Download presentation

1
**Construction Geometry**

Right Rectangular Prisms Surface Area Volume

2
Rectangular Prisms Right rectangular prisms are 3 dimensional rectangles. We often think of them as closed boxes or, in construction, examples would be rectangular concrete slabs.

3
Rectangular Prisms A right prism has bases which meet the lateral faces at right angles. A right rectangular prism has bases which are rectangles and form right angles with the other faces.

4
Surface Area Surface area can be thought of as the amount of wrapping paper, with no overlap, needed to cover a box.

5
**Surface Area Split into 3 separate rectangles. Front/back sides**

Top/bottom sides Right/left sides Find the areas of each (LxW) and double. Sum the areas. 10” 8” 8 “ 6” 10 in 8 in A = 80 sq in 10 in 6 in A= 60 sq in 6 in 8 in A = 48 sq in

6
**Surface Area 2(80) = 160 sq. in. 2(60) = 120 sq. in 2(48) = 96 sq. in**

10” 8” 8 “ 6” 10 in 8 in A = 80 sq in 10 in 6 in A= 60 sq in 6in 8 in A = 48 sq in

7
Cube A cube is a right rectangular prism. All its sides are congruent squares. All 6 faces have the same area. So the surface area of a cube = x (area of one face). Face = (4 x 4) = 16 ft2 Surface area = 6(16) = 96 ft2 4 ft

8
Surface Area The surface area of a rectangular prism can be found using a formula. SA= 2(LW + LH + WH) This formula is found on the Math Reference Sheet.

9
**Formula for a rectangular prism SA = 2(LW+ LH + WH)**

Surface Area Formula for a rectangular prism SA = 2(LW+ LH + WH) Width Height Length

10
Practice #1 Determine the surface area of the right rectangular prism using the formula. SA = 2(LW+ LH + WH) 2 mm 10 mm 5 mm

11
**Practice #1 SA = 2(LW + LH + WH) 2(10x2 + 10x5 + 2x5) 2(20 + 50 + 10)**

2(80) SA = 160 mm2 2 mm 10 mm 5 mm

12
Application Building wrap is commonly used in construction on exterior walls.

13
Application Exterior wrapping protects the structure from exterior water and air penetration. Interior space

14
Application But it also allows moisture from inside the building to escape. Exterior space moisture inside

15
Practice #2 Determine how much moisture wrap is needed for this structure. 10’ 12’ 22’

16
**Practice #2 2(10x12) = 240 2(10x22) = 440 1(12x22) = 264 SA = 944 ft2**

17
Volume Volume is the measure of the amount of space occupied by an object. Volume can also be thought of as the amount that an object can hold.

18
Volume Volume is the number of cubic units that a solid can hold. 1 cubic yard = cubic feet 27 26 25 24 23 22 3 feet 21 20 19 3 feet 3 feet

19
**Volume The volume of a rectangular prism has the formula:**

V = L*W*H This formula is found on the Math Reference Sheet.

20
Volume Volume is determined by the product of the 3 dimensions of a rectangular prism: height, length, width. Units for volume are “cubic” (cu) units or un3. V = (L x W x H) height width length

21
Practice #3 Determine the amount of concrete needed to replace this damaged slab. V = (L x W x H) 12’ 12’ 1’ thick

22
**Practice #3 V = (L x W x H) = (12 x 12 x 1) V = 144 ft3**

For cubic yards: 1 yd3 = 27 ft3 144 = 5⅓ yd3 27 12’ 12’ 1’

23
Application The footing is the most vital part of a foundation.

24
Application The foundation wall transfers weight to the footing.

25
Application The footing transfers the weight of the structure to the ground.

26
Application The foundation wall thickness is determined by the anticipated load of the structure. wall thickness

27
Application The heavier the load of the structure, the thicker the wall should be. wall thickness

28
Application The thickness of the footing is then determined by the wall thickness. X Foundation wall X footing 2X

29
**Application Steel reinforces the concrete.**

A footing should be poured in one piece for best results.

30
Practice #4 Determine the number of cubic feet of concrete needed for this footing. 52’ 2’ deep 1’ thick 22’

31
**Practice #4 52’ 2’ deep 1’ thick 22’**

Solve by adding the volumes of 4 separate sections OR outer section volume - inner section volume. 52’ 2’ deep 1’ thick 22’

32
**Practice #4 Volume (outer) = 52(22)(2) = 2288 ft3**

Volume (inner) = 50(20)(2) = 2000 ft3 52’ 50’ 2’ deep 1’ thick 22’ 20’

33
**Practice #4 Volume (outer) - Volume (inner) =**

2288 ft ft3 = 288 ft3 52’ 50’ 2’ deep 1’ thick 2288 ft3 2000 ft3 22’ 20’

34
Practice #5 Determine the volume and surface area for each of the cubes. 9’ 5’

35
**Practice #5 Volume = 5’ x 5’ x 5’ = 125 ft3**

Surface area = (5x5) x 6 = 150 ft2 Volume = 9’ x 9’ x 9’ = 729 ft3 Surface area = (9x9) x 6 = 486 ft2 9’ 5’

36
**Practice & Assessment Materials**

You are now ready for the practice problems for this lesson. After completion and review, take the assessment for this lesson.

Similar presentations

Presentation is loading. Please wait....

OK

Area, Perimeter, and Volume

Area, Perimeter, and Volume

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on area of plane figures worksheets Ppt on shipping corporation of india Mis ppt on nokia cell Ppt online ticket Ppt on regional transport office delhi Ppt on introduction to object-oriented programming definition Ppt on conservation of momentum example Ppt on unity in diversity studio Ppt on childhood obesity in india Ppt on acid-base indicators examples