# 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

**Solution:**

A frustum of a cone with h as height, l as the slant height, r_{1} and r_{2} radii of the ends where r_{1} > r_{2}

To Prove:

(i) CSA of the frustum of the cone = πl (r_{1} + r_{2})

(ii) TSA of the frustum of the cone = πl (r_{1} + r_{2}) + πr_{1}² + πr_{2}²

where r_{1}, r_{2}, h and l are the radii height and slant height of the frustum of the cone respectively.

Construction:-Extended side BC and AD of the frustum of cone to meet at O

Proof:

The frustum of a cone can be viewed as a difference of two right circular cones OAB and OCD. Let h_{1} and l_{1} be the height and slant height of cone OAB and h2 and l2 be the height and slant height of cone OCD respectively.

In ΔAPO and ΔDQO

ΔAPO = ΔDQO = 90° (Since both cones are right circular cones)

∠AOP = ∠DOQ (Common)

Therefore, ΔAPO ∼ ΔDQO (AA criterion of similarity)

AP/DQ = AO/DO = OP/OQ (Corresponding sides of similar triangles are proportional)

⇒ r_{1}/r_{2} = l_{1}/l_{2} = h_{1}/h_{2}

⇒ r_{1}/r_{2} = l_{1}/l_{2} or ⇒ r_{2}/r_{1} = l_{2}/l_{1}

Subtracting 1 from both sides we get

r_{1 }/ r_{2 }- 1 = l_{1 }/ l_{2 }- 1

(r_{1 - }r_{2})/r_{2} = (l_{1 - }l_{2}) / l_{2}

(r_{1} - r_{2})/r_{2} = l/l_{2}

l_{2} = l r_{2}/(r_{1} - r_{2}) (i)

or ⇒ r_{2 }/ r_{1} = l_{2 }/ l_{1}

Subtracting 1 from both sides we get

r_{2 }/ r_{1} - 1 = l_{2 }/ l_{1} - 1

(r_{2} - r_{1}) / r_{1} = (l2 - l1)l_{1}

(r_{1} - r_{2}) / r_{1} = (l_{1} - l_{2})l_{1}

(r_{1} - r_{2}) / r_{1 }= l_{ }/ l_{1}

l_{1} =lr_{1}/(r_{1} - r_{2}) (ii)

(i) CSA of frustum of cone = CSA of cone OAB - CSA of cone OCD

= πr_{1}l_{1} - πr_{2}l_{2}

= (r_{1}l_{1} - r_{2}l_{2})

= π [(r_{1} × lr_{1}/(r_{1} - r_{2}) - r_{2} × lr_{2}/(r_{1} - r_{2})] [Using (i) and (ii)]

= π [(lr_{1}² - lr_{2}²)/(r_{1} - r_{2})]

= π [l(r_{1}² - r_{2}²)/(r_{1} - r_{2})]

= π [l(r_{1} - r_{2})(r_{1} + r_{2})/(r_{1} - r_{2})] [a² - b² = (a - b)(a + b)]

= l (r_{1} + r_{2})

TSA of frustum of cone = CSA of frustum + Area of lower circular end + Area of top circular end

= πl (r_{1} + r_{2}) + πr_{1}² + πr_{1}²

Therefore, CSA of the frustum of the cone = πl (r_{1} + r_{2})

TSA of the frustum of the cone = πl (r_{1}+ r_{2}) + πr_{1}² + πr_{2}²

Hence Proved.

**Video Solution:**

## 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

### NCERT Solutions for Class 10 Maths - Chapter 13 Exercise 13.5 Question 6 :

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

The formula for the curved surface area and total surface area of the frustum of a cone are πl(r1+r2) and πl(r1+r2)+πr1^2+πr2^2 respectively where r1,r2, h and l are the radii height and slant height of the frustum of the cone respectively