Monday, May 19, 2008

Elearn Geometry Problem 69



See complete Problem 69
Square Inscribed in a Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

6 comments:

  1. Let the vertices of square are PQRS and BM perpendicular is drawn to BC. area triangle APM=(1/2).s.AM
    area of triangle SMC=(1/2).s.MC
    area of triangle PMS=(1/2).s.s
    area of triangle BPS=(1/2).s.(h-s)
    area of whole triangle=(1/2).b.h
    now
    1/2).s.AM+(1/2).s.MC+(1/2).s.s+(1/2).s.(h-s)=(1/2).b.h
    and AM + MC=b
    which on simplifying prove the relation

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  2. With P, Q, R and S vertices of the square and k be the height of the triangle ΔSBR, we have ΔABC ~ ΔSBR. Therefore,

    From h/k = b/s, we have:
    s = bk/h = b(h-s)/h = (bh-bs)/h
    and sh + bs = bh or s = bh/(b+h)

    QED

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  3. Let PQRS be the square.
    Area of triangle ABC= Area of triangle BSR + Area of Trapezium ACRS.
    (1/2)bh = (1/2)s*(h-s) + (1/2)(b+s)* (h-s)
    bh = (h-s)(2s+b)
    on simplification we get the result

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  4. Area of Tr. ABC is comprised of the areas of 3 Tr.s and the square.

    So s^2 + 1/2s(h-s) + 1/2s(b-s) = 1/2 bh

    From which the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Or from similar triangles

    (h-s)/h .= s/b and the result follows

    ReplyDelete