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805+ Level|   Geometry|         
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Bunuel
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We need to know the relationship among the similar triangles and how equally distanced lines divide the area of a triangle. I tried to explian is a photo i am attaching.
My answer is D
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photo_2021-08-19_11-24-08.jpg
photo_2021-08-19_11-24-08.jpg [ 93.99 KiB | Viewed 2963 times ]

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We have:
AB = 25
BC = 20
CA = 15
s = (AB + BC + CA)/2 = 30

Using Heron's Formula, the area of triangle ABC is √s(s-25)(s-20)(s-15) = √30*5*10*15 = 150

We have:
(1) Area of triangle ADE is AD/AC * AE/AB * 150 = 1/3*1/5*150 = 10
(2) Area of triangle CHI is CI/CA * CH/CB * 150 = 1/3*1/4*150 = 12.5
(3) Area of triangle BFG is BF/BA * BG/BC * 150 = 1/5*1/4*150 = 7.5

From (1), (2) and (3) we have:
Area of the shaded region = 150 - 10 - 12.5 - 7.5 = 120

The answer is D
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Screen Shot 2021-08-19 at 18.53.13.png
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