SOLUTION: In a right angle triangle ABC side AC is 4cm shorter than the hypotenuse and side BC is also 4cm shorter than the hypotenuse. Find the dimensions of the triangle

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Question 1205024: In a right angle triangle ABC side AC is 4cm shorter than the hypotenuse and side BC is also 4cm shorter than the hypotenuse. Find the dimensions of the triangle
Found 2 solutions by ikleyn, mananth:
Answer by ikleyn(52781) About Me  (Show Source):
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In a right angle triangle ABC side AC is 4cm shorter than the hypotenuse and side BC
is also 4cm shorter than the hypotenuse. Find the dimensions of the triangle
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x^2 + x^2 = (x+4)^2    <<<---=== Pythagorea equation;  x is the common length of any of the two its legs.

2x^2 = x^2 + 8x + 16

x^2 - 8x - 16 = 0


x = %288+%2B-+sqrt%28%28-8%29%5E2+-+4%2A1%2A%28-16%29%29%29%2F2 = %288+%2B-+sqrt%28128%29%29%2F2 = %288+%2B-+8%2Asqrt%282%29%29%2F2 = 4+%2B-+4%2Asqrt%282%29.


Only positive root is the solution to the problem.


ANSWER.  The triangle is isosceles right angled triangle.

         Its legs are 4%2Asqrt%282%29%2B4  cm long.  Its hypotenuse is  %284%2Asqrt%282%29%2B4%29%2Asqrt%282%29 = 8%2B4%2Asqrt%282%29 cm long.

Solved.

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Warning to a reader


        The solution by @mananth, giving two possible solutions for the hypotenuse, is NOT precisely correct.

        Only one (greater) of the two values of the hypotenuse length is admittable;
        the second (lesser) value is NOT admittable, since it leads to negative value of the leg length.



Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!

In a right angle triangle ABC side AC is 4cm shorter than the hypotenuse and side BC is also 4cm shorter than the hypotenuse. Find the dimensions of the triangle
Since it is a right angled triangle we apply pythagoras theorem
let hypotenuse be x
AC= x-4
BC = x-4
(x-4)^2+(x-4)^2= x^2
2(x-4)^2=x^2
2(x^2-8x+16)=x^2
2x^2-16x+32=x^2
subtract x^2 from both sides
x^2-16x+32=0
add 64 to both sides to solve by comp;eting the square metod
(x^2-16x+64)+32=64
(x-8)^2= 64-32
(x-8)^2= 32
take square root
x-8= +/- sqrt(32)
x= 8+/-sqrt(32) The hypotenuse
sides are 4 less than the hypotenuse