NMTC 2015

Geometry Level 3

ABCD is a square ,From the diagonal BD,a length BX is cut off equal to BA .From X,a straight line XY is drawn perpendicular to BD to meet AD at Y.Then AB+AY=

B D 2 \frac { BD }{ \sqrt { 2 } } 2 \sqrt { 2 } BD 3 \sqrt { 3 } BD BD

Bala Vidyadharan by Bala vidyadharan , 20, India

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1 solution

Akshay Yadav
Aug 24, 2015

A problem from Bhskara level of NMTC 2015,

Let us assume the side of square as 'x',

BD = 2 0.5 x 2^{0.5}x and BX = x,

DX = 2 0.5 x − x 2^{0.5}x - x

XY = DX,

DY = ( 2 − 2 0.5 ) x (2 - 2^{0.5})x

Now,

AB + AY = AB + DA - DY

= 1 x + 1 x − 2 x + 2 0.5 x 1x+1x-2x+2^{0.5}x

= 2 0.5 x 2^{0.5}x = BD

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abhishekrocks sahoo - 5 years, 9 months ago

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