SOLUTION: A right triangle has area 19 and the sum of its two legs is 20. Compute the length of the hypotenuse.

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Question 389474: A right triangle has area 19 and the sum of its two legs is 20. Compute the length of the hypotenuse.
Answer by MathLover1(20849) About Me  (Show Source):
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A right triangle has area 19 and the sum of its two legs is 20. Compute the length of the hypotenuse.

given:
area A=19+
and the sum of its two legs a%2Bb is 20
find: the length of the hypotenuse c
The sides a, b, and c of a right triangle satisfy the Pythagorean theorem, so
c%5E2+=+a%5E2+%2B+b%5E2
the area is A=%281%2F2%29ab+
now, find a and b
A=%281%2F2%29ab+.....................1
a%2Bb=+20...........................2.........solve for a...a=+20-b. and substitute in 1
2A=ab+.....................1
2%2A19=ab+.....................1
38=%2820-b%29b+.....................1
38=20b-b%5E2+.....................1
b%5E2+-20b+%2B+38=0.....................1.solve using quadratic formula
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve x%5E2-20%2Ax%2B38=0 ( notice a=1, b=-20, and c=38)





x+=+%28--20+%2B-+sqrt%28+%28-20%29%5E2-4%2A1%2A38+%29%29%2F%282%2A1%29 Plug in a=1, b=-20, and c=38




x+=+%2820+%2B-+sqrt%28+%28-20%29%5E2-4%2A1%2A38+%29%29%2F%282%2A1%29 Negate -20 to get 20




x+=+%2820+%2B-+sqrt%28+400-4%2A1%2A38+%29%29%2F%282%2A1%29 Square -20 to get 400 (note: remember when you square -20, you must square the negative as well. This is because %28-20%29%5E2=-20%2A-20=400.)




x+=+%2820+%2B-+sqrt%28+400%2B-152+%29%29%2F%282%2A1%29 Multiply -4%2A38%2A1 to get -152




x+=+%2820+%2B-+sqrt%28+248+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%2820+%2B-+2%2Asqrt%2862%29%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%2820+%2B-+2%2Asqrt%2862%29%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


x+=+%2820+%2B+2%2Asqrt%2862%29%29%2F2 or x+=+%2820+-+2%2Asqrt%2862%29%29%2F2



Now break up the fraction



x=%2B20%2F2%2B2%2Asqrt%2862%29%2F2 or x=%2B20%2F2-2%2Asqrt%2862%29%2F2



Simplify



x=10%2Bsqrt%2862%29 or x=10-sqrt%2862%29



So the solutions are:

x=10%2Bsqrt%2862%29 or x=10-sqrt%2862%29





b1=10+%2B+sqrt%2862%29
b1=10+%2B+7.9
b1=17.9


or b2=10-7.9
b2=2.1
find a

a1=+20+-+b
a1=+20+-+17.9
a1=+2.1

a2=+20+-+b
a2=+20+-+2.1
a2=+17.9



now, find c for a1=+2.1 and b1=17.9

c%5E2+=+a%5E2+%2B+b%5E2

c%5E2+=+%282.1%29%5E2+%2B+%2817.9%29%5E2

c%5E2+=+4.41+%2B+320.41...round it

c%5E2+=+4+%2B+320
c%5E2+=+324

c=+18.........