Simplicity is the most complex thing . Part 3

Geometry Level 5

A B C ABC is a triangle in which A D AD is the bisector of angle A A such that ∠ A D B = 12 0 ∘ \angle ADB = 120 ^\circ . A B : A C = 3 : 1 AB : AC = 3 :1 and A D = 10 cm AD = 10 \text{ cm} . C E CE is the median on side A B AB of triangle A B C ABC . O O is the centroid of the triangle A B C ABC . Through O O a line O F ∥ E A OF \parallel EA is drawn. Find the area of triangle O F C OFC in cm 2 \text{cm}^2 .

Round your answer to 2 decimal places.


The answer is 25.66.

D H by D H , 18, India

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1 solution

D H
Aug 5, 2016

Draw AG perpendicular on line BC . ( Note : It will meet BC in the part DC and not in BD.)

Let AB = 3 x x , AC = x x .

AD is the angle bisector of angle A.

.'. BD : DC = AB: AC 

      BD : DC = 3:1

Let BD = 3 y y and DC = y y

In right triangle AGD ,

A G A D \frac{AG}{AD} =sin 60

=> A G 10 c m \frac{AG}{10 cm} = √ 3 2 \frac{√3}{2}

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

=> AG = 5√3 cm

Also, D G A D \frac{DG}{AD} = cos 60

=> D G 10 c m \frac{DG}{10cm} = 1 2 \frac{1}{2}

=> DG = 5 cm

In right triangle AGB,

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

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

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

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

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

Similarly , in triangle AGC ,

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

=> x x ^2 - (DC - DG )^2 = (5√3)^2

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

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

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

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

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

120 y y = 800

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

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

Now, BC = BD + DC

= 3 y y + y y

=4 y y

= 4 * 20 3 \frac{20}{3}

= 80 3 \frac{80}{3}

.'. Area of △ ABC = 1 2 \frac{1}{2} * AG * BC

= 1 2 \frac{1}{2} * 5√3 cm * 80 3 \frac{80}{3}

= 200 √ 3 c m 2 3 \frac{200√3 cm^2}{3}

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

.'. Ar . △ ACE = 1 2 \frac{1}{2} Area of △ ABC

= 200 √ 3 c m 2 3 ∗ 2 \frac{200√3 cm^2}{3*2}

= 100 √ 3 c m 2 3 \frac{100√3 cm^2}{3}

Now , in △ ACE and △ OFC ,,

∠ C = ∠ C -( Common )

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

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

.'. A r . △ A C E A r . △ O F C \frac{Ar. △ ACE }{ Ar . △ OFC} = C O 2 O F 2 \frac{CO^2}{OF^2} ( In similar triangles ratio of area of triangles = ratio of square of . corresponding sides of the triangle )

=> A r . △ A C E A r . △ O F C \frac{Ar. △ ACE }{ Ar . △ OFC} = C E C O \frac{CE}{CO} ^2 -----------3

We know that centroid divides median in ratio 2:1

.'. CO : OE = 2 : 1

=> OE: CO =1 : 2

=> O E C O \frac{OE}{CO} + 1 = 1 2 \frac{1}{2} +1

=> O E + C O C O \frac{OE+CO}{CO} = 3 2 \frac{3}{2}

=> C E C O \frac{CE}{CO} = 3 2 \frac{3}{2}

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

=> A r . △ A C E A r . △ O F C \frac{Ar. △ ACE }{ Ar . △ OFC} = 3 2 \frac{3}{2} ^2

=> A r . △ A C E A r . △ O F C \frac{Ar. △ ACE }{ Ar . △ OFC} = 9 4 \frac{9}{4}

=> ( 100 √ 3 / 3 ) c m 2 A r . △ O F C \frac{(100√3 /3 )cm^2 }{ Ar . △ OFC} = 9 4 \frac{9}{4}

=> Ar. △ OFC = 100 √ 3 ∗ 4 c m 2 3 ∗ 9 \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.

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