Question

Q9. A cone is cut by a plane parallel to the base ans upper part is removed. If the C.S.A. of the remainder is 15/16 of the C.S.A. of whole cone, find the ratio of the line segments to which the cone's height is divided by the plane.​

Answer 1

$\large\underline{\sf{Solution-}}$

↝ A cone is cut by a plane parallel to the base and upper part is removed.

↝ It is given that the C.S.A. of the remainder is 15/16 of the C.S.A. of whole cone

So,

$\rm :\implies\:CSA_{(small cone)} = \dfrac{1}{16}CSA_{(big cone)} - - (i)$

Let us suppose that

↝ Dimensions of small cone be :-

• Radius of small cone = r units

• Height of small cone = h units

• Slant height of small cone = l units

and

↝ Dimensions of big cone :-

• Radius of big cone = R units

• Height of big cone = H units

• Slant height of big cone = L units

Now,

Consider,

• In triangle AOB and triangle ACD,

$\rm :\longmapsto\: \angle \: AOB = \angle \:ACD \: \: \{ \: each \: 90 \degree \}$

$\rm :\longmapsto\: \angle \: ABO = \angle \:ADC \: \: \{ \:corresponding \: angles \}$

$\rm :\implies\: \triangle \: AOB \: \sim \: \triangle \:ACD \: \: \: \: \: \{AA \: similarity \}$

$\rm :\implies\:\dfrac{r}{R} = \dfrac{l}{L} = \dfrac{h}{H} - - - (1)$

We know,

↝ Curved Surface Area of cone is

$\boxed{ \red{ \bf \: CSA_{(cone)} = \: \pi \: rl}}$

where,

• r is radius of cone

• l is Slant height of cone.

According to equation (i), we have

$\rm :\implies\:CSA_{(small cone)} = \dfrac{1}{16}CSA_{(big cone)}$

$\rm :\longmapsto\: \cancel\pi \: rl = \dfrac{1}{16} \cancel\pi \: RL$

$\rm :\longmapsto\:\dfrac{r}{R} \times \dfrac{l}{L} = \dfrac{1}{16}$

$\rm :\longmapsto\:\dfrac{h}{H} \times \dfrac{h}{H} = \dfrac{1}{16} \: \: \: \: \: \: \: \: \{ \: using \: (1) \}$

$\rm :\longmapsto\: {\bigg(\dfrac{h}{H} \bigg) }^{2} = {\bigg(\dfrac{1}{4} \bigg) }^{2}$

$\bf\implies \:\dfrac{h}{H} = \dfrac{1}{4}$

$\bf\implies \:H = 4h$

Now,

↝ We have to find the ratio of

$\rm :\longmapsto\:\dfrac{Height_{(small cone)}}{Height_{(remaining \: solid)}} = \dfrac{h}{H - h}$

$\rm :\longmapsto\:\dfrac{Height_{(small cone)}}{Height_{(remaining \: solid)}} = \dfrac{h}{4h - h}$

$\rm :\longmapsto\:\dfrac{Height_{(small cone)}}{Height_{(remaining \: solid)}} = \dfrac{h}{3h}$

$\rm :\longmapsto\:\dfrac{Height_{(small cone)}}{Height_{(remaining \: solid)}} = \dfrac{1}{3}$

Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

Answer 2

Figure:-

Given:-

• CSA of frustum = 15/16 CSA big cone.

To find:-

• find the ratio of the line segments to which the cone's height is divided by the plane.?

Solution:-

• CSA of big cone = πRL
• CSA of small cone = πrl

CSA of frustum = πRL - πrl

• h/H = r/R = l/L = 1/K
• h/(H - h)

We know that,

=> πRL - πrl = 15/16 πRL

=> 1 - πrl/πRL = 15/16

=> 1 - rl/RL = 15/16

=> -rl/RL = 15/16 - 1

=> r/l × R/L = 1 - 15/16

=> r/l × R/L = 1/16

=> 1/k × 1/k = 1/16⠀ [h/H = r/R = l/L = 1/K]

=> 1/k² = 1/16

=> 1/k = 1/4

Now

h/(H - h)

=> 1/(4 - 1)

=> 1/3

Hence, the ratio of the line segments to the cone's height is 1/3 in the plane.

Some Important:-

• Diameter of a Circle D = 2 × r
• Circumference of a Circle C = 2 × π × r
• Area of a Circle A = π × r2

Where,

• r denotes the radius of the circle.
• d indicates the diameter of the circle.
• c indicates circumference of the circle.

Curved Surface Area and Total Surface Area of the Frustum

• The curved surface area of the frustum of the cone = π(R+r)l1

• The total surface area of the frustum of the cone = π l1 (R+r) +πR2 +πr2

• The slant height (l1) in both the cases shall be = √[H2 +(R-r)2]