SOLUTION: The hypotenuse of a right triangle is twice as long as one of the legs and 10 inches longer than the other. What are the lengths of the sides of the triangle? So far I know that

Algebra ->  Triangles -> SOLUTION: The hypotenuse of a right triangle is twice as long as one of the legs and 10 inches longer than the other. What are the lengths of the sides of the triangle? So far I know that      Log On


   

Question 156850: The hypotenuse of a right triangle is twice as long as one of the legs and 10 inches longer than the other. What are the lengths of the sides of the triangle?
So far I know that i should use the Pythagorean theorem.
I've drawn the triangle and know:
1st Leg: h/2
2nd Leg: h-10
Hypotenuse: h
(h-10)^2 + (h/2)^2 = h^2
4(h^2-20h-100+ (h^2/4))=h^2
4h^2-80h-400+h^2=h^2
5h^2-80h-400=h^2
Found 2 solutions by jim_thompson5910, nerdybill:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You have the right equations set up, but your work is a little off.

%28h-10%29%5E2+%2B+%28h%2F2%29%5E2+=+h%5E2 Start with the given equation


h%5E2-20h%2B100+%2B+%28h%2F2%29%5E2+=+h%5E2 FOIL. Note: the third term is +100 (not -100)


h%5E2-20h%2B100+%2B+h%5E2%2F4+=+h%5E2 Square h%2F2 to get h%5E2%2F4


4h%5E2-80h%2B400+%2B+h%5E2+=+4h%5E2 Multiply EVERY term (including the terms on the right side) by the LCD 4 to eliminate the fraction



4h%5E2-80h%2B400%2Bh%5E2-4h%5E2=0 Subtract 4h%5E2 from both sides


h%5E2-80h%2B400=0 Combine like terms.


Notice we have a quadratic equation in the form of ah%5E2%2Bbh%2Bc where a=1, b=-80, and c=400


Let's use the quadratic formula to solve for h


h+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


h+=+%28-%28-80%29+%2B-+sqrt%28+%28-80%29%5E2-4%281%29%28400%29+%29%29%2F%282%281%29%29 Plug in a=1, b=-80, and c=400


h+=+%2880+%2B-+sqrt%28+%28-80%29%5E2-4%281%29%28400%29+%29%29%2F%282%281%29%29 Negate -80 to get 80.


h+=+%2880+%2B-+sqrt%28+6400-4%281%29%28400%29+%29%29%2F%282%281%29%29 Square -80 to get 6400.


h+=+%2880+%2B-+sqrt%28+6400-1600+%29%29%2F%282%281%29%29 Multiply 4%281%29%28400%29 to get 1600


h+=+%2880+%2B-+sqrt%28+4800+%29%29%2F%282%281%29%29 Subtract 1600 from 6400 to get 4800


h+=+%2880+%2B-+sqrt%28+4800+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


h+=+%2880+%2B-+40%2Asqrt%283%29%29%2F%282%29 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


h+=+%2880%29%2F%282%29+%2B-+%2840%2Asqrt%283%29%29%2F%282%29 Break up the fraction.


h+=+40+%2B-+20%2Asqrt%283%29 Reduce.


h+=+40%2B20%2Asqrt%283%29 or h+=+40-20%2Asqrt%283%29 Break up the expression.


h=74.641 or h=5.359 Now approximate the values of "h"


So the possible hypotenuses are h=74.641 or h=5.359


However, if you plug h=5.359 into h-10, you'll get a negative answer. So the only solution is h=74.641


So the length of the hypotenuse is approximately 74.641 units


So the first leg is 74.641%2F2=37.321 units long and the second leg is 74.641-10=64.641 units long

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
The hypotenuse of a right triangle is twice as long as one of the legs and 10 inches longer than the other. What are the lengths of the sides of the triangle?
.
So far I know that i should use the Pythagorean theorem.
I've drawn the triangle and know:
1st Leg: h/2
2nd Leg: h-10
Hypotenuse: h
.
(h-10)^2 + (h/2)^2 = h^2 <<-- formula looks good!
4(h^2-20h-100+ (h^2/4))=h^2 <<--why didn't you multiply the right side by 4?
.
Taking it from the top:
(h-10)^2 + (h/2)^2 = h^2
(h^2-20h+100) + (h^2/4) = h^2
4(h^2-20h+100) + 4(h^2/4) = 4h^2
4(h^2-20h+100) + h^2 = 4h^2
4h^2-80h+400 + h^2 = 4h^2
5h^2-80h+400 = 4h^2
h^2-80h+400 = 0
.
Can't factor so you must use the quadratic equation.
Doing so results in:
h = {74.64, 5.36}
If we pick h=5.36, our 2nd leg would be negative -- throw out solution.
Conclusion:
h = 74.64
1st Leg: h/2 = 37.32 inches
2nd Leg: h-10 = 64.64 inches
.
Reference: calculations for the quadratic
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-80x%2B400+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-80%29%5E2-4%2A1%2A400=4800.

Discriminant d=4800 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--80%2B-sqrt%28+4800+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-80%29%2Bsqrt%28+4800+%29%29%2F2%5C1+=+74.6410161513776
x%5B2%5D+=+%28-%28-80%29-sqrt%28+4800+%29%29%2F2%5C1+=+5.35898384862245

Quadratic expression 1x%5E2%2B-80x%2B400 can be factored:
1x%5E2%2B-80x%2B400+=+1%28x-74.6410161513776%29%2A%28x-5.35898384862245%29
Again, the answer is: 74.6410161513776, 5.35898384862245. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-80%2Ax%2B400+%29