SOLUTION: 1. Find the length of a shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotemuse is 10 miles less than twice the shorter leg.
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Question 77881: 1. Find the length of a shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotemuse is 10 miles less than twice the shorter leg.
2. The sum of two numbers is twentyfive and the sum of their square is three hundred and five. Find the number. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! 1. Find the length of a shorter leg of a right triangle 1. Find the length of a shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg.
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Using pythag: a^2 + b^2 = c^2
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Let x = a: the shorter leg
:
"longer leg is 10 miles more than the shorter leg"
b = (x+10)
:
"the hypotenuse is 10 miles less than twice the shorter leg."
c = (2x - 10)
:
a^2 + b^2 = c^2 is:
x^2 + (x+10)^2 = (2x-10)^2
:
x^2 + (x^2 + 20x + 100) = 4x^2 - 40x + 100
:
2x^2 + 20x + 100 = 4x^2 - 40x + 100
:
0 = 4x^2 - 2x^2 - 40x - 20x + 100 - 100
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2x^2 - 60x = 0; a simple equation
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Factors to:
2x(x - 30) = 0
:
2x = 0
and
x = 30 the length of the shorter side
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Check; a^2 + b^2 = c^2:
30^2 + 40^2 = (60-10)^2
900 + 1600 = 50^2 = 2500
:
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2. "The sum of two numbers is twenty five"
x + y = 25
y = (25 - x)
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"the sum of their squares is three hundred and five"
x^2 + y^2 = 305
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Find the number.
Substitute (25-x) for y in the 2nd equation:
x^2 + (25-x)^2 = 305
:
x^2 + 625 - 50x + x^2 = 305
:
2x^2 - 50x + 625 - 305 = 0
:
2x^2 - 50x + 320 = 0
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There is no solution: Look at the discriminant of this equation:
-b^2 - 4 * a * c
-50^2 - 4 * 2 * 320
2500 - 2560 = -60; <0 means x has no real solutions
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If we graph the two equations:
y = 25-x
y = Sqrt(305 - x^2)
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No point of intersection
