Chapter 60, Problem 31A

Solve these exercises.

The area of the irregularly shaped sheet metal piece in the following figure on the left is to be determined. The longest diagonal is drawn on the figure as shown in the figure on the right. Perpendiculars are drawn to the diagonal from each of the other vertices. The perpendicular segments are measured as shown. From the measurements, the areas of each of the common polygons are computed. This is one method often used to compute areas of irregular figures. Compute the area of the sheet metal piece.

To determine

The total area of the sheet metal piece.

Answer to Problem 31A

The total area of the sheet metal piece is $2743.5{\text{cm}}^{\text{2}}$.

Explanation of Solution

Given information:

The height of the triangle shown in part 1 is $1\text{8}\text{cm}$, the base of the triangle shown in part 1 is $13\text{cm}$, the height of the triangle shown in part 2 is $23\text{cm}$, the base of the triangle shown in part 2 is $25\text{cm}$, the base of the triangle shown in part 3 is $20\text{cm}$, the height of the part 4 is $12\text{cm}$, the base of the part 4 is $20\text{cm}$, the height of the triangle shown in part 5 is $12\text{cm}$, the base of the triangle shown in part 5 is $62\text{cm}$, the height of the part 6 is $21\text{cm}$, the base of the part 6 is $70\text{cm}$, the height of the triangle shown in part 7 is $21\text{cm}$ and the base of the triangle shown in part 8 is $70\text{cm}$.

Draw the schematic diagram for the sheet metal piece.

  

       Figure -(1)

Write the expression for the area of the part 1.

 $A_1=\frac{1}{2}{b}_1{h}_1$ ...... (I)

Here, the height of the triangle shown in part 1 is $h_1$ and the base of the triangle shown in part 1 is $b_1$.

Write the expression for the area of the part 2.

 $A_2=\frac{1}{2}{b}_2{h}_2$ ...... (II)

Here, the height of the triangle shown in part 2 is $h_2$ and the base of the triangle shown in part 2 is $b_2$.

Write the expression for the area of the part 3.

 $A_3=\frac{1}{2}{b}_3{h}_3$ ...... (III)

Here, the height of the triangle shown in part 3 is $h_3$ and the base of the triangle shown in part 3 is $b_3$.

Write the expression for the area of the part 4.

 $A_4=b_4{h}_4$ ...... (IV)

Here, the height of the part 4 is $h_4$ and the base of the part 4 is $b_4$.

Write the expression for the area of the part 5.

 $A_5=\frac{1}{2}{b}_5{h}_5$ ...... (V)

Here, the height of the triangle shown in part 5 is $h_5$ and the base of the triangle shown in part 5 is $b_5$.

Write the expression for the area of the part 6.

 $A_6=b_6{h}_6$ ...... (VI)

Here, the height of the part 6 is $h_6$ and the base of the part 6 is $b_6$.

Write the expression for the area of the part 7.

 $A_7=\frac{1}{2}{b}_7{h}_7$ ...... (VII)

Here, the height of the triangle shown in part 7 is $h_7$ and the base of the triangle shown in part 7 is $b_7$.

Write the expression for the area of the part 8.

 $A_8=\frac{1}{2}{b}_8{h}_8$ ...... (VIII)

Here, the height of the triangle shown in part 8 is $h_8$ and the base of the triangle shown in part 8 is $b_8$.

Write the expression for the total area of the sheet metal piece.

 $A=A_1+A_2+A_3+A_4+A_5+A_6+A_7+A_8$ ...... (IX)

Write the expression for the height $h_3$ of the part 3.

 $h_3=h_2-h_4$ ...... (X)

Write the expression for the base $b_7$ of the part 7.

 $b_7=\left(b_2+b_3+b_5\right)-\left(b_1+b_6\right)$ ...... (XI)

Write the expression for the height $h_8$ of the part 8.

 $h_8=h_7-h_1$ ...... (XII)

Calculation:

Substitute $23\text{cm}$ for $h_2$ and $12\text{cm}$ for $h_4$ in Equation (X).

 $\begin{array}{c}{h}_3=23\text{cm}-12\text{cm}\\ =11\text{cm}\end{array}$

Substitute $21\text{cm}$ for $h_7$ and $18\text{cm}$ for $h_1$ in Equation (XII).

 $\begin{array}{c}{h}_8=21\text{cm}-18\text{cm}\\ =3\text{cm}\end{array}$

Substitute $25\text{cm}$ for $b_2$, $20\text{cm}$ for $b_3$, $62\text{cm}$ for $b_5$, $13\text{cm}$ for $b_1$ and $70\text{cm}$ for $b_6$ in Equation (XI).

 $\begin{array}{c}{b}_7=\left(25\text{cm}+20\text{cm}+62\text{cm}\right)-\left(13\text{cm}+70\text{cm}\right)\\ =\left(107\text{cm}\right)-\left(83\text{cm}\right)\\ =24\text{cm}\end{array}$

Substitute $13\text{cm}$ for $b_1$ and $18\text{cm}$ for $h_1$ in Equation (I).

 $\begin{array}{c}{A}_1=\frac{1}{2}\left(13\text{cm}\right)\left(18\text{cm}\right)\\ =\frac{1}{2}\left(234{\text{cm}}^2\right)\\ =117{\text{cm}}^2\end{array}$

Substitute $25\text{cm}$ for $b_2$ and $23\text{cm}$ for $h_2$ in Equation (II).

 $\begin{array}{c}{A}_2=\frac{1}{2}\left(25\text{cm}\right)\left(23\text{cm}\right)\\ =\frac{1}{2}\left(575{\text{cm}}^2\right)\\ =287.5{\text{cm}}^2\end{array}$

Substitute $20\text{cm}$ for $b_3$ and $11\text{cm}$ for $h_3$ in Equation (III).

 $\begin{array}{c}{A}_3=\frac{1}{2}\left(20\text{cm}\right)\left(11\text{cm}\right)\\ =\frac{1}{2}\left(220{\text{cm}}^2\right)\\ =110{\text{cm}}^2\end{array}$

Substitute $20\text{cm}$ for $b_4$ and $12\text{cm}$ for $h_4$ in Equation (IV).

 $\begin{array}{c}{A}_3=\left(20\text{cm}\right)\left(12\text{cm}\right)\\ =240{\text{cm}}^2\end{array}$

Substitute $62\text{cm}$ for $b_5$ and $12\text{cm}$ for $h_5$ in Equation (V).

 $\begin{array}{c}{A}_5=\frac{1}{2}\left(62\text{cm}\right)\left(12\text{cm}\right)\\ =\frac{1}{2}\left(744{\text{cm}}^2\right)\\ =372{\text{cm}}^2\end{array}$

Substitute $70\text{cm}$ for $b_6$ and $18\text{cm}$ for $h_6$ in Equation (VI).

 $\begin{array}{c}{A}_6=\left(70\text{cm}\right)\left(18\text{cm}\right)\\ =1260{\text{cm}}^2\end{array}$

Substitute $24\text{cm}$ for $b_7$ and $21\text{cm}$ for $h_7$ in Equation (VII).

 $\begin{array}{c}{A}_7=\frac{1}{2}\left(24\text{cm}\right)\left(21\text{cm}\right)\\ =\frac{1}{2}\left(504{\text{cm}}^2\right)\\ =252{\text{cm}}^2\end{array}$

Substitute $70\text{cm}$ for $b_8$ and $3\text{cm}$ for $h_8$ in Equation (VIII).

 $\begin{array}{c}{A}_8=\frac{1}{2}\left(70\text{cm}\right)\left(3\text{cm}\right)\\ =\frac{1}{2}\left(210{\text{cm}}^2\right)\\ =105{\text{cm}}^2\end{array}$

Substitute $117{\text{cm}}^{\text{2}}$ for $A_1$, $287.5{\text{cm}}^{\text{2}}$ for $A_2$, $110{\text{cm}}^{\text{2}}$ for $A_3$, $240{\text{cm}}^{\text{2}}$ for $A_4$, $372{\text{cm}}^{\text{2}}$ for $A_5$, $1260{\text{cm}}^{\text{2}}$ for $A_6$, $252{\text{cm}}^{\text{2}}$ for $A_7$ and $105{\text{cm}}^{\text{2}}$ for $A_8$ in Equation (IX).

 $\begin{array}{c}A=\left[\begin{array}{l}117{\text{cm}}^{\text{2}}+287.5{\text{cm}}^{\text{2}}+110{\text{cm}}^{\text{2}}+240{\text{cm}}^{\text{2}}+\\ 372{\text{cm}}^{\text{2}}+1260{\text{cm}}^{\text{2}}+252{\text{cm}}^{\text{2}}+105{\text{cm}}^{\text{2}}\end{array}\right]\\ =2743.5{\text{cm}}^{\text{2}}\end{array}$

Conclusion:

The total area of the sheet metal piece is $2743.5{\text{cm}}^{\text{2}}$.