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.

## Monday, May 19, 2008

### Geometry Problem 102

Labels:
area,
equilateral,
interior point,
triangle

Subscribe to:
Post Comments (Atom)

Let C be (0,0), B:(a,0) Hence C:(a/2,V3a/2 where V = square root of the number following.

ReplyDeleteLet 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

Using the result of problem 101 we can say that angle CDB = 150 deg. since AD^2 = BD^2 + CD^2

ReplyDeleteAnd 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

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

ReplyDeleteI'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

ReplyDeleteGive me your Opinion.

It's a good solution!

ReplyDeleteDraw DE perpendicular to BC

ReplyDeletefind 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

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

ReplyDeleted = √(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

With the help of problem 101, <CDB=150 as of 3-4-5 combination

ReplyDeleteCB^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