Q. 74.2( 5 Votes )

# On a circular tab

Answer :

radius of the circle = 30 cm

Area of the circle = π r^{2}

= 3.14 × 30^{2}

= 2826 cm^{2}

PQR is an equilateral triangle

∠ QPR = 60^{O}

∠ QOR = 2 × ∠ QPR = 120^{o}

In OQR triangle,

let OS is perpendicular to QR

In triangle OQR as OQ = OR

∠ QOS = ∠ QPR = 60^{O}

In triangle OSQ ∠S = 90^{o}

sin 60^{o} =

=

QS = 15 √3 cm

Now,

QR = 2 QS = 2 (15 √3) = 30√3 cm

Area of equilateral triangle = × QR^{2}

= × (30√3)^{2}

= 1167.75 CM^{2}

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

= 2826 – 1167.75

= 1658.25 CM^{2}

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