Chapter 11.1, Problem 8PSB

a.

To determine

To calculate:Themissing dimension of given figure.

Answer to Problem 8PSB

The missing dimension are $w=5$ , x = $10$ , y = $25$ and z = $12.5$ .

Explanation of Solution

Given information:

Area of rectangle: $A=100$

In Figure 1, other side of $w=20.$

In Figure 2, other side of $x=10.$

In Figure 3, other side of $y=4.$

In Figure 4, other side of $z=8.$

Formula used:

Area of rectangle: $A=w\times l$

w = width of rectangle,

l = length of rectangle.

Calculation:

  

In Figure 1, $l=20.$

  $\begin{array}{l}A=w\times l\\ 100=w\times 20\\ \therefore w=\frac{100}{20}=5\end{array}$

  

In Figure 2, $l=10.$

  $\begin{array}{l}A=x\times l\\ 100=x\times 10\\ \therefore x=\frac{100}{10}=10\end{array}$

  

In Figure 3, $w=4.$

  $\begin{array}{l}A=4\times y\\ 100=4\times y\\ \therefore y=\frac{100}{4}=25\end{array}$

  

In Figure 4, $w=8.$

  $\begin{array}{l}A=8\times z\\ 100=8\times z\\ \therefore z=\frac{100}{8}=12.5\end{array}$

b.

To determine

To calculate: The length offencing needed to surround each figure.

Answer to Problem 8PSB

The length of fencing needed to surround Figure 1 is $50$ , Figure 2 is $40$ , Figure 3 is $58$ and

Figure 4 is $41$ .

Explanation of Solution

Given information:

Area of rectangle: $A=100$

In Figure 1, other side of $w=20.$

In Figure 2, other side of $x=10.$

In Figure 3, other side of $y=4.$

In Figure 4, other side of $z=8.$

Formula used:

Perimeter is sum of all sides.

Perimeter of rectangle: $P=b+b+h+h$

where b = breadth of rectangle,

h = height of rectangle.

Calculation:

  

In Figure 1, $l=20.$

  $\begin{array}{l}A=w\times l\\ 100=w\times 20\\ \therefore w=\frac{100}{20}=5\end{array}$

  $b=20,h=5.$

  $\begin{array}{l}P=b+b+h+h\\ P=20+20+5+5\\ P=50\end{array}$

  

In Figure 2, $l=10.$

  $\begin{array}{l}A=x\times l\\ 100=x\times 10\\ \therefore x=\frac{100}{10}=10\end{array}$

  $b=10,h=10.$

  $\begin{array}{l}P=b+b+h+h\\ P=10+10+10+10\\ P=40\end{array}$

  

In Figure 3, $w=4.$

  $\begin{array}{l}A=4\times y\\ 100=4\times y\\ \therefore y=\frac{100}{4}=25\end{array}$

  $b=25,h=4.$

  $\begin{array}{l}P=b+b+h+h\\ P=25+25+4+4\\ P=58\end{array}$

  

In Figure 4, $w=8.$

  $\begin{array}{l}A=8\times z\\ 100=8\times z\\ \therefore z=\frac{100}{8}=12.5\end{array}$

  $\begin{array}{l}b=8,\\ h=\mathrm{12.5.}\end{array}$

  $\begin{array}{l}P=b+b+h+h\\ P=8+8+12.5+12.5\\ P=41\end{array}$

c.

To determine

To find: The figure with a shortest perimeter.

Answer to Problem 8PSB

The figure with shortest perimeter is Figure 2.

Explanation of Solution

Given information:

Area of rectangle: $A=100$

In Figure 1, other side of $w=20.$

In Figure 2, other side of $x=10.$

In Figure 3, other side of $y=4.$

In Figure 4, other side of $z=8.$

Formula used:

Perimeter is sum of all sides.

Perimeter of rectangle: $P=b+b+h+h$

whereb = breadth of rectangle,

h = height of rectangle.

  

In Figure 1, $l=20.$

  $\begin{array}{l}A=w\times l\\ 100=w\times 20\\ \therefore w=\frac{100}{20}=5\end{array}$

  $b=20,h=5.$

  $\begin{array}{l}P=b+b+h+h\\ P=20+20+5+5\\ P=50\end{array}$

  

In Figure 2, $l=10.$

  $\begin{array}{l}A=x\times l\\ 100=x\times 10\\ \therefore x=\frac{100}{10}=10\end{array}$

  $b=10,h=10.$

  $\begin{array}{l}P=b+b+h+h\\ P=10+10+10+10\\ P=40\end{array}$

  

In Figure 3, $w=4.$

  $\begin{array}{l}A=4\times y\\ 100=4\times y\\ \therefore y=\frac{100}{4}=25\end{array}$

  $b=25,h=4.$

  $\begin{array}{l}P=b+b+h+h\\ P=25+25+4+4\\ P=58\end{array}$

  

In Figure 4, $w=8.$

  $\begin{array}{l}A=8\times z\\ 100=8\times z\\ \therefore z=\frac{100}{8}=12.5\end{array}$

  $b=8,h=\mathrm{12.5.}$

  $\begin{array}{l}P=b+b+h+h\\ P=8+8+12.5+12.5\\ P=41\end{array}$

The shortest perimeter is 40.

The figure with shortest perimeter is Figure 2.

d.

To determine

To find: A rectangle that encloses the maximum possible area with the shortest possible perimeter.

Answer to Problem 8PSB

A rectangle that encloses the maximum possible area with the shortest possible perimeter must be a square.

Explanation of Solution

Given information:

A rectangle with maximum possible area and shortest possible perimeter.

Formula used:

Area of rectangle: $A=w\ast l$

w = width of rectangle,

l = length of rectangle.

Perimeter is sum of all sides.

Perimeter of rectangle: $P=b+b+h+h$

b = Breadth of rectangle,

h = Height of rectangle.

The rectangle having the maximum possible area for a shortest possible perimeter is square because the length and width are the same.

  

Area of square is given by product of length and width.

In square, all sides are equal.

The magnitude of area will be larger because both values of length and width are same.

In case of other parallelogram, the length and width are different. This will result into area of certain value which is not as great as square.