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5 votes
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.

What is the ratio of the perimeter of ΔHKO to the perimeter of ΔFGO ? Express your-example-1
What is the ratio of the perimeter of ΔHKO to the perimeter of ΔFGO ? Express your-example-1
What is the ratio of the perimeter of ΔHKO to the perimeter of ΔFGO ? Express your-example-2
User Julivico
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4.9k points

1 Answer

6 votes

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

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In summary,

Answer for the first box: 12

Answer for the second box: 48

Answer for the third box: 1/4

User Issa
by
4.7k points