[Maths Class Notes] on Surface Area of a Prism Formula Pdf for Exam

$LSA = {text{perimeter}},{text{of}},{text{base}} times h$

For a regular right prism, we have the following equations.

TSA = 2 × BA + LSA = 2 × BA + n area of one rectangular face = 2 × [frac{n}{2}] × a × b + n × h × b =nb(a + h)

The following diagram shows the case of a triangular prism.

 

 

$TSA = 2 times BA + LSA = 2 times frac{1}{2} times b times h + l times left( {a + b + c} right) = bh + lleft( {a + b + c} right)$ 

In the case of a regular triangular prism, $a = b = c$. Therefore, we have the following formula for its surface area.

$TSA = bh + 3lb = bleft( {h + 3l} right)$ 

Let’s look at an example to see how to apply these formulae.

 

Question: The two bases of a triangular prism are equilateral triangles of side length 4 cm. If its length is 6 cm, then find its total surface area.

Solution:

$n = 3,,,b = 4,cm,,,h = 6,cm$ 

$BA = frac{{sqrt 3 }}{4}{b^2} = frac{{sqrt 3 }}{4}{4^2} = 4sqrt 3 ,c{m^2}$ 

 

TSA = 2 × BA + n × area of one rectangular face = 2 × 4[sqrt{3}] + 3 × 4 × 6 = 72 + 8[sqrt{3}] = 8(9 + [sqrt{3}]) [cm^{2}]

 

Why don’t you try using these formulae to solve the following problem?

 

Question: A regular hexagonal prism is 12 cm long and the perimeter of its hexagonal base is 30 cm. Find its total surface area.

 

Options:

(a) $15left( {12 + 15sqrt 3 } right),c{m^2}$ 

(b) $15left( {24 + 5sqrt 3 } right),c{m^2}$

(c) $12left( {25 + 5sqrt 3 } right),c{m^2}$

(d) $10left( {24 + 5sqrt 3 } right),c{m^2}$

 

Answer: (b)

 

Solution:

$n = 6,,,h = 12,cm,,,perimeter,of,base = 30,cm$ 

${text{Perimeter}},{text{of}},{text{base}} = n times b Rightarrow 6 times b = 30 Rightarrow b = 5,cm$ 

$BA = frac{{3sqrt 3 }}{2}{b^2} = frac{{3sqrt 3 times {5^2}}}{2} = frac{{75sqrt 3 }}{2},c{m^2}$ 

$LSA = {text{perimeter}},{text{of}},{text{base}} times h = 30 times 12 = 360,c{m^2}$ 

[TSA = 2 times BA + LSA = 2 times frac{{75sqrt 3 }}{2} + 360 = 360 + 75sqrt 3 = 15left( {24 + 5sqrt 3 } right),c{m^2}]