1. Geometry Construction Math
This presentation illustrates the use of geometry in construction.
Construction Math

2. Construction drawings are composed of various angles in order to represent the object that is being built. Therefore, it is important to be able to distinguish the different angles from one another.
Straight angles are flat lines, which measure to be 180 degrees
Two perpendicular lines form a right angle which measures to be 90 degrees. This is the most common angle in construction and is signified in drawings using the right angle symbol shown on the screen.
An acute angle measure between 0 and 90 degrees. The most common acute angles are 30, 45, and 60 degrees.
An obtuse angle is any angle that measures between 90 and 180 degrees.
Two angles are complementary when they add up to 90 degrees.
Two angles are supplementary when they add up to 180 degrees
Angles

3. Triangles are also important in construction drawings
Triangles are also important in construction drawings. Therefore, you need to become familiar with the various types. Remember that triangle are made up of 3 angles that makes up 180 degrees
Isosceles triangles have 2 equal sides and 2 equal angles. A line that bisects the isosceles triangle (meaning it cuts the triangle in half) from the base to the highest point forms two adjacent right triangles.
Scalene triangles have 3 different sides of unequal length and has 3 different angles.
Equilateral triangles have 3 equal sides and 3 equal angles
A right triangle has one 90 degree angle
An acute triangle is made up of all acute angles
And an obtuse triangle has one obtuse angle.
Think about it: Why can’t a triangle have more that one obtuse angle?
Types of Triangles

4. The Pythagorean theorem states that the sum of the squares of the two shorter sides is equal to the square of the longest side (also know as the hypotenuse). The formula is squared + b squared = c squared
The hypotenuse is always across from the 90 degree angle
Pythagorean Formula

5. Right triangles often follow the 3-4-5 rule
Right triangles often follow the rule. This means that one side is 3 units long, one side is 4 units long and the hypotenuse is 5 units long. This rule also applies to multiples of 3 4 and 5 as well. You will often see triangles or triangles.
3-4-5 Rule

6. Circles Formulas: Pi (π) = 3.14 Circumference = 2πr Diameter (d) = 2r
Radius (r) = 𝑑 2
Circles have their own special terminology.
Pi is a mathematical constant value of approximately 3.14 used to determine the area and the circumference of a circle.
The circumference of a circle is the length of the closed line that form a circle. Its formula is 2 pi r.
The diameter of a circle is the length of a straight line that crosses from one side of a circle through the center point to a point on the opposite side. The diameter is the longest straight line you can draw through a circle.
The radius of a circle is the length of a straight line from the center point of the circle to the outside of the circle. It is equal to half of the diameter
Circles

7. Area of Shapes Area of a Rectangle = l x w Area of a Square = 𝑙 2
Area of a Circle = π 𝑟 2
Area of a Triangle = 1 2 bh
Area is the measure of the surface of an object. Area is measured in units squared. For example, you could give an area in inches square which mean you multiplied the inches in width by the inches in height which gives you two dimensions.
The formula for the area of a rectangle is length times width.
The formula for the area of a square is also length times width but since each of the side are equal you could just square one side length. Therefore, its formula would be length squared
The formula for the area of a circle is pi times radius squared. Remember that pi is equal to 3.14.
And finally, the formula for the area of a triangle is one half base time height. The base is the side the triangle sits on and the height is the length of the triangle from the base to its highest point.
Area of Shapes

8. Volume of Shapes Volume of a Rectangular Prism = l x w x h
Volume of a Cube = 𝑙 3
Volume of a Cylinder = π 𝑟 2 h
Volume of a Triangular Prism = 1 2 bhd
Volume is measures in units cubed. For example, you could give an volume in inches cubed which mean you multiplied the inches in width by the inches in height and the inches in depth which gives you three dimensions.
The volume of a rectangular prism (a 3-d rectangle) would be length x width x height.
The volume of a cube (a 3d square) would also be length x width x height but each side should be the same length you can just cube one of the side lengths to find the volume.
The volume of cylinder is equal to pi r squared time height (or the area of the circle times the height of the cylinder)
The volume of a triangular prism is equal to on half base time height times depth (or the thickness of the prism)
Volume of Shapes

9. Geometry Practice Questions

10. Triangles & Area What is the area within this triangular gable fence?
8’-0”
32’-0”
Triangles & Area

11. Answer Area of a Triangle = 1 2 bh Height = 8’-0” Base = 32’-0”
x 8=128 sq.ft.
To find the area of a triangle, you should used the formula ½ base times height.
In this problem, the height of the triangle is 8 ft. and the base is 32 ft.
Therefore when you multiply ½ times 8 times 32, you get 128 sq. feet
Answer

12. What is the area of the hip roof shown if the length of one side is 22 feet and the common rafter length is 14 feet 6 inches?
14’-6”
Triangles & Area
22’-0”
22’-0”

13. Answer Area of a Triangle = 1 2 bh Base = 22’-0”
Height = 14’-6” = =
x =
x 4=638 sq. ft.
To find the area of this roof, you must realize 4 congruent triangles make up the roof.
Therefore, you should find the area of one of the triangles using the formula ½ base times height and multiply the area by 4.
In this case, the base is 22 ft. and the height is 14 feet 6 inches or 14.5 feet.
When you multiply ½ times 22 time 14.5 you get 159 ½ sq. feet /2 times 4 equals 638 sq. ft.
Answer

14. Find the number of square feet of form work required for the end of a concrete retaining wall illustrated below. (Hint: break the trapezoid into a triangle and a rectangle)
Write the answer as a fraction and a decimal
3’-0”
Triangles & Area
11’-3”
9’-6”

15. Answer Rectangle = l x w l = 11 3 12 = 11 1 4 ft. Triangle = 1 2 bh
w = 3 ft.
3 x = = 33.75
Triangle = 1 2 bh
b = – 3= = ft.
h = ft. = ft.
1 2 ( x ) = = 36.56
Total: = = sq. ft.
When you divide the trapezoid into a rectangle and a triangle,
Your rectangle has a length of 11 and 3 twelfths of a foot or 11 and 1 fourth of a foot. and a width of 3 ft.
Since area equals length times width, 11 and 1 fourth times 3 equals 33 and 3 fourths sq. ft.
Your triangle has a base of 9 and 6 twelfths of a foot minus 3 feet or 6 and a half feet. And a height of 11 and 3 twelfths ft. or 11 and 1 fourth ft.
Using ½ base times height, you get ½ times 6 and a half times 11 and a forth which gives you 36 and 9 sixteenths sq. feet.
To get the total square footage, add the area of your rectangle to the area of your triangle. 33 and 3 quarters plus 36 and 9 sixteenths equals 70 and 5 sixteenths or 70 and thirty one hundredths of a square foot.
Answer

16. Determine the floor area in the semicircular bay window shown
Determine the floor area in the semicircular bay window shown. Round your answer to the nearest thousandth.
3’-6”
Circles & Area

17. Answer Area of a circle = π 𝑟 2 r = 3 6 12 = 3.5 3.14 x 3.5 2 = 38.465
÷ 2 = sq. ft.
To find the area inside the bay window, you should use the formula pi 𝑟 2 .
The radius of the circle is 3 and six twelfths inches or 3 and a half inches
When you multiply pi times 3 and a half squared you get 38 and 4 hundred sixty 5 thousands inches.
Since the bay window only is a half circle, you divide 38 and 4 hundred sixty 5 thousands by 2 and find out that the area inside the bay window is 19 and 2 hundred thirty three thousandths square feet.
Answer

18. A rough floor is laid under the space occupied by the gymnasium track shown. Determine the area occupied by the track. Do not include the space inside the track.
5’-0”
20’-0”
35’-0”
Circles & Area

19. Answer Area of a circle = π 𝑟 2 Area of a Rectangle = l x w
r (whole) = 20 ft.
r (inner) = 20 – 5 = 15 ft.
l (whole) = 35 ft. / w (whole) = 40 ft.
l (inner) = 35 ft. / w (inner) = 30 ft.
Area (whole) = (3.14 x ) + (35 x 40) = 2656 sq. ft.
Area (inner) = (3.14 x ) + (35 x 30) = sq. ft.
Area of track = = sq. ft.
To determine the area of the track, first you must realize that a track is composed of a rectangle and a circle.
Next, you determine the radius, length and width of both the whole track including the field and just the field.
For the whole track, the radius of the circular parts of the track is 20 ft., the length of the rectangular part is 35 ft. and the width is 40 feet (the diameter of the circle).
For the inner part of the track, the radius of the circular parts of the track is 5 ft. less than 20 which equals 15 ft., the length of the rectangular part is 35 ft. and the width is 30 feet.
Now you need to determine the area of the inner part of the track and subtract that value from the area of the whole track.
The area of the whole track equals pi times 20 squared plus 35 times 40 giving 2,656 sq. feet.
The area of the inner part of the track equals pi times 15 squared plus 35 times 30 giving 1,756 and a half sq. feet
The area of the track equals which would give you sq. ft.
Answer

20. How many more cubic feet are in container A than in container B?
Dimensions:
2’ x 2’ x 2’
Dimensions:
4’ x 4’ x 4’ B A
Volume

21. Answer Volume of a Cube = 𝑙 3 Volume (A) = 4 3 = 64 ft. 3
Volume (B) = = 8 ft. 3
Difference = = 56 ft. 3
Since volume of a cube is l cubed.
The volume of cube a is 64 and the volume of cube b is 8.
The difference between the volume of the 2 cubes is 56 ft. cubed.
Answer

22. What is the capacity in cubic feet of the circular silo shown
What is the capacity in cubic feet of the circular silo shown? Put your answer as a decimal. (π = )
18’-0”
42’-6”
Volume 4”

23. Answer Volume of a Cylinder = π 𝑟 2 h r = 9 ft.
h = 42’-6” – 4” = 42’-2”= ft.
x x = 10, ft. 3
To find the volume of a cylinder, used the formula pi r squared times height.
The radius of the cylinder is 8 ft. and the height is 4 inches less than 42’ 6 inches or 42 and 1 thousand 6 hundred 67 ten thousandths feet.
Therefore the area of the cylinder is pi times 9 squared times 42 and 1 thousand 6 hundred 67 ten thousandths which equals 10,730 and thirteen hundredths feet cubed.
Answer

24. Find the length of the stringer required for the stairway if
Rise = 12’-0” and run = 16’-0”
Rise = 5’-2” and run = 7’-6”
Rise = 2’-7” and run = 3’-9”
Rise
(Altitude)
Right Triangles
Run (Base)

25. Answers 𝑎 2 + 𝑏 2 = 𝑐 2 or 𝑎 2 + 𝑏 2 = c 12 2 + 16 2 = 20 ft.
To solve for the length of the stringer, used the Pythagorean Theorem.
For part a. the square root of 12 squared plus 16 squared equals 20 ft.
For part b. the square root of 5 and 2 twelfths squared plus 7 and 6 twelfths squared equals 9 and 11 hundredths of a foot
For part c. the square root of 2 and 7 twelfths squared plus 3 and 9 twelfths squared equals 4 and 55 hundredths of a foot
Answers

26. What is the volume of the cupboard under the staircase shown?
Hypotenuse = 12’-5”
Rise = 9’- 2”
Width = 3’-6”
Right Triangles

27. 𝑎 2 + 𝑏 2 = 𝑐 2 or 𝑐 2 − 𝑎 2 = b
Volume of a Triangular Prism = 1 2 bhd
b = ( ) 2 − ( ) 2 = 8.38 ft.
h = ft.
d = ft.
Volume = x 8.38 x x = ft. 3
To solve for the volume of the cupboard under the staircase, you need to determine the base of the triangle.
To do so, you should use the Pythagorean theorem. Therefore, the square root of 12 and 5 twelfths squared minus 9 and 2 twelfths squared equals 8.38 ft.
You know the height is 9 and 2 twelfths feet and the depth is 3 and 6 twelfths feet.
Therefore, one half times 8.38 times 9 and 2 twelfths feet times 3 and 6 twelfths which equals 134 and forty three hundredths feet cubed.
Answer