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Expert-verified Found in: Page 478 ### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279 # Refer to rectangular solids with dimensions l, w and h. Complete the following table. $\begin{array}{cc}l& 6\\ w& 3\\ h& \\ \text{L.A.}& \\ \text{T.A.}& \\ V& 54\end{array}$

$\begin{array}{cc}l& 6\\ w& 3\\ h& 3\\ \text{L.A.}& 54\\ \text{T.A.}& 90\\ V& 54\end{array}$

See the step by step solution

## Step 1. Given information.

The length and width of a rectangle are 6 and 3 respectively. And the volume is 54.

## Step 2. Write the concept.

The lateral area of the rectangle is given by:

$\text{L.A.}=\text{Perimeter}×\text{Height}$

The total area of the rectangle is given by:

$\text{T.A.}=\text{L.A.+}2\left(\text{Base}\right)$

The volume of rectangle is given by:

$\text{V}=l×w×h$

## Step 3. Determine the value.

Firstly, find the third dimension of the rectangle using the volume.

$\begin{array}{c}54=6×3×h\\ 54=18×h\\ h=\frac{54}{18}\\ h=3\text{\hspace{0.17em}\hspace{0.17em}unit}\end{array}$

Now.

$\begin{array}{c}\text{L.A.}=2\left(l+w\right)×h\\ =2\left(6+3\right)×3\\ =2\left(9\right)×3\\ =18×3\\ =54{\text{\hspace{0.17em}\hspace{0.17em}unit}}^{2}\end{array}$

And

$\begin{array}{c}\text{T.A.}=54\text{+}2\left(6×3\right)\\ =54\text{+}2\left(18\right)\\ =54\text{+36}\\ =90{\text{\hspace{0.17em}\hspace{0.17em}unit}}^{2}\end{array}$ ### Want to see more solutions like these? 