Friday, May 24, 2013

Problem 881: Triangle, three Squares, Centers, Areas

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

Click the figure below to see the complete problem 881.

Online Geometry Problem 881: Triangle, three Squares, Centers, Areas

8 comments:

  1. BM=4 , BN= SQRT(23), MN=SQRT(67)
    In triangle MBN we have MN^2=MB^2+NB^2- 2.MB.NB.cos(B+90)
    But cos(B+90)=- sin(B)
    So sin(B)=(MN^2-MB^2-NB^2)/2.MB.NB
    =(67-16-23)/(2 * 4 .*SQRT(23)= 7/(2 . SQRT(23))
    Area (ABC)=1/2 . AB* BC* Sin(B) = 14
    BM=4 , BN= SQRT(23), MN=SQRT(67)
    In triangle MBN we have MN^2=MB^2+NB^2- 2.MB.NB.cos(B+90)
    But cos(B+90)=- sin(B)
    So sin(B)=(MN^2-MB^2-NB^2)/2.MB.NB
    =(67-16-23)/(2 * 4 .*SQRT(23)= 7/(2 . SQRT(23))
    Area (ABC)=1/2 . AB* BC* Sin(B) = 14

    ReplyDelete
  2. Join BM, BN.

    BM^2 = BD^2 / 2 = 16
    BN^2 = BF^2 / 2 = 23
    MN^2 = 67

    MN^2 = BM^2 + BN^2 − 2×BM×BN×cos(∠B+90°)
    67 = 16 + 23 + 2×BM×BN×sin∠B
    BM×BN×sin∠B = 14

    S(ABC) = 1/2×BA×BC×sin∠B
    = 1/2×(√2×BM)×(√2×BN)×sin∠B
    = BM×BN×sin∠B
    = 14

    ReplyDelete
  3. BD=AB=sqrt(32), BF=BC=sqrt(46), MN=sqrt(67)
    BM=sqrt(16)=4, BN=sqrt(23)
    By cosine law,
    MN^2 = BM^2 + BN^2 - 2*BM*BN*cosMBN
    MN^2 = BM^2 + BN^2 - 2*BM*BN*cos(90+ABC)
    MN^2 = BM^2 + BN^2 + 2*BM*BN*sinABC
    MN^2 = BM^2 + BN^2 + 2*1/sqrt2*1/sqrt2*(AB*BC*sinABC)
    67 - 16 - 23 = AB*BC*sinABC
    28 = AB*BC*sinABC
    14 = Area

    ReplyDelete
  4. Prove that the center of the square MNPQ coincides with the midpoint of AC.

    ReplyDelete
    Replies
    1. this is very easy.
      Because MSP, is, diagonal of the square MNPQ ,and angle MSN=90 ,angle SMN=angle MNS = 45 ,angle MNP=90 ,then, angle SNP=45 ,so, angle NPS =45 .Therefore SP=SN=SM so,,S midpoint of the MP

      Delete
  5. What is S?. Kindly clarify.
    If you mean S is the point where diagonals of the square MNPQ meet, it is trivial to see that SP = SN = SM.
    If you mean S is the point where a diagonal, say MP, meets AC, you need to prove that S is the midpoint of the side AC of the triangle ABC.

    ReplyDelete
  6. Please read my solution.S, is midpoint of the AC

    ReplyDelete