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Sagot :
Similar Triangles
Similar triangles have the same proportions of sides, but they have different side lengths.
To solve for missing sides in similar triangles, we can set up a proportion.
For instance, let's say that side a in Triangle A corresponds with side b in Triangle B. Let's say that side h in Triangle A also corresponds with side k in Triangle B. Then, it would be true that:
- [tex]\dfrac{a}{b}=\dfrac{h}{k}[/tex]
We need to make sure of a couple things:
- The numerators and denominators of fractions are corresponding
- The numerators describe one triangle, and the denominators describe another (can't switch, otherwise the calculations will get messed up)
Solving the Question
We're given two triangles (do you see it?).
- Triangle ABC
- Triangle ADE
These two triangles are similar.
We must solve for the length of side BC in Triangle ABC.
- We're given the length of DE, the corresponding side in Triangle ADE.
- We're also given the lengths of bottom sides, 20 units and 30 + 20 = 50 units.
Set up a proportion:
[tex]\dfrac{x}{15}=\dfrac{50}{20}\\\\\dfrac{x}{15}=\dfrac{5}{2}\\\\x=\dfrac{5}{2}*15\\\\x=\dfrac{75}{2}\\\\x=37.5[/tex]
Therefore, the unknown length is 37.5 units.
Answer
37.5 units
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