a² + b² = (26)²
a+b = 34
a = 34 - b
(34-b)² + b² = (26)²
(34)² -68b +b² + b² = (26)²
1156 - 68b + 2b² = 676
2b² -68b + 480 = 0
b² -34b + 240 = 0
(b-24)(b-10) = 0
Sides a and b are 10 and 24 meters long
2007-10-16 06:42:33
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answer #1
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answered by MamaMia © 7
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a² + b² = c²
In this case:
b = 34 - a
So:
a² + (34 - a)² = 26² { Substitute for b }
a² + 1156 - 68a + a² = 676 { Expand (34 - a)² }
2a² - 68a + 480 = 0 { Combine like terms }
a² - 34a + 240 = 0 { Divide everything by 2 }
(a - 24)(a - 10) = 0 { Quadratic Formula }
a = 24 or 10 { Possible values for a }
b = 10 or 24 { Possible values for b }
So, the two sides of your triangle are 10 meters and 24 meters.
2007-10-16 06:42:25
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answer #2
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answered by Dave 6
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Using the pythagorean theorem
a² + b² = c²
Let x be one leg, the other 34 - x
x² + (34 - x)² = 26²
x² + 1156 - 68x +x² = 676
2x² - 68x+ 480 = 0
divide thru by 2
x² - 34x +240 = 0
(x - 10)(x - 24) = 0
x = 10, x = 24
The two sides are 24 and 10
.
2007-10-16 06:47:03
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answer #3
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answered by Robert L 7
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Let one of the legs be x m long, so the other one is (34 - x) m.
Since it`s a right angled triangle, then using Pythagoras` Theorem:
x^2 + (34 - x)^2 = 26^2
You now solve this equation for x.
Hope this helps, Twiggy.
2007-10-16 06:44:58
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answer #4
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answered by Twiggy 7
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let 2 legs be a &b
sum of 2 legs is a+b=34
b=34-a
let hypotenuse be c
c=26
c^2 = a^2+ b^2
(26)^2 = a^2+(34-a)^2
676=a^2+1156+a^2-68a
676-1156=2a^2-68a
-480=2a^2-68a
-240 =a^2-34a
0=a^2-34a+240
0=a^2-24a-10a+240
0=a(a-24)-10(a-24)
0=(a-24)(a-10)
a= 24 or 10
if a =24
then
b= 34-a
b=34-24
b=10
if a =10
then
b=34-a
b=34-10
b=24
the length of 2 legs = 10 & 24
2007-10-16 06:54:52
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answer #5
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answered by Siva 5
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a - squared + b - squared = C - Squared.
axa + bxb = 26x26
axa + bxb = 676
What 2 numbers add up to 34 that when squared add up to 676.
the sides are 10m and 24 m
10x10 = 100
24x24 = 576
100+576 = 676
2007-10-16 06:49:08
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answer #6
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answered by bigfrank1255 3
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You have been given two equations:
1) a+b=34
2) c=26
Now just remember that the Pythagorean Theorem gives you a third equation. With three equations and three unknowns you should have enough information for a solution.
2007-10-16 06:45:33
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answer #7
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answered by K N 123 3
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