Asked by Lindsay
A right triangle has an area of 84 ftand a hypotenuse 25 ft long. What are the lengths of its two sides?
Ive gotten this far:
25^2= a^2+b^2
a=25-b
84= ab/2
168= 25b-b^2
b^2-25b+168=0
I then use the quad. formlua to solve for b... but I get an error. Can someone help me?
Ive gotten this far:
25^2= a^2+b^2
a=25-b
84= ab/2
168= 25b-b^2
b^2-25b+168=0
I then use the quad. formlua to solve for b... but I get an error. Can someone help me?
Answers
Answered by
MathMate
25^2= a^2+b^2... so far so good.
"a=25-b" ... problem stars here!
Start with
a²+b²=25²
and
(1/2)ab=84
2ab=4*84=336
So
a²+b²+2ab = 625+336 = 961
(a+b)²=961
a+b=√961=31
ab=2*84=168
a(31-a)=168
a²-31a+168=0
Solve quadratic in a to get
a=24, or a=7
So the two sides are 24 and 7
"a=25-b" ... problem stars here!
Start with
a²+b²=25²
and
(1/2)ab=84
2ab=4*84=336
So
a²+b²+2ab = 625+336 = 961
(a+b)²=961
a+b=√961=31
ab=2*84=168
a(31-a)=168
a²-31a+168=0
Solve quadratic in a to get
a=24, or a=7
So the two sides are 24 and 7
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