Question

a cone of radius 10 cm is cut into two part of a plan through the mindpiont. Of its vertical axis parallel to the base find the ratio of the volume of the smaller cone to the frustum of the cone​

Answer 1

AnswEr :

$\:\bullet\:\sf\ Radius \: of \: big \: cone = 10cm$

$\:\bullet\:\sf\ Height \: of \: big \: cone = H$

$\rule{100}{1}$

$\:\bullet$ It is given that cone is divided through a plane, at midpoint which shows that big cone is divided into a "small cone" and a "frustum".

$\:\bullet$Also, the cone id divided from midpoint, so height of small cone and frustum will be equal and radius of small cone is half of big cone.

$\underline{\dag\:\textsf{Let's \: head \: to \: question \: now:}}$

$\blacksquare\underline{\textsf{ Volume \: of \: frustum \: formed :}}$

$\:\bullet\:\sf\ Height \: of \: frustum(h)= \frac{H}{2}$

$\:\bullet\:\sf\ Radius \: of \: upper \: end(r_1) = 5cm$

$\:\bullet\:\sf\ Radius \: of \: lower \: end(r_2)= 10cm$

$\rule{100}{1}$

$\normalsize\star{\boxed{\sf{Volume \: of \: frustum = \dfrac{1}{3} \pi h[r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}] }}}$

$\normalsize\ : \implies\sf\ V_{f} = \frac{1}{3} \times\ \pi \times\ \frac{H}{2} \times\ [(10)^2 + (5)^2 + 10 \times\ 5] \\ \\ \normalsize\ : \implies\sf\ V_{f} = \frac{1}{3} \times\ \pi \times\ \frac{H}{2} \times\ [100 + 25 + 50] \\ \\ \normalsize\ : \implies\sf\ V_{f} = \frac{1}{3} \times\ \pi \times\ \frac{H}{2} \times\ [175]$

$\normalsize\ : \implies{\underline{\boxed{\sf \green{ V_{f} = \frac{1}{3} \times\ \pi \times\ \frac{H}{2} \times\ [175] }}}}$

$\rule{100}2$

$\blacksquare\underline{\textsf{ Volume \: of \: small \: cone \: formed :}}$

$\:\bullet\:\sf\ Height \: of \: cone(h)= \frac{H}{2}$

$\:\bullet\:\sf\ Radius \: of \: small \: cone \: end(r) = 5cm$

$\rule{100}{1}$

$\normalsize\star{\boxed{\sf{Volume \: of \: cone_{s} = \dfrac{1}{3} \pi r^2h }}}$

$\normalsize\ : \implies\sf\ V_{c} = \frac{1}{3} \times\ \pi \times\ 5 \times\ \frac{H}{2} \\ \\ \normalsize\ : \implies\sf\ V_{c} = \frac{1}{3} \times\ \pi \times\ 25 \times\ \frac{H}{2}$

$\normalsize\ : \implies{\underline{\boxed{\sf \green{V_{c} = \frac{1}{3} \times\ \pi \times\ 25 \times\ \frac{H}{2} }}}}$

$\rule{100}2$

$\blacksquare\underline{\textsf{ Ratio \: of \: volumes \: of \: smaller \: cone \: and \: frustum :}}$

$\normalsize\star{\boxed{\sf{Ratio \: of \: volumes = Cone_{volume} : Frustum_{volume} }}}$

$\normalsize\ : \implies\sf\ R_{v} = \frac{Volume \: of \: cone}{Volume \: of \: frustum} \\ \\ \normalsize\ : \implies\sf\ R_{v} = \frac{\cancel{\frac{1}{3}} \times\ \cancel{\pi} \times\ 25 \times\ \cancel{\frac{H}{2}} }{\cancel{\frac{1}{3}} \times\ \cancel{\pi} \times\ \cancel{\frac{H}{2}} \times\ 175} \\ \\ \normalsize\ : \implies\sf\ R_{v} = \frac{\cancel{25}}{\cancel{175}} \\ \\ \normalsize\ : \implies\sf\ R_{v} = \frac{1}{7} \\ \\ \normalsize\ : \implies\sf\ R_{v} = 1 : 7$

$\normalsize\ : \implies{\underline{\boxed{\sf \purple{Ratio \: of \: volumes \: of \: smaller \: cone \: and \: frustum \: = 1 :7}}}}$

Answer 2

Answer:

★ Ratio will be 1:7 ★

Step-by-step explanation:

Given:

• Radius (BC) of bigger cone is 10 cm
• Height of bigger cone is OB (2h)

To Find:

• Ratio of volume of smaller cone to the frustum of the cone

Solution: When we cut the cone into two parts by the plane through the mid point of its vertical axis parallel to its base then we will get a smaller cone and a frustum. ( See figure)

→ 'r' = AD is the radius of base of the smaller cone ←

→ 'h' = OA is the height of smaller cone = AB = Height of the frustum = 1/2 of OB

→ '2h' = OB ( OA = OB)

Now, In ∆OAD & ∆OBC we have,

→ ∠A = ∠A = 90°

→ ∠OAD = ∠OBC [ Corresponding angles because AD is parallel to BC ]

∴ ∆OAD ~ ∆ OBC by AA similarity

∵The corresponding sides of two similar triangles are proportional to each other

∴ OA / OB = AD / BC

→ h / 2h = r / R

→ r / R = 1 / 2

→ r = 1 / 2 x R

→ r = 1 / 2 x 10 .....( Radius is 10 cm)

→ r = 5 cm

† Volume of cone = 1/3 π(r)²h

∴ Volume of smaller cone = 1/3π(5)²h............(1)

∴ Volume of bigger cone = 1/3π(10)²2h.........(2)

† Volume of frustum will be = Volume of bigger cone – volume of smaller cone

∴ Volume of frustum = 1/3π(10)²2h – 1/3π(5)²h

$\small\implies{\sf }$ 1/3πh ( 200 – 25 )

$\small\implies{\sf }$ 1/3πh x 175 .........(3)

★ From equation (1) and (3) we will get

The ratio of the volume of the smaller cone to the volume of the frustum ★

→ Volume of smaller cone / Volume of frustum

$\small\implies{\sf }$ 1/3π(5)²h / 1/3πh(175)

→ After calculation and cancelling all similar terms, we will get

$\small\implies{\sf }$ 25 / 175

$\small\implies{\sf }$ 1 / 7

Hence, The ratio of the volume of the smaller cone to the frustum of the cone will be 1:7