Answer:
[tex]\textsf{D)}\quad \sqrt{(a+b)^2+c^2}[/tex]
Step-by-step explanation:
To prove that the diagonals of an isosceles trapezoid are congruent, we can use the distance formula:
[tex]\boxed{\begin{array}{l}\underline{\sf Distance \;Formula}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\;d\;\textsf{is the distance between two points.} \\\phantom{ww}\bullet\;\;\textsf{$(x_1,y_1)$ and $(x_2,y_2)$ are the two points.}\end{array}}[/tex]
The diagonals of the given isosceles trapezoid JKLM are JL and KM.
The coordinates of the vertices are J(-b, c), K(b, c), L(a, 0), and M(-a, 0).
Substitute the coordinates of J and L into the distance formula:
[tex]\overline{JL}=\sqrt{(x_L-x_J)^2+(y_L-y_J)^2}\\\\\overline{JL}=\sqrt{(a-(-b))^2+(0-c)^2}\\\\\overline{JL}=\sqrt{(a+b)^2+(-c)^2}\\\\\overline{JL}=\sqrt{(a+b)^2+c^2}[/tex]
Substitute the coordinates of K and M into the distance formula:
[tex]\overline{KM}=\sqrt{(x_M-x_K)^2+(y_M-y_K)^2}\\\\\overline{KM}=\sqrt{(-a-b)^2+(0-c)^2}\\\\\overline{KM}=\sqrt{(-(a+b))^2+(-c)^2}\\\\\overline{KM}=\sqrt{(a+b)^2+c^2[/tex]
Therefore, we have proved that JL = KM.
