Answer
405k+ views
Hint: Here we will use the formula of CSA (Curved surface area) of the cylinder for calculating the radius of the cylinder which equals to as below:
\[{\text{Curved surface area}} = 2\pi rh\]
Where \[r\] is known as the radius of the base of the cylinder, \[h\] is the height of the cylinder.
Complete step-by-step solution:
Step 1: From the below diagram we can see that the height of the cylinder is given which is \[14{\text{ cm}}\] and \[r\] is the radius of the base of the cylinder.
\[{\text{Curved surface area}} = 88{\text{ c}}{{\text{m}}^2}\]
Step 2: Now, by using the formula of the curved surface area of the cylinder and by substituting the values of \[h = 14{\text{ cm}}\]and \[\pi = \dfrac{{22}}{7}\] , and \[{\text{Curved surface area}} = 88{\text{ c}}{{\text{m}}^2}\], we get:
\[88{\text{ c}}{{\text{m}}^2} = 2 \times \dfrac{{22}}{7} \times r \times 14{\text{ cm}}\] ……………. (1)
By dividing \[14\] with \[7\] into the RHS side of the expression (1), we get:
\[ \Rightarrow 88{\text{ c}}{{\text{m}}^2} = 2 \times 22 \times r \times 2{\text{ cm}}\]
By bringing all the terms of the RHS side into the LHS side in the expression \[88 = 2 \times 22 \times r \times 2\] , we get:
\[ \Rightarrow \dfrac{{88{\text{ c}}{{\text{m}}^2}}}{{2 \times 22 \times 2{\text{ cm}}}} = r\]
By dividing into the LHS side we get:
\[ \Rightarrow 1{\text{ cm}} = r\]
\[\therefore \] The radius of the base of the cylinder is equal to \[1{\text{ cm}}\]
Note: Students need to remember some basic formulas for finding surface areas of the cylinder because there are two types of area, one is a Curved surface area and the second is the Total surface area. The formulas for finding both is given below:
\[{\text{Curved surface area}} = 2\pi rh\]
The total surface area of the cylinder\[ = 2\pi r\left( {r + h} \right)\], where \[r\] is known as the radius of the base of the cylinder, \[h\] is the height of the cylinder.
Similarly, the formula for finding the volume of the cylinder is as below:
\[{\text{Volume of the cylinder}} = \pi {r^2}h\]
\[{\text{Curved surface area}} = 2\pi rh\]
Where \[r\] is known as the radius of the base of the cylinder, \[h\] is the height of the cylinder.
Complete step-by-step solution:
Step 1: From the below diagram we can see that the height of the cylinder is given which is \[14{\text{ cm}}\] and \[r\] is the radius of the base of the cylinder.
\[{\text{Curved surface area}} = 88{\text{ c}}{{\text{m}}^2}\]
![seo images](https://www.vedantu.com/question-sets/66c1cc61-8d35-4986-b1c7-6ba62b4ac62c6551895491319750601.png)
Step 2: Now, by using the formula of the curved surface area of the cylinder and by substituting the values of \[h = 14{\text{ cm}}\]and \[\pi = \dfrac{{22}}{7}\] , and \[{\text{Curved surface area}} = 88{\text{ c}}{{\text{m}}^2}\], we get:
\[88{\text{ c}}{{\text{m}}^2} = 2 \times \dfrac{{22}}{7} \times r \times 14{\text{ cm}}\] ……………. (1)
By dividing \[14\] with \[7\] into the RHS side of the expression (1), we get:
\[ \Rightarrow 88{\text{ c}}{{\text{m}}^2} = 2 \times 22 \times r \times 2{\text{ cm}}\]
By bringing all the terms of the RHS side into the LHS side in the expression \[88 = 2 \times 22 \times r \times 2\] , we get:
\[ \Rightarrow \dfrac{{88{\text{ c}}{{\text{m}}^2}}}{{2 \times 22 \times 2{\text{ cm}}}} = r\]
By dividing into the LHS side we get:
\[ \Rightarrow 1{\text{ cm}} = r\]
\[\therefore \] The radius of the base of the cylinder is equal to \[1{\text{ cm}}\]
Note: Students need to remember some basic formulas for finding surface areas of the cylinder because there are two types of area, one is a Curved surface area and the second is the Total surface area. The formulas for finding both is given below:
\[{\text{Curved surface area}} = 2\pi rh\]
The total surface area of the cylinder\[ = 2\pi r\left( {r + h} \right)\], where \[r\] is known as the radius of the base of the cylinder, \[h\] is the height of the cylinder.
Similarly, the formula for finding the volume of the cylinder is as below:
\[{\text{Volume of the cylinder}} = \pi {r^2}h\]
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)