A conical flask of height 24 cm and curved surface area is 550 CM square. find the volume.
Curved Surface area of a Cone = pi r L, where r is the radius of the base & L is its slant height.
Height(h) of the cone = 24cm.
So, in right triangle slant height L= √( h² + r²) = √(24² + r²)
= √(576+r²)
Pi r L = 550
=> pi * r * (√(24² + r²) = 550
=> pi² * r² * ( 24² + r²) = 550² ( On squaring)
=> 576pi²r² + pi²r^4 = 550²
=> pi² r^4 + 576pi² r² - 550² =0
Let r² be x
=> pi²x² + 576pi²x - 550² =0
=> 22/7 *22/7 *x² + 576*22/7*22/7 -2²*5²*5²*11²= 0 Now divide each term by 11²
=> 4/49x² + 576*4/49 *x - 2500=0
= 4x² + 2304x - 122500 =0 ( by multiplying each term by 49)
=> x² + 576x - 30625 = 0
=> x = (-576 +,-√454276)/2
x = (-576 + 674)/2 ( -ve value is ruled out)
=> x = 98/2 = 49
=> r² = 49
=> r = 7
So, Volume of Cone = 1/3* pi* r² * h
=> Volm = 1/3 * 22/7 * 7*7*24
= Volm = 22*7*8
=> 1232 cu cm
Curved Surface area of a Cone = pi r L, where r is the radius of the base & L is its slant height.
Height(h) of the cone = 24cm.
So, in right triangle slant height L= √( h² + r²) = √(24² + r²)
= √(576+r²)
Pi r L = 550
=> pi * r * (√(24² + r²) = 550
=> pi² * r² * ( 24² + r²) = 550² ( On squaring)
=> 576pi²r² + pi²r^4 = 550²
=> pi² r^4 + 576pi² r² - 550² =0
Let r² be x
=> pi²x² + 576pi²x - 550² =0
=> 22/7 *22/7 *x² + 576*22/7*22/7 -2²*5²*5²*11²= 0 Now divide each term by 11²
=> 4/49x² + 576*4/49 *x - 2500=0
= 4x² + 2304x - 122500 =0 ( by multiplying each term by 49)
=> x² + 576x - 30625 = 0
=> x = (-576 +,-√454276)/2
x = (-576 + 674)/2 ( -ve value is ruled out)
=> x = 98/2 = 49
=> r² = 49
=> r = 7
So, Volume of Cone = 1/3* pi* r² * h
=> Volm = 1/3 * 22/7 * 7*7*24
= Volm = 22*7*8
=> 1232 cu cm
