Question

In the diagram below of right triangle ACB, altitude CD is drawn to
hypotenuse A
If AB = 36 and AC = 12, what is the length of AD?

Answer 1

B
D

C A

\[\text{BC}=36\text{ Sin}[\text{ArcCos}[12/36]]=24 \sqrt{2} \]\[\text{area} = \frac{1}{2}\text{BC } 12 =\text{ }\frac{1}{2}24 \sqrt{2}\text{ } 12 =144 \sqrt{2} \]\[\text{area} = \frac{1}{2}\text{CD } \text{AB} \]\[144 \sqrt{2} = \frac{1}{2}\text{CD } 36 \]\[\text{CD}=8 \sqrt{2}= 11.3137 \]

Answer 2

AD^2 = AC^2 - DC^2
AD^2 = 12^2 - (8 sqrt(2))^2
AD = 4

Answer 3

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Answer 4

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Answer 5

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Answer 6

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Answer 7

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So I know the length of

Answer 8

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Answer 9

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Answer 10

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Answer 11

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Answer 12

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Answer 13

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Answer 14

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Answer 15

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Answer 16

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Answer 17

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Answer 18

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Answer 19

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Answer 20

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Answer 21

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Answer 22

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Answer 23

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