The hypotenuse of a right triangle is 2.6 units long. The longer leg is 1.6 units longer than the shorter leg.

Question

Find the lengths of the sides of the triangle.

Answer 1

Let's call the short leg s, for short. (I always recmomend using letters that stand for what you're looking for, rather than x and y, becaseu when you get to the end, you know what you got. Also letters like x and y tend to make it look harder than it is)
s^2 + (s+1.6)^2 = 2.6^2
Expand out the (s+1..6)^2 = (s+1.6)(s+1.6) FOIL it out...
s^2 + 3.2s + 2.56
evaluate 2.6^2 = 6.76
Insert this into the main expression and subtract 6.76 from each side so you can set this to zero to use the quadratic forumla...
s^2 + s^2 + 3.2s + 2.56 - 6.76 = 0
Combine terms and rearrange to standard form
2s^2 + 3.2s -4.2 = 0
Put this in the quadratic formula and there you go.
This will give you two values. Discard the negative one, since you can't have a negative length. The positive one is the length of the short leg. The longer leg is 1.6 + s and the hypotenuse is 2.6

Answer 2

x^2 + y^2 = c^2

x^2 + (x + 1.6)^2 = 2.6^2
x^2 + ((x + 1.6)(x + 1.6)) = 6.76
x^2 + (x^2 + 1.6x + 1.6x + 2.56) = 6.76
x^2 + (x^2 + 3.2x + 2.56) = 6.76
x^2 + x^2 + 3.2x + 2.56 = 6.76
2x^2 + 3.2x + 2.56 = 6.76
2x^2 + 3.2x - 4.2 = 0
2(x^2 + 1.6x - 2.1) = 0
x^2 + 1.6x - 2.1 = 0
(10x^2 + 16x - 21)/10 = 0
10x^2 + 16x - 21 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-16 ± sqrt(16^2 - 4(10)(-21)))/(2(10))
x = (-16 ± sqrt(256 + 840))/20
x = (-16 ± sqrt(1096))/20
x = (-16 ± sqrt(4 * 274))/20
x = (-16 ± 2sqrt(274))/20
x = (1/10)(-8 ± sqrt(274))

since you can't have a negative length

x = (1/10)(-8 + sqrt(274))

y = (-4/5) + (sqrt(274)/10) + (8/5)
y = (4/5) + (sqrt(274)/10)
y = (1/10)(8 + sqrt(274))

The Lengths are (1/10)(8 + sqrt(274)) and (1/10)(-8 + sqrt(274))

Answer 3

Let the shorter leg be x

the longer leg=x+1.6

X^2+(x+1.6)^2=2.6^2
2x^2+3.2X+2.56=6.76
2X^2+3.2X+2.56-6.76=0
2x^2+3.2X-4.2=0
X^2+1.6X-2.1=0
x=0.8555
Calculate other length

Answer 4

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Answer 5

x ^2 + 1.6x ^2 = 2.6^2

x^2+x^2+3.2x+2.56=2.6^2

x^2+1.6x-2.1=0

Lengths: .855, 2.455, 2.6

Answer 6

x^2+(x+1.6)^2=2.6^2
2x^2+3.2x+2.56=6.76
2x^2+3.2x-4.20=0
20x^2+32x-42=0
x=[-32+/-rt(32^2+3360)]/40
=[-32+/-rt(4384)]/40
=-32+/-66.21 /40
=34.21/40
=0.855
say 0.86
the longer length is 1.86
shorter side is 0.86

Answer 7

h=2.6
b=a+1.6

2.6^2=a^2+(a+1.6)^2
6.76=a^2+a^2+3.2a+2.56 subtract 6.76 from both sides & combine like terms
0=2a^2+3.2a-4.2 divide both sides by 2
a^2+1.6a-2.1=0
a=(-1.6+/-√(1.6^2+4*2.1))/2
a=-0.8+/-.5√(2.56+8.4)
a=-.8+/-0.5√10.96
a=-0.8+/-√2.74 since a>0
a=√2.74-0.8
b=a+1.6=√2.74+0.8

a=√2.74-0.8
b=√2.74+0.8
h=2.6

Answer 8

a=2,6
b= c + 1,6
c= ?
2.6² = (c+1.6)² + c²
6.76 = c² + 3,2c + 2,56 + c²
2c² + 3,2c + 2,56 - 6,76 = 0
2c² + 3,2c - 4,2 = 0
c² + 1,6c -2,1 = 0
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