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Otrada [13]
3 years ago
6

Triangle ABC is similar to triangle DEF. The length of

Mathematics
2 answers:
inysia [295]3 years ago
4 0

<u>Answer:</u>

The length of EF = 15 cm.

<u>Step-by-step explanation:</u>

We are given that the two triangles, ABC and DEF, are similar to each other,

Given that the length of AC = 12 cm, BC = 18 cm and DF = 10 cm, we are to find the length of EF.

For this, we can simply use the ratio method.

\frac { EF } { BC } = \frac { DF } { AC }

\frac { EF } { 18 } = \frac { 10 } { 12 }

E F = \frac { 10 } { 12 } \times 18

EF = 15 cm

zavuch27 [327]3 years ago
3 0

Answer: EF=15cm

Step-by-step explanation:

You know that the triangle ABC and the triangle DEF are similar.

Therefore, if the lenght of AC is 12 centimeters and the lenght of DF is 10 centimeters, then you can find the ratio as following:

ratio=\frac{DF}{AC}\\\\ratio=\frac{10cm}{12cm}\\\\ratio=0.8333

Then, the calculte the length of EF, you must multiply the lenght BC of the triangle ABC by the ratio obtained above.

Therefore, the lenght EF is the following:

EF=0.8333(18cm)\\EF=15cm

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sine = oppsite / hypotenuse
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In that case, we can essentially add 180° to our current 177° to the same effect. That means that (-2, 177°) = (2, 357°).

Note that since our angle is in Quadrant IV, our cosine will be positive, but our sine will be negative. (as derived from the unit circle) We don't have to worry about this since our calculator figures this for us, but you should pay attention to it if you are converting from Cartesian to polar.

cosine = adjacent / hypotenuse
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sine = opposite / hypotenuse
sinθ = y/r
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So (-2, 177°) ⇒ (1.99725906951, -0.10467191248).

Now we must use the distance formula with our two points.
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
d\approx\sqrt{(1.99725906951-3.84504678375)^2+(-0.10467191248-1.10254942327)^2}
d\approx\sqrt{-1.84778771^2+-1.20722134^2}
d\approx\sqrt{3.41431942+1.45738336}
d\approx\sqrt{4.87170278}
\boxed{d\approx2.20719342}

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