Chapter 59, Problem 64A

To determine

The area of the template.

Answer to Problem 64A

Area $\text{A =}\text{73030}\text{.95}{\text{mm}}^{\text{2}}\text{.}$

Explanation of Solution

Concept used:

The template is divided into two trapeziums and a rectangle as shown.

The area of two trapeziums and the area of rectangle are added to find the required area.

Calculation:

Area of trapezium (1), $A_1=\frac{1}{2}\times h\left(b_1+b_2\right)$

Here, height h = 215 − 100 = 115 mm, base b₁= 310 mm and

Base (b₂) is calculated as:

Since the groove makes an angle of 28° then, the angle made by the triangle is,

  ${90}^{°}-{28}^{°}={62}^{°}$

  $\begin{array}{l}\tan {62}^{°}=\frac{115}{b}\\ b=\frac{115}{\tan {62}^{°}}\\ b=\frac{115}{1.880}\\ b=61.2mm\end{array}$

Thus, from symmetry the base of another triangle is also 61.2 mm.

Now, the base b₂ of the trapezium is:

  $\begin{array}{l}\Rightarrow 310-\left(61.2+61.2\right)\\ \Rightarrow 310-122.4\\ b_2=187.6mm\end{array}$

Substituting the above values in the formula;

  $\begin{array}{l}{A}_1=\frac{1}{2}\times 115\left(310+187.6\right)\\ A_1=\frac{1}{2}\times 115\times 497.6\\ A_1=28612m{m}^2\_\_\_\_\_\_\_\_\_(1)\end{array}$

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

Here, height h = 85 mm,

Base b₁is calculated as:

Since the groove makes an angle 28°, then the $\angle GCE={90}^{°}-{28}^{°}={62}^{°}$

Hence,

  $\begin{array}{l}\sin {62}^{°}=\frac{190}{b_1}\\ b_1=\frac{190}{\sin {62}^{°}}\\ b_1=\frac{190}{0.8829}\\ b_1=215.2mm\end{array}$

Again, From the geometry of the figure the line BCE makes an angle of 90° and $\angle ACB={62}^{°}$ as calculated above thus, $\angle ACE={90}^{°}-{62}^{°}={28}^{°}$

Now, the combined $\angle ACD={28}^{°}+{28}^{°}={56}^{°}$

  $\begin{array}{l}\tan {56}^{°}=\frac{85}{b}\\ b=\frac{85}{\tan {56}^{°}}\\ b=\frac{85}{1.4825}\\ b=57.33mm\end{array}$

Thus, from symmetry the base of another triangle is also 57.33 mm.

Now, the base b₂ of the trapezium is:

  $\begin{array}{l}\Rightarrow 215.2-\left(57.33+57.33\right)\\ \Rightarrow 215.2-114.66\\ b_2=100.54mm\end{array}$

Substituting the above values in the formula;

  $\begin{array}{l}{A}_2=\frac{1}{2}\times 85\left(215.2+100.54\right)\\ A_2=\frac{1}{2}\times 85\times 315.74\\ A_2=13418.95m{m}^2\_\_\_\_\_\_\_\_\_(2)\end{array}$

Area of rectangle ${\text{A}}_3=l\times w$

Here, length l = 310 mm and Width w = 100 mm

By substituting the above values in the formula;

  $\begin{array}{l}{\text{A}}_{\text{3}}\text{= 310 × 100}\\ {\text{A}}_{\text{3}}\text{= 31000}{\text{mm}}^{\text{2}}\_\_\_\_\_\_\_\_\_\_\_(3)\end{array}$

Now, Adding the equations (1), (2), and (3) to calculate the required area.

Thus,

  $\begin{array}{l}A=A_1+A_2+A_3\\ A=28612+13418.95+31000\\ A=73030.95m{m}^2\end{array}$

Conclusion:

Thus, the area of the given template is $\text{A =}\text{73030}\text{.95}{\text{mm}}^{\text{2}}\text{.}$