# Triangle KLB

It is given equilateral triangle ABC. From point L which is the midpoint of the side BC of the triangle it is drwn perpendicular to the side AB. Intersection of perpendicular and the side AB is point K. How many % of the area of the triangle ABC is area of a triangle KLB?

Result

p =  12.5 %

#### Solution:

$a=1 \ \\ S_{1}=\sqrt{ 3 }/4 \cdot \ a^2=\sqrt{ 3 }/4 \cdot \ 1^2 \doteq 0.433 \ \\ LB=a/2=1/2=\dfrac{ 1 }{ 2 }=0.5 \ \\ KL=LB \cdot \ \sin 60 ^\circ =\sin π/3=0.43301 \ \\ KB=LB \cdot \ \cos 60 ^\circ =\cos π/3=0.25 \ \\ S_{2}=KL \cdot \ KB / 2=0.433 \cdot \ 0.25 / 2 \doteq 0.0541 \ \\ p=100 \cdot \ S_{2}/S_{1}=100 \cdot \ 0.0541/0.433=\dfrac{ 25 }{ 2 }=12.5=12.5 \%$

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