a sector of a circle radius 9cm and central angle of 120 it is rolled up so that the two bounding radii joined together to form a cone find the slant height of the cone, radius of the base of the cone, volume of the cone , and the T.S.A of the cone?
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When a sector of a circle is rolled to form a cone:
The slant height of the cone = radius of the circle = 9cm
The base of the cone forms a circle equal in length to the length of the arc
= 120/360 x 2x 22/7 x 9 = 18.86 cm
The radius of the base is found by equating the circumference of the base to 18.86cm
2x22/7 x r = 18.86
r = 18.86/2 x 7/22 = 3 cm
Height of cone is obtained by pythagoras theorem
h² = 9² - 3²
h² = 81-9 = 72
h = 8.49 cm
Volume of cone = 1/3 base area x height
= 1/3 x 22/7 x 3² x height
= 1/3 x 22/7 x 3² x 8.49
= 80 cm³
Surface area of the cone = area of sector
= 120/360 x 22/7 x 9²
= 84.86 cm²
The slant height of the cone = radius of the circle = 9cm
The base of the cone forms a circle equal in length to the length of the arc
= 120/360 x 2x 22/7 x 9 = 18.86 cm
The radius of the base is found by equating the circumference of the base to 18.86cm
2x22/7 x r = 18.86
r = 18.86/2 x 7/22 = 3 cm
Height of cone is obtained by pythagoras theorem
h² = 9² - 3²
h² = 81-9 = 72
h = 8.49 cm
Volume of cone = 1/3 base area x height
= 1/3 x 22/7 x 3² x height
= 1/3 x 22/7 x 3² x 8.49
= 80 cm³
Surface area of the cone = area of sector
= 120/360 x 22/7 x 9²
= 84.86 cm²
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