A point O is taken inside an equilateral ΔABC. If OM⊥AC, OL⊥BC and ON⊥AB such that OL = 14 cm, OM = 10 cm and ON = 6 cm, find the area of ΔABC.
300√3 cm2
Let each side of ΔABC be 'a' cm.
Then Area (ΔABC)=Area (ΔOBC)+Area (ΔAOC)+Area (ΔAOB)
We know area of a triangle with base 'b' and height 'h' id given by '12×base×height'
So, Area (ΔABC)=12a(OL+OM+ON)
=12a(14+10+6)
=15a
Given, ΔABC is equailateral.
Area of an equilateral triangle with side 'a' is given by √34a2.
⇒√34a2=15a
⇒a=15×4√3
⇒a=60×√33=20√3
∴Area (ΔABC)=15×20√3=300√3 cm2