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Question

# Question 19 Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90∘ at the centre.

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Solution

## Let the radius of the circle be r. ∴ OA=OB = r cm Given that, length of chord of a circle, AB = 5cm And central angle of the sector AOBA(θ)=90∘ Now, in ΔAOB (AB)2=(OA)2+(OB)2 [ by Pythagoras theorem] (5)2=r2+r2 ⇒2r2=25 ∴ r=5√2cm [Since, the perpendicular drawn from the centre to the chord of a circle divides the chord into two equal parts] By Pythagoras theorem, in ΔADO (OA)2=OD2+AD2 ⇒OD2=OA2−AD2 =(5√2)2−(52)2=252−254 =50−254=254 ⇒OD=52cm ∴Area of an isosceles ΔAOB = \frac{1}{2}\times Base (=AB)\times Height(=OD)\) =12×5×52=254cm2 Now, area of sector AOBA =πr2360∘×θ=π×(5√2)2360∘×90∘ =π×252×4=25π8cm2 ∴ Area of minor segment = Area of sector AOBA - Area of an isosceles ΔAOB =(25π8−254)cm2 Now, area of the circle =πr2=π(5√2)=25π2cm2 ∴ Area of major segment = Area of circle - Area of minor segment =25π2−(25π8−254) =25π8(4−1)+254 =(75π8+254)cm2 ∴ Difference of the areas of two segments of a circle = [Area of minor segment] =∣∣(75π8+254)−(25π4−254)∣∣ =∣∣(75π8+254)−(25π8−254)∣∣ =∣∣75π−25π8+504∣∣=∣∣50π8+504∣∣ =(25π4+252)cm2 Hence, the required difference of the areas of two segments is (25π4+252)cm2

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