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Question

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The diagram shows a section of a bridge between the points A and K. The length of line segment  is 640 meters. ∆ABC, ∆CDF, and ∆FJK are similar, and 2AC = CF = 2FK. The first pillar, , is 20 meters tall.

The area of ∆CDF is square meters.

Answer 1

Hi Amauri,

Because your triangles are similar, you can set up a proportion to calculate the area of the triangle CDF.

Remember that the area of a triangle is calculated by taking 1/2 of the Height (altitude) x Base

Since 2AC = CF, we know that 2BG = DH, since BG is 20, we know that DH has to be 40. Now to calculate the length of the base CH.

The full length of the bridge is 640 meters, and we know that if we add up all of the lengths it will equal that 640 meters. so let's call the lengths of AC and FK, 2x for simplicity, CF would then be x because of the proportions of these similar triangles. So mathematically we find that 2x + x + 2x = 640 and x would then be 5x=640 (divide by 5) and get x = 128.

So, the base of the triangle we're interested in is 128 and it's height or altitude is 40. Plug that into your formula

A = 1/2 bh

A = 1/2(128)(40)

A = 2,560 square meters