Question

question 3
jannette says \\( \triangle a b c \sim \triangle d e f \\) because \\( \triangle a b c \\) s sides form a pythagorean triple and \\( \triangle d e f \\) s side lengths are multiples of \\( \triangle a b c \\) s side lengths. is she correct? explain your reasoning.

Explanation:

Step1: Calculate the length of \( AC \)

Use the Pythagorean theorem \( a^{2}+b^{2}=c^{2} \) for \( \triangle ABC \), where \( a = 20 \), \( b=21 \).
\( AC=\sqrt{20^{2}+21^{2}}=\sqrt{400 + 441}=\sqrt{841}=29 \)

Step2: Check the ratio of corresponding sides

For \( \triangle ABC \) and \( \triangle DEF \), the ratio of \( DE\) to \( AB\) is \( \frac{40}{20}=2 \).
If the triangles are similar, the other sides should also have a ratio of \( 2 \).
If \( AC = 29 \), then the corresponding side of \( AC \) in \( \triangle DEF\) (let's assume it's \( DF = 58 \)) has a ratio \( \frac{58}{29}=2 \).
Let's find the length of \( EF \) using the Pythagorean theorem for \( \triangle DEF\), \( EF=\sqrt{58^{2}-40^{2}}=\sqrt{(58 + 40)(58 - 40)}=\sqrt{98\times18}=\sqrt{1764}=42 \), and \( \frac{42}{21}=2 \)

Answer:

Yes, Jannette is correct. Since the ratios of the corresponding sides of \( \triangle ABC \) and \( \triangle DEF\) (\( \frac{DE}{AB}=\frac{EF}{BC}=\frac{DF}{AC} = 2 \)) are equal, by the Side - Side - Side (SSS) similarity criterion, \( \triangle ABC\sim\triangle DEF \)