CAT Practice : Coordinate Geometry

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Geometry - Triangles

Basic Geometry

Q.43:
In the above figure, ΔABC is right angled and AC = 100 cm. Also, AD = DE = EF = FC. Find the value of:
BD2 + BE2 + BF2 (in cm2)
1. 10,000
2. 5,000
3. 8,750
4. 12,500

Choice (C). 8,750

Detailed Solution

E is the midpoint of hypotenuse AC so, BE = $\frac{1}{2} * AC = 50$ cm
Now, applying Apollonius’s theorem on ΔABE
=) AB2 + BE2 = 2 x (BD2 + AD2)
=) AB2 + 502 = 2 x (BD2 + 252)……….(1) (As, AD = DE = EF = FC = 25 cm)
Similarly, in ΔBEC,
=) BC2 + BE2 = 2 x (BF2 + EF2)
=) BC2 + 502 = 2 x (BF2 + 252)………..(2)
=) AB2 + BC2 + 2 x 502 = 2 x (BD2 + BF2 + 2 x 252)
=) AC2 + 2 x 502 = 2 x (BD2 + BF2) + 4 x 252 (As, AB2 + BC2 = AC2)
=) 1002 + 2 X 502 – 502 = 2 x (BD2 + BF2)
=) 1002+ 502 = 2 x (BD2 + BF2)
=) $\frac{(10,000 + 2,500)}{2} = BD^2 + BF^2 = 6250$
=) BD2 + BF2 + BE2 = 6250 + 502 = 6250 + 2500 = 8750 cm2

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Geometry is probably the most vital topic as far as CAT preparation is concerned. Geometry sets the stage for Trigonometry, Cogeo and Mensuration as well.