271 mensuration notes oh1

8/22/2019 271 Mensuration Notes OH1

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271 Mensuration Notes OH1 1 16/01/2010

Mensuration

IntroductionMensuration is the process of measuring and calculating with measurements.Mensuration deals wi th the determination of length, area, or volume

Measurement Types

The basic measurement types include: distances (including perimeters and circumferences) areas volumes

Quantities such as area and volume are often calculated from measured distancessuch as length, width, height, diameter, circumference, perimeter or other distances.

Measured ShapesMeasured shapes can be classified in various ways, including:

two-dimensional (flat) surfaces, such as rectangles, triangles, trapezoids and other polygons circles, sectors and segments of circles.

three-dimensional (solid) shapes, such as prisms and cylinders spheres cones and pyramids

Another classif ication of shapes is :

regularshapes (such as the shapes identified above) irregularshapes, which are shapes bound by irregular edges such as a pond, ortrench.

Two Dimensional Shapes - Perimeter and AreaMany problems in architecture, construct ion, surveying, mechanical design and civi lengineering involve two dimensional shapes.

Mensuration calculations for two dimensional shapes include:

distances and areas.

Distance - UnitsDistance calculations result in linearunits.Typical units fordistances include:

cm centimetres

km kilometres

m metres

ft feet


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Area - UnitsArea calculations result in square units.Typical units forareas include:

cm2 square centimetres

m2 squ are metres

in2 square inches

ft2 square feet

This section considers perimeter and area for:

triangles, quadrilaterals, and circu lar shapes

PolygonsPolygons are closed plane figures bound by three or more line segments.

Common polygons include those with:

Three sides Triangle

Four sides Quadrilateral

Five sides: Pentagon

Six sides: Hexagon

Eight sides: Octagon

Twelve sides: Dodecagon


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Do not copy into your notes

Source: regular polygons - Wikipedia

For regular polygons with side s=1 this produces the following table:

Sides Name Approximate area

3 equilateral triangle 0.433

4 square 1.000

5 regularpentagon 1.720

6 regularhexagon 2.598

7 regularheptagon 3.634

8 regularoctagon 4.828

9 regularnonagon 6.18210 regulardecagon 7.694

11 regularhendecagon 9.366

12 regulardodecagon 11.196

13 regulartriskaidecagon 13.186

14 regulartetradecagon 15.335

15 regularpentadecagon 17.642

16 regularhexadecagon 20.109

17 regularheptadecagon 22.735

18 regularoctadecagon 25.52119 regularenneadecagon 28.465

20 regularicosagon 31.569

100 regularhectagon 795.513

1000 regularchiliagon 79577.210

10000 regularmyriagon 7957746.893

1,000,000 regularmegagon 79,577,471,545.685


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DefinitionA general po lygon may have unequal s ide lengths and unequal angles.DefinitionA regular po lygon has equal side lengths and equal angles.

regular po lygons

This section considers mainly

triangles and quadrilaterals.

Triangles

DefinitionA t riang le is a c losed three s ided f igure, in which each s ide is a st raight l inesegment and includes the following types:

Right Triangles - contains a 90 degree angle

Oblique Triangles contains no 90 degree angle

Acute Triangles all angles are less than 90 degrees

Equilateral Triangles all sides and angles are of equal length

Isosceles Triangles two angles and two sides are equal

Obtuse Triangles contains one angle greater than 90 degrees

Scalene Triangles no equal sides and no equal angles


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Quadrilaterals

DefinitionA quadr ilateral is a c losed geometric figure (two d imens ional) with four s traightsides, and inc ludes the following types:

Rectangles and Squares all angles are 90 degrees

Parallelograms - each pair of opposite sides are parallel

Rhombuses

Every rhombus has two diagonals connecting opposite pairs of vertices and two pairs ofparallel sides. It follows that any rhombus has the following two properties:

1. Opposite angles of a rhombus have equal measure.

2. The two diagonals of a rhombus are perpendicular.

Trapezoids - one pair of opposite sides are parallel

General Quadrilaterals sides are of unequal length


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Triangles

DefinitionA t riang le is a c losed 3 s ided f igure, in which each s ide is a st raight l ine segment.

Notation

a one side of the triangle

b base side of the triangle

c third side of the triangle

h height of the triangle

s one half the perimeterof the triangle

P perimeter of the triangle

A area of the triangle

The height h is the perpendicular distance from any vertex to the opposite side.Any side may be considered the base of the triangle.

Triangles are commonly labelled with the sides in lower case and each angle in upper

case of the same letter as the opposite side (e.g. angle opposite side a ).


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Formulas The perimeter of a triangle is the sum of the lengths of the sides.

P a b c

The area of a triangle is one half the base times the height:

A 12

bh

Knowing the length of the 3 sides of the triangle, the area can be obtained fromHeron's Formula:

A ss as bs c

where s 1

2 a b c

Knowing the length of2 sides a and b and the included angle C of a triangle,the area is:

A 1

2absinC


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Perimeter Problems

Example 1Determine the perimeterof a triangular structure with sides of 2.5m , 4.7m , and

3.1m.

Solution

Analysis:

Calculation:

For P :

P a b c

4.7 2.5 3.1 10.3P = + + =

Conclusion:

The perimeterof a triangular st ructure is 10.3m.


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Example 2Determine the perimeter of the triangular steel plate.

Solution

Analysis:

Calculations:

For a : For P :

Its a right angle triangle, so use Pythagoras to solve for the length of the missingside

2 2 2

2 2 2

2 2

54 46

54 46 28.284

r x y

x

x

= +

= +

= =

28 54 46 128.28P = + + =

Conclusion:

The perimeter of the steel plate is 128.28cm.


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Example 3Determine the perimeter of the fabricated truss.

Solution

Analysis:

Calculations:

For a (the third side): For P :

2 2 2 2a b c bcCosA= +

( )( )

2 2 2

2 2

2

22.4 14.5 2 22.4 14.5 cos130 1129.5648

1129.5648 33.6

a b c bcCosA

a

= +

= + =

= =

22.4 14.5 33.6 70.51P = + + =

Conclusion:

The perimeter of the fabricated truss is 70.51m.


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Area Problems

Example 1Determine the triangular section area of an industrial building wall.

Solution1

Analysis:

Calculation:

1

2Area bh=

1(28.5)(6.0) 85.5

2Area = =

Conclusion:

The area of the wall is 85.5 m2.


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Example 2

Determine the area of the illustrated triangular structure.

Solution 2Analysis:

Calculations:

For s : For :

( )

( ) ( )( )

( ) ( ) ( )

1 3.1 4.7 2.5 5.152

5.15 5.15 3.1 5.15 4.7 5.15 2.5

3.548

s m

Area s s a s b s c

= + + =

=

=

=

Conclusion:

The area of the triangular st ructure is 3.548 m2.


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Example 3Calculate the enclosed floor area.

Solution 3Analysis:

Calculations:

1sin

2Area ab C=

( ) ( )1

163 178 sin125 118832

oArea = =

Conclusion:

The enclosed floor area is 41.188 10 cm2


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Example 4

Determine the floor area of a commercial display.

Analysis :

A

1

2bh

(requires the base and height) A ss as bs c (requires three sides)

A 12

absinC(requires two sides and the included angle)

Which formula can be used in this example?

For C:

1

4.83 6.25

38.4 sin

sin 0.080376

sin 0.080376

53.49o

sin C

C

C

C

=

=

=

=

For B:

180 (38.4 53.49) 88.109oB = + =


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Forthe area :

( ) ( )

1sin

2

14.83 6.25 sin88.109 15.086

2

Area ac B=

= =

Conclusion:

The floor area of a commercial display is 15.086 m2.


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Quadrilaterals

The next sections explore mensuration calculations involving the followingquadrilateral shapes.

Rectangle and Square

Definition

A rectangle is a quadr ilateral where each in ter ior angle is 90

and the oppositesides are equal in length.

DefinitionA square is a rectangle where all sides are equal in leng th.

Parallelogram and Rhombus

DefinitionA parallelogram is a quadrilateral with two pai rs o f opposite sides that are equaland parallel.

DefinitionA rhombus is a parallelogram where all sides are of equal length.The diagonals of a rhombus intersect at right angles.

Trapezoid

DefinitionA trapezoid is a quadr ilateral with one pair of paral lel sides .


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General Quadrilateral

DefinitionA general quadr ilateral is a c losed geometric f igure wi th four st raight sides.

Rectangles and Squares

Rectangles and squares are quadrilaterals with specific characteristics.

RectanglesDefinition

A rectangle is a quadri lateral where each interior angle is 90

and the oppositesides are equal in length.

SquaresDefinition

A square is a rectangle where al l s ides are equal in length.

Notationl

length of the rectanglew width of the rectangle or square

d diagonal distance

P perimeterof the square or rectangle

area of the square or rectangle

Note that the length and width for a square are equal and are both labelled w.


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FormulasWith reference to the diagram:

The perimeter is the sum of the lengths of the sides.

rectangle squ are

P 2l 2w P 2w 2w

P 2l w P 4w

The area is the length times the width.

rectangle sq uare

A l w A w2

The diagonal can be determined from the Pythagorean Theorem:

rectangle square

d2 l2 w

2 d2 w2 w

2

d l2 w

2 d 2w2


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Example 1Determine the perimeterand area of a rectangular garden plot of width 12.4m and

length 5.2m .

Solution

Analysis

Calculations:

For P :

2 12.4 2 5.2 35.2P = + =

For :

( ) ( )12.4 5.2 64.48A = =

Conclusion:

The rectangular garden plot has a perimeterof 35.2 m and an area of 64.48 m2.


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Example 2Determine the perimeterof the foundation layout as shown.

Analysis:

Calculations:

For a and b :

9.2 2.1 4.8 2.3

9.5 7.25 2.25

m

m

=

=

For P :

2.1 7.25 9.2 14.5 4.8 9.5 2.3 2.25 51.9P = + + + + + + + =

Conclusion:

The perimeterof the foundation layout is 51.9 m.Stopped Tuesday Jan 12


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Example 3Determine the area of the steel plate.

Solution

4.5 7.2 15.1 3.4c c cm+ + = =

( ) ( )12

13.4 5.2

2

8.84

A

cm

=

=

( )( )22

4.5 5.2

23.4

A

cm

=

=

( )( )32

17.2 5.2

2

18.72

A

cm

=

=

8.84 23.4 18.72 50.96A = + + =

Conclusion:

The area of the steel plate is 50.96 cm2.


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Example 4Determine the perimeterand the area of the retail floor d isplay space.The floor area is square with a diagonal of 48.5 m.

Analysis:

Calculation:From the Pythagorean Theorem:

2 2 2

2 2

22

48.5

2 48.5

48.51176.125

2

34.29

w w

w

w

w

+ =

=

= =

=

4 34.29 137.18

34.29 34.29 1176.1

perimeter

area

= =

= =

Conclusion:

The retail floor display space has a perimeter 137.2 m of and an area of 1176.1 m2.


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Parallelograms and Rhombuses

Parallelograms and rhombuses are quadrilaterals with specific characteristics.

Parallelogram

A paral lelogram is a quadr ilateral with two pairs of oppos ite sides that areequal and parallel.

Rhombus

A rhombus is a parallelogram where all sides are o f equal length.The diagonals of a rhombus intersect at right angles.

Parallelograms

Rhombus

a and b are the lengths of the sides.

h height (the perpendicular distance between a pair of parallel sides).

area

P perimeter

Formulas The perimeter is the sum of the edge lengths.

parallelogram rhombus

P 2a 2b P 4b

The area is the base length times the height.

parallelogram rhombus

A b h A b h


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Example 1A paral lelogram shaped building lot which covers an area of 500 m2.Determine the perimeter of the lot.

For h :

130 90 40o = =

cos40 24.8

18.998

o h

h m

=

=

For b :

( )500 18.998

50026.31918.998

Area bh

b

b m

=

=

= =

For P :

( ) ( )2 26.319 2 24.8 102.24P m= + =

Conclusion:

The perimeter of the lot is 102.24 m.


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Example 2Determine the perimeter of the rhombus shaped pendant.

For b :

2 2 21.25 2.0 5.5625

5.5625 2.3585

w

w

= + =

= =

For P :

( )4 2.3585 9.434P = =

Conclusion:

The perimeter of the pendant is 9.434 cm.


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Trapezoids

Trapezoid

A t rapezoid is a quadrilateral with one pai r o f paral lel sides .

a and b are the two parallel sidesc and d are the other sides

h height of the trapezoid (perpendicular distance between the parallel sides).

area

P perimeter

The perimeter is the sum of the lengths of the sides.

P a b c d

The area is the average of the lengths of the two parallel sides times the height .

A a b2

h


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Example 1Determine the area of the steel plate.

Analysis:

Calculations:

2

a bA h

+ =

4.5 15.1 5.2 50.962

A + = =

Conclusion:

The area of the steel plate is 50.96 cm2.


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Example 2Calculate the perimeterand area for this cross section of an industrial building roofstructure.


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Calculations:

For e : Consider the similar triangles:

2 1

10

2 10

5

e

e

e

=

=

=

2 2 210.0 5.0

11.18

c

c m

= +

=

For : Consider the similar triangles:

2 3

10

15

f

f

=

=


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2 2 210.0 15.0

18.028

d

d m

= +

=

For P :

5 8.2 15 28.2b = + + = For :

2a bA h+ =

8.2 28.210 182

2A

+ = =

11.18 8.2 18.0 28.2 65.58P = + + + =

Conclusion:

The industrial building roof structure has a perimeter of 65.6 m and an area of 182m2.


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Example 3Calculate the area of the trapezoid shaped panel with dimensions as indicated.

Solution 3Analysis:

Consider combining the two triangles into a single triangle.

Knowing all three sides of the triangle provides different alternatives.

an algebraic alternative a trigonometric alternative

Calculate triangle from Heron'sformula

Calculate an angle by the cosine law

Calculate h from

1

2bh

Calculate h by right trigonometry

Calculate rectangle from

lw

Calculate trap . from

a b

2

h

Sum the two areas

Calculations:Calculations are shown for both alternatives.


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The trigonometric alternative:

For angle C (by the cosine law) :

( ) ( )

2 2 2

2 2 2

2 cos

11.1 15.4 17.0 2 15.4 17.0 cos

cos 0.76958

39.684o

c a b ab C

C

C

C

= +

= +

=

=

For h (by right triangle trigonometry) :

sin 39.68415.4

15.4sin 39.684 9.8337

o h

h cm

=

= =

For trapezoid :

221.5 38.5 9.8337 295.02

Area cm+

= =

The area of the trapezoid is 295.0 cm2.St d T d J 13

271 mensuration notes oh1

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