Question

A hollow cylinder having external radius 8 cm and height 10 cm has a total surface area of 338 pai cm square. Find the thickness of the hollow metalic cylinder​

Answer 1

$<b>$External radius = R = 8 cm

internal radius = r

Height = 10 cm

TSA = 338 π cm²

TSA = Area of external cylinder + Area of internal cylinder + twice Area of ring

= 2πRh + 2πrh + 2π(R² - r²)

2π(Rh + rh + R² - r²) = 138 π

Rh + rh + R² - r² =169

(10 × 8)+(r × 10) + 8² - r² = 169

r² - 10r + 25 = 0

(r - 5)² = 0

r = 5

Thickness of metal = R-r = (8 - 5) cm = 3cm

Answer 2

Solution:-

Given:-

External Radius = 8cm.

Height = 10cm.

Total Surface Area = 338π cm²

Let Internal Radius be "x" cm.

TSA = Outer Surface Area + Inner Surface Area + 2(Area of thickness of Cylinder).

=> 2πrh + 2πRh + 2π(R² -r²)

=> 338π = 20πr + 160π + 128π - 2πr²

=> 2πr² - 20πr + 50π =0

=> r² -10r + 25 =0

=>(r -5)²=0

=> r =5 cm

Hence,

Area of Thickness = Outer Radius - Inner Radius

=> Area of Thickness = 8 - 5

=> Area of Thickness = 3cm.

Hence,

Area of Thickness of Metallic Cylinder is 3cm.