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sweet [91]
3 years ago
5

In triangle $ABC,$ point $D$ is on $\overline{AC}$ such that $AD = 3CD = 12$. If $\angle ABC = \angle BDA = 90^\circ$, then what

is $BD$?

Mathematics
1 answer:
Neporo4naja [7]3 years ago
8 0

Answer:

BD=4\sqrt{3}\ units

Step-by-step explanation:

we know that

AD=12\ units

3CD=12 ----> CD=4\ units

see the attached figure to better understand the problem

Triangles ABD and BCD are similar by AA Similarity Theorem

Remember that

If two figures are similar, then the ratio of its corresponding sides is proportional

so

\frac{BD}{AD}=\frac{CD}{BD}

substitute the given values

\frac{BD}{12}=\frac{4}{BD}

BD^2=48

BD=\sqrt{48}\ units

simplify

BD=4\sqrt{3}\ units

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Answer:

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Step-by-step explanation:

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Now, in the given triplets:

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Hence,  {4, 8, 12} is NOT a triplet.

(b) {6, 8, 10}

Here, (6)^{2}  + (8)^{2}   = 36 + 64  = 100\\\implies H = \sqrt{100}  =  10

So, third side of the triangle  10

Hence,  {6, 8, 10} is  a triplet.

(c) {6, 8, 15}

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So, third side of the triangle  10  ≠ 15

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(d) {5, 7, 13}

Here, (5)^{2}  + (7)^{2}   = 25 + 49  = 74\\\implies H = \sqrt{74}  =  8.60

So, third side of the triangle  8.60  ≠ 13

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4 0
3 years ago
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