ABCD is a square. P is the midpoint of
AB. The line passing through A and perpendicular to DP intersects the diagonal at Q and BC at R.

(A) 1/2 (B) √3/2 (C)
√2 (D) 1 (E) None of these

If AB =2 then PR = _______?

In ∆ADM, as ÐAMD = 90

ÐADP + ÐDAM = 90

^{0},ÐADP + ÐDAM = 90

^{0 }
------(1)

And also as ÐDAP = 90

^{0}, this can be split into
ÐMAP + ÐDAM = 90

^{0}
------(2)

From (1) and (2), ÐADP = ÐMAP
ÐDAP = 90

^{0}= ÐRBP
By A-A-A similarity, ∆ADP ~ ∆ARB

AB/AP = BR/AP => 1 = BR/1 => BR = 1

Now on right angle ∆PBR, applying Pythagoras
theorem,

PR

^{2}= PB^{2}+ BR^{2}= 1 = 1 = 2
PR = √2

Answer (C)

## No comments:

## Post a Comment