# Please solve the NCERT question No. 6 Exercise-13.5 (p-258)

### Asked by Mdivya | 12th Mar, 2009, 11:23: PM

Expert Answer:

--------------------- +P --------------------

| /|\ |

| / | \ |

| / | \ S-s |H-h

| / | \ |

| / | \ |

| /*****|**r**\ |

| ** C+-----*+D--------------

H| S / *********** \ |

| / | \ |

| / | \ |

| / | \ s |

| / | \ |h

| / | \ |

| / | \ |

| / **********|********** \ |

| /***** | R *****\ |

-----* A+----------------+B----

****** ******

*********************

We know R, r, and h, but not H, the total height of the cone from

which the frustum was cut. If we can find it, then the volume of the

frustum will be the volume of the whole cone, pi R^2 H/3, minus the

volume of the cone we cut off the top, pi r^2 (H-h)/3.

The triangles PAB and PCD are similar, so we can write the equation

AB CD R r

-- = -- or - = ---

PA PC H H-h

Cross-multiplying [that is, multiplying both sides by H(H-h)], we get

R(H-h) = rH

We can distribute the left side and collect H terms, then divide:

RH - Rh = rH

RH - rH = Rh

(R-r)H = Rh

Rh

H = ---

R-r

Now let's write the volume formula and substitute this formula for H:

pi pi

V = -- R^2 H - -- r^2 (H-h)

3 3

pi

= -- (R^2 H - r^2 H + r^2 h)

3

pi

= -- [(R^2 - r^2) H + r^2 h]

3

pi Rh

= -- [(R^2 - r^2) --- + r^2 h]

3 R-r

pi R

= -- [(R^2 - r^2) --- + r^2] h

3 R-r

We can write R^2 - r^2 as (R - r)(R + r) and cancel:

pi

= --- [(R + r) R + r^2] h

3

pi

= --- [R^2 + Rr + r^2] h

3

That's the formula.

Now let's work on the lateral surface area. The formula for a complete

cone is:

A = pi R S

where R is the radius and S is the slant height of the whole cone. For

the frustum, we will subtract the area of the cut-off cone (whose

slant height is S-s) from the whole:

A = pi R S - pi r (S-s)

= pi (RS - rS + rs)

= pi ((R-r)S + rs)

By the same similar triangles as before, we can write

AB CD R r

-- = -- or - = ---

PB PD S S-s

Again solving for S,

R(S-s) = rS

RS - Rs = rS

RS - rS = Rs

(R-r)S = Rs

Rs

S = ---

R-r

Now the area is

Rs

A = pi ((R-r)--- + rs) = pi (Rs + rs) = pi(R+r)s

R-r

| /|\ |

| / | \ |

| / | \ S-s |H-h

| / | \ |

| / | \ |

| /*****|**r**\ |

| ** C+-----*+D--------------

H| S / *********** \ |

| / | \ |

| / | \ |

| / | \ s |

| / | \ |h

| / | \ |

| / | \ |

| / **********|********** \ |

| /***** | R *****\ |

-----* A+----------------+B----

****** ******

*********************

We know R, r, and h, but not H, the total height of the cone from

which the frustum was cut. If we can find it, then the volume of the

frustum will be the volume of the whole cone, pi R^2 H/3, minus the

volume of the cone we cut off the top, pi r^2 (H-h)/3.

The triangles PAB and PCD are similar, so we can write the equation

AB CD R r

-- = -- or - = ---

PA PC H H-h

Cross-multiplying [that is, multiplying both sides by H(H-h)], we get

R(H-h) = rH

We can distribute the left side and collect H terms, then divide:

RH - Rh = rH

RH - rH = Rh

(R-r)H = Rh

Rh

H = ---

R-r

Now let's write the volume formula and substitute this formula for H:

pi pi

V = -- R^2 H - -- r^2 (H-h)

3 3

pi

= -- (R^2 H - r^2 H + r^2 h)

3

pi

= -- [(R^2 - r^2) H + r^2 h]

3

pi Rh

= -- [(R^2 - r^2) --- + r^2 h]

3 R-r

pi R

= -- [(R^2 - r^2) --- + r^2] h

3 R-r

We can write R^2 - r^2 as (R - r)(R + r) and cancel:

pi

= --- [(R + r) R + r^2] h

3

pi

= --- [R^2 + Rr + r^2] h

3

That's the formula.

Now let's work on the lateral surface area. The formula for a complete

cone is:

A = pi R S

where R is the radius and S is the slant height of the whole cone. For

the frustum, we will subtract the area of the cut-off cone (whose

slant height is S-s) from the whole:

A = pi R S - pi r (S-s)

= pi (RS - rS + rs)

= pi ((R-r)S + rs)

By the same similar triangles as before, we can write

AB CD R r

-- = -- or - = ---

PB PD S S-s

Again solving for S,

R(S-s) = rS

RS - Rs = rS

RS - rS = Rs

(R-r)S = Rs

Rs

S = ---

R-r

Now the area is

Rs

A = pi ((R-r)--- + rs) = pi (Rs + rs) = pi(R+r)s

R-r

### Answered by | 13th Mar, 2009, 01:55: AM

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