Find the lateral surface area and total surface area of an equilateral triangle-based prism with height 8 cm and base 4 cm.
Answer
Verified
478.5k+ views
Hint: First of all, draw the diagram of the equilateral triangle-based prism to have a clear idea of what we have to find. Then use the formula for finding L.S.A and T.S.A of the equilateral triangle-based prism.
Complete step-by-step solution:
Given that the base of the prism is an equilateral triangle with side 4 cm as shown in the given figure:
Let \[p\] be the perimeter of the base of the prism i.e., the perimeter of equilateral \[\Delta ABC\].
So, \[p = 4 + 4 + 4 = 12{\text{ cm}}\]
Given height of the prism \[h = 8\;{\text{cm}}\]
We know that the lateral surface area of the triangle-based prism is given by \[p \times h\] where \[p\] is the perimeter of the base triangle and \[h\] is the height of the prism.
Therefore, L.S.A of the prism \[ = p \times h = 12 \times 8 = 96{\text{ c}}{{\text{m}}^2}\]
Now, consider the area of equilateral \[\Delta ABC\].
We know that the area of the equilateral triangle with side \[a\] is given by \[\dfrac{{{a^2}}}{4}\].
So, area of \[\Delta ABC = \dfrac{{\sqrt 3 {{\left( 4 \right)}^2}}}{4} = 4\sqrt 3 \]
We know that the total surface area of the triangle based prism with base area \[A\], perimeter \[p\] and \[h\] is the height of the prism is given by \[2 \times A + p \times h\].
Therefore, the total surface area of the prism \[ = 2 \times 4\sqrt 3 + 12 \times 8\]
\[
= 8\sqrt 3 + 96 \\
= 8\left( {1.73} \right) + 96 \\
= 13.84 + 96 \\
= 109.84{\text{ c}}{{\text{m}}^2} \]
Thus, the L.S.A of the prism is \[96{\text{ c}}{{\text{m}}^2}\] and the T.S.A of the prism is \[{\text{109}}{\text{.84 c}}{{\text{m}}^2}\].
Note: The lateral surface area of the triangle-based prism is given by \[p \times h\] where \[p\] is the perimeter of the base triangle and \[h\] is the height of the prism. The total surface area of the triangle based prism with base area \[A\], perimeter \[p\] and \[h\] is the height of the prism is given by \[2 \times A + p \times h\].
Complete step-by-step solution:
Given that the base of the prism is an equilateral triangle with side 4 cm as shown in the given figure:
Let \[p\] be the perimeter of the base of the prism i.e., the perimeter of equilateral \[\Delta ABC\].
So, \[p = 4 + 4 + 4 = 12{\text{ cm}}\]
Given height of the prism \[h = 8\;{\text{cm}}\]
We know that the lateral surface area of the triangle-based prism is given by \[p \times h\] where \[p\] is the perimeter of the base triangle and \[h\] is the height of the prism.
Therefore, L.S.A of the prism \[ = p \times h = 12 \times 8 = 96{\text{ c}}{{\text{m}}^2}\]
Now, consider the area of equilateral \[\Delta ABC\].
We know that the area of the equilateral triangle with side \[a\] is given by \[\dfrac{{{a^2}}}{4}\].
So, area of \[\Delta ABC = \dfrac{{\sqrt 3 {{\left( 4 \right)}^2}}}{4} = 4\sqrt 3 \]
We know that the total surface area of the triangle based prism with base area \[A\], perimeter \[p\] and \[h\] is the height of the prism is given by \[2 \times A + p \times h\].
Therefore, the total surface area of the prism \[ = 2 \times 4\sqrt 3 + 12 \times 8\]
\[
= 8\sqrt 3 + 96 \\
= 8\left( {1.73} \right) + 96 \\
= 13.84 + 96 \\
= 109.84{\text{ c}}{{\text{m}}^2} \]
Thus, the L.S.A of the prism is \[96{\text{ c}}{{\text{m}}^2}\] and the T.S.A of the prism is \[{\text{109}}{\text{.84 c}}{{\text{m}}^2}\].
Note: The lateral surface area of the triangle-based prism is given by \[p \times h\] where \[p\] is the perimeter of the base triangle and \[h\] is the height of the prism. The total surface area of the triangle based prism with base area \[A\], perimeter \[p\] and \[h\] is the height of the prism is given by \[2 \times A + p \times h\].
Recently Updated Pages
A uniform rod of length l and mass m is free to rotate class 10 physics CBSE
Solve the following pairs of linear equations by elimination class 10 maths CBSE
What could be the possible ones digits of the square class 10 maths CBSE
Where was the Great Bath found A Harappa B Mohenjodaro class 10 social science CBSE
PQ is a tangent to a circle with centre O at the point class 10 maths CBSE
The measures of two adjacent sides of a parallelogram class 10 maths CBSE
Trending doubts
Imagine that you have the opportunity to interview class 10 english CBSE
Find the area of the minor segment of a circle of radius class 10 maths CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Frogs can live both on land and in water name the adaptations class 10 biology CBSE
Fill in the blank One of the students absent yesterday class 10 english CBSE
Write a letter to the Principal of your school requesting class 10 english CBSE