Question 566959
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Missed it by <i><b>that</b></i> much (holding his thumb and index finger just a little bit apart).


Your equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1}{2}\right)(\underline{2}b)(2b\ +\ 4)\ =\ 35]


with the problem spot underlined.  If you are going to use *[tex \Large b] to represent the measure of the base, then *[tex \Large b] is what goes in the equation to represent the base.  You put in *[tex \Large 2b].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1}{2}\right)(b)(2b\ +\ 4)\ =\ 35]


is what you want.


Multiply *[tex \Large 2b\ +\ 4] by *[tex \Large \frac{1}{2}].  Distribute across the parentheses.  Add -35 to both sides.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2\ +\ 2b\ -\ 35\ =\ 0]


Factor


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (b\ +\ 7)(b\ -\ 5)\ =\ 0]


So *[tex \Large b\ =\ -7] or *[tex \Large b\ =\ 5].  Well, -7 for the measure of the base of a triangle is absurd, so toss that result out.  The base is 5 and the height must be 14.  And 5 times 14 is 70.  70 divided by 2 is 35.  Answer checks.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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