The hypotenuse of a right triangle has length 20cm.
a² + b² = 20²
a² + b² = 400
The sum of the lengths of the other two sides is 28cm.
a + b = 28
So we have this system of equations:
ìa² + b² = 400
í
îa + b = 28
Solve the second equation for one of the letters, say b
a + b = 28
b = 28 - a
Substitute (20 - a) for x in the first equation:
a² + b² = 400
a² + (28 - a)² = 400
a² + (28 - a)(28 - a) = 400
a² + 784 - 56a + a² = 400
2a² - 56a + 784 = 400
Subtract 400 from both sides to get 0 on the right
2a² - 56a + 384 = 0
Divie every term by 2
a² - 28a + 192 = 0
Factor as
(a - 16)(a - 12) = 0
a - 16 = 0 a - 12 = 0
a = 16 a = 12
b = 28 - a b = 28 - a
b = 28 - 16 b = 28 - 12
b = 12 b = 16
It looks as though there are two solutions,
but there is really only one, as it is only
a matter of which side is chosen to have length
represented by the letter "a" and which is to
have length represented by letter "b".
Regardless, the other two sides are 12 and 16
Edwin
