DEF is a right triangle with hypotenuse DF. DE = 6 and EF =8. If Y lies on DF and EY bisects angle DEF, find DY.
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Because the sides are 6,8, we know its in the form of a 3,4,5 triangle. So the hypotenuse is 10.
Given the fact that line EY bisects angle DEF, and that Y lies on DF, we know that line EY and DF are perpendicular.
So that makes a 45,45,90 triangle, EDY.
Use the 1,1, sqrt(2),
(sqrt(2)/6) = (1/x)
so it would be 6/(sqrt(2))
or aprrox.. 4.243
or
Given the fact that line EY bisects angle DEF, and that Y lies on DF, we know that line EY and DF are perpendicular.
So that makes a 45,45,90 triangle, EDY.
Use the 1,1, sqrt(2),
(sqrt(2)/6) = (1/x)
so it would be 6/(sqrt(2))
or aprrox.. 4.243
or
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The answer is five.
If DEF is a right triangle then use the Pythagorean Theorem to find DF which is 10. If Y bisects DF then divide DF by 2 and get 5.
EDIT - I know, but in angle DEF, E is the vertex and if it bisects E then it bisects DF. EY bisects DEF meaning that the line would hit directly in the middle of DF. DY=YF and they are both 5 which make up 10.
If DEF is a right triangle then use the Pythagorean Theorem to find DF which is 10. If Y bisects DF then divide DF by 2 and get 5.
EDIT - I know, but in angle DEF, E is the vertex and if it bisects E then it bisects DF. EY bisects DEF meaning that the line would hit directly in the middle of DF. DY=YF and they are both 5 which make up 10.
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Who cares? Just go jack off to your dead grandmother.