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 S1=3/4 a2=3/4 120.433 LB=a/2=1/2=12=0.5 KL=LB sin60=sinπ/3=0.43301 KB=LB cos60=cosπ/3=0.25 S2=KL KB/2=0.433 0.25/20.0541 p=100 S2/S1=100 0.0541/0.433=252=12.5=12.5%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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Our percentage calculator will help you quickly calculate various typical tasks with percentages.
Pythagorean theorem is the base for the right triangle calculator.
See also our trigonometric triangle calculator.

 

 

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