
A formula for the surface area (A) of the rectangular solid shown below is A = 2lw + 2lh + 2wh where l represents length; w, width; and h, height. By doubling each of the dimensions (l, w, and h), the surface area will be multiplied by what factor?
A). 2
B). 4
C). 6
D). 8
Answer
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Hint: In these types of questions use the given information to form the new equation and compare to the given equation and also remember to use the new l, w, and h as $2l$, $2w$, and $2h$, using this information can help you to reach towards the solution of the problem.
Complete step-by-step answer:
According to the given information we have rectangular solid whose surface area is $A = 2lw + 2lh + 2wh$
Now taking $A = 2lw + 2lh + 2wh$ as equation 1
Now we know that all the dimensions of the given rectangular solid are doubled
Therefore, the new dimensions are length = $2l$, width = $2w$, and height = $2h$
We know that the formula of surface area of rectangular solid is given by $A = 2 (wl + hl + hw)$ here w is the width of the rectangular solid, l is the length, and h is the height of the rectangular solid
Substituting the given values in the formula of the surface area of rectangular solid we get
$A’ = 2[ (2l) (2w) + (2l) (2h) + (2w) (2h)]$
\[ \Rightarrow \] $A’ = 2[ 4lw + 4lh + 4wh]$
\[ \Rightarrow \] $A’ = 4[ 2lw + 2lh + 2wh]$ taking this equation as equation 2
Comparing equation 1 and equation 2 we get
\[ \Rightarrow \] $A’ = 4[A]$
Therefore, we can say that the new surface area of the rectangular solid is 4 times the initial area of the rectangular solid.
Hence option B is the correct option.
Note: In the above solution we came across the term “rectangular solid” which can be defined as a three dimensional shape or an object which consists of 6 faces here all the angles formed in the rectangular solid are right angled the reason it is named as a rectangular is due to the shape of all the faces of the rectangular solid are rectangular in shape
Complete step-by-step answer:
According to the given information we have rectangular solid whose surface area is $A = 2lw + 2lh + 2wh$
Now taking $A = 2lw + 2lh + 2wh$ as equation 1
Now we know that all the dimensions of the given rectangular solid are doubled
Therefore, the new dimensions are length = $2l$, width = $2w$, and height = $2h$
We know that the formula of surface area of rectangular solid is given by $A = 2 (wl + hl + hw)$ here w is the width of the rectangular solid, l is the length, and h is the height of the rectangular solid
Substituting the given values in the formula of the surface area of rectangular solid we get
$A’ = 2[ (2l) (2w) + (2l) (2h) + (2w) (2h)]$
\[ \Rightarrow \] $A’ = 2[ 4lw + 4lh + 4wh]$
\[ \Rightarrow \] $A’ = 4[ 2lw + 2lh + 2wh]$ taking this equation as equation 2
Comparing equation 1 and equation 2 we get
\[ \Rightarrow \] $A’ = 4[A]$
Therefore, we can say that the new surface area of the rectangular solid is 4 times the initial area of the rectangular solid.
Hence option B is the correct option.
Note: In the above solution we came across the term “rectangular solid” which can be defined as a three dimensional shape or an object which consists of 6 faces here all the angles formed in the rectangular solid are right angled the reason it is named as a rectangular is due to the shape of all the faces of the rectangular solid are rectangular in shape
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