Math, asked by bodarohith1432, 5 months ago

There are 8 different lines, 7 different circles and 6 different equilateral triangles, then maximum number of points on ntersections of


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Answered by amitnrw
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Given : 8 different lines, 7 different circles and 6 different equilateral triangles,  

To Find :  maximum number of points of intersections  

Solution:

8 Lines

7 circles

6 Equilateral Triangles

8 Lines can intersect with each other in ⁸C₂ = 28  

1 Line can intersect a circle at 2 points hence  8 x 7 x 2 = 112  

A line can intersect an equilateral triangle at 2 points = 8 x 6 x 2 = 96  

A Circle  can intersect another circle at 2 point  =  2 x ⁷C₂ = 42  

A triangle can intersect another triangle at 6 points = 6 x ⁶C₂ = 90

A triangle and circle can intersect at 6 points => 6 x 6 x 7 = 252

Maximum point of intersections = 28 + 112 + 96 + 42 + 90 + 252

= 620

maximum number of points on intersections  = 620

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