Please open the attachment for the problem description. Thanks in advance
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krishnasty
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not able to download the attachment..can you please repost it?
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Appreciation in thanks please!!
Appreciation in thanks please!!
All three triangles are similar right triangles because they each share a 90 degree plus one other angle. Therefore they have the same three angles. Triangle ABC is a nifty little 3:4:5 number so AB = 5. The other two triangles have to be in the ratio 3:4:5.
The side AC of the right-hand triangle corresponds to side CB of the left hand triangle so the ration is 4:3. That means the ratio of DC:AC must also be 4:3. Since we know AC is 4, DC must be 4*4/s = 16/3. (It's a lot easier to follow if you draw out the triangles and line up the equal angles.)
The side AC of the right-hand triangle corresponds to side CB of the left hand triangle so the ration is 4:3. That means the ratio of DC:AC must also be 4:3. Since we know AC is 4, DC must be 4*4/s = 16/3. (It's a lot easier to follow if you draw out the triangles and line up the equal angles.)
Tani Wolff
In triangle ABC to the right, if BC = 3 and AC = 4, then what is the length of segment CD?
3
15/4
5
16/3
20/3
3
15/4
5
16/3
20/3
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Hi,
Tani Wolff has already provided a solution. I will post another method
Let CD = x
In right triangle ABC, BC = 3 and AC = 4. So, AB = 5
In right triangle ACD, AD^2 = 4^2 + x^2 = 16+x^2
In right triangle ABD, BD^2 = AB^2 + AD^2 => (3+x)^2 = 5^2 + 16 + x^2
So, x^2+ 6x + 9 = 41 + x^2 =>6x = 32 => x = 16/3
Hence, D
Tani Wolff has already provided a solution. I will post another method
Let CD = x
In right triangle ABC, BC = 3 and AC = 4. So, AB = 5
In right triangle ACD, AD^2 = 4^2 + x^2 = 16+x^2
In right triangle ABD, BD^2 = AB^2 + AD^2 => (3+x)^2 = 5^2 + 16 + x^2
So, x^2+ 6x + 9 = 41 + x^2 =>6x = 32 => x = 16/3
Hence, D
Cheers!
Things are not what they appear to be... nor are they otherwise
Things are not what they appear to be... nor are they otherwise
