Chapter 59, Problem 78A

To determine

The area of the metal plate.

Answer to Problem 78A

Area of stamping ${\text{A}}_s\text{ =}\mathrm{193.605.}$

Explanation of Solution

Concept used:

The metal sheet is divided into three trapeziums and a rectangle as shown.

  

The area of three trapeziums is added and the area of rectangle is subtracted from the metal plate to find the required area.

Calculation:

Area of whole rectangle $\text{A}=l\times w$

Here, length l = 46 ft and Width w = 42 ft

By substituting the above values in the formula;

  $\begin{array}{l}\text{A= 46 × 42}\\ \text{A= 1932}\text{ft}\_\_\_\_\_\_\_\_\_\_\_(1)\end{array}$

Area of trapezium (2), $A_2=\frac{1}{2}\times h\left(b_1+b_2\right)$

Here, height h = 6.50 ft, base b₁= 9 ft and base b₂= 14 ft

  $\begin{array}{l}{A}_2=\frac{1}{2}\times 6.50\left(9+14\right)\\ A_2=\frac{1}{2}\times 6.50\times 23\\ A_2=74.75f{t}^2\_\_\_\_\_\_\_\_\_(2)\end{array}$

Since the dimensions of trapezium (2) and (3) are same thus, from symmetry

The area of trapezium (3), $A_3=74.75f{t}^2\_\_\_\_\_\_\_\_\_(3)$

Again, Area of trapezium (4), $A_4=\frac{1}{2}\times h\left(b_1+b_2\right)$

Here, height h = 6.50 ft, base b₁= 11 ft and base b₂= 16 ft

  $\begin{array}{l}{A}_4=\frac{1}{2}\times 6.50\left(11+16\right)\\ A_4=\frac{1}{2}\times 6.50\times 27\\ A_4=87.75f{t}^2\_\_\_\_\_\_\_\_\_(4)\end{array}$

And,

The area of rectangle, ${\text{A}}_r=l\times w$

Here, length l = 12.50 ft and

Width w = 42 − 28 = 14 ft

By substituting the above values in the formula;

  $\begin{array}{l}{\text{A}}_{\text{r}}\text{= 12}\text{.50×14}\\ {\text{A}}_{\text{r}}\text{= 175}\text{ft}\_\_\_\_\_\_\_\_\_\_\_(5)\end{array}$

Now, Adding the equations (1), (2), (3) and (4) and subtract equation (5) to calculate the required area of metal plate.

Thus,

  $\begin{array}{l}A=A_1+A_2+A_3+A_4-A_r\\ A=1932+74.75+74.75+87.75-175\\ A=1994.25ft\end{array}$

Conclusion:

Thus, the area of stamping is ${\text{A}}_s\text{ =}\mathrm{193.605.}$