Ccgps Geometry Day 21 Homework

Presentation on theme: "Daily Check Find x. 1. 2. 3. 4. (2x + 8)º 3 (6x – 16)º 3 3 2 x 4."— Presentation transcript:

1 Daily Check Find x. 1. 2. 3. 4. (2x + 8)º 3 (6x – 16)º 3 3 2 x 4

2 Review Homework

3 CCGPS Geometry Day 27 (9-13-13) UNIT QUESTION: What special properties are found with the parts of a circle? Standard: MMC9-12.G.C.1-5,G.GMD.1-3 Today’s Question: How do we find the volume of cylinders, prisms, cones, and pyramids? Standard: MMC9-12.G.GMD1 and 3

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5 VOLUME = the number of cubic units contained in its interior VOLUME has cubic units Cm 3, ft 3, units 3

6 The volume of a cube is side 3 4 ft Volume = lengthwidthheight Volume = s 3 Volume = 4 3 Volume = 64 ft 3 In a CUBE they are all the same

7 When a prism is NOT a cube… B (area of the BASE) The formula for B will depend on the shape of the base. B (area of the BASE) The formula for B will depend on the shape of the base.

8 Volume Formula for Prisms & Cylinders V = Bh

9 B = area of the BASE h = HEIGHT Volume of a Prism V = 12 in 3 4 in 3 in 2 in

10 Volume of a Cylinder 7 cm 5 cm V = 769.7 cm 3 Round to the nearest tenth.

11 WS 1 – 5

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13 How many cones will it take to fill the cylinder? They have the same size base. They also have the same height.

14 Volume Formula for Cones & Pyramids

15 8 ft 12 ft 10 ft Find the volume.

16 2 cm 3 cm What is the volume of this cone? Round to the nearest tenth. 3.6 cm

17 WS 6 – 8

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20 r Radius of a Sphere

21 If you cut a sphere right down the middle you would create two congruent halves called HEMISPHERES. You can think of the globe. The equator cuts the earth into the northern and southern hemisphere.

22 Look at the cross section formed when you cut a sphere in half. What shape is it? A circle!!! This is called the GREAT CIRCLE of the sphere.

23 Surface Area of a Sphere

24 8 in Surface Area of a Sphere (round to the nearest hundredths)

25 10 cm Surface Area of a Sphere (round to the nearest hundredths)

26 25 in The circumference of a great circle of a sphere is 25 inches. Find the surface area of the sphere. (Round to the nearest tenths.)

27 A sphere is inscribed in a cube of volume 27 cubic meters. What is the surface area of the sphere? Give an exact answer and an answer rounded to the nearest hundredth.

28 Volume of a Sphere

29 2 cm Volume of a Sphere (round to the nearest hundredths)

30 10 cm Volume of a Sphere (round to the nearest hundredths)

31 Volume of a Sphere A sphere is inside a cube. The cube has a volume of 27 cm 3. Find volume of the sphere. Round to the nearest hundredths.

32 WS 9 – 16

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Presentation on theme: "Warm up Find the missing side.. Skills Check CCGPS Geometry Applications of Right Triangle Trigonometry."— Presentation transcript:

1 Warm up Find the missing side.

2 Skills Check

3 CCGPS Geometry Applications of Right Triangle Trigonometry

4 Use the 3 ratios – sin, cos and tan to solve application problems. Solving Word Problems Choose the easiest ratio(s) to use based on what information you are given in the problem.

5 Draw a Picture When solving math problems, it can be very helpful to draw a picture of the situation if none is given. Here is an example. Find the missing sides and angles for Triangle FRY. Given that angle Y is the right angle, YR = 68, and FR = 88. 68 88 r The picture helps to visualize what we know and what we want to find!

6 1. From a point 80m from the base of a tower, the angle of elevation is 28˚. How tall is the tower? 80 28 ˚ x Using the 28˚ angle as a reference, we know opposite and adjacent sides. Use tan 28˚ = 80 (tan 28˚) = x 80 (.5317) = x x ≈ 42.5 About 43 m tan

7 2. A ladder that is 20 ft is leaning against the side of a building. If the angle formed between the ladder and ground is 75˚, how far is the bottom of the ladder from the base of the building? ladder building 20 x 75˚ Using the 75˚ angle as a reference, we know hypotenuse and adjacent side. Use cos 75˚ = 20 (cos 75˚) = x 20 (.2588) = x x ≈ 5.2 About 5 ft. cos

8 3. When the sun is 62˚ above the horizon, a building casts a shadow 18m long. How tall is the building? x 18 shadow 62˚ Using the 62˚ angle as a reference, we know opposite and adjacent side. Use tan 62˚ = 18 (tan 62˚) = x 18 (1.8807) = x x ≈ 33.9 About 34 m tan

9 4. A kite is flying at an angle of elevation of about 55˚. Ignoring the sag in the string, find the height of the kite if 85m of string have been let out. string 85 x 55˚ kite Using the 55˚ angle as a reference, we know hypotenuse and opposite side. Use sin 55˚ = 85 (sin 55˚) = x 85 (.8192) = x x ≈ 69.6 About 70 m sin

10 5. A 5.50 foot person standing 10 feet from a street light casts a 14 foot shadow. What is the height of the streetlight? 5.5 14 shadow x˚ tan x˚ = x° ≈ 21.45° About 9.4 ft. 10

11 Depression and Elevation If a person on the ground looks up to the top of a building, the angle formed between the line of sight and the horizontal is called the angle of elevation. If a person standing on the top of a building looks down at a car on the ground, the angle formed between the line of sight and the horizontal is called the angle of depression. horizontal line of sight horizontal angle of elevation angle of depression

12 6. The angle of depression from the top of a tower to a boulder on the ground is 38º. If the tower is 25m high, how far from the base of the tower is the boulder? 25 x angle of depression 38º Using the 38˚ angle as a reference, we know opposite and adjacent side. Use tan 38˚ = 25/x (.7813) = 25/x X = 25/.7813 x ≈ 32.0 About 32 m tan Alternate Interior Angles are congruent