Learning Outcome 1 - Ac 1 (Analytical Methods)

7/28/2019 Learning Outcome 1 - Ac 1 (Analytical Methods)

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LOC 1 AC 1 : Construction Planning and Determination

(1) Application of Simple Equation

(a) Problem Solving Sequence Read the problem carefully at least twice.

If possible, draw a diagram. This will often help you to visualize themathematical relationship needed to write the equation.

Choose a symbol to represent the unknown quantity in the problem and write

what it represents.

Write an equation which express the information given in the problem and

which involves the unknown.

Solve the equation from point 4.

Check your solution both in the equation from point 4 and in the original

problem itself.

(b) Examples:(i) You need to tile the floor of a rectangular room with a wooden outer border

of 6 in. The floor of the room is 10 ft by 8 ft 2 in. How many rows of 4 in.by 4 in. tiles are needed to fit across the length of the room ?(Answer: 27 rows of tiles)

(ii) An interior wall measures 30 ft 4 in. long. It is to be divided by 10 evenlyspaced posts; each post is 4 in. by 4in. (Posts are to be located in the

corners.) What is the distance between the posts?(Answer: 36 in.)

(c) Classroom Exercise:(i) A set of eight built-in bookshelves is to be constructed in a room. The floor-

to-ceiling clearance is 8 ft 2 in. Each shelf is 1 in. thick. An equal space isto be left between shelves. What space should there be between each shelfand the next? (There is no shelf against the ceiling and no shelf on the floor)(Answer: 10 in. where there are 10 spaces + 8 pieces of shelves)

(ii) A given type of concrete contains twice as much sand as cement and 1.5times as much gravel as sand. How many yd3 of each must be used to make9 yd3 of concrete? Assume no loss of volume in mixing.(Answer: 1.5 yd3 of cement, 3 yd3 of sand, and 4.5 yd3 of gravel)

(iii) One side of a rectangular yard is bounded by the side of a house. The otherthree sides are to be fenced with 345 ft of fencing. The length of fenceopposite the house is 15ft less than either of the other two sides. Find thelength and width of the yard.

(Answer: 105 ft + 120 ft + 120 ft)


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(2) Application of Ratio and Proportion

(a) Types of measurement Ratio

The ratio between 2 numbers, a and b, is the first number divided by the secondnumber.

ProportionA proportion states that two ratios or two rates are equal,

d

c

b

a=

Percent and Proportion

Proportion can be written as100

R

B

P= where R is the rate written as a percent.

Direct Variation

When two quantities, y and x, change so that their ratios are constant, that is

2

2

1

1

x

y

x

y= , they are to said to vary directly. This relationship between the two

quantities is called direct variation.

Inverse Variation

If two quantities, y and x, change so that they product is constant, that is, if

2211 xyxy = , they are said to vary inversely. This relationship between the twoquantities is called inverse variation.

(b) Examples / Classroom Exercise:(i) A construction crew uses 4 buckets of cement and 12 buckets of sand to mix

a supply of concrete. What is the ratio of cement to sand?(Answer: 1 : 3)

(ii) The pitch of a roof is the ratio of the rise to the run of a rafter. The pitch ofthe roof is 2 : 7. Find the rise if the run is 21 ft.(Answer: 6 ft, which is the rise)

(iii) A concrete mix is composed of 1 part cement, 2.5 parts sand, and 4 partsgravel by volume. What is the percent by volume in the dry mix of (a)

cement, (b) sand, and (c) gravel)(Answer: (a) 13.3%, (b) 33.3% and (c) 53.3%)

(iv) A 400-lb force applied to the small piston of a hydraulic press produces a3600-lb force by the large piston. Find its mechanical advantage.(Answer: 9 : 1)

(v) The number of workers needed to complete a particular job is inverselyproportional to the number of hours that they work. If 12 electricians can

complete a job in 72 h, how long will it take 8 electricians to complete thesame job? Assume that each person works at the same rate no matter howmany persons are assigned to the job.(Answer: 108 h)


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(3) Application of Linear Equations Systems

(a) Types of Solution Sequence Solving Pairs of Linear Equations by Addition

1) If necessary, multiply both sides of one or both equations by a number (ornumbers) so that the numerical coefficients of one of the variables are negativesof each other.

2) Add the two equations from step 1 to obtain an equation containing onevariable.

3) Solve the equation from step 2 for the one remaining variable.4) Solve for the second variable by substituting the solution from step 3 in either

of the original equations.5) Check your solution by substituting the ordered pair in the original equation not

chosen in step 4.

Solving Applications Involving Equations with Two Variables

1) Choose a different variable for each of the two unknowns you are asked to find.Write what each variable represents.

2) Write the problems as two equations using both variables. To obtain these twoequations, look for two different relationships that express the two unknownquantities in equation form.

3) Solve this resulting system of equations using the methods given.4) Answer the question or questions asked in the problem.5) Check your answers using the original problem.

Solving Pairs of Linear Equations by Substitution

1) From either of the two given equations, solve for one variable in terms of theother.

2) Substitute the result from step 1 into the remaining equation. Note that this stepeliminates one variable.

3) Solve the equation from step 2 fro the remaining variable.4) Solve for the second variable by substituting the solution from step 3 into the

equation resulting from step 1.5) Check your solution by substituting the ordered pair in the original equation not

used in step 1.

(b) Examples / Classroom Exercise:(i) In a concrete mix, there is four times as much gravel as concrete. The total

volume is 20ft3. How much of each is in the mix? If x = the amount ofconcrete and y = the amount of gravel.(Answer: x = 4, y = 16)

(ii) A contractor finds a bill for $ 225 for 720 ceiling tiles. She knows that therewere two types of tiles used; one selling at 25 cents a tile and the other at 40cents a tile. How many of each type were used?(Answer: x = 420 and y = 300)

(iii) Enclose a rectangular yard with a fence so that the length is twice the width.The length of the 80-ft house is used to enclose part of one side of the yard. If580 ft of fencing are used, what are the dimensions of the yard?(Answer: length = 220 ft and width = 110 ft)


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(4) Application of Geometry

(a) Types of Geometry Form

Quadrilaterals

A parallelogram is a quadrilateral with opposite sides parallel. For example, ifside AB and CD are parallel, whereas AD and BC are parallel. Thus, polygon

ABCD is a parallelogram.Summary of Formulas for Area and Perimeter of Quadrilaterals

Quadrilateral Area Perimeter

Rectangle bh )(2 ba +

Square 2b b4

Parallelogram bh )(2 ba +

Rhombus bh b4

Trapezoidh

ba

+

2

dcba +++

Triangles

Triangles may be classified or named by the relative lengths of their sides. ThePythagorean theorem relates the lengths of the sides of any right triangle.

Summary of Formulas for Triangles

Pythagorean Theorem Area (1) Area (2)222

bac += bh2

1 ))()(( csbsass

Similar Polygons

Polygons with the same shape are called similar polygons. Polygons are similarwhen the corresponding angles are equal. When two polygons are similar, thelengths of the corresponding sides are proportional.

Circles

A circle is a plane curve consisting of all points at a given distance (called theradius, r) from a fixed point in the plane, called the centre. The diameter, d, ofthe circle is a line segment through the centre of the circle with endpoints on thecircle. Note that the length of the diameter equals the length of two radii, thatis, d = 2r. The circumference of a circle is the distance around the circle.

Summary of Formulas for Triangles

Circumference AreardC 2==

4

2

2 drA

==

Radian Measures (Refer to LOC 2 AC 3)

PrismsA prism is a solid whose sides are parallelograms and whose bases are one pair of

polygons that are parallel and congruent (same size and shape).

Summary of the formulas : The lateral surface area of a prism is the sum of the areasof the lateral faces of the prism. The total surface area of a prim is the sum of theareas of the lateral faces and the areas of the bases. Volume of prim is themultiplication of the area of base surface with height.


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Cylinders

A cylinder is a geometric solid with a curved lateral surface. The axis of acylinder is the line segment between the centres of the bases. The altitude, h, isthe shortest (perpendicular) distance between the bases. If the axis is

perpendicular to the bases, the cylinder is called a right circular cylinder and theaxis is the same length as the altitude.

Summary of Formulas for CylindersVolume Lateral Surface Area

hrV 2= rhA 2=

Pyramids and Cones

A pyramid is a geometric solid whose base is a polygon and whose lateral facesare triangles with a common vertex. The common vertex is called the apex ofthe pyramid. If a pyramid has a base that is a triangle, then it is called atriangular pyramid. If a pyramid has a base that is a hexagon, then it is called ahexagonal pyramid. In general, a pyramid is named by the shape of its base.

Summary of Formulas for Pyramid and Cones

Volume Lateral Surface Area

BhV3

1= for Pyramids

hrV 2

3

1= for Cones

rsA = where s is the slant height of the cone

Spheres

A sphere is a geometric solid formed by a closed curved surface with all pointson the surface the same distance from a given point, the centre. The givendistance from a point on the surface to the centre is called the radius.

Summary of Formulas for Spheres

Volume Lateral Surface Area3

4

3rV =

24 rA =

(b) Examples:

(i) Find the amount of sheathing needed for the roof, (page 375). How manysquares of shingles must be purchase? (1 square = 100 ft2)(Answer: 7 square, which is 12 ft x 28 ft x 2 nearest)

(ii) Find the length of the brace needed in the diagram attached (page 383).(Answer: 33.4 inches, including the 1 inch edge each side)

(iii) An inclined ramp is to be built so that it reaches a height of 6 ft over a 15 ftrun. Braces are placed every 5 ft. Find the height of braces x and y. (page

390)(Answer: x = 2 ft, y = 4 ft)


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(iv) A circular hole for an air duct 4.8 ft in diameter is cut from a rectangularpiece of dry wall 16 ft by 36 ft.(a) Find the area of the dry wall that is cut out.

(Answer: 18 ft2, 22/7 x 4.82/2)(b) Find the area of the dry wall that is left to be finished.

(Answer: 558 ft2, 36 ft x 16 ft (a))

(v) Find the volume of the wagon box shown (page 409)(Answer: 300 ft2 , (2 + 8) x 4 x 15)

(vi) How many square feet of sheet metal are needed to form a trough (page414)(Answer: 18.88 ft2, [(22/7 x 5)/12 x 14] + [22/7 x (5/12)2])

(vii) The volume of a right circular cone is 238 cm3. The radius of its base is5.82 cm. Find its height.

(Answer: 6.71cm)

(c) Classroom Exercises:(i) The Smith family plan to paint their home (page 383). The area of the

openings not to be painted is 325 ft2. The cost per square foot is RM 0.85.What will be the cost of painting the house?(Answer: RM 927.35, (48x9x2 + 24x9x2 + 1/2x24x5 325) x 0.85)

(ii) Find the length of the rafter for a roof (page 384).

(Answer: 14.5 ft for each side, 13 ft + 1.5 ft suspended edge)

(iii) A 6 ft by 8 ft bookcase is to be built. It has horizontal shelves every foot.A support is to be notched in the shelves diagonally from one corner to theopposite corner (page 391). At what point should each of the shelves benotched? That is, find lengths A, B, C, D and E, and how long is thecrosspiece?(Answer: A = 1.33 ft, B = 2.67 ft, C = 4 ft, D = 5.33 ft, E = 6.67 ft, 10ft)

(iv) Find the smallest circular duct that will rest on the ceiling joists withouttouching the ceiling.

(Answer: 7.08 in, R2 = (R 6)2 + 72)

(v) (1) What is the area of the lateral surface to be painted for figure (page 409)if windows takes 2 m2?(Answer: 2490 m2)

(2) What is the area of roof to be covered with shingles?(Answer: 1230 m2)

(3) What is the volume of concrete needed to be poured a floor 16 cm deep?(Answer: 188 m2)

(4) What is total surface area of the figure?(Answer: 4895 = 4900 m2, take out 0.16 m from floor)

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Learning Outcome 1 - Ac 1 (Analytical Methods)