Thursday, March 12, 2009

Problem 268: Right Triangle, Catheti and Altitude

Proposed Problem
Problem 268. Right Triangle, Catheti and Altitude.

See complete Problem 268 at:
gogeometry.com/problem/p268_right_triangle_catheti_altitude_hypotenuse_square.htm

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 3:06 PM
Labels: , , ,

4 comments:

  1. AjitMay 21, 2009 at 12:19 AM

    We know that a^2 = BH*BA, b^2 = AH*BA and h^2=AH*BH. So 1/a^2 + /b^2 = 1/BH*BA + 1/AH*BA = 1/BA*[1/BH+1/AH] = 1/c * [AH+BH=c)/AH*BH = 1/(AH-BH) = 1/h^2. Hence, 1/a^2 + /b^2 = 1/h^2
    Ajit: ajitathle@gmail.com

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  2. AnonymousAugust 31, 2009 at 10:06 AM

    By JCA
    1/a^2+1/b^2=(a^2+b^2)/(a^2*b^2)=c^2/(a^2*b^2)
    =1/h^2*(c^2*h^2)/(a^2*b^2)
    =1/h^2*(4[ABC]^2)/(4*[ABC]^2]=1/h^2

    ReplyDelete
  3. Geek37March 25, 2020 at 4:13 AM

    https://www.youtube.com/watch?v=Pxuw-ro45ZA

    ReplyDelete
  4. rv.littlemanDecember 9, 2020 at 1:55 PM

    See the drawing

    Define c=AB
    S=[ABC]=a.b/2=c.h/2 => a.b=c.h
    c=a.b/h => c^2=a^2.b^2/h^2
    Pythagoras => a^2+b^2=c^2
    => a^2+b^2=a^2.b^2/h^2
    dividing by a^2.b^2 => 1/b^2 + 1/a^2 = 1/h^2

    ReplyDelete

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