Simultaneous solution question

[evanr1234]

Hi Everyone,

I can not solve the following problem:

"the hypotenuse of a right triangle is 41 ft long and the area of the triangle is 180 ft^2, find the length of the two legs.

I understand how the equations are:

1) x^2 + y^2 = 41^2

and

2) 1/2(xy) = 180

But I can not get the simultaneous solution. The answers are 9 and 40.

Any help is greatly appreciated.

 

[Cubist]

I'd start by making y the subject of equation 2, and then substitute it into equation 1. What do you end up with?

 

[evanr1234]

I'd start by making y the subject of equation 2, and then substitute it into equation 1. What do you end up with?

360^2/x^2 + x^2 = 41^2

then

129,600 + x^4 = 1681(x^2) (I multiplied everything by x^2)

I also can get 129,600 = (1681 - x^2)/x^2

then

129,601 = 1681/x^2 but am getting very low numbers for x

 

[Dr.Peterson]

360^2/x^2 + x^2 = 41^2

then

129,600 + x^4 = 1681(x^2) (I multiplied everything by x^2)

I also can get 129,600 = (1681 - x^2)/x^2

then

129,601 = 1681/x^2 but am getting very low numbers for x

Your equation 129,600 + x^4 = 1681(x^2) is correct. Replace x^2 with u and solve the resulting quadratic equation, then find x. It works.

Your equation 129,600 = (1681 - x^2)/x^2 is wrong. How did you get it? It looks like you divided by x^2 when you meant to multiply.

 

[pka]

You want to solve \(\displaystyle =x^4-41^2x^2+360^2=0\)
SpoilerSEE HERE

 

[MarkFL]

As an alternate method, you could write:

[MATH]z=x+y[/MATH]
[MATH]z^2=x^y+2xy+y^2=41^2+720=40^2+2(40)(9)+9^2=(40+9)^2=49^2[/MATH]
Hence:

[MATH]x+y=49[/MATH]
[MATH]x+\frac{360}{x}=49[/MATH]
[MATH]x^2-49x+360=0[/MATH]
[MATH](x-40)(x-9)=0[/MATH]
And so:

[MATH](x,y)\in\{(9,40),(40,9)\}[/MATH]