What is the ratio of the perimeter of δhko to the perimeter of δfgo

?

express your answer as a simplified fraction.

enter your answer in the box.

I think you meant to say the ratio of the areas (not perimeters). Also, HKO and FGO are not showing up, so it seems like a different problem. I'm going to answer the question you posted in the image attachments.

The triangle ABC has area 12 since

A = b*h/2 = 4*6/2 = 24/2 = 12

Triangle CEF has area 48 because

A = b*h/2 = 8*12/2 = 96/2 = 48

The ratio of the areas is found by dividing the area of ABC over the area of CEF (as the last box instructs) so we have 12/48 = 1/4

Therefore, the area of ABC is 1/4 that of the area of CEF

In summary,

$4

step-by-step explanation:

$30×5=$150

$150 minus $130=$20

$20÷5=$4.

hope to and hope that my answer is brainliest.(follow me )