Question 62.
In a square ABCD, 2 quadrants of a circle are drawn and named as ACD and BCD intersecting at point O. Find the sum of the areas of region AOD and region BOC.
Answer 62.
Let A and D be left vertices of square ABCD of side length "a".
Let DC be x-axis and DA y-axis with D as origin.
Eqn. of circle with center D and radius "a" is
x^2 + y^2 = a^2
By symmetry, x-coordinate of O is a/2
=> Area DOC
= 2 ∫(a/2 to a) √(a^2 - x^2) dx
= [x√(a^2 - x^2) + a^2 *arcsin(x/a)] (x=a/2 to a)
= [0 + a^2 * π/2] - [(a/2)*(√3a/2) + a^2 * (π/6)]
= (π/3)a^2 - a^2(√3/4)
=> Required area
= 2 * [area of quarter-circle - area DOC]
= (π/2)a^2 - 2 * [(π/3)a^2 - (√3/4)a^2]
= (√3/2 - π/6) a^2.
LINK to YA!
In a square ABCD, 2 quadrants of a circle are drawn and named as ACD and BCD intersecting at point O. Find the sum of the areas of region AOD and region BOC.
Answer 62.
Let A and D be left vertices of square ABCD of side length "a".
Let DC be x-axis and DA y-axis with D as origin.
Eqn. of circle with center D and radius "a" is
x^2 + y^2 = a^2
By symmetry, x-coordinate of O is a/2
=> Area DOC
= 2 ∫(a/2 to a) √(a^2 - x^2) dx
= [x√(a^2 - x^2) + a^2 *arcsin(x/a)] (x=a/2 to a)
= [0 + a^2 * π/2] - [(a/2)*(√3a/2) + a^2 * (π/6)]
= (π/3)a^2 - a^2(√3/4)
=> Required area
= 2 * [area of quarter-circle - area DOC]
= (π/2)a^2 - 2 * [(π/3)a^2 - (√3/4)a^2]
= (√3/2 - π/6) a^2.
LINK to YA!
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