Respuesta :

Check the picture below, notice the tickmarks.

for a chord perpendicular to the diameter chord, their intersection makes two equal segements and those twin segments make twin arcs.

[tex]\bf 2x+5=3x-2\implies 5=x-2\implies \boxed{7=x} \\\\\\ y+8=5y\implies 8=4y\implies \cfrac{8}{4}=y\implies \boxed{2=y} \\\\[-0.35em] ~\dotfill\\\\ \widehat{AB}=2(7)+5\implies \widehat{AB}=19~\hfill \overline{DC}=5(2)\implies \overline{DC}=10[/tex]

Ver imagen jdoe0001
1m1c7h

Answer for arc AB:

since we know FB is the perpendicular bisector of AC then we know AB is the same length as BC so: 2x + 5 = 3x - 2 put everything on the left side

2x + 5 + 2 = 3x -2 + 2    2x + 7 = 3x put the variables on one side

2x - 2x + 7 = 3x - 2x   7 = x  plug it in

2(7) + 5 = 19

AB = 19

Answer for segment DC:

since we know FB is the perpendicular bisector of AC then we know AD is the same as DC.

y + 8 = 5y put the variable on one side

y - y + 8 = 5y - y   8 = 4y simplify

8/4 = 4y/4

2 = y plug it in

5(2) = 10

DC = 10

Otras preguntas

Q&A Education