Monday, May 19, 2008

Geometry Problem 102

Try to use elementary geometry (Euclid's Elements).




See complete Problem 102
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 10:18 PM
Labels: , , ,

8 comments:

  1. AjitMarch 26, 2009 at 2:39 AM

    Let C be (0,0), B:(a,0) Hence C:(a/2,V3a/2 where V = square root of the number following.
    Let D be (x,y). Then we can say:(x^2+y^2=9,
    (x-a)^2+y^2=16 and (x-a/2)^2+(y-(a/2)(3)^(1/2))^2=25 which can be solved to obtain
    x=2.865957009,y=0.8867301846 and a=[(25+12(3)^(1/2)]^(1/2) = 6.766432568
    Area of the triangle =((a^2)/4)(3)^(1/2)
    = 9+(25/4)(3)^(1/2)
    = 19.82531755 sq units.
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. AjitApril 7, 2009 at 3:13 AM

    Using the result of problem 101 we can say that angle CDB = 150 deg. since AD^2 = BD^2 + CD^2
    And thus from Tr.BDC we've:
    BC^2 = 5^2-2*3*4cos(150)
    = 25 + 24V3/2 ~ 45.784609691
    Area of Tr. ABC ~ (V3/4)*(45.784609691)
    ~ 19.82531755 sq units as b4.
    Ajit: ajitathle@gmail.com

    ReplyDelete
  3. TouskaSeptember 23, 2009 at 1:33 AM

    to solve this problem you have to rotate the triangle by angle 60 where the side AC will be on AB

    ReplyDelete
  4. TouskaSeptember 23, 2009 at 8:17 AM

    I'm going to give you a good solution ,rotate the triangle by angle 60 where the side AC will be on AB and AD will be AE and AB will have AF as image . Now ADE is equilateral triangle(AD=AE and DAE=60) ,and DBE is right triangle (Converse of pythagora's) ,calcutate the areas of these two triangle to get the area of ADBE which is equale to the part ACBD of the triangle now let us to calculate the area of the triangle CDB,now the angle CBF = 120 and DBE =90 then DBC+EBF=30 then the angle CDB=150 because the angle EBF=DCB (use the same fashion to solve problem 101) finally use the formula of the area 1/2*DC*DB*sin (150) to get the area of DCB add the three results of the three tiangles .QED
    Give me your Opinion.

    ReplyDelete
  5. Antonio GutierrezSeptember 23, 2009 at 11:03 AM

    It's a good solution!

    ReplyDelete
  6. c .t . e. oDecember 11, 2009 at 1:04 PM

    Draw DE perpendicular to BC
    find a side of tr

    DE'2=16-x'2 ( x EB )
    DE'2=9-y'2 (y CE )
    16-x'2 = 9 - y'2
    7 = x'2 - y'2
    7 = (x+y)(x-y)
    7*1 = (x+y)(x-y)
    or
    3,5*2=(x+y)(x-y) where x+y give a side of tr
    How can i continue

    ReplyDelete
  7. Ignacio Larrosa CañestroApril 9, 2018 at 4:01 AM

    a=AD, b = BD, c = CD, d = AB = BC = CA. Then a^2 = b^2 + c^2 and

    d = √(a^2 + √3bc) (https://goo.gl/Hj5iWc)
    S = (√3/4)(a^2 + √3bc)= (3bc + √3a^2)/4 = (36+25√3)/4 ~= 19.82531754

    See also https://goo.gl/eJiQJo

    ReplyDelete
  8. MarcoDecember 25, 2022 at 3:00 AM

    With the help of problem 101, <CDB=150 as of 3-4-5 combination
    CB^2=3^2+4^2-2(3)(4)cos150
    CB^2=25+12sqrt(3)
    Area of ABC=(CB^2)sin60/2
    =(25+12sqrt3)*sqrt(3)/4
    =(25sqrt3+36)/4

    ReplyDelete

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