Mathematics For Machine Technology
Mathematics For Machine Technology
8th Edition
ISBN: 9781337798310
Author: Peterson, John.
Publisher: Cengage Learning,
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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.
Chapter 60, Problem 31A, Solve these exercises. The area of the irregularly shaped sheet metal piece in the following figure

Expert Solution & Answer
Check Mark
To determine

The total area of the sheet metal piece.

Answer to Problem 31A

The total area of the sheet metal piece is 2743.5cm2.

Explanation of Solution

Given information:

The height of the triangle shown in part 1 is 18cm, the base of the triangle shown in part 1 is 13cm, the height of the triangle shown in part 2 is 23cm, the base of the triangle shown in part 2 is 25cm, the base of the triangle shown in part 3 is 20cm, the height of the part 4 is 12cm, the base of the part 4 is 20cm, the height of the triangle shown in part 5 is 12cm, the base of the triangle shown in part 5 is 62cm, the height of the part 6 is 21cm, the base of the part 6 is 70cm, the height of the triangle shown in part 7 is 21cm and the base of the triangle shown in part 8 is 70cm.

Draw the schematic diagram for the sheet metal piece.

  Mathematics For Machine Technology, Chapter 60, Problem 31A

       Figure -(1)

Write the expression for the area of the part 1.

A1=12b1h1 ...... (I)

Here, the height of the triangle shown in part 1 is h1 and the base of the triangle shown in part 1 is b1.

Write the expression for the area of the part 2.

A2=12b2h2 ...... (II)

Here, the height of the triangle shown in part 2 is h2 and the base of the triangle shown in part 2 is b2.

Write the expression for the area of the part 3.

A3=12b3h3 ...... (III)

Here, the height of the triangle shown in part 3 is h3 and the base of the triangle shown in part 3 is b3.

Write the expression for the area of the part 4.

A4=b4h4 ...... (IV)

Here, the height of the part 4 is h4 and the base of the part 4 is b4.

Write the expression for the area of the part 5.

A5=12b5h5 ...... (V)

Here, the height of the triangle shown in part 5 is h5 and the base of the triangle shown in part 5 is b5.

Write the expression for the area of the part 6.

A6=b6h6 ...... (VI)

Here, the height of the part 6 is h6 and the base of the part 6 is b6.

Write the expression for the area of the part 7.

A7=12b7h7 ...... (VII)

Here, the height of the triangle shown in part 7 is h7 and the base of the triangle shown in part 7 is b7.

Write the expression for the area of the part 8.

A8=12b8h8 ...... (VIII)

Here, the height of the triangle shown in part 8 is h8 and the base of the triangle shown in part 8 is b8.

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

A=A1+A2+A3+A4+A5+A6+A7+A8 ...... (IX)

Write the expression for the height h3 of the part 3.

h3=h2h4 ...... (X)

Write the expression for the base b7 of the part 7.

b7=(b2+b3+b5)(b1+b6) ...... (XI)

Write the expression for the height h8 of the part 8.

h8=h7h1 ...... (XII)

Calculation:

Substitute 23cm for h2 and 12cm for h4 in Equation (X).

h3=23cm12cm=11cm

Substitute 21cm for h7 and 18cm for h1 in Equation (XII).

h8=21cm18cm=3cm

Substitute 25cm for b2, 20cm for b3, 62cm for b5, 13cm for b1 and 70cm for b6 in Equation (XI).

b7=(25cm+20cm+62cm)(13cm+70cm)=(107cm)(83cm)=24cm

Substitute 13cm for b1 and 18cm for h1 in Equation (I).

A1=12(13cm)(18cm)=12(234 cm2)=117cm2

Substitute 25cm for b2 and 23cm for h2 in Equation (II).

A2=12(25cm)(23cm)=12(575 cm2)=287.5cm2

Substitute 20cm for b3 and 11cm for h3 in Equation (III).

A3=12(20cm)(11cm)=12(220 cm2)=110cm2

Substitute 20cm for b4 and 12cm for h4 in Equation (IV).

A3=(20cm)(12cm)=240cm2

Substitute 62cm for b5 and 12cm for h5 in Equation (V).

A5=12(62cm)(12cm)=12(744 cm2)=372cm2

Substitute 70cm for b6 and 18cm for h6 in Equation (VI).

A6=(70cm)(18cm)=1260cm2

Substitute 24cm for b7 and 21cm for h7 in Equation (VII).

A7=12(24cm)(21cm)=12(504 cm2)=252cm2

Substitute 70cm for b8 and 3cm for h8 in Equation (VIII).

A8=12(70cm)(3cm)=12(210 cm2)=105cm2

Substitute 117cm2 for A1, 287.5cm2 for A2, 110cm2 for A3, 240cm2 for A4, 372cm2 for A5, 1260cm2 for A6, 252cm2 for A7 and 105cm2 for A8 in Equation (IX).

A=[117 cm 2+287.5 cm 2+110 cm 2+240 cm 2+372 cm 2+1260 cm 2+252 cm 2+105 cm 2]=2743.5cm2

Conclusion:

The total area of the sheet metal piece is 2743.5cm2.

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Chapter 60 Solutions

Mathematics For Machine Technology

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