A Geometry Problem . T^3 Tilted, Turned And Twisted Maths

Draw AG perpendicular on line BC . ( Note : It will meet BC in the part DC and not in BD.)
Let AB = 3 $x$ , AC = $x$ ,

(by angle bisector theorem we know that bisector of any angle of a triangle divides the base side in the same ratio as the remaining sides .'. BD : CD= 3:1 Let BD = 3 $y$ and DC = $y$

In right triangle AGD ,

=> $\frac{AG}{AD}$ =sin 60

=> $\frac{AG}{10 cm}$ = $\frac{√3}{2}$

=> AG = $\frac{10√3 cm}{2}$

=> AG = 5√3 cm

Also, $\frac{DG}{AD}$ = cos 60

=> $\frac{DG}{10cm}$ = $\frac{1}{2}$

=> DG = 5 cm

In right triangle AGB,

AB^2 - BG^2 = AG^2 (by Pythagoras theorem )

=> (3 $x$ )^2 - (BD + DG )^2 = (5√3)^2

=> 9 $x$ ^2 - (3 $y$ +5 )^2 = 75

=> 9 $x$ ^2 - 9 $y$ ^2 - 30 $y$ - 25 = 75

=> 9 $x$ ^2 - 9 $y$ ^2 - 30 $y$ =100 ------------------------------ eq n (1)

Similarly , in triangle AGC ,

AC^2 - GC^2 = AG^2 (by Pythagoras theorem )

=> $x$ ^2 - (DC - DG )^2 = (5√3)^2

=> $x$ ^2 - ( $y$ - 5 )^2 = 75

=> $x$ ^2 - $y$ ^2 + 10 $y$ - 25 = 75

=> $x$ ^2 - $y$ ^2 + 10 $y$ = 100

=> 9 $x$ ^2 - 9 $y$ ^2+ 90 $y$ = 900 ------------------------------------ eq n (2) (Multiplying both sides of eq n by 9)

Subtracting equation (1) from equation (2) , we get :

120 $y$ = 800

=> $y$ = $\frac{800}{120}$

=> $y$ = $\frac{20}{3}$

.'. BC = BD + DC
= 3 $y$ + $y$
= 4 $y$
= 4 * $\frac{20}{3}$
= 4 * $\frac{20}{3}$
= $\frac{80}{3}$

.'. Area of △ ABC = $\frac{1}{2}$ * AG * BC
= $\frac{1}{2}$ * 5√3 cm * $\frac{80}{3}$
= $\frac{200√3 cm^2}{3}$

'.' We know that median of a triangle divides a triangle into two triangles of equal area.

.'. Ar . △ ECA = $\frac{1}{2}$ Area of △ ABC = $\frac{200√3 cm^2}{3*2}$
= $\frac{100√3 cm^2}{3}$

Now , in △ ECA and △ OFC ,,

∠ C = ∠ C ( Common )

& ∠A = ∠ CFO ( '.' EA || OF .'. alt. int. ∠s are equal )

.'. △ ECA ~ △ OFC --(by AA similarity criterion)

.'. $\frac{Ar. △ ECA}{ Ar . △ OFC}$ = $\frac{CO^2}{OF^2}$ ( In similar triangles ratio of area of triangles = ratio of .. square of corresponding sides of the triangle ) => $\frac{Ar. △ ECA }{ Ar . △ OFC}$ = $\frac{CE}{CO}$ ^2 ------------ eq n (3)

We know that centroid divides median in ratio 2:1

.'. CO : OE = 2 : 1

=> OE: CO =1 : 2

=> $\frac{OE}{CO}$ + 1 = $\frac{1}{2}$ +1
. => $\frac{OE+CO}{CO}$ = $\frac{3}{2}$

=> $\frac{CE}{CO}$ = $\frac{3}{2}$

Putting CE/CO = 3/2 in equation 3 , we gets :

=> $\frac{Ar. △ ECA }{ Ar . △ OFC}$ = $\frac{3}{2}$ ^2

=> $\frac{Ar. △ ECA }{ Ar . △ OFC}$ = $\frac{9}{4}$

=> $\frac{(100√3 /3 )cm^2 }{ Ar . △ OFC}$ = $\frac{9}{4}$ ('.' Area of △ ECA = $\frac{100√3 cm^2}{3}$

=> Ar. △ OFC = $\frac{100√3*4cm^2}{3*9}$
= 25.660
* This question has been TILTED using GEOMETRY, TURNED from its original path using ALGEBRA and finally TWISTED . using TRIGONOMETRY . *