Q7 of 36 Page 253

On a circular table top of radius 30 cm a design is formed leaving an equilateral triangle inscribed in a circle. Find the area of the design. (π = 3.14)

radius of the circle = 30 cm


Area of the circle = π r2



= 3.14 × 302


= 2826 cm2


PQR is an equilateral triangle


∠ QPR = 60O


∠ QOR = 2 × ∠ QPR = 120o


In OQR triangle,


let OS is perpendicular to QR


In triangle OQR as OQ = OR


∠ QOS = ∠ QPR = 60O


In triangle OSQ ∠S = 90o


sin 60o =


=


QS = 15 √3 cm


Now,


QR = 2 QS = 2 (15 √3) = 30√3 cm


Area of equilateral triangle = × QR2


= × (30√3)2


= 1167.75 CM2


Area of design = Area of circle - area of triangle PQR


= 2826 – 1167.75


= 1658.25 CM2


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figure 13.25


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