqrt{10})^2$$

Step2: Calculate squared values

$$x^2 + 81 = 250$$

Step3: Solve for $x^2$

$$x^2 = 250 - 81 = 169$$

Step4: Find square root of $x^2$

$$x = \sqrt{169} = 13$$
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Step1: Use Pythagorean theorem for XY

Let $XY = x$, legs $XZ=13$, $YZ=10$.
$$x^2 = 13^2 + 10^2$$

Step2: Calculate squared values

$$x^2 = 169 + 100 = 269$$

Step3: Find square root of $x^2$

$$x = \sqrt{269}$$
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Step1: Identify 45-45-90 triangle properties

$\triangle GHJ$ is isosceles right triangle, hypotenuse $GH=12$. Legs $GJ=HJ$, hypotenuse $= \text{leg} \times \sqrt{2}$.

Step2: Solve for leg length

Let $GJ=HJ=x$.
$$x = \frac{12}{\sqrt{2}} = 6\sqrt{2}$$
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Step1: Identify 30-60-90 triangle properties

$\triangle XYZ$: hypotenuse $XZ=18$. Shorter leg $YZ = \frac{1}{2} \times \text{hypotenuse}$, longer leg $XY = \text{shorter leg} \times \sqrt{3}$.

Step2: Calculate YZ

$$YZ = \frac{1}{2} \times 18 = 9$$

Step3: Calculate XY

$$XY = 9 \times \sqrt{3} = 9\sqrt{3}$$
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Step1: Find RT from 30-60-90 triangle

In $\triangle QRT$, $\angle Q=30^\circ$, $QT=10$. $RT = QT \times \tan(30^\circ)$
$$RT = 10 \times \frac{\sqrt{3}}{3} = \frac{10\sqrt{3}}{3}$$

Step2: Find TS from 45-45-90 triangle

$\triangle RTS$ is isosceles right triangle, $RT=TS=\frac{10\sqrt{3}}{3}$

Step3: Calculate QS

$$QS = QT + TS = 10 + \frac{10\sqrt{3}}{3} = \frac{30 + 10\sqrt{3}}{3}$$
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17. Converse of Pythagorean Theorem Proof:

Assume $\triangle ABC$ has $a^2 + b^2 = c^2$. Construct a right triangle $\triangle A'B'C'$ with legs $a,b$ and hypotenuse $c'$. By Pythagoras, $c'^2=a^2+b^2=c^2$, so $c'=c$. By SSS congruence, $\triangle ABC \cong \triangle A'B'C'$, so $\angle C$ is a right angle, proving the converse.

18. Two-Column Proof of 45-45-90 Theorem:

Statement	Reason
$JK=KL$	Definition of isosceles triangle
Let $JK=KL=x$	Assign variable to legs
$JL^2 = x^2 + x^2 = 2x^2$	Pythagorean Theorem
$JL = x\sqrt{2}$	Square root property
Hypotenuse = leg $\times \sqrt{2}$	Simplify

19. Paragraph Proof of 30-60-90 Theorem:

Take $\triangle ABC$ with $\angle A=30^\circ$, $\angle B=60^\circ$, $\angle C=90^\circ$. Construct $\triangle ABD$ congruent to $\triangle ABC$ with $D$ on opposite side of $AB$ from $C$. $\triangle ACD$ is equilateral, so $AC=AD=CD=AB/2$, meaning the shorter leg $BC = \frac{1}{2}AB$. Using Pythagoras, $AC = \sqrt{AB^2 - BC^2} = \sqrt{AB^2 - (AB/2)^2} = \frac{AB\sqrt{3}}{2}$, so longer leg = shorter leg $\times \sqrt{3}$.

Answer:

1. $13$
2. $\sqrt{269}$
3. (See paragraph proof above)
4. (See two-column proof above)
5. (See paragraph proof above)
6. $GJ = 6\sqrt{2}$, $HJ = 6\sqrt{2}$
7. $XY = 9\sqrt{3}$, $YZ = 9$
8. $\frac{30 + 10\sqrt{3}}{3}$