Question

this is a 2 - page document!
directions: identify the similar triangles in the diagram, then sketch them so the corresponding sides and angles have the same orientation.
1.
2.
directions: solve for x.
3.
4.
5.
6.

Explanation:

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For Problems 1 & 2 (Similar Triangles):

Problem 1:

Step1: Identify right triangles

The diagram has 3 right triangles: $\triangle JKL$, $\triangle JKM$, $\triangle KML$ (all have right angles: $\angle K$, $\angle M$, $\angle K$ respectively).

Step2: Prove similarity

All triangles share common acute angles, so by AA (Angle-Angle) similarity:
$\triangle JKL \sim \triangle JKM \sim \triangle KML$

Step3: Sketch with matching orientation

Align right angles and corresponding vertices:

1. $\triangle JKL$: Right angle at $K$, vertices $J-K-L$
2. $\triangle JKM$: Right angle at $M$, vertices $J-M-K$ (rotate to match $J-K-L$ orientation)
3. $\triangle KML$: Right angle at $M$, vertices $K-M-L$ (rotate to match $J-K-L$ orientation)

Problem 2:

Step1: Identify right triangles

The diagram has 3 right triangles: $\triangle WYZ$, $\triangle WXZ$, $\triangle XYZ$ (right angles at $Z$, $X$, $Z$ respectively).

Step2: Prove similarity

All triangles share common acute angles, so by AA similarity:
$\triangle WYZ \sim \triangle WXZ \sim \triangle XYZ$

Step3: Sketch with matching orientation

Align right angles and corresponding vertices:

1. $\triangle WYZ$: Right angle at $Z$, vertices $W-Z-Y$
2. $\triangle WXZ$: Right angle at $X$, vertices $W-X-Z$ (rotate to match $W-Z-Y$ orientation)
3. $\triangle XYZ$: Right angle at $Z$, vertices $X-Z-Y$ (rotate to match $W-Z-Y$ orientation)

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For Problems 3-6 (Solve for $x$):

Problem 3:

Step1: Use right triangle altitude rule

The altitude $x$ to hypotenuse: $x = \frac{\text{product of legs}}{\text{hypotenuse}}$
<Expression>
$x = \frac{6 \times 8}{10}$
</Expression>

Step2: Calculate the value

<Expression>
$x = \frac{48}{10} = 4.8$
</Expression>

Problem 4:

Step1: Split hypotenuse into segments

Let segments be $a$ and $b$, so $a + b = 29$. Use $\frac{a}{20} = \frac{20}{29}$ and $\frac{b}{21} = \frac{21}{29}$ to find $a=\frac{400}{29}$, $b=\frac{441}{29}$.

Step2: Use altitude formula

Altitude $x = \sqrt{a \times b}$
<Expression>
$x = \sqrt{\frac{400}{29} \times \frac{441}{29}} = \frac{\sqrt{400 \times 441}}{29}$
</Expression>

Step3: Simplify the expression

<Expression>
$x = \frac{20 \times 21}{29} = \frac{420}{29} \approx 14.48$
</Expression>

Problem 5:

Step1: Use right triangle altitude rule

Altitude $x = \frac{\text{product of legs}}{\text{hypotenuse}}$
<Expression>
$x = \frac{20 \times 48}{52}$
</Expression>

Step2: Simplify the fraction

<Expression>
$x = \frac{960}{52} = \frac{240}{13} \approx 18.46$
</Expression>

Problem 6:

Step1: Use right triangle altitude rule

Altitude $x = \frac{\text{product of legs}}{\text{hypotenuse}}$
<Expression>
$x = \frac{13.2 \times 22.4}{26}$
</Expression>

Step2: Calculate the value

<Expression>
$x = \frac{295.68}{26} \approx 11.37$
</Expression>

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Answer:

1. Similar triangles: $\triangle JKL \sim \triangle JKM \sim \triangle KML$ (sketched with aligned right angles and corresponding vertices)
2. Similar triangles: $\triangle WYZ \sim \triangle WXZ \sim \triangle XYZ$ (sketched with aligned right angles and corresponding vertices)
3. $x = 4.8$
4. $x = \frac{420}{29} \approx 14.48$
5. $x = \frac{240}{13} \approx 18.46$
6. $x \approx 11.37$