Thursday, March 17, 2016

Geometry Problem 1200: Square, Right Triangle, Perpendicular, 90 Degrees, Equal Areas

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1200.

Geometry Problem 1200: Square, Right Triangle, Perpendicular, 90 Degrees, Equal Areas, Online College School
at

4 comments:

  1. Peter TranMarch 17, 2016 at 6:25 PM

    http://s22.postimg.org/yor6ykls1/pro_1200.png

    Let α= ∡ (CBM)= ∡ (PCB)= ∡ (BMA)
    And β= ∡ (NDC)= ∡ (DCQ)= ∡ (NAD)

    Triangle BCP congruent to tri. BAM …( hypoth- leg congruence)
    And Triangle DAN congruent to tri. DCQ …( hypoth- leg congruence)
    So AM=CP and AN= CQ
    S(MAN)= ½. MA.NA.sin(MAN)= ½.MA.NA.sin(90- α- β)
    S(CPQ)= ½.CP.CQ.sin(PCQ)= ½.CP.CQ.sin(90+ α +β)
    Note that ∡ (MAN) supplement to ∡ (PCQ) so sin(MAN)= sin(PCQ) =>S(MAN)=S(CPQ)

    ReplyDelete
  2. Sumith PeirisMarch 17, 2016 at 8:34 PM

    Triangles BCP and ABM are congruent ASA and triangles ADN and CDQ are also congruent ASA.

    Hence <s PCQ and AMN can easily be shown to be supplementary.

    So altitudes QX and NY of Tr.s PCQ and AMN respectively are equal since Tr.s QCX and NMY are congruent ASA

    So using area = 1/2 base X height

    S(CPQ) = S(AMN)

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. AnonymousJanuary 4, 2018 at 12:20 PM

      Hi there . Was wondering how you prove that sides ND=DQ and CQ=NA ? You need to prove this to substantiate your congruency theory , or am I incorrect?

      Delete
  3. APOSTOLIS MANOLOUDISMarch 18, 2016 at 2:07 AM

    Is AM and AN perpendicular CP and CQ respectively.Therefore tri. ABM= tri. CBP or
    tri. AND=tri. CDQ (CP=AM ,CQ=AN, <PCB=<BAM, <DCQ=<NAD ) .But <PCQ +<MAN=180.
    Therefore (PCQ)/(MAN)=(CP.CQ)/(AM.AN)=1

    ReplyDelete

Share your solution or comment below! Your input is valuable and may be shared with the community.