SOLUTION: I need help solving this equation: The hypotenuse of a right isosceles triangle is one meter greater than either of it's sides. Find the length of both.

Algebra ->  Pythagorean-theorem -> SOLUTION: I need help solving this equation: The hypotenuse of a right isosceles triangle is one meter greater than either of it's sides. Find the length of both.      Log On


   

Question 612307: I need help solving this equation: The hypotenuse of a right isosceles triangle is one meter greater than either of it's sides. Find the length of both.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
A "right isosceles" triangle is a right triangle with two congruent legs.
Let x = the length of each of these legs
And if the hypotenuse is 1 meter greater than the sides, then it must be: x+1

With these expressions we can use the Pythagorean equation to solve the problem. (Remember to put the hypotenuse by itself.)
x%5E2+%2B+x%5E2+=+%28x%2B1%29%5E2

Now we simplify. (Be careful to use FOIL or the %28a%2Bb%29%5E2+=+a%5E2%2B2ab%2Bb%5E2 to square the right side:
2x%5E2+=+x%5E2+%2B2x+%2B+1
Since this is a quadratic equation and since quadratic equations are a little easier to solve when the squared terms has a positive coefficient, I'm going to subtract the entire right side from both sides:
x%5E2+-+2x+-+1+=+0
Next we factor or use the quadratic formula. This will not factor so we must use the formula:
x+=+%28-%28-2%29+%2B-+sqrt%28%28-2%29%5E2+-+4%281%29%28-1%29%29%29%2F2%281%29
Simplifying...
x+=+%28-%28-2%29+%2B-+sqrt%284+-+4%281%29%28-1%29%29%29%2F2%281%29
x+=+%28-%28-2%29+%2B-+sqrt%284+%2B+4%29%29%2F2%281%29
x+=+%28-%28-2%29+%2B-+sqrt%288%29%29%2F2%281%29
x+=+%282+%2B-+sqrt%288%29%29%2F2
x+=+%282+%2B-+sqrt%284%2A2%29%29%2F2
x+=+%282+%2B-+sqrt%284%29%2Asqrt%282%29%29%2F2
x+=+%282+%2B-+2%2Asqrt%282%29%29%2F2
x+=+%282%281+%2B-+sqrt%282%29%29%29%2F2
x+=+%28cross%282%29%281+%2B-+sqrt%282%29%29%29%2Fcross%282%29
x+=+1+%2B-+sqrt%282%29
So
x+=+1+%2B+sqrt%282%29 or x+=+1+-+sqrt%282%29
Since sqrt%282%29+%3E+1 the second x is negative. So the seoncd x is negative. Since x is the length of a side of a triangle and since we cannot have negative lengths, we must reject that solution. So
x+=+1+%2B+sqrt%282%29
is the only solution to the equation and, therefore, the length of the two legs. The length of the hypotenuse, x+1, would then be x+=+1+%2B+sqrt%282%29+%2B+1+=+2+%2B+sqrt%282%29

Another way to solve this is to recognize and take advantage of the fact that right isosceles triangles are 45-45-90 triangles. And the sides of 45-45-90 triangle have a fixed, known relationship to each other: If x if the length of a leg, then x%2Asqrt%282%29 will always be the hypotenuse. But we already have the expression of x+1 for the hypotenuse. Therefore these expression must be equal:
x%2Asqrt%282%29+=+x+%2B+1
We can solve this for x. Subtracting x from each side we get:
x%2Asqrt%282%29+-+x+=+1
Factoring out x:
x%2A%28sqrt%282%29+-+1%29+=+1
Dividing by %28sqrt%282%29+-+1%29:
x+=+1%2F%28sqrt%282%29+-+1%29
Rationalizing the denominator:
x+=+%281%2F%28sqrt%282%29+-+1%29%29%28%28sqrt%282%29+%2B+1%29%2F%28sqrt%282%29%2B1%29%29
x+=+%28sqrt%282%29+%2B+1%29%2F%28%28sqrt%282%29%29%5E2+-+1%5E2%29
x+=+%28sqrt%282%29+%2B+1%29%2F%282+-+1%29
x+=+%28sqrt%282%29+%2B+1%29%2F1
x+=+sqrt%282%29+%2B+1
and x%2B1+=+sqrt%282%29%2B2
which is the solution we got earlier.