Question

The TSA of a hollow cylinder open at both ends of external radius 8 cm and height 10 cm is 338π cm². If 'r' is the inner radius then write an equation in 'r' and use it to find the thickness of the Cylinder.

Answer 1

Here in this question,

TSA = 338 pi.

Now,

=> T.S.A = CSA of outer Cylinder + C.S.A of Inner Cylinder + 2 x Area of base ring.

$= > 338 \pi = 2 \pi \times r_{1} \times h + 2 \pi \times r_{2} \times h + 2 \pi( r_{1}^{2} - r_{2}^{2} )$

$= > 338 \pi = 2 \pi( r_{1}h + r_{2}h + ( r_{1}^{2} - r_{2}^{2} ))$

$= > \frac{338 \pi}{2 \pi} = (8 \times 10 + 10 \times r_{2} + ( {8}^{2} - r_{2}^{2} ))$

$= > 169 = 80 + 10 r_{2} + 64 - r_{2}^{2}$

$= > 169 - 144 - 10 r_{2} + r_{2}^{2} = 0$

$= > 25 - 10 r_{2} + r_{2}^{2} = 0$

$= > r_{2}^{2} - 10 r_{2} + 25 = 0$

$= > r_{2}^{2} - 5 r_{2} - 5 r_{2} + 25 = 0$

$= > ( r_{2} - 5)( r_{2} - 5) = 0$

$= > r_{2} - 5 = 0$

$= > r_{2} = 5$

Now,

Thickness of Cylinder = External Radius -Internal Radius

$= > t = r_{1} - r_{2}$

$= > t =( 8 - 5) \: cm \: = 3 \: cm \:$


Be Brainly

Answer 2

Answer:

SA of the hollow cylinder = 338π cm²

↠ 2π(8)10 + 2π(r)10 + 2[π(8)² – π(r)²] = 338π

↠ 160 + 20r + 2(64 – r²) = 338

↠ – 2r² + 20r – 50 = 0

↠ r² – 10r + 25 = 0

↠ (r – 5)² = 0

↠ r = 5

∴ Thickness of the metal = 8 – r

= 8 – 5 = 3 cm