Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3:1 then the ratio of their

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Let h be the height, l the slant height and r1 and r2 the radii of the circular bases of the frustum ABB’ A’ shown in Fig. such that r1 > r2.
Let the height of the cone VAB be h1 and its slant height be i.e., VO = h1 and VA = VB = l1
∴ VA’ = VA – AA’ = l1– l
and VO’ = VO – OO’ = h1– hHere, ΔVOA ~ ΔVO‘A’

Now,Height of the cone VA‘B’

Let S denote the curved surface area of the frustum of cone. Then,S = Lateral (curved) surface area of cone VAB

- Curved surface area of cone VA‘B’

Curved surface area of the frustum
= π(r1 + r2)lTotal surface area of the frustum= Lateral (curved) surface area+ Surface area of circular bases

= π (r1 + r2) I + πr12 + πr22
= π {(r1 + r2) l + r12 + r22}.

Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Find the ratio of their volumes. 

Given that, let height →h say 

Height of `1^(st)` cone = h 

Height of `2^(nd)`cone = 3h 

Let the ratio of radii be r 

∴ Radius of `1^(st)`  cone=3r  

Radius of` 2 ^(nd)` cone = r 

∴ ratio of volume =` V_1/V_2` 

⇒ `V_1/V_2=(1/3pir_1^2h_1)/(1/3pir_2^2h_2)=(r_1^2h_1)/(r_2^2h_2)` 

=`((3r)^2xxh)/(r^2xx3h)`

`=(9r^2h)/(3r^2h)` 

= 3/1 

⇒  `v_1/v^2=3/1`

Concept: Surface Area of a Right Circular Cone

  Is there an error in this question or solution?

>

Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3:1.Show that their volumes are in the ratio 3 : 1.

Solution

Let the heights be 1x and 3x respectively and their radii be 3y and 1y .

ratio=volume of cone1 volume of cone 2=13π(3y)2x13π(1y)23x=9y2×xy2×3x=31

Mathematics

Secondary School Mathematics IX

Standard IX