jugoslav

D is on side AC of Triangle ABC. We know [ABC] = 120, AD = 8, DC = 4. What is [ABD]?

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Let BD be x

8x + 4x = 240

x = 20

[ABD] = 8x / 2 = 80 u²

BD = sqrt(20² + 15²) = 25

AD = sqrt( 25² - 7²) = 24

The perimeter of quadrilateral ABCD = 7 + 15 + 20 + 24 = 66

2x+8 +x = 66

That number is 46.66666667

let diagonal  be            d = 2

square, area                 A_s = 2 u²

rectangle, area             A_r = √3

ratio               R = A_r / A_s = √3 / 2 = 0.866025403

R² = 0.75  ==> 3/4 

In triangle ABC,  AB= AC=5 and BC=7. Let O be the circumcenter of triangle ABC. Find the area of triangle OBC.

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Let M be a midpoint on AB.

Angle BAO ==>    β = arcsin(3.5 / 5) ==>  44.427004º

Area of triangle ABC ==>     A₁ [sqrt(5² - 3.5²) * 3.5 ==>   A₁ = 12.49749975 u²

Area of polygon ABOC ==>  A₂ = 2[(tanβ * 2.5) * 2.5]       A₂ = 12.25245073 u²

Δ OBC area ==>             A = A₁ - A₂ = 0.24504902 u²

Angle ACB = 180 - 142

Y'Z' = 37.10795063

That's nice, but I like my method too.

Let's denote a side of a square with a letter  a

Tangent of angle BAC = 1.33333333

Tangent of angle BCA = 0.75

So, we have     (a / 1.3333333) + a + (a / 0.75) = 5  ==>   a = 1.621621622     or     60 / 37

This is an interesting answer: 

Using similar triangles twice, and  calling the length of the side of the square x,

in the triangle CYZ,  CY/x = 4/3, so CY = 4x/3,

in the triangle WXA,  x/XA = 3/4, so XA = 3x/4.

The hypotenuse of the triangle is

CY + YX + XA  = 4x/3 + x + 3x/4 = 37x/12 = 5,

so x = 60/37.

(Solved by Guest -  Nov 13, 2015)