Chapter 18, Problem 48A

A drawing has a scale of $\frac{1\text{'}\text{'}}{4}=3\text{'}$ . Determine the value of each missing value in the table.

To determine

(a)

The actual length.

Answer to Problem 48A

The actual length is $13{\frac{1}{2}}^{\prime }$.

Explanation of Solution

Given:

The given relation is ${\frac{1}{4}}^{\prime \text{}\prime }=3^{\prime }$.

Concept used:

The unitary method is used in which the value of single is evaluated first then the required value is found by multiple it by single unit value.

Calculation:

The given scale length is $1{\frac{1}{8}}^{\prime \text{}\prime }$.

  $\begin{array}{c}1{\frac{1}{8}}^{\prime \text{}\prime }={\frac{9}{8}}^{\prime \text{}\prime }\\ =\frac{9}{8}\times 12\left({\frac{1}{4}}^{\prime \text{}\prime }=3^{\prime }\Rightarrow 1^{″}=12^{\prime }\right)\\ =13\frac{1}{2}\end{array}$

The actual length is $13{\frac{1}{2}}^{\prime }$.

Conclusion:

The actual length is $13{\frac{1}{2}}^{\prime }$.

To determine

(b)

The actual length.

Answer to Problem 48A

The actual length is $5{\frac{1}{4}}^{\prime }$.

Explanation of Solution

Given:

The given relation is ${\frac{1}{4}}^{\prime \text{}\prime }=3^{\prime }$.

Concept used:

The unitary method is used in which the value of single is evaluated first then the required value is found by multiple it by single unit value.

Calculation:

The given scale length is ${\frac{7}{16}}^{\prime \text{}\prime }$.

  $\begin{array}{c}\frac{7}{16}=\frac{7}{16}\times 12\left({\frac{1}{4}}^{\prime \text{}\prime }=3^{\prime }\Rightarrow 1^{″}=12^{\prime }\right)\\ =5{\frac{1}{4}}^{\prime }\end{array}$

The actual length is $5{\frac{1}{4}}^{\prime }$.

Conclusion:

The actual length is $5{\frac{1}{4}}^{\prime }$.

To determine

(c)

The scale length.

Answer to Problem 48A

The scale length is $9{\frac{1}{2}}^{\prime \text{}\prime }$.

Explanation of Solution

Given:

The given relation is ${\frac{1}{4}}^{\prime \text{}\prime }=3^{\prime }$.

Concept used:

The unitary method is used in which the value of single is evaluated first then the required value is found by multiple it by single unit value.

Calculation:

The given actual length is $6^{\prime }{9}^{″}$.

  $\begin{array}{c}{6}^{\prime }{9}^{″}=6\times \frac{1}{12}\left(9^{″}\right)\left({\frac{1}{4}}^{\prime \text{}\prime }=3^{\prime }\Rightarrow 1^{\prime }={\frac{1}{12}}^{\prime \text{}\prime }\right)\\ =9{\frac{1}{2}}^{\prime \text{}\prime }\end{array}$

The scale length is $9{\frac{1}{2}}^{\prime \text{}\prime }$.

Conclusion:

The scale length is $9{\frac{1}{2}}^{\prime \text{}\prime }$.

To determine

(d)

The scale length.

Answer to Problem 48A

The scale length is $1{\frac{13}{192}}^{\prime \text{}\prime }$.

Explanation of Solution

Given:

The given relation is ${\frac{1}{4}}^{\prime \text{}\prime }=3^{\prime }$.

Concept used:

The unitary method is used in which the value of single is evaluated first then the required value is found by multiple it by single unit value.

Calculation:

The given actual length is $16^{\prime }1{\frac{1}{16}}^{\prime \text{}\prime }$.

  $\begin{array}{c}16^{\prime }1{\frac{1}{16}}^{\prime \text{}\prime }=\frac{1}{16\times 12}\left(1{\frac{1}{16}}^{\prime \text{}\prime }\right)\left({\frac{1}{4}}^{\prime \text{}\prime }=3^{\prime }\Rightarrow 1^{\prime }={\frac{1}{12}}^{\prime \text{}\prime }\right)\\ ={\frac{1}{192}}^{\prime \text{}\prime }+{\frac{17}{16}}^{\prime \text{}\prime }\\ =\frac{205}{192}\\ =1{\frac{13}{192}}^{\prime \text{}\prime }\end{array}$

The scale length is $1{\frac{13}{192}}^{\prime \text{}\prime }$.

Conclusion:

The scale length is $1{\frac{13}{192}}^{\prime \text{}\prime }$.