Check the picture below.
now, we're making an assumption that, the two blue shaded region are equal in shape, and thus if that's so, that area above the 14 is 6 and below it is also 6, 14 + 6 + 6 = 26.
so hmm if we simply get the area of the trapezoid and subtract the area of the yellow triangle and the area of the cyan triangle, what's leftover is what we didn't subtract, namely the shaded region.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h~~=height\\ a,b=\stackrel{parallel~sides}{bases~\hfill }\\[-0.5em] \hrulefill\\ h=15\\ a=14\\ b=26 \end{cases}\implies A=\cfrac{15(14+26)}{2}\implies A=300 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\Large Areas}}{\stackrel{trapezoid}{300}~~ - ~~\stackrel{yellow~triangle}{\cfrac{1}{2}(26)(9)}~~ - ~~\stackrel{cyan~triangle}{\cfrac{1}{2}(15)(6)}} \\\\\\ 300~~ - ~~117~~ - ~~45\implies 138\qquad \textit{blue shaded area}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h~~%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases~%5Chfill%20%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D15%5C%5C%20a%3D14%5C%5C%20b%3D26%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B15%2814%2B26%29%7D%7B2%7D%5Cimplies%20A%3D300%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5CLarge%20Areas%7D%7D%7B%5Cstackrel%7Btrapezoid%7D%7B300%7D~~%20-%20~~%5Cstackrel%7Byellow~triangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%2826%29%289%29%7D~~%20-%20~~%5Cstackrel%7Bcyan~triangle%7D%7B%5Ccfrac%7B1%7D%7B2%7D%2815%29%286%29%7D%7D%20%5C%5C%5C%5C%5C%5C%20300~~%20-%20~~117~~%20-%20~~45%5Cimplies%20138%5Cqquad%20%5Ctextit%7Bblue%20shaded%20area%7D)
Well, by doing the work and trying to solve this equation, i have came to the terms that it is a no solution problem
B is your answer!
Hope I helped!
Contact me for more!
FLVS teacher,
~Amber Fish
Answer:
is that a question?
Step-by-step explanation:
Answer:
When dividing rational numbers, multiply by the "right" reciprocal. In this case, the "right" reciprocal means to take the reciprocal of the fraction on the right-hand side of the division operator.
