Question

ABC forms a equilateral triangle whose side length is a cms ,D is another point which equidistant from A,B,C which is 2cms.Find side a of ABC?

Answer 1

Answer:

Given : △ABC is equilateral triangle of side 2

3

cms.

Let P be any interior point of △ABC such that x,y,z are the distances of P from the sides of a triangle.

Now, A(△ABC)=A(△APB)+A(△APC)+A(△BPC)

=

2

1

×PE×AB+

2

1

×PG×AC+

2

1

×PF×BC

=

2

1

×PE×2

3

+

2

1

×PG×2

3

+

2

1

×PF×2

3

=(x+y+z)

3

∴

4

3

×(2

3

)

2

=

3

(x+y+z)

∴ x+y+z=3cms