Question 324836
a) I'm assuming that 'x' is the missing leg of the triangle. If so, then you are correct.


b) Let x = hypotenuse. So {{{(4a)^2+(3a)^2=x^2}}}. Solve for 'x' to get {{{x=sqrt((4a)^2+(3a)^2)=sqrt(16a^2+9a^2)=sqrt(25a^2)=5a}}}


So the hypotenuse is 5a units long. So x = 5a



c) So you're saying that the triangle has side lengths of 13, 13, and 10. You then bisect the line of 10 units long to get 5. So you end up with two triangles with side lengths of x, 5, and 13. If that's the case, then you are correct. The height of the triangle is 12 units.



d) If you bisect a line and the half is 's' units long, then the equilateral triangle will have a side length of 2s units. So if you look at the picture, this means that the hypotenuse is 2s units long.



So by the pythagorean theorem, this means that {{{x^2+s^2=(2s)^2}}}. Solve for 'x' to get {{{x=sqrt(4s^2-s^2)=sqrt(3s^2)=s*sqrt(3)}}}



So {{{x=s*sqrt(3)}}}. Note: this should look familiar if you've studied 30-60-90 triangles.