Question

please find the area of the shaded region and show calculation:

Answer 1

Construction:-

Draw seg PO, OQ and OR which are

perpendicular to AB, BC and AC. PQR are

the midpoint of of AB, BC and AC.

Draw seg BO. such that ∠PBO = ∠QBO

and ∠POB = ∠QOB.

Given:-

BC = 5.8cm

Solution:-

By construction

BQ = QC .......(1)

BP = AP ........(2)

AR = CR ........(3)

BC = BQ + QC

5.8 = BQ + BQ .....{BQ = QC}

5.8 = 2BQ

BQ = 2.9 cm .....(4)

now,

By Tangent Segment theorem

•°• BP = BQ ....(5)

•°• PA = AR .....(6)

•°• RC = QC .....(7)

From 1, 2, 3, 5, 6, 7

BQ = QC = BP = AP = AR = CR

•°• AB = BC = AC .....(8)

FROM THIS IT PROVES THAT △ABC IS

EQUILATERAL TRIANGLE.

•°• ∠ABC = ∠ACB = ∠BAC = 60° ........(9).........{angles of equilateral △ is 60°}

•°• ∠PBO + ∠QBO = 60°

•°• ∠PBO = ∠QBO = 60/2 = 30° ...... (10)....{BY construction}

now,

Area of equilateral △ ABC = √(3)/4 × side²

•°• A(△ABC) = √(3)/4 × BC²

= √(3)/4 × (5.8)²

= √(3)/4 × 33.64

= 8.41√(3)

= 8.41 × 1.73

= 14.5493

•°• A(△ABC) = 14.54 cm² ......(11)

now,

□ POQB is kite ........(°•°BQ = BP and PO = OQ)

sum of all angles of Quadrilateral is 360°

•°• ∠PBQ + ∠POQ +∠BPO + ∠BAO = 360°

•°• 60° + ∠POQ + 90° + 90° = 360° ....{from 9 and Tangent Theorem}

•°• ∠POQ = 360° - 240°

•°• ∠POQ = 120° ......(12)

•°• ∠POB +∠QOB = 120

•°• ∠POB = ∠QOB = 120/2 = 60° .......(13)....{ By construction}

now,

We will Find the radius of circle.

In △OQB is right angled Triangle

as ∠OQB = 90° .......{Tangent theorem}

∠QBO = 30° ....... {FROM 10}

∠QOB = 60° .......{From 11}

tan30° = OQ/BQ

1/√(3) = r/2.9

r = 2.9/√(3)

r = 2.9/√(3) cm ......(14)

now,

Area of circle = πr²

= 22/7 × (2.9/√(3))²

= 22/7 × 8.41/3

= 22/7 × 2.80

= 61.6/7

= 8.8 cm²

•°• Area of circle = 8.8cm² ........(15)

now,

Area of Shaded region = Area of triangle - Area of circle

= 14.54 - 8.8

= 5.74 cm²

Area of Shaded region = 5.74cm²

Note: { Answer is approximate due to Decimal. √(3) = 1.73}