Radius of the circle = R = 5 c m , R=5cm, R = 5 c m , angle of the sector = θ = 216 ° . \theta=216\degree. θ = 216°.
Length of the circular arc cut off from the circle
L = 2 π R × θ 360 ° = 2 π ( 5 ) 216 ° 360 ° = 6 π ( c m ) L=2\pi R\times{\theta \over 360\degree}=2\pi(5){216\degree \over 360\degree}=6\pi(cm) L = 2 π R × 360° θ = 2 π ( 5 ) 360° 216° = 6 π ( c m ) When the sector is cut and its bounding radii is bent to form a cone, slant height of the cone, l = R = 5 c m . l=R=5cm. l = R = 5 c m .
Let r r r and h h h be the radius and height of the cone formed.
Circumference of the base of the cone = 2 π r = 6 π . =2\pi r=6\pi. = 2 π r = 6 π .
r = 3 c m . r=3cm. r = 3 c m . The base radius is 3 c m . 3cm. 3 c m .
Height of the cone
h = l 2 − r 2 = 5 2 − 3 2 = 4 ( c m ) h=\sqrt{l^2-r^2}=\sqrt{5^2-3^2}=4(cm) h = l 2 − r 2 = 5 2 − 3 2 = 4 ( c m ) Let φ \varphi φ be a vertical angle. Then
sin ( φ / 2 ) = r l = 3 5 = 0.6 \sin(\varphi/2)={r \over l}={3\over 5}=0.6 sin ( φ /2 ) = l r = 5 3 = 0.6
φ = 2 arcsin ( 0.6 ) ≈ 73.74 ° \varphi=2\arcsin(0.6)\approx73.74\degree φ = 2 arcsin ( 0.6 ) ≈ 73.74° Find the curved surface area
π r l = π ( 3 ) ( 5 ) = 15 π ( c m 2 ) ≈ 47.12 ( c m 2 ) \pi rl=\pi(3)(5)=15\pi (cm^2)\approx47.12(cm^2) π r l = π ( 3 ) ( 5 ) = 15 π ( c m 2 ) ≈ 47.12 ( c m 2 )
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