# Find the Area of the Triangle Inside Three Circles

## Geometry Math Problem: Find the area of the triangle

From figure, We have three co-centric circle with center *O*, here *OP *= 1 cm, *OQ *= 1 cm and* OR *=* *3* *cm. ABC is a triangle, *AB* and* AC* are tangents of the smallest circle. Then what is the area of the blue triangle?

## Solution

From this figure

P is the midpoint of chord *AB* because *OP* and *AB* are perpendicular to each other so, *AP *= *BP*

*AS *=* AP *(Tangent of the circle from Point *A*)

Now we can connect *AE* and *AG* so we can form two triangles

From triangle *BOP *

*OB *= 2 cm

*OP *= 1 cm

*BP² *=* OB*² –* OP*² = 2² – 1² = 4 – 1 = 3

⇒ *BP *= √3 cm

From triangle *OCS*

*OC* = 3* *cm

*OS *= 1* *cm

*CS*²* *=* OC*² – *OS*²* *

⇒ *CS*² = 3² – 1² = 9 – 1 = 8

so, *CS* = 2√2 cm

Then From* ∆ABC*

*AB *= *AP + BP *

⇒ AB = √3 +√3 = 2√3 cm

*AC *=* AS *+* CS*

⇒ AC = √3 + 2√2* *cm

Connect *B* and* T*

Here we form a new triangle* ABT* with inner circle radius = 1 cm and outer circle radius *= *2* *cm, that is, *Outer circle Radius *=* *2 × *Inner circle Radius *so we can understand *∆ ABT *is an equilateral triangle

or,* AB, BT,* and *AT* are chords of circle *ABT*. These chords have the same distance from the centre so* AB *= *BT *= *AT* so we can say *∆ ABT *is an equilateral triangle

That is *∠F *= *60°*

Then Area of Triangle = *½ *×* FE *×* FG *× sin* F* = ½ × 2√3 × (√3 + 2√2) × (√3/2)

⇒ **Area of Triangle = 3√2 + ½ × 3√3 cm²**