A regulation baseball (hardball) has a great circle circumference of 9 inches; a regulation softball has a great circle circumference of 12 inches. a. Find the volumes of the two types of balls. b. Find the surface areas of the two types of balls?

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Answer 1 · 2018-06-13T02:35:49

$Volume of baseball V_{h} = 12.31 cub inch$ $color(blue)("Volume of baseball " V_h = 12.31 " cub inch"$

$Volume of softball V_{2} = 29.18 cub inch$ $color(blue)("Volume of softball " V_2 = 29.18 " cub inch"$

$Surface Area of baseball A_{h} = 25.78 sq inch$ $color(green)("Surface Area of baseball " A_h = 25.78 " sq inch"$

$Surface Area of softball A_{s} = 45.84 sq inch$ $color(green)("Surface Area of softball " A_s = 45.84 " sq inch"$

$Volume of Sphere V = (\frac{4}{3}) π r^{3}$ $color(crimson)("Volume of Sphere " V = (4/3) pi r^3$

$Surface Area of Sphere A_{s} = 4 π r^{2}$ $color(crimson)("Surface Area of Sphere " A_s = 4 pi r^2$

$Circumference of circle C = 2 π r$ $color(crimson)("Circumference of circle " C = 2 pi r$

$Area of circle A_{c} = π r^{2}$ $color(crimson)("Area of circle "A_c = pi r^2$

$Given : C_{h} = 2 π r_{h} = 2 π \cdot 9 inch$ $"Given : " C_h = 2 pi r_h = 2pi * 9 " inch"$

$:. r_h = 9 / 2pi$

$V_h = (4/3) pi (r_h)^3 = (4/3) * pi * (9/(2pi))^3$

$V_h = 12.31 " cub inch"$

$A_h = 4 pi( r_h)^2 = 4* pi * (9/(2pi))^2 = 25.78 " sq inch"$

$"Given : " C_s = 2 pi r_h = 2pi * 12 " inch"$

$:. r_s = 12 / 2pi$

$V_s = (4/3) pi (r_s)^3 = (4/3) * pi * (12/(2pi))^3$

$V_s = 29.18 " cub inch"$

$A_s = 4 pi( r_s)^2 = 4* pi * (12/(2pi))^2 = 45.84 " sq inch"$