Chapter 1.6, Problem 29PPS

a.

To determine

To find perimeter of rectangle

Answer to Problem 29PPS

  Perimeter is 14 ft

Explanation of Solution

Given information:

  $length=4ft$

  $breadth=3ft$

Formula used:

  Perimeter= $2(lengh+breadth)$

Calculation:

Substituting values,

  Perimeter= $2(4+3)$

Solving,

  Perimeter= $2(7)$

Hence, perimeter is 14 ft.

b.

To determine

To find the area

Answer to Problem 29PPS

area is $21f{t}^2$

Explanation of Solution

Given information:

  $length=4ft$

  $breadth=3ft$

Formula used:

  Area= $length\times breadth$

Calculation:

Substituting values in formula,

  Area= $7\times 3$

Solving,

  Area= $21$

Hence, area is $21f{t}^2$

c.

To determine

To find the effect on area and perimeter.

Answer to Problem 29PPS

  Perimeter is doubled.

  Area is increased 4 times.

Explanation of Solution

Given information:

  $\begin{array}{l}length=4ft\\ breadth=3ft\end{array}$

Length and breadth are doubled.

Calculation:

  Area₁= $length\times breadth$

Now,

  Area₂= $2length\times 2breadth$

So,  area= $4(length\times breadth)$

  $\frac{are{a}_1^{}}{are{a}_2^{}}=\frac{length\times breadth}{4(length\times breadth)}$

Solving,

  $\frac{are{a}_1^{}}{are{a}_2^{}}=\frac{1}{4}$

Hence, $4are{a}_1^{}=are{a}_2^{}$

Hence, area is increased 4 times.

Now,

  Perimeter₁= $2(l+b)$

And,

perimeter₂= $2(2l+2b)$

Simplifying,

  Perimeter= $4(l+b)$

Now,

  $\frac{p_1^{}}{p_2^{}}=\frac{2(l+b)}{4(l+b)}$

Solving,

  $\frac{p_1^{}}{p_2^{}}=\frac{2}{4}$

So,

  $2p_1^{}=p_2^{}$

Hence, perimeter is doubled.

d.

To determine

To find the effect on area and perimeter.

Answer to Problem 29PPS

The perimeter is halved.

  Area is decreased 4 times.

Explanation of Solution

Given information:

  $\begin{array}{l}length=4ft\\ breadth=3ft\end{array}$

Length and breadth are halved.

Calculation:

Area₁ is $length\times breadth$

  Area₂is $\frac{length}{2}\times \frac{breadth}{2}$

Solving,

  Area₂= $\frac{length\times breadth}{4}$

Now,

  $\frac{are{a}_1^{}}{are{a}_2^{}}=\frac{length\times breadth}{\frac{length\times breadth}{4}}$

So,

  $\frac{are{a}_1^{}}{are{a}_2^{}}=\frac{4(length\times breadth)}{length\times breadth}$

Now,

  $are{a}_1^{}=4are{a}_2^{}$

Hence, area is decreased 4 times.

  $perimete{r}_1^{}=2(l+b)$

And $perimete{r}_2^{}=2(\frac{l+b}{2})$

So,

  $perimete{r}_2^{}=l+b$

Comparing both perimeter, the perimeter is halved.