Q. 174.0( 15 Votes )

# A solid cone of b

Answer :

Let ‘H’ be the height of cone ‘R’ be the radius of base of cone.

R = 10 cm

When the cone is cut at midpoint of H by a plane parallel to its base we get a cone of height H/2 and a frustum also of height H/2

Let the radius of the base of the cone which we got after cutting and the radius of upper circle of frustum be ‘r’ as shown in figure

From figure consider ΔABC and ΔADE

∠ABC = ∠ADE = 90˚

∠BAC = ∠DAE …(common angle)

as two angles are equal by AA criteria we can say that

ΔABC∼ΔADE

∴ r = 5 cm

Let V_{c} be volume of the cone and V_{f} be the volume of frustum

Volume of cone = (1/3)π(radius)^{2}(height) cm^{3}

∴ V_{c} = (1/3) × π × r^{2} × (H/2) cm^{3}

= (1/3) × π × 5^{2} × (H/2) cm^{3}

= (1/3) × π × 25 × (H/2) cm^{3}

Volume of frustum = (1/3)πh(R^{2} + r^{2} + Rr) cm^{3}

∴ V_{f} = (1/3) × π × (H/2)(10^{2} + 5^{2} + 10 × 5) cm^{3}

= (1/3) × π × (H/2) × 175 cm^{3}

∴ The ratio of volumes of two parts after cutting = V_{c}:V_{f} = 1:7

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