Question

Please solve this geometry (circumcircle)
I will mark as Brainlist​

Answer 1

Answer:

1: each side of eq. ∆ ABC = 6 cm

let's draw an altitude AD on side BC.

now area of ∆ ABC = √3/4(side)²

= √3/4(6)²

= √3/4*36

= 9√3 cm²

now area of ∆ ABD = 1/2 area of ∆ ABC

=> 1/2*b*h = 9√3/2

=> 1/2*3*h = 9√3/2

=> h = (9√3/2)*(2/3)

=> h = 3√3

so, altitude of ∆ = 3√3 cm

alternatively :

By using Pythagoras theorem,

AB² = BD² + AD²

6² = 3² + AD²

AD² = 36 - 9

AD² = 27

AD = 3√3

2:

Polygon with 3 equal sides is called an

equilateral Triangle .

Let ABC is an equilateral triangle whose

side length is 5.2 cm

AB = BC = CA = 5.2 cm

Steps to Construction :

1 ) Draw AB = 5.2 cm line segment.

2 ) Take A and B are centers with equal

radius 5.2 cm draw two intersecting arcs.

C is the intersecting point.

3 ) Join A and B to C .

Therefore ,

Required equilateral ∆ABC formed.