Question

100.

Answer 1

Check the picture below.

so our bases are "d" and "d+a+b", and a height of "c".

$\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(x+y)}{2}~~ \begin{cases} x,y=\stackrel{parallel~sides}{bases}\\ h~~~=height\\ \cline{1-1} x=d\\ y=d+a+b\\ h=c\\ A=(d+b)c \end{cases}\implies (d+b)c=\cfrac{c[d+(d+a+b)]}{2} \\\\\\ 2(d+b)c=c[d+(d+a+b)]\implies 2dc+2bc=c(2d+a+b)$

$\bf ~~\begin{matrix} 2dc \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+2bc=~~\begin{matrix} 2dc \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+ac+bc\implies 2bc=ac+bc\implies 2bc=c(a+b) \\\\\\ \cfrac{2b~~\begin{matrix} c \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} c \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}=a+b\implies 2b=a+b\implies 2b-b=a\implies \blacktriangleright b=a \blacktriangleleft$