Question

Find the area of the shaded region ??​

Answer 1

Answer:

the area of the shaded region ≈ 51.3 cm²

Step-by-step explanation:

**In triangle ABC:

∵ circle (A), circle (b) circle (d) are equidistant and internally tangential to the same circle

∵ triangle ABC are joining between the circles centers

∴ triangle ABC is equilateral

∵ the half of (BC) =2 cm

∴ BC = 4 cm

∵ triangle ABC is equilateral

∴ AB = BC = CA = 4 cm

*let the triangle's hight be named (AM)

∵ AM = √(( MC² + AC² )

∴ AM = √( 2² +4² ) = 2√5

∵ triangle (ABC ) Area = 1/2 base x hight

∴ triangle (ABC ) Area = 1/2 BC x AM

∴ triangle (ABC ) Area = 1/2 x 4 x 2√5

∴ triangle (ABC ) Area = 4√5 cm² --------------> (1)

**In triangle ABC:

*let the mid. of (AB) = D , mid. of (BC) = E , mid. of (AC) = F

∵ circle (A), circle (B) circle (C) are equidistant and internally tangential to the same circle

∵ triangle ABC are joining between the circles centers

∴ the circles cut triangle (ABC ) at its mid points D, E, F

∵ triangle ABC is equilateral

∴ AD = DB = BE = EC = CF = FA = 2 cm

**In triangle DEF :

∵ the lines DE, EF, FD join between two mid points of triangle ABC

∵ triangle ABC is equilateral

∴ DE = EF = FD = 1/2 base of triangle ABC

∴ triangle DEF is equilateral

∴ DE = EF = FD = 1/2 x4

∴ DE = EF = FD = 2 cm

*let the triangle's hight be named (EO)

∵ EO = √( OE² + OF² )

∴ EO = √( 2² +1² ) = √5

∵ triangle ( DEF ) Area = 1/2 base x hight

∴ triangle ( DEF ) Area = 1/2 DE x EO

∴ triangle ( DEF ) Area = 1/2 x 2 x √5

∴ triangle ( DEF ) Area = √5 cm² --------------> (2)

**In Circle A :

∵ the radius = 2 cm

∵ Area of circle A = π r²

∴ Area of circle A = π 2²

∴ Area of circle A = 4π cm²

∵ circle (A), circle (B) circle (C) are equidistant and internally tangential to the same circle

∴ Area of circle A = Area of circle B = Area of circle C = 4π cm² --------------> (3)

**In Circle O ( the circum circle = the biggest circle ) :

∵ circle (A), circle (B) circle (C) are equidistant and internally tangential to the circle O

∴ Circle O's radius = any circle's diameter + X

∵ X = the distance between the triangle ( DEF )'s center and triangle ( DEF )'s base

∵ triangle DEF is equilateral

∴ X = 1/2 DEF's hight

∴ X = 1/2 X √5

∴ X = ( √5 ) / 2

∴ Circle O's radius = 4 +( ( √5 ) / 2 )

∵ Area of circle O = π ( 4 +( ( √5 ) / 2 ) )²

∴ Area of circle O ≈ 26.19 π cm² --------------> (4)

** The area of the shaded region :

∵ the area of the shaded region = Area of circle O - ( Area of circle A + Area of circle B + Area of circle C ) + Area of triangle ABC - Area of triangle ( DEF )

from (1), (2), (3) and (4) we get that :

the area of the shaded region = 26.19 π - ( 4π + 4π + 4π ) + 4√5 -√5

∴ the area of the shaded region ≈ 51.3 cm²

# sorry for the long answer