Triangle ABC is isosceles with right angle at B. There is a point D in the interior ...?

Question

Triangle ABC is isosceles with right angle at B. There is a point D in the interior of triangle ABC such that |DA| = 5, |DB| = 3 and |DC| = 4. Find the area of the triangle ABC.

to the first one who answered, how did you arrive at cos theta = (x^2 - 16) / 6x

to the first one who answered, how did you arrive at cos theta = (x^2 - 16) / 6x

Answer 1

I think there is a easier way to solve this problem than developing equations that required extensive use of reverse sin & cos. I need to dig through my Trig formulas to figure out the numbers, but I doubt simply Geometry equations can solved this. I will post the edit as soon as I get a hold of the formulas.
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Edit: The formula I was thinking of is Heron's area formula; given 3 sides of a triangle, a, b, c to find the area = √[s(s-a)(s-b)(s-c)], where s = (a+b+c)/2. Using the lines from point D that seperate the right triangle into 3 smaller triangles, I was able to come up with an algebraic equation with one unknown, but the equation was too difficult to solve, with square roots that just won't go away. I think I might need a good week to think about this some more. Had use trig, geometric proofs, and inequalities. Have not found one working yet.


XR

Answer 2

For convenience let angle ADB = x, angle BDC = y, and angle CDA = z.
Using the law of cosines we get:
AB^2 = 5^2+3^2 -15cos x = 34-15cos x <-- Eq 1
BC^2 = 3^2 + 4^2 -12 cos y = 25- 12 cos y <-- Eq 2
AC^2 = 4^2+5^2 -20 cos z = 41 -20cos z <-- Eq 3
Since AB =BC, AB^2 = BC^2 so EQ 1 = EQ 2, giving:
34-15cos x = 25-12cos y <-- Eq 4
Since AC^2 =2 BC^2 we get:
41-20cos z = 2(25-12 cos y) <-- Eq 5
Since AC^2 = AB^2 +BC^2 we get:
41-20 cos z = 34-15cosx + 25-12cos y <-- Eq 6

You can now solve Eq 4,5,6 simultaneosly and find cos x which will then allow you to find AB
The area of triangle is then AB^2/2

Answer 3

(sdlei, you asked me a question in additional details which I saw only after 2 days. I have given the required explanation now. I could have posted the explanation earlier had you sent me an email.)

Let AB = BC = x
Let angle DBA = θ
So angle DBC = 90° - θ
Using cosine rule of trigonometry for triangle DBA,
cos θ = (BA^2 + BD^2 - AD^2) / 2*BD*BA
= (x^2 + (3)^2 - (5)^2) / 2*3*x
= (x^2 - 16) / 6x
Using cosine rule for triangle DBC,
cos (90° - θ) = (BC^2 + BD^2 - CD^2) / 2 * BC * BD
=> sin θ = (x^2 + 9 - 16) / 6x = (x^2 - 7) / 6x

cos^2 θ + sin^2 θ = 1
=> (x^2 - 16)^2 / (6x)^2 + (x^2 - 7)^2 / (6x)^2
=> (x^2 - 16)^2 + (x^2 - 7)^2 = 36x^2
=> 2x^4 - 82x^2 + 305 = 0
=> x^2 = (1/4) [ 82 ± √{(82)^2 - 2440} ]
=> x^2 = (1/4) [ 82 ± 65.45 ]
=> x^2 = 36.86 or 4.14
=> area = x^2 = 36.86 sq. units. (4.14 if point outside)

CORRECTION:
Thank you year 2012 for drawing my attention to my mistake.
As it is an isosceles triangle area is 18.43 sq. units ( 2.07 if point outside)
I had solved a similar problem earlier on Nick's mathematical puzzles which was appreciated by Mr. Nick by acknowledging on his site in which the figure was a square.

Answer 4

I solve it by the drawing ( using compass ) C B = 6unit
......................................... AB = 6unit and AC =8.485 unit
area =1/2 x 6 x 6 = 1/2 x 36 = 18 unit ^2 approx .
I do,nt know how to trig it sorry .

Answer 5

AB should be3>=x>=6
BC should be3>=x>=6
AC should be3>=x>=9

Answer 6

I agree with my above answer.