Chapter Nine: Polygons
Index:
A: Interior & Exterior Angles
B: Perimeter & Area
C: Volume
D: Volume (Day 2)
E: Volume (Day 3) and Density
F: Volume (Day 3) and Density Day 2
G: Cross Sections and Multiple Choice Practice
Chapter 9 Polygons
•Perimeter: the continuous line forming the boundary of a closed geometric figure.
Any polygon: P = sum of all sides
Circle: C = πd = 2πr
•Rectangle: P = 2l+2w
• Square: P = 4 side
•Area: The amount of space inside the boundary of a flat (2-dimensional) object such as a triangle or circle.
•Triangle: A = 12 bh
•Rectangle: A = lw
•Square: A = (side)²
• Circle: A= π r²
•Trapezoid: A= 12 h (b1+b2)
•Parallelogram: A = b h
Volume:
Volume is described as the amount of space inside a figure. It is used in three-dimensional objects and can be represented as the amount of space an object takes up. Volume always uses cubic units (or units³).Volume Formulas: Where B is the area of the shape.
General Prisms : V =Bh
•Cone: V = 43 πr² h
•Sphere : V = 43 πr³
•Cylinder: V = 𝜋 r² h
•Pyramid: V = 13 Bh = 13 l x w x h
Geometry Interior/Exterior Angles 9A
Introduction to Polygons
A Polygon is a closed figure that is formed by 3 or more coplanar line segments. The line segments are the sides of the polygon and the endpoints are the vertices. Polygons are named by its vertices starting with any vertex and going in order in either direction. Polygons are classified by their number of sides.
***Regular Polygons are both Equilateral and Equiangular***
**7 sides can also be a Septagon**
**9 sides can also be a Novagon**
A Diagonal of a polygon is a line segment connecting one vertex to any other non consecutive vertex. (When connecting the diagonals, notice that triangles are formed!)
Types of Angles:
Interior Angles: the angles inside the polygon, each angle is less than 180o
Exterior Angles: the angles formed by extending one of the two adjacent sides
***** Special Relationship: Each interior and adjacent exterior angle is supplementary
To find the measure of an angle of a given polygon we utilize the following formulas (
Practice:
1.) If the measure of one exterior angle of a regular polygon is 72°, find the number of sides that the polygon has.
••The measure of each exterior angle of a regular polygon is always the same, and it's given by the formula:
•
••Given that one exterior angle is 72^\circ , we can set up the equation:
••
••To find the number of sides, we can rearrange the equation:
••
••
••So, the regular polygon has 5 sides.
2.) What is the measure of an interior and an exterior angle of a regular 46-gon. Round to the nearest tenth if necessary.
•To find the measure of an interior angle of a regular polygon, we use the formula:
••
••where n is the number of sides.
••For a regular 46-gon, n = 46 . Substituting this into the formula:
••
•
•
••To find the measure of an exterior angle, we use the fact that the sum of an interior angle and its corresponding exterior angle is 180^\circ in a regular polygon. Therefore, the exterior angle is:
••
•
•
••So, the measure of the interior angle is approximately 175.7^\circ , and the measure of the exterior angle is approximately 4.3^\circ .
3.) If the measure of one interior angle of a regular polygon is 144°, find the number of sides that the polygon has. Classify this polygon.
The number of sides of a regular polygon can be determined using the formula for the measure of each interior angle:
Where n is the number of sides of the polygon.
Given that the measure of one interior angle of the regular polygon is 144^\circ , we can set up the equation:
Solving for n :
So, the regular polygon has 10 sides. Since it has 10 sides, it is classified as a decagon.
4.) If the sum of the interior angles of a regular polygon is 2160°, how many sides does the polygon have?
•The sum of the interior angles of a polygon with n sides can be calculated using the formula:
•
•Given that the sum of the interior angles of the regular polygon is 2160^\circ , we can set up the equation:
•
•Solving for n :
•
•
•
•
•
•So, the regular polygon has 14 sides.
5.) Using the provided diagram, find the value of x.
a.)
In a polygon with n sides, the sum of the interior angles is given by the formula:
Given that the polygon has 4 sides, we can substitute n = 4 into the formula:
Now, we know that the sum of the interior angles of this polygon is 360°.
The interior angles of the polygon are given as 60°, x , x , x . So, we can write the equation:
Simplifying:
Subtracting 60° from both sides:
Dividing both sides by 3:
So, the value of x is 100^\circ .
b.)
In any polygon, the sum of the exterior angles is always 360^\circ. Since the polygon has 4 sides, the sum of its exterior angles is 360^\circ.
Given that the exterior angles are x, 2x, 3x, and 4x, the sum of these angles is:
To find the value of x, divide both sides of the equation by 10:
So, the value of x is 36^\circ.
c.)•Number of sides = n =5
•42°+a=180°
•a=180°-42° = 138°
•b = 180°-80°= 100°
•X°+100°+110°+2x°+138°=180°(5-2)
•3x + 348 = 540°
•3x = 192°
•X = 64°
2x = 128°
Geometry Interior/Exterior Angles 9A HW
Directions: Complete each of the following in the space provided. No work = no credit.
1.) Using hexagon HIJKLM
a.) Name all angles that are consecutive to ∠H ____________ and _______________________
b.) Name two diagonals. _____________ and _________________
c.) Name two consecutive sides._______ and ________________
Consecutive Angles Definition: When two parallel lines are crossed over by a transversal, there are two sets of angles formed on each of the parallel lines.
There are two types of consecutive angles according to their position with relation to parallel lines and the transversal. i.e.,
Consecutive interior angles & Consecutive exterior angles
Consecutive Interior Angles
Consecutive interior angles are angles that are within a straight line from one corner to the other.
These consecutive angles lie on the interior region of the two parallel lines and on the same side of the transversal. They are also known as same side interior angles or co-interior angles.
Consecutive Exterior Angles
These consecutive angles lie on the outside or exterior region of the two parallel lines and on the same side of the transversal.
In hexagon HIJKLM:
a.)Angles consecutive to angle H: 1. Angle I 2. Angle M
Angles consecutive to angle I: 1. Angle H 2. Angle J
b.) - Diagonals: HJ , IK ,MJ & HK
c.)- Consecutive sides: 1. Side HI 2. Side IJ, LK & KJ
2.) If the measure of one interior angle of a regular polygon is 150° , how many sides does the polygon have?
The measure of one interior angle of a regular polygon with n sides is given by the formula:
Given that the measure of one interior angle is 150^\circ, we can set up the equation:
Now, let's solve for n:
So, the polygon has 12 sides.
3.) Find the measure of an interior and an exterior angle of a regular polygon who has 18 sides.
•To find the measure of an interior angle of a regular polygon with ° n sides, you can use the formula:
••Interior angle} = \frac{180°(n - 2)}{n}
••Given that the polygon has 18 sides, plug in ° n = 18 into the formula:
••Interior angle} = \frac{180°(18 - 2)}{18}
• = \frac{180°16}{18}
• = \frac{2880°}{18} = 160°
••So, the measure of an interior angle of the regular 18-sided polygon is ° 160°.
••To find the measure of an exterior angle, you can use the fact that the sum of an interior angle and its corresponding exterior angle is always ° 180°. Therefore, the measure of an exterior angle of the polygon is ° 180°- 160°= 20°.
4.) If the measure of one exterior angle of a regular polygon is 24° , find the number of sides that the polygon has.
The sum of the exterior angles of any polygon is 360^\circ . Therefore, if the measure of one exterior angle of a regular polygon is 24^\circ , then the number of sides of the polygon can be found using the formula:
Substituting the given value, we get:
So, the regular polygon has 15 sides.
5.) If the sum of the interior angles of a regular polygon is 3060°, how many sides does the polygon have?
•The sum of the interior angles of a polygon can be calculated using the formula:
•Sum of interior angles} = (n - 2) 180°
•Where ° n is the number of sides of the polygon.
•Given that the sum of the interior angles of the regular polygon is ° 3060°, we can set up the equation:
•° (n - 2) 180°= 3060°
•Solving for ° n :
•° n - 2 = \frac{3060°}{180°}
•° n - 2 = 17
•° n = 17 + 2
•° n = 19
•So, the regular polygon has 19 sides.
6.) Using the diagram find the value of x.
•N=5
•98+104+120+90+x=180(n-2)
•412 + x = 180(5-2)
•412+x = 180 x 3
•412 + x = 540
•X = 540 - 412
•X = 128°
7.) Find the value of x.
•X + 47 + 102 + 50 + 95 =360
•X + 294 = 360
•X = 360-294
•X = 66°
8.)Find the value of m.
•3m+2m+m+2+100 = 360°
•6m+102=360
•6m = 360-102
•6m = 258
•M=258/6
•M = 43°
9.) Find the value of x and the measure of each angle.
N= 6
2x-50+x+40+80+x+20+x+150=180(6-2)
5x=720-240
X=480/5
X = 96°
X+20 = 96 + 20 = 116°
X+40 = 96 +40 =136°
2x-50 = 2(96)-50 = =192-50 = 142°
10.) Using the regular hexagon find both x and y.
N=6
•One interior angle = y= (180(n-2))/n
= (180(6-2))/6
= (180(4))/6 = 120°
•X = 180 – y
•X = 180-120
•X = 60°
Review Section:
11.) a.) What can be concluded about the relationship between line l and plane P?
b.) What can be concluded about the relationship between planes P and Q?
c.) What can be concluded about the relationship between lines l and m ?
d.) What can be concluded about segments AB and plane Q?
a) Perpendicular
b) Parallel
c) Parallel
d)Perpendicular
12.) If the rectangle below is continuously rotated about side w, which solid figure is formed? [1] pyramid
[2] rectangular prism
[3] cone
[4] cylinder
If the rectangle is continuously rotated about side w , the solid figure formed is a [4] cylinder.
13.) A sequence of transformations maps rectangle ABCD onto rectangle A”B”C”D”, as shown in the diagram. Which sequence of transformations maps ABCD onto A’B’C’D’ and then maps A’B’C’D’ onto A”B”C”D”? [1] a reflection followed by a rotation
[2] a reflection followed by a translation
[3] a translation followed by a rotation
[4] a translation followed by a reflection
Reflection means the object is mirrored from a base. So that’s from ABCD to A’B’C’D’ it reflected by x-axis
From A’B’C’D’ to A”B”C”D”, the rectangle rotate 90° counter clockwise. So reflection and rotation. Answer [1]
Homework Answers
Geometry Interior/Exterior Angles 9A HW
1.) a.) <M and <I
b.) MJ and HK
c.) LK and KJ
2.) n = 12 sides
3.) interior = 160 exterior = 20
4.) n = 15 sides
5.) n = 19 sides
6.) x = 128
7.) x = 66
8.) m = 43
9.) x = 96
10.) x = 60 y = 120
11.) a.) perpendicular
b.) parallel
c.) parallel
d.) perpendicular
12.) [4]
13.)[1]
Geometry Perimeter/Area 9B
Let’s Review before we start:
Perimeter: the continuous line forming the boundary of a closed geometric figure.
Any polygon: P = sum of all sides.
•Circle: C= 2πr
•Rectangle: P = 2l+2w
• Square: P = 4 side
•Area:
•The amount of space inside the boundary of a flat (2-dimensional) object such as a triangle or circle.
•Triangle: A = 1/2 bh
•Rectangle: A = lw
•Square: A = (side)²–
• Circle: π r^2
•Trapezoid: A= 1/2 h (b1+b2)
•Parallelogram: A = b h
• Perimeter & Area Formulas all must be memorized!
•
Practice:
1.) A rectangle has a width of 5mm and a length of 13mm. Find the perimeter.
To find the perimeter of a rectangle, you add up all the sides. Since opposite sides of a rectangle are equal:
Perimeter = 2 (length + width)
Given:
Width = 5 mm
Length = 13 mm
Substitute the values into the formula:
Perimeter = 2 (13 + 5)
= 2 x 18
= 36 mm
So, the perimeter of the rectangle is 36 mm.
2.) The perimeter of a pentagon is 30. The sides are represented by (y + 3), (2y + 7), (3y – 10), 15 and 3. What is the value of y?
To find the value of y , you need to add up all the sides of the pentagon and set the sum equal to the given perimeter.
Given:
Perimeter = 30
Sides: (y + 3) , (2y + 7) , (3y - 10) , 15, and 3
So, the perimeter of the pentagon is:
Simplify the expression:
Combine like terms:
Subtract 18 from both sides:
Divide both sides by 6:
So, the value of y is 2.
3.) What is the diameter of a circle whose area is 36𝜋?
To find the diameter of a circle given its area, you can use the formula for the area of a circle:
Where:- \pi is a constant (approximately 3.14159)
- r is the radius of the circle
Given that the area of the circle is 36\pi , we can set up the equation:
To solve for r , divide both sides by \pi :
Now, the diameter of the circle is twice the radius, so:
So, the diameter of the circle is 12 units.
4.) If the area of a circle is 49𝜋, what is the circumference in terms of 𝜋 ?
The formula for the area (A) of a circle is given by:
where r is the radius of the circle.
Given that the area of the circle is 49\pi, we can set up the equation:
Divide both sides by \pi :
Now, the circumference (C) of a circle is given by:
Substitute the value of r:
So, the circumference of the circle is 14\pi.
5.) Keira has a square poster that she is framing and placing on her wall. The poster has a diagonal 58 cm long and fits exactly inside the frame. The width of the frame around the picture is 4 cm. Determine and state the total area of the poster and frame to the nearest tenth of a square centimeter.
To find the area of the poster and frame, we first need to find the dimensions of the poster.
Given that the poster has a diagonal of 58 cm and is a square, we can use the Pythagorean theorem to find the length of each side (s) of the poster.
The Pythagorean theorem states that for a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b):
In this case, the diagonal of the square poster is the hypotenuse (c), and the sides of the square are the other two sides (s).
So, we have:
Now, the total width of the poster and frame is 41 + 4 + 4 = 49 cm (adding the width of the frame on both sides of the poster).
Similarly, the total height of the poster and frame is also 49 cm.
So, the total area of the poster and frame is:
Therefore, the total area of the poster and frame is 2401 cm².
6.) A walking path at a local park is modeled on the grid below, where the length of each grid square is 10 feet. The town needs to submit paperwork to pave the walking path. Determine and state, to the nearest square foot, the area of the walking path.
Area of the walking path = Area of 2 rectangle shaped walking path + Area of 2 semi circled walking path
Area of 2 rectangle shaped walking path = 2 x 90 x 10 = 1800 Square feet
Area of 2 semi circled walking path = 22/7 x 30 x 30 - 22/7 x 20 x 20 = 22/7 x (900-400) = 22/7 x 500
Area of the walking path = 1800 + 1571.42 = 3371 Square foot
Geometry Perimeter/Area 9B HW
Directions: Complete each of the following in the space provided. No work = no credit.
1.) If the circumference of a circle is 10𝜋 inches, what is the area in square inches of the circle?
The circumference (C) of a circle is related to its radius (r) and diameter (d) by the formula:
Given that the circumference is 10\pi inches, we can set up the equation:
Divide both sides by 2\pi:
Now, we can use the formula for the area (A) of a circle, which is:
Substitute the value of r:
So, the area of the circle is 25\pi square inches.
2.) A side of a regular pentagon is 16m. Find the side of a square having the same perimeter.
The perimeter P of a regular pentagon is given by the formula:
where s is the length of one side of the pentagon.
Given that the side of the regular pentagon is 16 m, we can calculate its perimeter:
Now, we need to find the side length of a square S with the same perimeter. In a square, all sides are equal, so we can divide the perimeter by 4 to find the length of one side:
Therefore, the side length of the square having the same perimeter is 20 meters.
3.) What is the base of a triangle whose area is 135 cm2 and whose height is 18cm?
The area A of a triangle can be calculated using the formula:
Given that the area of the triangle is 135 sq.cm and the height is 18 cm, we can rearrange the formula to solve for the base b:
Substituting the given values:
Therefore, the base of the triangle is 15 \, cm}.
4.) A countertop for a kitchen is modeled with the dimensions shown below. An 18-inch by 21-inch rectangle will be removed for the installation of the sink. What is the area of the top of the installed countertop, to the nearest square foot?
Area of rectangle 1 + Area of rectangle 2 – Area of sink =
Review Section:
5.) If ∆ABC ≅ ∆JKL ≅ ∆ RST , the BC must be congruent to:
(1) JL (2) JK (3) ST (4) RS
6.) What is the difference between the sum of the measures of the interior angles of a regular pentagon and the sum of the measures of the exterior angles of a regular pentagon?
(1) 36 (2) 72 (3) 108 (4) 180
The sum of the measures of the interior angles of a polygon can be found using the formula (n - 2) \times 180^\circ , where n is the number of sides of the polygon. For a regular pentagon (5 sides), the sum of the interior angles is (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ .
The sum of the measures of the exterior angles of any polygon, including a regular pentagon, is always 360^\circ .
Therefore, the difference between the sum of the measures of the interior angles and the sum of the measures of the exterior angles of a regular pentagon is 540^\circ - 360^\circ = 180^\circ .
Hence, the correct answer is option (4) 180.
7.) As shown in the diagram, AB and CD intersect at E, and AC // BD. Given , which equation is true?
∠AEC = ∠BED
∠EAC =∠EBD
AE = BE
EC = ED
AC = BD
Answer (2) : AE/BE = AC/BD
8.) Triangle RST and XYZ are drawn below. If RS =6, ST = 14, XY =9, YZ = 21 and ∠S ≅ ∠Y is ∆RST~ ∆XYZ? Justify your answer.
To determine if triangles RST and XYZ are similar, we need to check if their corresponding sides are proportional and if their corresponding angles are congruent.
Given:
- RS = 6 , ST = 14 , XY = 9 , YZ = 21
- ∠S and ∠Y are congruent
We can first check if the ratios of corresponding sides are equal:
- \frac{RS}{XY} = \frac{6}{9} = \frac{2}{3}
- \frac{ST}{YZ} = \frac{14}{21} = \frac{2}{3}
The ratios of corresponding sides are equal, indicating that the sides of the triangles are proportional.
Next, we verify if the corresponding angles are congruent:
- Given: ∠S and ∠Y are congruent
Since both the ratios of corresponding sides and the congruence of corresponding angles hold true, by the Side-Angle-Side (SAS) similarity criterion, we can conclude that triangles RST and XYZ are similar.
9.) Construct an equilateral triangle inscribed in circle T shown below. [leave all construction marks]
To construct an equilateral triangle inscribed in circle T , follow these steps:
1. Draw circle T with its center at point O .
2. Choose any point on the circumference of circle T and label it A .
3. Draw the radius of circle T from the center O to point A . Label the point where the radius intersects the circle as B .
4. Draw another radius from the center O to a point on the opposite side of the circle and label it C .
5. Draw a straight line segment connecting points B and C . This segment will pass through point A .
6. Label the point where the line segment intersects the circumference of the circle as D .
7. Draw the line segment AD . This line segment is the third side of the equilateral triangle.
8. Finally, draw the line segments BD and CD to complete the equilateral triangle \triangle ABC inscribed in circle T .
Make sure to use a compass and ruler to ensure accuracy in constructing the equilateral triangle and to leave all construction marks visible.
Homework Answers
Geometry Perimeter/Area 9B HW
1.)
2.) x = 20
3.) x = 15
4.) 19
5.) [3]
6.) [4]
7.) [2]
8.) Yes by SAS ~
9.) Construction
Geometry Volume 9C
Volume:
Volume is described as the amount of space inside a figure. It is used in three-dimensional objects and can be represented as the amount of space an object takes up. Volume always uses cubic units (or units3). It is important not to mix up units when solving volume, area, and perimeter problems.
Volume:
Volume is described as the amount of space inside a figure. It is used in three-dimensional objects and can be represented as the amount of space an object takes up. Volume always uses cubic units (or units³).Volume Formulas: Where B is the area of the shape.
General Prisms : V =Bh
•Cone: V = 43 πr² h
•Sphere : V = 43 πr³
•Cylinder: V = 𝜋 r² h
•Pyramid: V = 13 Bh = 13 l x w x h
Examples:
1.) The height of a cylinder is 15 in and the diameter of its base is 12 in. In cubic inches, find the volume of the cylinder , V = 𝜋𝑟² ℎ, . Leave the answer in terms of pi.
The volume V of a cylinder is given by the formula V = 𝜋 r² h , where r is the radius of the base and h is the height of the cylinder.
Given that the diameter of the base is 12 inches, the radius r is half of the diameter, so r =12/2 = 6 inches.
Also, the height h of the cylinder is 15 inches.
Now, we can substitute these values into the formula to find the volume:
V = 𝜋 x(6)² x15
V = 𝜋 x36 x15
V = 540𝜋
So, the volume of the cylinder is 540𝜋 cubic inches.
2.) A paper container in the shape of a right circular cone has a radius of 3 inches and a height of 8 inches. Determine and state the number of cubic inches in the volume, Cone: V = 1/3 πr^2 h , of the cone, in terms of π
The volume V of a cone is given by the formula V = \frac{1}{3} \pi r^2 h , where r is the radius of the base and h is the height of the cone.
Given that the radius r is 3 inches and the height h is 8 inches, we can substitute these values into the formula:
So, the volume of the cone is cubic inches.
3.) A regular pyramid has a height of 12 cm and a square base. If the volume of the pyramid, V = 1/3 Bh is 256 cubic centimeters, how many centimeters are in the length of one side of its base?
The volume V of a pyramid is given by the formula V = \frac{1}{3} Bh , where B is the area of the base and h is the height of the pyramid.
Given that the volume V is 256 cubic centimeters and the height h is 12 cm, we can substitute these values into the formula:
To solve for the area of the base B , we multiply both sides by \frac{3}{12} :
Since the base of the pyramid is square, and the area of a square is equal to the side length squared, we have:
To find the side length s of the square base, we take the square root of both sides:
So, the length of one side of the square base is 8 cm.
4.) A fish tank with a rectangular base has a volume of 3,360 cubic inches. The length and width of the tank are 14 inches and 12 inches, respectively. Find the height, in inches, of the tank.
To find the height (h) of the fish tank, we can use the formula for the volume of a rectangular prism: V = l \times w \times h, where l is the length, w is the width, and h is the height.
Given that the volume (V) of the tank is 3,360 cubic inches, and the length (l) and width (w) are 14 inches and 12 inches, respectively, we can rearrange the formula to solve for the height (h):
Plugging in the given values, we get:
So, the height of the fish tank is 20 inches.
5.) Determine the volume, V= Bh = = ½ b x h x l of the prism below. It is a regular prism with a right triangular base is 8 cm and height is 2.5 cm and length is 5.3 cm.
To find the volume (V) of the prism, we'll use the formula for the volume of a prism: V = Bh, where B is the area of the base and h is the height of the prism.
The base of the prism is a right triangle with legs of length 8 cm and 2.5 cm. The area (B) of a right triangle is given by the formula B = \frac{1}{2} \times base} \times height}. So, the area of the base is:
Given that the height (h) of the prism is 5.3 cm, we can now calculate the volume using the formula:
So, the volume of the prism is 53 \, cm}^3.
6.) The diameter of a sphere is 12 inches. What is the volume, Sphere : V = 4/3 πr^3 , of the sphere to the nearest cubic inch?
To find the volume (V) of the sphere, we'll use the formula for the volume of a sphere: V = \frac{4}{3} \pi r^3, where r is the radius of the sphere.
Given that the diameter of the sphere is 12 inches, the radius (r) is half of that, so r = \frac{12}{2} = 6 inches.
Now, we can substitute this value into the formula:
So, the volume of the sphere is 288 \pi cubic inches. = 905 cubic inches
7.) The height of a soda can is 8 in and the diameter of its base is 4 in. In cubic inches, find the volume of the can of soda, Cylinder: V = 𝜋𝑟^2 ℎ . Leave answer in terms of pi.
To find the volume (V) of the soda can, we'll use the formula for the volume of a cylinder: V = \pi r^2 h, where r is the radius of the base and h is the height.
Given that the diameter of the base is 4 inches, the radius (r) is half of that, so r = \frac{4}{2} = 2 inches.
The height (h) of the can is given as 8 inches.
Now, we can substitute these values into the formula:
So, the volume of the soda can is 32 \pi cubic inches.
Geometry Volume 9C HW
1.) As shown in the diagram, a regular pyramid has a square base whose side measures 6 inches. If the altitude of the pyramid measures 12 inches, its volume, in cubic inches is: Pyramid: V = 1/3 𝐵ℎ
[1] 72 [2]144 [3] 288 [4] 432
To find the volume of the pyramid, we'll use the formula for the volume (V) of a pyramid: V = \frac{1}{3}Bh, where B is the area of the base and h is the height.
Since the base is a square with side length 6 inches, the area of the base (B) is 6 \times 6 = 36 square inches}.
Given that the altitude (height) of the pyramid is 12 inches, we can substitute the values into the volume formula:
So, the correct answer is option [2], 144 cubic inches.
2.) A fish tank in the shape of a regular prism has dimensions of 14 inches, 16 inches and 10 inches. The tank contains 1680 cubic inches of water. What percent of the fish tank is empty? V=Bh
[1] 10 [2] 25 [3] 50 [4] 75
To find the volume of the fish tank, we use the formula for the volume (V) of a rectangular prism: V = Bh, where B is the area of the base and h is the height.
Given that the dimensions of the tank are 14 inches, 16 inches, and 10 inches, we can calculate its volume:
Now, to find the percentage of the tank that is empty, we subtract the volume of water from the total volume of the tank, and then divide by the total volume, and multiply by 100 to express the result as a percentage.
So, the correct answer is option [2], 25%.
3.) The Great Pyramid of Giza was constructed as a regular pyramid with a square base. It was built with an approximate volume of 2,592,276 cubic meters and a height of 146.5 meters. What was the length of one side of its base, to the nearest meter? Pyramid: V = 1/3 𝐵ℎ
[1] 73 [2] 77 [3] 133 [4] 230
The volume (V) of a regular pyramid is given by the formula V = \frac{1}{3}Bh, where B represents the area of the base and h is the height of the pyramid.
Given that the volume of the Great Pyramid of Giza is approximately 2,592,276 cubic meters and its height is 146.5 meters, we can use this information to find the area of the base, and then determine the length of one side of the square base.
We know that V = \frac{1}{3}Bh, so we can rearrange the formula to solve for the area of the base (B):
Substituting the given values:
Since the base of the Great Pyramid of Giza is a square, we can find the length of one side (s) by taking the square root of the area:
Rounded to the nearest meter, the length of one side of the base is approximately 230 meters. So, the correct answer is option [4].
4.) The volume of a rectangular prism is 144 cubic inches. The height of the prism is 8 inches. Which measurements in inches, could be the dimensions of the base?
[1] 3.3 by 5 [2] 2.5 by 7.2 [3] 12 by 8 [4] 9 by 9
The volume V of a rectangular prism is given by the formula V = Bh, where B represents the area of the base and h is the height.
Given that the volume of the prism is 144 cubic inches and the height is 8 inches, we can use this information to find the possible dimensions of the base.
We know that V = Bh, so we can rearrange the formula to solve for the area of the base B:
Substituting the given values:
So, the area of the base is 18 square inches.
Now, we need to find which set of dimensions gives an area of 18 square inches. We check each option:
1. Option [1]: 3.3 \times 5 = 16.5 square inches (not 18 square inches)
2. Option [2]: 2.5 \times 7.2 = 18 square inches (matches)
3. Option [3]: 12 \times 8 = 96 square inches (not 18 square inches)
4. Option [4]: 9 \times 9 = 81 square inches (not 18 square inches)
So, the only option that gives an area of 18 square inches is option [2]: 2.5 inches by 7.2 inches. Therefore, option [2] is correct.
5.) Walter wants to make 100 candles in the shape of a cone for his new candle business. The mold shown below will be used to make the candles. Each mold will have a height of 8 inches and a diameter of 3 inches. To the nearest cubic inch, what will the total volume of 100 candles? Cone: V = 1/3 πr^2 h
To find the volume of one cone-shaped candle, we use the formula for the volume of a cone:
Given that the height ( h ) of each candle is 8 inches and the diameter ( d ) is 3 inches, we first need to find the radius ( r ).
The radius is half of the diameter, so r = \frac{d}{2} = \frac{3}{2} = 1.5 inches.
Now, we can substitute the values into the formula:
So, the volume of one candle is 6 \pi cubic inches.
To find the total volume of 100 candles, we multiply the volume of one candle by 100:
Therefore, the total volume of 100 candles, to the nearest cubic inch, is 600 \pi cubic inches.
600 x 3.14 = 1885 in³
6.) A cylinder has a height of 7 cm and a base with a diameter of 10 cm. Determine the volume, in cubic centimeters, of the cylinder in terms of 𝜋, ? Cylinder: V = 𝜋𝑟^2 ℎ
To find the volume V of the cylinder, we'll use the formula for the volume of a cylinder:
Given that the diameter d of the base of the cylinder is 10 cm, we can find the radius r by dividing the diameter by 2:
And the height h of the cylinder is given as 7 cm.
Now, we can substitute the values of r and h into the volume formula and calculate the volume:
So, the volume of the cylinder is 175 \pi cubic centimeters.
7.) The diameter of a sphere is 15 inches. What is the volume of the sphere to the nearest tenth of a cubic inch? V = 4/3 𝜋𝑟^3
To find the volume V of the sphere, we'll use the formula for the volume of a sphere:
Given that the diameter d of the sphere is 15 inches, we can find the radius r by dividing the diameter by 2:
Now, we can substitute the radius into the volume formula and calculate the volume:
Rounded to the nearest tenth of a cubic inch, the volume of the sphere is approximately 1767.1 cubic inches.
Review Section:
8.) In ∆ ABC, M∠A=X, M∠B = 2X+2 and M∠C=3X+4 . What type of triangle is ∆ ABC ?To determine the type of triangle \triangle ABC , we need to examine the measures of its angles.
Given:
- m A = X
- m∠ B = 2X + 2
- m∠ C = 3X + 4
We know that the sum of the interior angles of a triangle is 180^\circ . Therefore, we can write the equation:
Solving for X :
Now that we have found the value of X , we can find the measures of the angles:
In this case, since m∠ C = 91^\circ , which is greater than 90^\circ , \triangle ABC is an obtuse scalene triangle.
9.) Construct a square inscribed in circle T shown below. [leave all construction marks]
To construct a square inscribed in circle T , follow these steps:
1. Draw the circle T with its center at point O .
2. Draw a diameter AB of circle T .
3. Bisect AB at point M . This will be the midpoint of \overline{AB} .
4. From point M , draw a perpendicular line to \overline{AB} intersecting the circle at points C and D .
5. Connect points A and C and points B and D .
6. ABCD is the inscribed square.
This construction ensures that each side of the square is tangent to the circle and that the vertices of the square lie on the circumference of the circle.
Homework Answers
Geometry Volume 9C HW
1.) [2]
2.) [2]
3.) [4]
4.) [2]
5.) 1885 in³
6.) 175𝜋
7.) 1767.1 in³
8.) Obtuse scalene triangle
9.) Construction
Geometry Volume (Day 2) 9D
1.) Given a rectangular pyramid with a height of 14 cm and the base measurements of 13 cm and 6 cm, find the volume Pyramid: V = 1/3 Bh
•To find the volume V of a rectangular pyramid, you can use the formula:
•
•Where B is the area of the base and h is the height of the pyramid.
•For a rectangular pyramid, the base is a rectangle, so the area of the base B is given by:
•
•Given that the length is 13 cm and the width is 6 cm, we can calculate the area of the base:
•
•Now, we can use the volume formula to find the volume of the pyramid:
•
•
•
•So, the volume of the rectangular pyramid is 364 cm³.
2.) Given a right circular cone with a height of 11 and a slant height of √137. find the length of the radius of the base. Using the radius, then find the volume of the cone to the nearest tenth. Cone: V = 1/3 πr^2 h
To find the length of the radius r of the base of the cone, we can use the Pythagorean theorem. The slant height l , height h , and radius r form a right triangle where l is the hypotenuse.
According to the Pythagorean theorem:
Given that l = 137 and h = 11 , we can solve for r :
Now that we have found the radius r = 4 , we can use it to find the volume V of the cone using the formula:
Substituting the given values, we get:
So, the length of the radius of the base is 4 units, and the volume of the cone is approximately 184.7 cubic units.
3.) Given a square pyramid with a slant height of 13 and a base side length of 10, find the volume of the pyramid. Pyramid: V = 1/3 Bh
To find the volume V of the square pyramid, we can use the formula:
where B is the area of the base and h is the height of the pyramid.
For a square base, the area B is given by the formula for the area of a square: B = s^2 , where s is the length of a side of the square base.
Given that the base side length s = 10 and the slant height l = 13 , we can use the Pythagorean theorem to find the height h of the pyramid. In this case, the slant height l , the height h , and half the diagonal of the square base form a right triangle.
According to the Pythagorean theorem:
Now that we have found the height h = 12 , we can use it to find the volume V of the pyramid:
So, the volume of the square pyramid is 400 cubic units.
4.) Given a right circular cone with a slant height of √185 and a diameter of 16, find the volume of the cone in terms of pi.
To find the volume V of the right circular cone in terms of \pi , we can use the formula:
where r is the radius of the base and h is the height of the cone.
Given that the diameter of the cone is 16 , the radius r is half of the diameter, so r = \frac{16}{2} = 8 units.
We are also given the slant height l = \sqrt{185} units.
To find the height h , we can use the Pythagorean theorem. The height h , radius r , and slant height l form a right triangle.
According to the Pythagorean theorem:
Substituting the given values, we get:
Now that we have found the height h = 11 units, we can use it to find the volume V of the cone:
So, the volume of the cone in terms of \pi is \frac{704}{3} \pi cubic units.
5.) Two prisms with equal altitudes have equal volumes. The base of one prism is a square with a side length of 5 inches. The base of the second prism is a rectangle with a side length of 10 inches. Determine and state, in inches, the measure of the width of the rectangle.
Since the volumes of the two prisms are equal, and they have equal altitudes, we can set up the equation:
For a prism, the volume V is given by the formula:
where B is the area of the base and h is the height.
For the first prism, with a square base of side length 5 inches, the area of the base B_1 is:
For the second prism, with a rectangular base of width w inches and length 10 inches, the area of the base B_2 is:
Since the volumes of the prisms are equal, we have:
Given that the heights h_1 and h_2 are equal, we can cancel them out:
So, we have:
Solving for w :
So, the width of the rectangular base of the second prism is 2.5 inches.
6.) A sphere is inscribed inside a cube with edges of 6 cm. In cubic centimeters, what is the volume of the sphere, in terms of pi.
To find the volume of the inscribed sphere, we first need to find its radius, which is half the length of the side of the cube. Since the cube has edges of 6 cm, the radius of the sphere is cm.
The formula for the volume of a sphere is:
Substituting the radius cm into the formula:
So, the volume of the sphere, in terms of , is cubic centimeters.
7.) In the diagram below, a right circular cone with a radius of 3 inches has a slant height of 5 inches, and a right cylinder with a radius of 4 inches has a height of 6 inches. Determine and state the number of full cones of water needed to completely fill the cylinder with water.
To find the number of full cones needed to completely fill the cylinder with water, we first need to find the volume of the cylinder and the volume of one cone.
The formula for the volume of a cylinder is:
Substituting the given values, where inches and inches:
The formula for the volume of a cone is:
Substituting the given values, where inches and h=4 inches:
Now, to find the number of cones needed to fill the cylinder, we divide the volume of the cylinder by the volume of one cone:
Number of cones = 8
Since we can't have a fraction of a cone, we round up to the nearest whole number. Therefore, we need 8 full cones to completely fill the cylinder with water.
Geometry Volume (Day 2) 9D HW
1.) The diameter of a basketball is approximately 9.5 inches and the diameter of a tennis ball is approximately 2.5 inches. The volume of the basketball is about how many times greater than the volume of the tennis ball? Sphere : V = 4/3 πr^3
[1] 3591 [2] 65 [3] 55 [4] 4
To find out how many times greater the volume of the basketball is compared to the volume of the tennis ball, we can compare their volumes using the formula for the volume of a sphere:
First, let's calculate the volume of the basketball using its diameter of approximately 9.5 inches (which gives a radius of inches):
Next, let's calculate the volume of the tennis ball using its diameter of approximately 2.5 inches (which gives a radius of inches):
Now, we can find out how many times greater the volume of the basketball is compared to the volume of the tennis ball by dividing the volume of the basketball by the volume of the tennis ball:
Rounded to the nearest whole number, the volume of the basketball is about 55 times greater than the volume of the tennis ball. So, the correct option is [3] 55.
2.) The diameter of a sphere is 15 inches. What is the volume of the sphere, to the nearest tenth of an inch?
[1] 706.9 [2] 1767.1 [3] 2827.4 [4] 14,137.2
To find the volume of the sphere, we can use the formula:
where is the radius of the sphere.
Given that the diameter of the sphere is 15 inches, we can find the radius by dividing the diameter by 2:
Now, we can plug this value into the formula for volume:
Rounded to the nearest tenth, the volume of the sphere is approximately 1767.1 cubic inches.
So, the correct option is [2] 1767.1.
3.) A sphere inscribed inside a cube with edges of 6 cm. In cubic centimeters, what is the volume of the sphere in terms of pi?
[1] 12𝜋 [2]36𝜋 [3] 48 𝜋 [4] 288𝜋
Given that the radius of the sphere is 3 cm (half the edge length of the cube), we can use the formula for the volume of a sphere:
Substituting cm, we get:
So, the volume of the sphere in terms of is .
Therefore, the correct answer is [2] .
4.) Tennis balls are sold in cylindrical cans with the balls stacked one on top of the other. A tennis ball has a diameter of 6.7cm. To the nearest cubic centimeter, what is the minimum volume of the can that holds a stack of 4 tennis balls? (Hint: The radius of the ball is the radius of the cylinder. The diameter of each ball is involved in the height of the cylinder)
[1] 236 [2] 282 [3] 564 [4] 945
To find the volume of a cylinder, you can use the formula:
Where:
- is the volume of the cylinder
- is the radius of the cylinder
- is the height of the cylinder
Given that the diameter of the cylinder is 6.7 cm, we first need to find the radius:
Now, we can use the given height of the cylinder, which is 26.8 cm, along with the calculated radius to find the volume:
Therefore, the volume of the cylinder is approximately .
5.) Given a square pyramid with a slant height of √193 and a base side length of 14, find the volume.
To find the volume of a square pyramid, you can use the formula:
Where:
- is the area of the base
- is the height of the pyramid
For a square pyramid, the area of the base () is given by the formula , where is the side length of the square base.
Given that the base side length is , and the slant height () is , we can find the height () using the Pythagorean theorem:
Now, we can find the volume using the formula for the volume of a pyramid:
Therefore, the volume of the square pyramid is cubic units.
6.) Using the fact that the volume of a sphere is represented by V = 4/3 πr^3 , find the volume of the hemisphere if the diameter is 18 feet in terms of π
The volume of a hemisphere can be calculated using the formula for the volume of a sphere and then dividing by 2, because a hemisphere is half of a sphere.
Given the diameter of the hemisphere is 18 feet, the radius () is half of the diameter, so feet.
Now, we can use the formula for the volume of a sphere:
Substituting the value of , we get:
VOLUME OF HEMISPHERE = 972PI/2 = 486PI
Therefore, the volume of the hemisphere is cubic feet.
6.) 486 𝜋
7.) Given a right circular cone with a slant height of and a diameter of 30, find the volume of the cone in terms of π.
To find the volume of a cone, we use the formula:
where is the radius of the base and is the height of the cone.
Given that the diameter of the cone is 30 units, the radius is half of the diameter, so units.
To find the height , we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (slant height ) is equal to the sum of the squares of the lengths of the other two sides (radius and height ).
So, the volume of the cone in terms of is cubic units.
Review Section:
8.) In the diagram∆ DEF is the image of ∆ABC after a clockwise rotation of 180° and a dilation. AC = 4.5, AB = 3, BC = 5.5, FD = 9, DE = 6, FE = 11, Which relationship must always be
Answer: △ABC and △DEF are similar and, in similar triangles corresponding angles are equal, So, ratio of corresponding angles are also equal.
[4] (𝑚∠𝐵)/(𝑚∠𝐸 )=(𝑚∠𝐶)/(𝑚∠𝐹)
9.) To find the distance across a pond from point B to point C, a surveyor drew the diagram. The measurements he made are indicated on his diagram. Use the surveyor’s information to determine and state the distance from point B to point C, to the nearest yard.
∆AED~∆ACB
So 𝐴𝐸/𝐴𝐶= 𝐸𝐷/(𝐶𝐵 )
CB = (𝐸𝐷 𝑋 𝐴𝐶)/𝐴𝐸= (120 𝑋 (230+85))/230
CB = (12 𝑋 315)/23
CB = 164.34 yards
Homework Answers
Geometry Volume (Day 2) 9D HW
1.) [3]
2.) [2]
3.) [2]
4.) [4]
5.)784
6.) 486 𝜋
7.) 1500𝜋
8.) [4]
9.) 164 yards
Geometry Volume (Day 3) & Density 9E
Cavalieri’s Principle:
If, in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are
1) The diagram below shows two figures. Figure A is a right triangular prism and figure B is an oblique triangular prism. The base of figure A has a height of 5 and a length of 8 and the height of prism A is 14. The base of figure B has a height of 8 and a length of 5 and the height of prism B is 14.
Use Cavalieri’s Principle to explain why the volumes of these two triangular prisms are equal. According to Cavalieri’s Principle, the volumes of these two triangular prisms are equal because they have the same _, __, and _.
Density:Density and volume are related. Density is described as the mass of the substance per unit volume. In other words, it is how much “stuff” (matter that makes up mass) that can fit inside its space (volume) . Formula : Density = mass/volume
:
Less Dense More Dense
1) A hemispherical tank is filled with water and has a diameter of 10 feet. If water weighs 62.4 pounds per cubic foot, what is the total weight of the water in a full tank, to the nearest pound?
To find the total weight of the water in the tank, we first need to find the volume of the water in the hemispherical tank.
The volume of a hemisphere can be calculated using the formula:
where is the radius of the hemisphere.
Given that the diameter of the tank is 10 feet, the radius is half of the diameter, so feet.
Plugging in the value of into the formula, we get:
Now, we know the volume of the water in the tank. To find the weight of the water, we multiply the volume by the weight per cubic foot, which is 62.4 pounds:
Rounded to the nearest pound, the total weight of the water in the full tank is 16336 pounds.
2) A wooden cube has an edge length of 6 cm and a mass of 137.8 grams. Determine the density of the cube, to the nearest thousandth. State which type of wood the cube is made of, using the density table below.
To find the density of the wooden cube, we use the formula:
Given that the edge length of the cube is 6 cm, the volume of the cube can be calculated using the formula for the volume of a cube:
Now, we can calculate the density:
To determine the type of wood the cube is made of, we compare its density to the densities provided in the table:
- Pine: 0.373 g/cm³
- Hemlock: 0.431 g/cm³
- Elm: 0.554 g/cm³
- Birch: 1.601 g/cm³
- Ash: 0.638 g/cm³
- Maple: 0.676 g/cm³
- Oak: 0.711 g/cm³
The density of the cube (0.638 g/cm³) is closest to the density of Ash (0.638 g/cm³). Therefore, the cube is most likely made of Ash wood.
More Difficult Volume Questions:
1) June 2015 Exam (manipulated):
The water tower in the picture below is modeled by the two dimensional figure beside it. The water tower is composed of a hemisphere, a cylinder, and a cone. Let C be the center of the hemisphere and let D be the center of the base of the cone.
a.) If AC = 8.5 feet, BF = 25 feet, ED = 9.1 feet, and m∠EFD = 47° , determine and state, to the nearest cubic foot, the volume of the water tower.
a.) To find the volume of the water tower, we need to find the volumes of the hemisphere, cylinder, and cone separately, and then add them together.
1. Volume of the hemisphere:
The formula for the volume of a hemisphere is , where is the radius.
Given that feet, the radius of the hemisphere is feet.
Therefore, the volume of the hemisphere is:
2. Volume of the cylinder:
The formula for the volume of a cylinder is , where is the radius and is the height.
Given that feet, which is also the height of the cylinder, and the radius is the same as that of the hemisphere ( feet).
Therefore, the volume of the cylinder is:
3. Volume of the cone:
The formula for the volume of a cone is , where is the radius and is the height.
Given that feet and , we need to calculate the radius and height of the cone.
Using trigonometry, we find that the radius of the cone is approximately feet and the height of the cone is approximately feet.
Therefore, the volume of the cone is:
Finally, the total volume of the water tower is the sum of the volumes of the hemisphere, cylinder, and cone:
b.) The water tower was constructed to hold a maximum of 400,000 pounds of water. If water weighs 62.4 pounds per cubic foot, can the water tower be filled to 85% of its volume and not exceed the weight limit? Justify your answer.
b.) To determine if the water tower can be filled to 85% of its volume without exceeding the weight limit, we first calculate 85% of the maximum weight capacity:
Then, we compare this value to the weight of the water tower when it's filled to 85% of its volume, calculated by multiplying the volume by the weight of water per cubic foot (62.4 pounds):
If the weight of the water tower when filled to 85% of its volume is less than or equal to the maximum weight capacity, then the water tower can be filled to 85% of its volume without exceeding the weight limit. Otherwise, it cannot.
Another method
The water tower is composed of three parts.
Volume of hemisphere = 1243 r³ = 1243 (8.5)³ = 1286.220392
Volume of Cylinder = r²h = (8.5)² x 25 = 5674.501731
EFD = 47 °
ED . EF sin 47° = DF. tan 47° = 9.12
Volume of cone = 13 r²h =13 (8.5)² x 9.1 = 688.50621
Volume total = 1286.220392 + 5674.501731 + 688.50621 = 7649.228333 = 7649 ft³
M = 7651.74 x 85 / 2 x 62.4 = 4050848.29 > 400000
So it is impossible. No, the water tower cannot be filled to 85% of its Volume.
Geometry Volume (Day 3) & Density 9E HW
1) Sue believes that the two cylinders shown in the diagram below have equal volumes.
Is Sue correct? Explain why. First cylinder radius is 5 m and height is 11.5 m and second cylinder slant height is 12m.
Volume of Cylinder1= r²h = x 5 x5 x 11.5 = 287.5
Volume of Cylinder1= r²h = x 5 x5 x 11.5 = 287.5
Yes, by Cavalieri’s Principle Sue is correct, the 2 cylinders have the same volume.
2) Molly wishes to make a lawn ornament in the form of a solid sphere. The clay being used to make the sphere weighs .075 pound per cubic inch. If the sphere’s radius is 4 inches, what is the weight of the sphere, to the nearest pound?
Volume of Sphere = 43r³ = 434³ = 268.0825731 in³
Total Weight = V( weight per in³) = V sphere (0.075) = 20.10619298 = 20 lbs.
3) Lou has a solid clay brick in the shape of a rectangular prism with a length of 8 inches, a width of 3.5 inches, and a height of 2.25 inches. If the clay weighs 1.055 oz/in3, how much does Lou's brick weigh, to the nearest ounce?
the volume of a brick will be (8 × 3.5 × 2.25) = 63 cubic inches.
weight of clay = weight of brick
1 cubic inches of clay or bricks weigh = 1.055 ounces
63 cubic inches of clay or bricks weigh = V( weight) = V brick (1.055) = 66.465 = 66 ounces.
Therefore, weight of Lou's brick is 66 ounces. ( nearest ounces )
4) The table below shows the population and land area, in square miles, of four counties in New York State at the turn of the century. (Hint: the formula is population divided by land area)
Which county had the greatest population density?
(1) Broome (2) Dutchess (3) Niagara (4) Saratoga
Broome : 200536 706.82 = 283.716
Dutches : 280150 801.59 = 349.493
Niagara : 219846 522.95 = 420.396
Saratoga : 200635 811.84 = 247.136
Niagara>Dutchess> Broome > Saratoga
Niagara had the greatest population density.
Review Section:
5) August 2016 Exam:
Answer
(2) l || p. If two exterior angles on the same side of a transversal are supplementary, then the lines are parallel. The sum of 112 and 68 is 180 degrees.
6) August 2016 Exam Triangle MNP is the image of Triangle JKL after a 120° counterclockwise rotation about point Q. I f the measure of angle L is 47° and the measure of angle N is 57°. Determine the measure of angle M. Explain how you arrived at your answer.
Solution:
Because angle K is angle N after the rotation and angle L becomes angle P and angle J becomes angle M.
M∼J
N∼K
P∼L
∠ L= ∠ P
∆ MNP≌ ∆JKL
Rotation is a rigid motion, so their measures will be the same. On ∆ MNP, m ∠P = 47° and m ∠N = 57°.
180 -47 – 57 =76°
m∠M = 76°
Homework Answers
Geometry Volume (Day 3) & Density 9EHW
**All of these require A LOT of work. Failure to show work will not result in full credit for this assignment** 1) Yes with work
2) 20 pounds
3) 66 ounces
4) (3)
5) (2)
6) with explanation
Geometry Volume (Day 3) & Density 9F
1) A silo is used to store grain that farm animals eat during the winter months. The top of the silo is a hemispherical with a radius of 8 feet. The cylindrical body of the silo shares the same radius as the hemisphere and has a height of 40 feet.
The truck hauling grain to the silo has a rectangular container attached to the back that is 8 feet in length, 5 feet in width, and 4 feet in height. Determine the number of truckloads of grain required to fill an empty silo.
The volume of silo: 12x 43 x 8³ + x 8² x 40 = 8704 3
The volume of truck is : 8×5×4=160
The number of truckloads: 8704 3 ÷ 160 ≈ 57
57 truckloads of grain required to fill an empty silo.. .
2) The frozen yogurt cone is 12 cm in height and has a diameter of 6 cm. A scoop of frozen yogurt is placed on the wide end of the cone. The scoop is a sphere with a diameter of 6 cm. If the scoop of frozen yogurt melts into the cone, will the cone overflow? Explain your reasoning.
Solution :
V cone = 13 r²h = 13 x 3² x 12 = 36 cm³
V sphere = 43 r³ =43 x 27 = 36
So V cone = V sphere
Both the cone and the scoop of frozen yogurt have the same volume, which is 36π cubic cm. Since the volume of the melted frozen yogurt perfectly matches the volume of the cone, there will be no overflow. The melted frozen yogurt will fill the cone completely, with no excess.
3) June 2016 Exam:
A water glass can be modeled by a truncated right cone (a cone which is cut parallel to its base) as shown below:
The diameter of the top of the glass is 3 inches, the diameter at the bottom of the glass is 2 inches, and the height of the glass is 5 inches.
The base with a diameter of 2 inches must be parallel to the base with a diameter of 3 inches in order to find the height of the cone. Explain why. Determine and state, in inches, the height of the larger cone.
Determine and state, to the nearest tenth of a cubic inch, the volume of the water glass.
ABCD is the glass. DC = 3 inches
EC = DC 2 = 32inches
AB = 2 inch
FB = AB 2 = 22 = 1 inch
Height of the larger cone = EG
Height of glass = EF = 5 inches
tan(EGC) = ECEG = FBFG
Let FG = x
ECEF + FG = FBFG
32(5 + x ) = 1x
3x = 2(5+x)
3x = 10 + 2x
X = 10
Height of the larger cone = EG = EF + FG = 5+10 = 15 inches
b) Volume of frustum = 3 h ( R² + r² + Rx r)
H = 5 inches, R = 32 inches , r = 22 = 1 inch
Volume = .3 5[ ( 32)² + 1² + 32x 1)
Volume = 227x 53 [ 34+ 1 + 32)
= 227x 53 x 194 = 24.88 inch³
Geometry Volume (Day 3) & Density 9F HW
1) A trophy has a rectangular prism as its base and a pyramid that lies on top of the prism. If the monument weighs 2.5 grams per cubic inch, then what is the weight of the monument to the nearest gram?
V = V prism + V pyramid 13 Bh
V = 2 x 6 x 10 + 6 x 10 x 9 x 13 = 120 + 180 = 300 in ³
m = Density x volume = 300 in ³ x 2.5 grams / in = 750 g
2) A storage tank is in the shape of a cylinder with a hemisphere on the top. The highest point on the inside of the storage tank is 13 meters above the floor of the storage tank, and the diameter inside the cylinder is 8 meters. Determine and state, to the nearest cubic meter, the total volume inside the storage tank.
Given:
Highest point in storage tank = 13m
Diameter = 8m
To find: Total volume inside the storage tank.
Solution:
Since, diameter = 8
Radius = d/2 = 8/2 = 4m
Highest side = 13m
Other side = 13 - 9 = 4cm
According to the cylinder formulas -
πr²h + 2/3πr³
= π ( 4 )²( 9) + 2/3π ( 4)³
= 22/7 × 16 × 9 + 2/3 × 22/7 × 64
= 586
Answer: The total volume inside storage tank is 586 m³
3) A plastic container is in the shape of a right, circular cone with a base radius of 12 inches and a total height of 18 inches. The container is filled with water to a depth of 15 inches as shown.
(a) What is the volume of the plastic container, to the nearest cubic inch?
(b) Determine the radius, r, of the cone of water sitting in the plastic container.
(c) How much volume, to the nearest cubic inch, remains unfilled in the plastic container?
Volume of plastic container = 13r²h = 13(12)² x 18 = 864 = 2714.53 cubic in
Since OEB & OFD are similar. OEOF = EBFD ⇒ 1518= r12 ⇒ r = 12 x 15 18=10 in
Volume of water inside container = 13(10)² x 15 =500
Volume remains unfilled = Volume of container - Volume of water = 864 - 500 = 364 = 1143.53
Review Section:
4) In the following diagram QR is the image of NP after a dilation centered at the origin with a scale factor of k. Which of the following would not be a correct calculation of k.
1) QRNP
2) OQQN
3) OROP
4) OQON
ANSWER
Scalefactor = QRNP = OQON = OROP
∆ OQR ∼∆ONP
Option 2 ) OQQN
5) Describe a sequence of transformations that would map ∆ABC onto ∆DEF. Explain why this shows that the triangles are congruent.
To map triangle ABC onto triangle DEF, we need to perform a series of transformations: translations, rotations, reflections, or combinations of these.
Given that we're trying to show congruence between the triangles, we should aim to perform transformations that preserve size and shape. Here's a sequence of transformations that would achieve this:
1. **Translation**: Translate triangle ABC so that point A coincides with point D. Since A(-8,-2) and D(6,4), we need to translate by adding 14 to the x-coordinate and adding 6 to the y-coordinate.
2. **Reflection**: Reflect the translated triangle across the line that passes through points D and E. This reflection will ensure that the corresponding vertices line up.
3. **Rotation**: Rotate the reflected triangle about point D until side AB aligns with side DE. Since AB is parallel to DE, no rotation is necessary, but if needed, we'd rotate until they align perfectly.
Let's now explain why this sequence shows that the triangles are congruent:
- **Translation** preserves the shape and size of the triangle, only shifting it in space. So, triangle ABC and its image after translation are congruent.
- **Reflection** preserves distances and angles, maintaining congruence. Reflecting the translated triangle across the line passing through points D and E ensures that the corresponding vertices are in the correct position.
- **Rotation**, if required, would also preserve congruence since it simply reorients the triangle without changing its shape or size.
Since each transformation individually preserves the size and shape of the triangle, the composition of these transformations ensures that triangle ABC maps onto triangle DEF in a way that maintains congruence. Therefore, the triangles are congruent.
Homework Answers
Geometry Volume (Day 3) & Density 9F HW
1) 750 grams
2) 586 cubic meters
3) (a) 2714 cubic inches (b) 10 inches (c) 1143 cubic inches
4) (2)
5) list transformations
Geometry Cross Sections & Multiple Choice 9G
Cross Section: A cross section is the face you get when you make one slice through an object. Below is a sample slice through a cube, showing one of the cross sections you can get.
The polygon formed by the slice is the cross section. The cross section cannot contain any piece of the original face; it all comes from "inside" the solid. In this picture, only the gray piece is a cross section. In short, a cross section is the 2D shape you get when cutting straight across a 3D object.
Here are some examples of cross sections:
Examples:
https://www.ixl.com/math/geometry/cross-sections-of-three-dimensional-figures
Warm Up:
A prism is a solid that has two congruent faces (the bases) with all other faces being parallelograms. Two examples of right prisms are shown below.
(a) Why do you believe these are called right prisms?
(b) If you sliced a prism with a horizontal plane, what would be true about the cross section that you would see?
(c) If you sliced a prism with a vertical plane, what would be true about the cross section that you would see?
(a) These prisms are called "right prisms" because the angles formed between the bases and the lateral faces are all right angles (90 degrees). This characteristic distinguishes them from oblique prisms, where these angles are not necessarily right angles.
(b) If you sliced a right prism with a horizontal plane, the cross section that you would see would be congruent to the bases of the prism. This is because a horizontal plane intersects the prism parallel to the bases, cutting through the lateral faces, but not changing their shape. Therefore, the cross section would be a congruent shape to the bases.
(c) If you sliced a right prism with a vertical plane, the cross section that you would see would be a parallelogram. This is because a vertical plane intersects the prism perpendicular to the bases but not necessarily parallel to the sides of the bases. As a result, the cross section would be a parallelogram, representing a slice through the prism that includes both the bases and parts of the lateral faces.
Examples:
1) Which figure can have the same cross section of a sphere?
ANSWER 2 CONE
2) Jennifer is trying to determine if there is any difference in shape for various horizontal cross sections of a cone. She works with two cones that are similar to the one pictured below. Jennifer takes a cross-section of the first cone that is parallel to the base of the cone. Next, she takes a cross section of the second cone that is slanted 5 degrees from parallel. What were her results?
(1) The cross sections were both circles.
(2) The cross sections were both ellipses.
(3) The first cross section was a circle and the second cross section was an ellipse.
(4) The first cross section was an ellipse and the second cross section was a circle.
Answer: (3) The first cross section was a circle and the second cross section was an ellipse.
The correct answer is:
Step 1: Horizontal cross-sections of a cone parallel to the base are circles.
When Jennifer takes a cross-section of the first cone that is parallel to the base of the cone, the resulting shape will always be a circle. This is because parallel cross-sections to the base of a cone are always circles.
Step 2: A slanted cross-section of a cone is an ellipse.
However, when she takes a cross-section of the second cone that is slanted 5 degrees from parallel, the resulting shape will be an ellipse. This is because the plane intersects the slanted cone at an angle, resulting in an elliptical cross section rather than a perfect circle.
3) The cross section of a regular pyramid contains the altitude of the pyramid. The shape of this cross section is a
(1) circle (2) square (3) triangle (4) rectangleThe correct answer is: (3) triangle
The cross section of a regular pyramid that contains its altitude is a triangle. This cross section will have a base that coincides with one of the sides of the pyramid's base, and its apex will be at the apex of the pyramid. Since a regular pyramid has a polygonal base and triangular lateral faces that converge at the apex, any plane containing the altitude will intersect the lateral faces, forming a triangle.
Let’s review some multiple choice questions now that we know more about 3D figures!
4) A student has a rectangular postcard that he folds in half lengthwise. Next, he rotates it continuously about the folded edge. Which three-dimensional object is generated by this rotation?
When the student folds the rectangular postcard in half lengthwise and then rotates it continuously about the folded edge, the three-dimensional object generated by this rotation is a **cylinder**.
Here's how the transformation occurs: Folding the postcard lengthwise creates a long, narrow strip. When this strip is rotated continuously about the folded edge, it sweeps out a cylindrical shape. The folded edge becomes the axis of rotation, and the resulting solid has circular cross-sections at every point along its length.
Therefore, the correct answer is a **cylinder**.
5) If the rectangle below is continuously rotated about side w, which solid figure is formed?
(1) pyramid (2) rectangular prism (3) cone (4) cylinderIf the rectangle below is continuously rotated about side w , the solid figure formed is a cylinder.
The process of rotating a rectangle about one of its sides creates a cylindrical shape. The resulting solid will have two circular bases (formed by the rotation of the ends of side w ) connected by a curved surface (formed by the rotation of the sides perpendicular to w ). This matches the definition of a cylinder: a solid figure with two parallel circular bases and a curved surface connecting them. Therefore, the correct answer is:
(4) cylinder
Geometry Cross Sections & Multiple Choice 9G HW
1) Describe the cross section:
(1) The cross section is a circle.
(2) The cross section is a cylinder.
(3) The cross section is a plane.
(4) The cross section is a parallelogram.
ANSWER : (1) The cross section is a circle.
2) William is drawing pictures of cross sections of the right circular cone below.
Which drawing can not be a cross section of a cone?
ANSWER: (1) NOT POSSIBLE
(2) FRONT VIEW
(3) CUT ALONG THE TOP AND BOTTOM DIAMETER. GET A HALF CONE, FROM THE BOTTOM
(4) VIEW FROM THE BOTTOM
3) The figure below is a square pyramid. Which of the following is NOT a cross section from the square pyramid?
(1) )
(3) 4)
ANSWER :
1 A cross section of a square pyramid can be a triangle, a trapezoid, or a square
2 However, a cross section of a square pyramid cannot be a circle, RECTANGLE
4) Which object is formed when right triangle shown below is rotated around leg ? (1) a pyramid with a square base
(2) an isosceles triangle
(3) a right triangle
(4) a cone
ANSWER: When the right triangle is rotated around one of its legs, it forms a cone. So, the correct option is (4) a cone.
5) If you rotated the shaded figure below about line m, which solid would result from the revolution?
(1) cylinder
(2) cone
(3) cube
(4) sphere
ANSWER: (1) cylinder
If the shaded figure is rotated about line m, it would form a cylinder. So, the correct option is (1) cylinder.
6) If a right triangle ABC, shown, was rotated around segment BC, then the solid produced would look like
(1) (3)
(2) 4)
ANSWER: OPTION (1) The solid generated by the rotation of a right triangle about one of its sides containing the right angle is called a right circular cone.
Review Section:
7) Segment CD is the perpendicular bisector of AB at E. Which pair of segments does not have to be congruent?
(1) AD, BD
(2) AC, BC
(3)AE, BE
(4) DE, CE
Hint: Draw a diagram!ANSWER : (4) DE, CE
8) In the diagram of right triangle ABC, CD intersects hypotenuse AB at D.
I f AD = 4 and DB = 6, which length of AC makes CD ┴AB?
(1) 2√6 (3) 2√15
(2) 2√10 (4)4√2
To find the length of AC that makes CD \perp AB , we can use the property of similar triangles.
Since CD \perp AB , triangles ADC and BDC are similar to triangle ABC because they share the same angles.
Let x be the length of AC .
Using the property of similar triangles, we can set up the following proportion:
Given that AD = 4 and DB = 6 , we can find BC using the fact that AB = AD + DB = 4 + 6 = 10 .
Thus, BC = 10 .
Now, we can find CD using the Pythagorean theorem in triangle BDC :
Now, substituting the values into our proportion:
Thus, AC = 20 .
However, we need to find the length of AC in terms of \sqrt{} . Using the Pythagorean theorem in triangle ADC :
Therefore, the correct answer is:
(2) 2\sqrt{10}
9) In the diagram below, has coordinates A(1,1), B(4,1), and C(4,5). Graph a label ∆A”B”C” , the image of ∆ABC after the translation five units to the right and two units up following by a reflection over the line Y=0.
To find the coordinates of the vertices of triangle \Delta A'B'C' , the image of ∆ABC after the described transformations, we first perform a translation of five units to the right and two units up, followed by a reflection over the line y = 0 .
Let's calculate the new coordinates:
1. Translation:
- A'(x, y) = (1 + 5, 1 + 2) = (6, 3)
- B'(x, y) = (4 + 5, 1 + 2) = (9, 3)
- C'(x, y) = (4 + 5, 5 + 2) = (9, 7)
2. Reflection over the line y = 0 :
- Reflecting over the line y = 0 simply means changing the sign of the y-coordinate.
- So, the new coordinates become:
- A''(6, -3)
- B''(9, -3)
- C''(9, -7)
Now, let's graph the labeled triangle \Delta A''B''C'' on the coordinate plane:
```
C'' (9, -7) B'' (9, -3)
*----------------*
/ \ / \
/ \ / \
/ \ / \
/_________\__________\
A'' (6, -3) B'' (9, -3)
```
Homework Answers
Geometry Cross Sections & Multiple Choice 9G HW
1) (1)
2) (1)
3) (1)
4) (4)
5) (1)
6) (1)
7) (4)
8) (2)
9) A” (6,-3) B” (9,-3) C” (9,-7) with graph
Chapter 9 Polygons Notes and Homework
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