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

 

  • Correct Answer
    Choice (C). 8,750

Detailed Solution

E is the midpoint of hypotenuse AC so, BE = 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)
Adding (1) and (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)
=)
=) BD2 + BF2 + BE2 = 6250 + 502 = 6250 + 2500 = 8750 cm2


Correct Answer: 8,750



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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.