In the 1st method, I found the radius of semi-circle by joining O and E. Then using Pythagoras Theorem in the triangle OEC (since the angle OCE is 90°). Cheers

Almost always if it is solvable by intersecting chord, it is solvable by Pythagoras as well.

I love the 2° method but the 1° is also beautiful.

A method 3: At 6:15, drop a perpendicular from E to OB. Label the intersection with CP as G and with OB as H. Note that EH and CP are perpendicular and, therefore, EPOH is a rectangle, so OP = GH and PG = OH.. ΔEGC is an isosceles right triangle with hypotenuse r, so length EG = EC = r/(√2). EH = EG + GH, but GH = OP = r, so EH = r + r/(√2) = r(1 + 1/(√2)). PG = CP - EC = r - r/(√2) = r(1 - 1/(√2)). However, PG = OH so OH = r(1 - 1/(√2)), Construct EO and note that it is a radius of the quarter circle, so has length 3. Consider right ΔEHO. Applying the Pythagorean theorem, (OH)² + (EH)² = (3)², (r(1 - 1/(√2))² + ( r(1 + 1/(√2)))² = 9, r²(1 - 2/√2 + 1/2) + r²(1 + 2/√2 + 1/2) = 9, 3r² = 9 and r² = 3. Compute area of semicircle from r² as at 14:20.

(3/2)π ?

interseco le equazioni x^2+y^2=9,y=-x+p..e trovo i 2 punti la cui distanza è 2r...e risulta 36-2q^2=4r^2,ma la retta y=-x+q passa per C(r,r),per cui q=2r...quindi 36-8r^2=4r^2...r^2=3..Ared=3π/2

Please also solve question of class10 or 11.

Solution: R = radius of the quarter circle = 3, r = radius of the semicircle, M = center of the quarter circle, N = center of the semicircle, A = Point of contact with the semicircle at the top left, B = Point of contact of the semicircle at the bottom right, C = corner of the semicircle on the right, Because of the symmetry: Pythagoras: MN² = AN²+BN² = r²+r² = 2r² And another Pythagoras: MN²+NA² = MA² ⟹ 2r²+r² = R² ⟹ 3r² = 3² |/3 ⟹ r² = 3 ⟹ Area of the red semicircle = 3π/2 ≈ 4,7124

Generally r=R/sqrt(3).😊