Triangle KLB

It is given equilateral triangle ABC. From point L, the midpoint of the side BC of the triangle, it is drawn perpendicular to the side AB. The intersection of the perpendicular and the side AB is point K. How many percents of the area of the triangle ABC is an area of a triangle KLB?

Correct answer:

p =  12.5 %

Step-by-step explanation:

a=1 S1=3/4 a2=3/4 120.433 LB=a/2=1/2=21=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=12.5%



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