I have a question but I am stuck on the factoring 1/2*b*(h+7)=22
Define 'factoring' - because that isn't something I'd look to factorise.
1/2*b*(h+7)=22 1/2(b^2+h7)=22 2*1/2(b^2+h7)=2*22 b^2+h7=44 b^2+h7-44=44-44 b^2+h7-44=0 (b- )(h+ )= this is where I am lost
u cant solve for b and h in same equation
Here I will give you the question. The base of a triangle is 7cm greater than the height. The area is 22 cm^2. I have to find the height and base of the triangle.
Define b = h+7 (or h = b-7) , and go from there.
i.e. you should be solving: \[\frac{1}{2} \cdot h \cdot (h+7) = 22 \text{ or } \frac{1}{2} \cdot b \cdot (b-7) \]
The last one =22 too :@
Post the quadratic equation you get when you follow through with INewton's suggestion.
My suggestion is to use the first one:\[(1/2) \times h \times(h + 7)=22 cm sq\]
1/2*h(h+7)=22 1/2(h^2+h7)=22 2*1/2(h^2+h7)=2*22 h^2+h7=44 h^2-h7=44=44-44 h^2+h7-44=0 h=9 b=35
\[h ^{2}=7h-44=0\] is correct. Which method did you use to get the values of h? quadratic formula, or factor method?
\[h ^{2}+7h-44=0\] I meant to type sri
which method did you use to get 9 for h??
i think factor
9 doesn't go into 44 so something went wrong. Factor is the way to go. That equation will factor beautifully. Do you need some guidance in factoring that equation?
yes please
Well we are in luck in that the coefficient of h^2 is 1. Also we need to find to factors of 44 which will when combined equal 7. We have a hint since 44 is negative, one of these factors will positive and the other negative. Do you follow so far?
It will start looking like this: (h- ?) (h +?) the ?*?=-44 and ?+?=-7
Any questions??? so far.
leahpl81872 can you think of two numbers when multiplied equal -44 and when added =+7 ? It is important cause one of these numbers is your answer!
11 & 4
Right on!! (h+11)(h-4)=0
which one do you think is the answer? the -11 or the 4 ???
4
very good, now remember the base was equal to the height + 7 cm, so what is the base?
11
Correcto, you seem to have a handle on this problem, good luck.
thank you very much
No problem, and INewton's input was helpful.
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