version - Mathematics in Education and Industry · The extension is symmetrical about a vertical...

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ContentsInitial Problem Statement 2 Narrative 3-9 Solutions 10-24 Appendix 25

Dormer Window ExtensionHow can an engineer determine lengths and angles from the drawings of three-dimensional objects?

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How can an engineer determine lengths and angles from the drawings of three-dimensional objects?

Before an extension or alteration can be made to

a building, an engineer must plan and design the

details. This is done using both two-dimensional

projections and three-dimensional drawings.

Once drawn, the representation allows design

lengths and angles to be determined which can

then be used in the construction.

Dormer Window ExtensionInitial Problem Statement

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Narrative IntroductionThe following drawings show the location and size of a planned dormer window extension to a house. The extension is symmetrical about a vertical centre line on the front view (front elevation). All dimensions are given in mm.

Figure 1

Discussion 1How could you produce such drawings and what might they be used for?

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Activity 1The architect has not yet stated the length x. Use trigonometry to calculate its value in mm to the nearest whole number.

Hint HintThe length x forms one side of a right angled triangle. The length of a second side is given on the diagram. If you know the value of one of the acute angles in the triangle you can determine x.

The two angles shown are equal as the top of the extension is parallel with the horizontal line drawn through the points where the roof meets the walls.

800 800

3100

x

9922

A

A

Figure 2

HintThe roof is symmetrical so the angle can be calculated from the half-width of the building and the pitch height of the roof.

Discussion 2Is there another method you could use to calculate x?

HintCan you see two similar triangles?

Activity 2Produce a three-dimensional sketch of what the frame for the extension will look like. Label your sketch with the known lengths.

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Some of the dimensions and angles can be found using more than one method. Identify alternative

calculations and use them to check your results.

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2. ConstructionIn order to construct the extension a hole of the appropriate size and shape needs to be made in the existing roof.

800

800

2500

2561

A

a

b

B

C

Figure 3

The construction will have some engineering tolerances built in so you can use 2561 mm as marked and take A = 32.0°.

Activity 3Sketch the shape of the hole required in the roof. Calculate the dimensions a and b to the nearest mm and the angles B and C to 1 decimal place.

HintUse the original two-dimensional drawings to help you calculate lengths and ratios.

Figure 4

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3. Effects of the extension on the buildingDiscussion 3Adding an extension changes the design of the house. What effects externally and internally might arise as a result of these changes? What might you need to calculate from the drawing to evaluate the magnitude of these effects?

Discussion 4Look at the diagram below. How many exterior plane surfaces are there in the design? How many uniquely shaped exterior plane surfaces are there?

800

800

2500

2561

32°

c

1960.148…

D

E d

Q

R V

W

P Figure 5

Activity 4Sketch the roof section QRVW. Calculate the lengths c and d giving results to the nearest mm value. Calculate the angles D and E, giving results to 1 d.p.

HintUse the original two-dimensional drawings to help you calculate lengths and ratios.

Figure 6

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4. Calculating the surface areaYou have discovered that in order to calculate the total surface area of the extension you need to calculate the area of three unique plane shapes; a roof section, the front and a triangular side. You have also calculated the dimensions of the sides of the plane shapes.

A roof section is shown below. All dimensions are shown in mm and have been rounded to appropriate values for construction allowing for engineering tolerances.

1484

2561

1280.5

1960

2561

1960

1280.5

1484

Figure 7

Figure 8

Discussion 5What would be an appropriate unit for the area?

Activity 5Make an order of magnitude estimate of the area.

Activity 6Establish an approximate upper and lower bound on the area.

Activity 7Calculate the area of the roof section using the above dimensions giving your result in appropriate units, as discussed above.

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Discussion 6What is the effect of rounding the lengths on the calculated area? Recall that the base was previously calculated lengths are 1484.082…, 1280.5 and 1960.148….

Activity 8Calculate the area of the front and side parts as shown below and give the total new external surface area.

800

2500

1600

Figure 9

800

1280.5

1510

Figure 10

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5. Checking your resultsAll of the calculations you have carried out in this problem can be automated using CAD software.

Activity 9Create a CAD model of the building and extension. Use it to check your results.

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Solutions Introduction

Discussion 1 solutionThe engineering drawings shown could be made using pencil and paper or, more likely, using a CAD system. They are often produced to show how a design will look from several viewing angles and are necessary to demonstrate compliance if any planning regulations are in place. They can also show how individual

components, the frame for example, fit together on site and can be used to estimate material usage.

Activity 1 solutionThe length x forms one side of a right angled triangle. The length of a second side is given on the diagram as 800 + 800 = 1600.

Figure 11

The two angles shown in the diagram below are equal as the top of the extension is parallel with the horizontal line drawn through the points where the roof meets the walls.

800 800

3100

x

9922

A

A

Figure 12

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tan-1 is the inverse tangent function. It is also sometimes

written as arctan or atan.

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As the roof is symmetrical the angle A is given by

tan A =( )

=

31009922 2

31004961

This gives the angle A as

tan

tan

.

A

A

A

=

=

=

−

31004961

31004961

32 00

1

…The value of x is give by

tan

tan

Ax

x

xA

=+

=

=

800 800

1600

1600

At this point you could use the angle you have just calculated (remembering to keep all the digits in your calculator) x

A=

=

= ( )

1600

160032 0

2561

tan

tan . …nearest whole number

Discussion 2 solutionHowever, you could save a bit of effort because you have already determined what tan A is – you didn’t need to evaluate the angle for this calculation! Recall that from your previous calculation you determined that

tan A =31004961

Substituting this directly into the calculation of x gives

x

A=

=

=×

=

=

1600

160031004961

1600 49613100

7 937 6003100

25

tan

661 nearest whole number( )

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The last method is an algebraic form of similar triangles. Look again at the diagram.

800 800

3100

x

9922

A

A

Similar triangles

Figure 13

The extension triangle is similar to the half-roof triangle. This means that

x

x

x

1600

99222

3100

1600496131001600 4961

31007937600

31002

=

=

=×

=

= 5561 nearest whole number( )

Activity 2 solutionThe two projections of the extension can be sketched in three-dimensions as shown below relative to the house

Figure 14

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A sketch of just the frame of the extension with known dimensions is shown below.

800

800

2500

2561

Figure 15

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2. ConstructionActivity 3 solutionThe hole in the roof has the following shape. Construction lines and points P, Q, R, S and T have been added to help with the calculation of lengths and angles.

2500

a

b

B

C

P

Q

R

T

S U

Figure 16

To calculate a consider the points P, Q and R in the three-dimensional and two-dimensional side views. Note that a = length PQ.

800

800

2500

2561

32°

a

b

B

C

P

Q

R

S

T

Figure 17

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Keep all the digits in your calculator from

the previous calculation when using SR = a

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800 800

2561

32°

32°

P, T

Q, S R

Figure 18

From the lower diagram you can calculate a = length PQ using trigonometric functions

sin

sin

32 800

80032

1510

�

�

=

=

= ( )

a

a

a mm, nearest whole value

There is an alternative method. As the vertical height from P and T to Q and S is equal to the vertical height from Q and S to R (both are 800 mm) then the point Q must bisect the line TR, i.e. PQ = TS = ½ TR.

The length of TR can be calculated using Pythagoras’s theorem

TR2 2 2

2 2

2561 800 800

2561 16006 558 721 2 560 0009118 721

= + +( )= += +=

Taking the square root of both sides

TR =

=

9118 7213019 722. …

The length a = length PQ = ½ TR so that a = = × = ( )1

212 3019 722 1510TR mm, nearest whole number. …

The two calculations give the same result.

To calculate the length b you can use Pythagoras’s theorem by noting that b2 2 2= +SR SQSQ is half the width of the figure, i.e. SQ = ½ × 2500 = 1250 (mm). From the previous calculation you have shown that SR = TS and that TS = PQ = a.

Substituting these values gives

b2 2 2

2 21509 861 12502 279 680 25 1562 5003 842180 25

= +

= += +=

SR SQ.

..

…

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b =

= ( )3 842180 25

1960

.

mm, nearest whole numberThe angle B can be calculated by noting that

B = +∠90� SQR

∠SQR can be calculated using, for example,

sin

.

..

∠ =

=

=

=

SQR SRb

ab1509 8611960 1480 770

……

… ∠ =

= ( )

−SQR1 d.p.

sin ..

1 0 77050 4

…

Of course you could also one of the other trigonometric relationships

cos∠ = =SQR SQb b

1250

or

tan∠ = =SQR SR

SQa

1250All three methods give the same value.

The angle B can be calculated from

B = +∠

= +

= ( )

9090 50 4140 4

�

� �

�

SQR

1 d.p..

.

To calculate the angle C you note that ∆RUQ is isosceles with ∠ = ∠SQR SUR.

From the properties of triangles

∠ +∠ + = ⇒ ×∠ + =SQR SUR SQRC C180 2 180� �

You have already calculated ∠SQR. Substituting the value (keeping all digits in your calculator from previous calculations) gives

2 50 378 180100 757 180

180 100 75779 2

× + =+ =

= −

= (

.

..

.

……

…

CCCC 1 d.p.))

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There is an alternative method of calculating C. The internal angles of the quadrilateral PQRT must sum to 360°. Two of the angles are right angles while the third is the angle B. The angle QRS = ½ C as ∆RUQ is isosceles. This gives

90 90 360

360 90 90

180180

2 180

12

12

12

� � �

� � �

�

�

�

+ + + =

+ = − −

=

= −

=

B C

B C

C B

C −−( )B

You have calculated B previously. Substituting its value (keeping all the digits in your calculator) gives

C B= −( )= −( )= ×

= ( )

2 180

2 180 140 378

2 39 62179 2

…

…

.

.. 1 d.p.

This result agrees with the previous method.

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3. Effects of the extension on the buildingDiscussion 3 solutionInternally the extension will increase the living space available. If the extra space is to be heated you need to ensure that the current heating system will have enough spare capacity to do this or consider what alternative or additional heating must be supplied. To evaluate this you need to calculate the volume of the extra space

produced. You may also need to calculate the surface area added, as heat will be lost through the walls, windows and roof.

Externally, the extension will change the area of the roof that must be used in the rain collection calculation in order to demonstrate compliance with the relevant British Standard. You must account for this to ensure that the existing guttering can accommodate any extra flow. To calculate the effects of this you need to calculate the areas of the roof and wall sections.

Discussion 4 solutionThere are five exterior plane surfaces to consider. These are the front, the two triangular sides and two sections for the roof. The back face of the object represents the hole that must be made in the roof. The two triangular side sections and the two roof sections are the same size and shape. This means that there are 3 unique exterior plane surface shapes.

Activity 4 solutionThe roof section QRVW has the following shape

R

Q

W V

2561

1960.148…

c

D

E

d

Figure 19

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The dimension d is calculated from the triangle section shown

800

800

2500

Q

W

R V

d

X

Figure 20

The point X is halfway along the base of the triangle. The length WX is therefore 2500/2 = 1250.

The distance XV is shown as 800. Using Pythagoras’s theorem

d 2 2 2

2 21250 8001562 500 640 0002 202 500

= +

= += +=

WX XV


d =

= ( )2 202 500

1484 mm, nearest whole numberThe distance c can be calculated by looking at the side view. In this diagram c = length WQ.

800 800

2561

32°

32°

P

Q R

W

V

Figure 21

There are several methods.

Method 1You have already calculated the length PQ = a = 1509.861…. The length WQ = c is then given using trigonometry as c

a=

== ×

=

length WQ

mm, nearest whole

cos. .32

1509 861 0 8481281

… …

number( )

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Method 2Triangles PRV and PQW are similar. The triangle PRV is obtained by enlarging the triangle PQW with a scale factor 2. You can see this as in triangle PQW the length of PW is 800 while in triangle PRV the length of PV is 1600. This means that the length WQ is half the length of VR, which is shown on the diagram, so that

c ==

= ×

=

length WQlength VR

mm, nearest whole numb

1212 2561

1281 eer( )To calculate the angles reconsider the diagram with the known and calculated lengths added.

R

Q

W V

2561

1960.148…

c=1280.5

D

E

d = 1484.082…

Y

Figure 22

You have just found that length WQ = c is half the length VR. This means you can construct the triangle QRY where

length RY = length YV = length WQ = c

Angle E can be calculated using, for example,

sin

.

.

..

E

d

=

=

=

=

YQQR

1960 1481484 0821960 1480 757

………

…

E =

= ( )

−sin ..

1 0 75749 2

…

1 d.p.

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Of course you could also use one of the other trigonometric relationships

cos.

..

E c= = =

RYQR 1960 148

1280 51960 148… …

or

tan ..

E dc

= = =YQRY

1484 0821280 5

…

All three results give the same value.

The angle D can be calculated from D = +∠90� YQR∠YQR is calculated from triangle QRY as

∠ +∠ +∠ =YQR QRY RYQ 180�

You know that ∠ =QRY E�

and ∠ =RYQ 90�

so that ∠ +∠ +∠ =

∠ + + =∠ = − −

= −= −

YQR QRY RYQYQR

YQR

18090 180

180 909090 49 2

EE

E. 111

40 788…

…= .This gives

D = +∠

= +

= ( )

9090 40 788130 8

…

YQR

1 d.p..

.There is an alternative method of calculating D. The internal angles of the quadrilateral WQRV must sum to 360°. Two of the angles are right angles while the third is the angle E. This gives

90 90 360360 90 90180

� � �

� � �

�

+ + + =

= − − −

= −

E DD ED E

You have calculated E previously. Substituting its value (keeping all the digits in your calculator) gives D E= −

= −

= ( )

180180 49 211130 8

….. 1 d.p.

This result agrees with the previous method.

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4. Calculating the surface areaDiscussion 5 solutionThe dimensions are given in mm. This is an easy unit to measure and can provide lengths that are accurate enough for general construction purposes without having to give values to many decimal places. However, in the design the lengths have an order of magnitude of ~1000. This means that an area will have a value

of order of magnitude ~millions. This is cumbersome to write and far too accurate for calculation purposes.

If the lengths were converted to cm this would still lead to large values for the area. The lengths are to an order of magnitude ~100 (in cm) so areas would be to an order of magnitude ~10 000 (in cm2).

Activity 5 solutionIf the lengths were converted to m this would lead to a reasonable unit for the area. The lengths are to an order of magnitude ~1 (in m) so areas would be to an order of magnitude ~1 (in m2). You might want to report the areas in m2 to 2 or 3 d.p. to prevent the reporting from being too coarse.

Activity 6 solutionAn approximate upper bound for the area can be made by considering a rectangle of sides 1.5 m and 2.5 m (1484 mm ~ 1.5 m and 2561 mm ~ 2.5 m). An approximate lower bound can be made by considering a rectangle of sides 1.5 m and 1.3 m (1484 mm ~ 1.5 m and 1280.5 mm ~ 1.3 m).

2.5

1.5

1.3

Actual area Upper bound Lower bound

Figure 23The bounds are

Upper = 2.5 × 1.5 = 3.75 m2

Lower = 1.3 × 1.5 = 1.95 m2

Notice that the order of magnitude for your bounds agrees with your previous estimate.

Activity 7 solutionThe actual area of the roof section is given by the trapezium of vertical sides 2561 and 1280.5 with a base of 1484. This gives an area of (remembering to convert mm to m to obtain an answer in m2).

Area = +( )×= × ×

= (

12

12

2 561 1 2805 1 484

3 8415 1 484

2 85

. . .

. .

. m 2 d.p.2 ))

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Discussion 6 solutionIf the lengths to full accuracy are used the area is calculated as Area = +( )×

= × ×

=

12

12

2 561 1 2805 1 484082

3 8415 1 4840822 85

. . .

. ..

……

0555090…Comparing this with the full digits from the previous calculation

Area = +( )×= × ×

=

12

12

2 561 1 2805 1 484

3 8415 1 4842 850393

. . .

. ..

it is seen that the values differ from the 4th decimal place. The error is less than one hundredth of one percent.

Activity 8 solution

800

2500

1600

Figure 24

The area of the front is given by the area of a rectangle of sides 800 and 2500 and a triangle of base 2500 and perpendicular height 800. This gives an area of (remembering to convert mm to m to obtain an answer in m2).

Area = × + × ×0 8 2 5 2 5 0 812. . . .

rectangle area triangle area

= +

= ( )2 1

3 2m

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The area of the side is given by the area of a triangle of base 1280.5 and perpendicular height 800. This gives an area of (remembering to convert mm to m to obtain an answer in m2).

Area = × ×

= ( )12 1 2805 0 8

0 51

. .

. m , 2 d.p.2

The total surface area is given by

Area = 2 × roof section area + 2 × triangle section area + front area Area = × + × +

= ( )2 2 85 2 0 51 3

9 72 2

. .

.

… …

m 2 d.p.

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Appendixmathematical coverageUse trigonometry and coordinate geometry to solve engineering problems• KnowandbeabletousePythagoras'theorem• Knowthemeaningsofsine,cosineandtangent• Beabletofindsine,cosineandtangentofanyangle• Beabletousesine,cosineandtangenttofindunknownsidesandanglesinright-angledtriangles• Beabletofindsides,anglesandareasinfiguresinvolvingmorethanonetriangle• Solvesimpleequationsinvolvingtrigonometricfunctions• Beabletointerpretdrawingsofthree-dimensionalobjects• Beabletodrawsimplethree-dimensionalobjects• Beabletocalculatelengthsandanglesinthree-dimensionalobjects

Use algebra to solve engineering problems• Be able to use appropriate units• Be able to convert from one set of units to another• Be able to comment on the expected order of magnitude of an answer and to decide if the

answer is reasonable• Be able to establish reasonable upper and lower bounds for the answer to a problem• Work with formulae, including those for the areas and perimeters of plane shapes, and the

surface areas and volumes of regular solids

version - Mathematics in Education and Industry · The extension is symmetrical about a vertical...

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