algebra II problem

[vduncan]

Hello - I am having a hard time with this - I have done it 5 different ways and would like to say that it is written wrong. But...it's probably just me.

Multiply out (a-b)^2. and use the result to answer the following question: If the height of a right triangle is 18 units larger than the base and the hypotenuse is 24, find the area.

 

[tkhunny]

Please demonstrate one of your methods.

 

[vduncan]

(a-b)^2=(a-b)(a-b)
a^2-ab-ab+b^2
a^2-2ab+b^2
566=a^2 + (a-18)^2
566=a^2 + (a^2-18a-18a+324)
566=2a^2-36a+324

a^2-2a(a-18)+(a-18)^2
a^2-2a^2+a^2-36a-18a-18a +324
-72a+324
a=4.5
a=4.5 so b=4.5+18=22.5

4.5^2+22.5^2=24^2.....NO!

 

[Denis]

Double check your "566" :shock:

WHY are you jumping from
566=2a^2-36a+324
to
a^2-2a(a-18)+(a-18)^2
??
All you need to do is solve
576 = 2a^2 - 36a + 324
for a

 

[tkhunny]

(a-b)^2=(a-b)(a-b)
a^2-ab-ab+b^2
a^2-2ab+b^2

Fine.

576=a^2 + (a-18)^2

This needs a definition before you jump in. Something like, "a = heaight of triangle" will do.

576=a^2 + (a^2-18a-18a+324)

I have to object to this step, just a little. You multiplied it nicely enough, but you didn't "use the result". Had you "used the result", you would just have written down the next step.

576=2a^2-36a+324

a^2-2a(a-18)+(a-18)^2

No idea what this is supposed to be.

The area is, of course, \(\displaystyle \frac{1}{2}a(a-18)\). Does that do us any good?

 

[vduncan]

Ok - seriously? I am such a dummy - a woman in my neighborhood asked if I could tutor her daughter in math - I have been tutoring ACT prep but it's been a while. I have a fever of 100.5 and that must have thrown me for a loop - I was trying to use a squared plus b squared equals c squared. I never went to the actual area formula - wow! Thanks and sorry for the waste of time.