Point A can also be slid down to point O. The area of the small quarter circle then becomes 0 and the diameter of the semicircle is then the same as the radius of the large quarter circle. The area of the large quarter circle minus the area of the semicircle is 4π - 2π = 2π.
Limit condition: r₁=R=2cm. ; r₂=0 A = ½πR² = 2π cm² (Solved √ ) With these modifications, the calculation is easier, and original conditions are meeting Other limit condition: r₁=0 , r₂=½R , R=4cm A = ¼πR² - ½πr₂² A = ¼π4² - ½π2² A = 4π - 2π = 2π cm² ( Solved √ )
Answer 2pi Let the radius of the quarter circle to the LEFT =n , Hence, the area = n^2 pi /4 Hence, the radius of the quarter to the RIGHT = 4-n , Hence, the area = ( n^2 + 16- 8n pi) /4 Hence, the diameter of the semi-circle on TOP = (4-n-n) or 4- 2n) Hence its radius is 2-n, Hence, the area of the semi-circle on Top = (n^2+4- 4n + 4pi)/2 Since the other two have a denominator of 4, Let's change this also to 4 by multiplying the numerator and denominator by 2 (2n^2 + 8-8n + 8pi)/4. Since the area of the shaded region is the Large quarter circle - the Large semi-circle + the small quarter circle, then ( n*2+ 16 -8n pi) /4 - ( 2n^2+8 -8n + 8pi)/4 + n^2 pi/4 pi/4 ( n^2 + 16-8n - 2n^2 -8 + 8n - 8pi + n^2 factor out pi/4 pi/4 ( 2n^2- 2n^2 -8n+ 8n+ 16-8) pi/4 ( 0 + 16-8) pi/4 ( 8) 8/4 pi 2 pi
Cómo no se da el valor de los radios del cuarto de círculo pequeño y del semicírculo, podemos desplazar verticalmente el punto de tangencia entre ambos hasta que el radio del semicírculo se anule y la zona roja se transforme en un semicírculo de diámetro 4 ud→ r=4/2=2→ Área roja =2²π/2=2π ud². Gracias y saludos.
Nice. I would emphasize the fact that the area is independent of r. But can we know it before making some calculations ?
Point A can also be slid down to point O. The area of the small quarter circle then becomes 0 and the diameter of the semicircle is then the same as the radius of the large quarter circle. The area of the large quarter circle minus the area of the semicircle is 4π - 2π = 2π.
Limit condition: r₁=R=2cm. ; r₂=0
A = ½πR² = 2π cm² (Solved √ )
With these modifications, the calculation is easier, and original conditions are meeting
Other limit condition:
r₁=0 , r₂=½R , R=4cm
A = ¼πR² - ½πr₂²
A = ¼π4² - ½π2²
A = 4π - 2π = 2π cm² ( Solved √ )
Answer 2pi
Let the radius of the quarter circle to the LEFT =n , Hence, the area = n^2 pi /4
Hence, the radius of the quarter to the RIGHT = 4-n , Hence, the area = ( n^2 + 16- 8n pi) /4
Hence, the diameter of the semi-circle on TOP = (4-n-n) or 4- 2n)
Hence its radius is 2-n, Hence, the area of the semi-circle on Top = (n^2+4- 4n + 4pi)/2
Since the other two have a denominator of 4, Let's change this also to 4 by multiplying the numerator
and denominator by 2 (2n^2 + 8-8n + 8pi)/4.
Since the area of the shaded region
is the Large quarter circle - the Large semi-circle + the small quarter circle, then
( n*2+ 16 -8n pi) /4 - ( 2n^2+8 -8n + 8pi)/4 + n^2 pi/4
pi/4 ( n^2 + 16-8n - 2n^2 -8 + 8n - 8pi + n^2 factor out pi/4
pi/4 ( 2n^2- 2n^2 -8n+ 8n+ 16-8)
pi/4 ( 0 + 16-8)
pi/4 ( 8)
8/4 pi
2 pi
calculated in 4 seconds.
Cómo no se da el valor de los radios del cuarto de círculo pequeño y del semicírculo, podemos desplazar verticalmente el punto de tangencia entre ambos hasta que el radio del semicírculo se anule y la zona roja se transforme en un semicírculo de diámetro 4 ud→ r=4/2=2→ Área roja =2²π/2=2π ud².
Gracias y saludos.
Answer 2pi
Background music is super annoying