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July 29, 2014

July 29, 2014

Posted by **Megan** on Saturday, December 4, 2010 at 12:20am.

- Math -
**hy**, Saturday, December 4, 2010 at 1:31am(X-14)^2+X^2 = 26^2

solve for x and x-14

- Math -
**Bosnian**, Saturday, December 4, 2010 at 8:59ama=shorter leg

b=longer leg

c=hypotenuse

a=b-14

c^2=a^2+b^2

26^2=(b-14)^2+b^2

676=b^2-2*b*14+14^2+b^2

676=2b^2-28b+196

676-196=2(b^2-14b)

480=2(b^2-14b) Divide with 2

240=b^2-14b

b^2-14b=240

b^2-14b+7^2=240+7^2

Becouse: (b-7)^2=b^2-14b+7^2

(b-7)^2=240+49

(b-7)^2=289

b-7=sqroot(289)

You have two solutions:

b-7=17 and

b-7=-17

Becouse sqroot(289)=+/- 17

First solution: b-7=17

b=17+7=24

Second solution:

b-7=-17

b=17+7=-10

leg of triangle can not be negative number so:

b=24

a=b-14=24-14=10

a=10 b=24

c=sqroot(a^2+b^2)

c=sqroot(10^2+24^2)

c=sqroot(100+576)

c=sqroot(676)

c=26

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