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A semicircle is subtended on hypotenuse of right angled isosceles triangle with side 2 units. Two sector of circles are drawn with center on the semicircle, one end at of side at end of semicircle and one common side. Both sector subtended with equal angles. Find sum of areas of sector of circles.
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(Shortcut Image Solution with Calculations is gives below in this post)
Step 1:
Give names to all vertices.
Step 2:
AC = BC = 2;
∠ACB = 90°;
⇒ AB = √8;
Let radius of sector of circles are m, n.
Step 3:
∠ADB = 90°;
⇒ AD² + BD² = AB²;
⇒ m² + n² = 8;
Also,
∠ADB = 90°;
⇒ ∠ADF + ∠BDE = 360 - 90 = 270°;
But ∠ADF = ∠BDE;
⇒ ∠ADF = ∠BDE = ²⁷⁰⁄₂ = 135°;
⇒ Sum of areas = ⅜π(m² + n²) = ⅜π(8) = 3π
Image Solution:
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