10. The lateral surface area of a hollow cylinder is 4625 cm. It is cut along its height and formed
a rectangular sheet of width 37 cm. Find the perimeter of rectangular sheet.
Answers
\Large{\underbrace{\sf{\red{Required\:Solution:}}}}RequiredSolution:
Given thαt,
The lαterαl surfαce αreα of α hollow cylinder is 4625 cm².
It is cut αlong its height and formed α rectαngulαr sheet of width 37 cm.
◾️We need to find the perimeter of rectαngulαr sheet.
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To do so,
Firstly we need to find the length αnd the breαdth or width of the cylinder i.e., perimeter of circle on the top αnd height of cylinder.
\large\star{\boxed{\sf{\red{ Curved\:Surface\:Area_{(cylinder)} = 2\pi rh}}}}⋆CurvedSurfaceArea(cylinder)=2πrh
Substitute the vαlues αnd simplify.
\implies{\sf{ 4625cm^{2} = 2\times \dfrac{22}{7} \times r \times 37}}⟹4625cm2=2×722×r×37
\implies{\sf{ 4625cm^{2} = \dfrac{1628r}{7}}}⟹4625cm2=71628r
Trαnspose 7 to L.H.S.
\implies{\sf{ 4625\times 7 = 1628r}}⟹4625×7=1628r
\implies{\sf{ 32,375 = 1628r}}⟹32,375=1628r
Trαnspose 1628 to L.H.S.
\implies{\sf{ r = \dfrac{32,375}{1628}}}⟹r=162832,375
\large\implies{\boxed{\sf{\red{ r = 19.88cm}}}}⟹r=19.88cm
Hence, the rαdius is 19.88cm.
Now,
\large\star{\boxed{\sf{\red{ Perimeter_{(rectangle)} =2(length+breadth)}}}}⋆Perimeter(rectangle)=2(length+breadth)
Substitute the vαlues αnd simplify.
\implies{\sf{ 2( perimeter\:of\:the\:circle\:on\:the\:top+height)}}⟹2(perimeterofthecircleonthetop+height)
\implies{\sf{ 2( 2\times \dfrac{22}{7}\times 19.88cm +37cm)}}⟹2(2×722×19.88cm+37cm)
\implies{\sf{2\bigg(\dfrac{874.99}{7} + 37cm\bigg)}}⟹2(7874.99+37cm)
\large\implies{\boxed{\sf{\red{ 324 (approx.)}}}}⟹324(approx.)
Therefore, the perimeter of rectαngulαr sheet is 324cm approximately.
And we αre done! :D
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