Question

If total surface area of a cylinder is 94.2 cm sq and it's height is 5 cm then find radius of it's base

Answer 1

Answer

The radius of its base is 2.11 cm (approx)

Explanation

Total surface area of a cylinder is given as

2πrh + 2πr² or 2πr(r + h)

Given, height (h) = 5 cm and T.S.A = 94.2 cm²

To find, the radius of the base

Let the base radius be r

Now,

2πr(h + r) = T.S.A

This gives,

2πr(5 + r) = 94.2 cm²

⇒ r(5 + r) = 94.2/2π

⇒ 5r + r² = 47.1/π

Taking π as 3.14, we get

⇒ 5r + r² = 47.1/3.14

⇒ 5r + r² = 15

⇒ r² + 5r - 15 = 0

Using Quadratic formula,

$\sf r = \frac{-5\pm \sqrt{5^2 -4(1)(-15)}}{2}\\\\r = \frac{-5\pm\sqrt{25+60}}{2}\\\\r = \frac{-5\pm\sqrt{85}}{2}$

Now, since r can not be negative since it's a length, we will neglect the negative value and thus, r equals

$\sf r = \frac{-5+\sqrt{85}}{2}\\\\r \approx 2.11\ cm$

Answer 2

Given :-

• TSA of cylinder = 94.2cm²
• Height of Cylinder = 5cm .

To Find :-

• Radius of Base .

Solution :-

we know That, TSA of cylinder is 2πr(h+r) . Lets Assume That, radius of Base is r cm.

Putting values :-

→ 2 * 3.14 * r * (5 + r) = 94.2

→ 2r(5 + r) = 30

→ r(5 + r) = 15

→ r² + 5r - 15 = 0

____________________

Now,

Sridharacharya formula for Solving Quadratic Equation ax² +bx + c = 0 says that,

→ x = [ -b±√(b²-4ac) / 2a ]

or,

→ x = [ - b± √D /2a ] where D(Discriminant)= b²-4ac.

_____________________

Comparing r² + 5r - 15 = 0 with ax² + bx + c = 0, we get,

→ a = 1

→ b = 5

→ c = (-15)

So ,

→ D = b² - 4ac

→ D = (5)² - 4*1*(-15)

→ D = 25 + 60

→ D = 85

So ,

→ r = [ - b± √D /2a ]

→ r = [ -5 ± √85 / 2 ]

{ putting √85 = 9.2 (Approx) } ,

→ r = { -5 - 9.2 } /2 or, { - 5 + 9.2 } /2

→ r = (-7.1cm) or, (2.1cm). { Negative value ≠ }.

ஃ Radius of Base of cylinder is 2.1cm (Approx).