CAT Geometry is an important topic with lots of weightage in the **CAT Exam**. There are quite a few ways to find the area of the triangle. But which one to use when the lengths of the sides are given? It Looks like something heroic might work. But this question is beautiful because it tests a wonderful idea from the triangle. If you get this one idea correct, then you're through. Will you get this right or, you'll say, 'Damn! I missed it".

Question 16 : From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is :

- 225√3
- \\frac{500}{√3})
- \\frac{275}{√3})
- \\frac{250}{√3})

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Given that from a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. Here GBC is the one third of the area of the triangle.

We can join AG and GD which is the median. Each of the shaded triangle has the same area and therefore the remaining area is two-thirds of ABC

The ratio of the sides are 8 : 5 : 7

Area of triangle = √(s(s−a)(s−b)(s−c))

where, semi perimeter (s) = \\frac{8 + 5 + 7}{2}) = 10

Area = √(10(10 − 8)(10 − 5)(10 − 7))

Area = √(10(2)(5)(3))

Area = 10√3

Area of ABC = 25 × 10 √3 (As 8 : 5 : 7 multiplied by 5 gives the sides of triangle ABC)

Therefore area of the remaining traingle = \\frac{2}{3}) × 250√3 = \\frac{500}{√3})

The question is **"From a triangle ABC with sides of lengths 40 ft, 25 ft and 35 ft, a triangular portion GBC is cut off where G is the centroid of ABC. The area, in sq ft, of the remaining portion of triangle ABC is :" **

Choice B is the correct answer.

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