Question

the lateral surface area of a hollow cylinder is 4224 sq cm. it is cut along its height and formed a rectangular sheet of width 48 cm.find the perimeter of the rectangular sheet.​

Answer 1

Given data : The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 48 cm.

To find : The perimeter of the rectangular sheet ?

Solution :

⟹ lateral surface area of the cylinder

= 4224 cm²

⟹ width/breadth of rectangular sheet

= 48 cm

Now according to given we know that, the lateral surface area of the cylinder is equal to area of the rectangular sheet.

⟹ lateral surface area of the cylinder = area of the rectangular sheet

⟹ 4224 = length * breadth

⟹ 4224 = length * 48

⟹ length = 4224/48

⟹ length = 88 cm

Now,

⟹ Perimeter of rectangular sheet

= 2 * (length * breadth)

⟹ Perimeter of rectangular sheet

= 2 * (88 * 48)

⟹ Perimeter of rectangular sheet

= 2 * 136

⟹ Perimeter of rectangular sheet

= 272 cm

Answer : Hence, perimeter of the rectangular sheet is 272 cm.

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Answer 2

Step-by-step explanation:

${\large{\bold{\rm{\underline{Let's\ understand\ the\ concept\ 1^{st}:-}}}}}$

★ This question says that the lateral surface area of a hollow cylinder is 4224cm². It is cut along its height and formed a rectangular sheet of width 48cm. As through the given data we know that the length of the rectangular sheet is not given. Also, the lateral surface area of the cylinder is equal to the area of the rectangular sheet, so through this data we will find out the length of the rectangular sheet. One more concept is used that is perimeter of rectangle. Since, the breadth of the rectangular sheet is already given and after having the length we easily find out the perimeter of the rectangular sheet. So let's do....!!!

${\large{\bold{\rm{\underline{Given\ that:-}}}}}$

$\sf {\bigstar Lateral\ surface\ area\ of\ hollow\ cylinder\ =\ 4424cm^2.}$

$\sf {\bigstar Breadth\ of\ rectangular\ sheet\ =\ 48cm.}$

${\large{\bold{\rm{\underline{To\ find:-}}}}}$

★ In this question we have to find the perimeter of the rectangular sheet ?

${\large{\bold{\rm{\underline{Using\ formula:-}}}}}$

$\star\;\underline{\boxed{\bf{\orange{Perimeter\ of\ rectangle\ =\ 2(length\ +\ breadth).}}}}$

▪︎Where,

• Perimeter = ?
• Length = ?
• Breadth = 48cm.

${\large{\bold{\rm{\underline{Solution:-}}}}}$

★ The perimeter of the rectangular sheet is 272cm.

${\large{\bold{\rm{\underline{Full\ solution:-}}}}}$

$\sf : \implies {Lateral\ surface\ area\ of\ cylinder\ =\ Area\ of\ rectangular\ sheet}$

$\sf : \implies {4224\ =\ length \times breadth}$

$\sf : \implies {4224\ =\ length \times 48}$

$\sf : \implies {\cancel \dfrac{4224}{48}\ =\ length}$

$\sf : \implies {88\ =\ length}$

$\bf : \implies {\blue{Length\ =\ 88cm.}}$

∴ Hence, the length of the rectangular sheet is 88cm.

~Since, we have to find out the perimeter of the rectangular sheet. As the breadth of the rectangular sheet is already given and now we also have the length, that is 88cm. So, by using the formula of perimeter of rectangle we will find out the perimeter of the rectangular sheet.

$\sf : \implies {Perimeter\ of\ rectangular\ sheet\ =\ 2(length\ +\ breadth)}$

$\sf : \implies {Perimeter\ of\ rectangular\ sheet\ =\ 2(88 \times 48)}$

$\sf : \implies {Perimeter\ of\ rectangular\ sheet\ =\ 2(136)}$

$\sf : \implies {Perimeter\ of\ rectangular\ sheet\ =\ 2 \times 136}$

$\bf : \implies {\green{Perimeter\ of\ rectangular\ sheet\ =\ 272cm.}}$

∴ Hence, the perimeter of the rectangular sheet is 272cm.

${\large{\bold{\bf{\underline{\underline{More\ to\ know:-}}}}}}$

● Diagram of rectangle:-

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large x cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large y cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

$\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto {Area\ of\ rectangle\ =\ Length \times Breadth.}$

$\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto {Perimeter\ of\ rectangle\ =\ 2(Length\ +\ Breadth).}$

● Diagram of cylinder:-

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{r}}\put(9,17.5){\sf{h}}\end{picture}$

$\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto {C.S.A\ of\ cylinder\ =\ 2{\pi}rh.}$

$\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto {T.S.A\ of\ cylinder\ =\ 2{\pi}r(r\ +\ h).}$

$\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \leadsto {Volume\ of\ cylinder\ =\ {\pi}r^2h.}$