Module #GM Geometry - Arizona State University · PDF fileIn a similar manner, we can...

MAT 142 College Mathematics Module #GM

GeometryTerri L. Miller & Elizabeth E. K. Jones

revised April 20, 2011

Perimeter and Area

We are going to start our study of geometry with two-dimensional figures. We will look atthe one-dimensional distance around the figure and the two-dimensional space covered bythe figure.

The perimeter of a shape is defined as the distance around the shape. Since we usuallydiscuss the perimeter of polygons (closed plane figures whose sides are straight line segment),we are able to calculate perimeter by just adding up the lengths of each of the sides. Whenwe talk about the perimeter of a circle, we call it by the special name of circumferenceSince we dont have straight sides to add up for the circumference (perimeter) of a circle, wehave a formula for calculating this.

Circumference (Perimeter) of a Circle

C = 2πr

r = radius

π =the number that is approximated by 3.141593

Example 1. Find the perimeter of the figure below.

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Solution:: It is tempting to just start adding of the numbers given together, but that willnot give us the perimeter. The reason that it will not is that this figure has SIX sides andwe are only given four numbers. We must first determine the lengths of the two sides thatare not labeled before we can find the perimeter. Lets look at the figure again to find thelengths of the other sides.

Since our figure has all right angles, we are able to determine the length of the sides whoselength is not currently printed. Lets start with the vertical sides. Looking at the imagebelow, we can see that the length indicated by the red bracket is the same as the length of

2 Geometry

the vertical side whose length is 4 units. This means that we can calculate the length of thegreen segment by subtracting 4 from 11. This means that the green segment is 7 units.

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Now that we have all the lengths of the sides, we can simply calculate the perimeter by addingthe lengths together to get Since perimeters are just the lengths of lines, the perimeter wouldbe 50 units.

The area of a shape is defined as the number of square units that cover a closed figure. Formost of the shape that we will be dealing with there is a formula for calculating the area.In some cases, our shapes will be made up of more than a single shape. In calculating thearea of such shapes, we can just add the area of each of the single shapes together.

We will start with the formula for the area of a rectangle. Recall that a rectangle is aquadrilateral with opposite sides parallel and right interior angles.

Area of a Rectangle

A = bh

b = base of the rectangle

h = the height of the rectangle

Example 2. Find the area of the figure below.

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MAT 142 - Module 3

Solution: This figure is not a single rectangle. It can, however, be broken up into tworectangles. We then will need to find the area of each of the rectangles and add themtogether to calculate the area of the whole figure.

There is more than one way to break this figure into rectangles. We will only illustrate onebelow.

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We have shown above that we can break the shape up into a red rectangle (figure on left) anda green rectangle (figure on right). We have the lengths of both sides of the red rectangle.It does not matter which one we call the base and which we call the height. The area of thered rectangle is A = bh = 4× 14 = 56.

We have to do a little more work to find the area of the green rectangle. We know that thelength of one of the sides is 8 units. We had to find the length of the other side of the greenrectangle when we calculated the perimeter in Example 1 above. Its length was 7 units.

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Thus the area of the green rectangle is A = bh = 8 × 7 = 56. Thus the area of the wholefigure is

area of red rectangle + area of green rectangle = 56 + 56 = 112.

In the process of calculating the area, we multiplied units times units. This will produce afinal reading of square units (or units squared). Thus the area of the figure is 112 squareunits. This fits well with the definition of area which is the number of square units that willcover a closed figure.

Our next formula will be for the area of a parallelogram. A parallelogram is a quadrilateralwith opposite sides parallel.

4 Geometry

Area of a Parallelogram

A = bh

b = base of the parallelogram

h = the height of the parallelogram

You will notice that this is the same as the formula for the area of a rectangle. A rectangleis just a special type of parallelogram. The height of a parallelogram is a segment thatconnects the top of the parallelogram and the base of the parallelogram and is perpendicularto both the top and the base. In the case of a rectangle, this is the same as one of the sidesof the rectangle that is perpendicular to the base.

Example 3. Find the area of the figure below.

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Solution: In this figure, the base of the parallelogram is 15 units and the height is 6 units.This mean that we only need to multiply to find the area of A = bh = 15 × 6 = 90 squareunits.

You should notice that we cannot find the perimeter of this figure since we do not have thelengths of all of the sides, and we have no way to figure out the lengths of the other twosides that are not given.

Our next formula will be for the area of a trapezoid.

Area of a Trapezoid

A = 12(b1 + b2)h

b1 = one base of the trapezoid

b2 = the other base of the trapezoid

h = the height of the trapezoid

A trapezoid is a quadrilateral that has one pair of sides which are parallel. These two sidesare called the bases of the trapezoid. The height of a trapezoid is a segment that connectsthe one base of the trapezoid and the other base of the trapezoid and is perpendicular toboth of the bases.

Example 4. Find the area of the figure below

MAT 142 - Module 5

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Solution: For this trapezoid, the bases are shown as the top and the bottom of the figure.The lengths of these sides are 45 and 121 units. It does not matter which of these we say isb1 and which is b2. The height of the trapezoid is 20 units. When we plug all this into theformula, we get A = 1

2(b1 + b2)h = 1

2(121 + 45)20 = 1660 square units.

Our next formulas will be for finding the area of a triangle (a three-sided polygon). Wewill have more than one formula for this since there are different situations that can comeup which will require different formulas

Triangle Area Formulas

Area of a Triangle

for a triangle with a base and a height

A = 12bh

b = the base of the triangle

h = the height of the triangle

Heron’s Formula

for a triangle with only sides

A =√s(s− a)(s− b)(s− c)

a = one side of the triangle

b = another side of the triangle

c = the third side of the triangle

s = 12(a+ b+ c)

The height of a triangle is the perpendicular distance from any vertex of a triangle to theside opposite that vertex. In other words the height of triangle is a segment that goes fromthe vertex of the triangle opposite the base to the base (or an extension of the base) that isperpendicular to the base (or an extension of the base). Notice that in this description ofthe height of a triangle, we had to include the words or an extension of the base. This isrequired because the height of a triangle does not always fall within the sides of the triangle.Another thing to note is that any side of the triangle can be a base. You want to pick thebase so that you will have the length of the base and also the length of the height to thatbase. The base does not need to be the bottom of the triangle.

6 Geometry

You will notice that we can still find the area of a triangle if we dont have its height. Thiscan be done in the case where we have the lengths of all the sides of the triangle. In thiscase, we would use Herons formula.


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Solution: Notice that in this figure has a dashed line that is shown to be perpendicular tothe side that is 8.2 units in length. This is how we indicate the height of the triangle (thedashed line) and the base of the triangle (the side that the dashed line is perpendicular to).That means we have both the height and the base of this triangle, so we can just plug thesenumbers into the formula to get

A =1

2bh =

1

2(8.2)(4.5) = 18.45square units.

Notice that the number 6 is given as the length of one of the sides of the triangle. This sideis not a height of the triangle since it is not perpendicular to another side of the triangle. Itis also not a base of the triangle, since there is no indication of the perpendicular distancebetween that side and the opposite vertex. This means that it is not used in the calculationof the area of the triangle.


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Solution: In this figure there are two dashed lines. One of them is extended from the sideof the triangle that has a length of 1.7. That dashed line is show to be perpendicular to thedashed line that has a length of 2.6. This is how we indicate that the dashed line that hasa length of 2.6 is the height of the triangle and the base of the triangle is the side of length1.7. The height here is outside of the triangle. Also, the dashed line that is extended fromthe base is not part of the triangle and its length is not relevant to finding the area of thetriagnle. Since we have both the height and the base of this triangle, we can just plug thesenumbers into the formula to get A = 1

2bh = 1

2(1.7)(2.6) = 21.21 square units.


MAT 142 - Module 7

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Solution: You should notice that we do not have a height for this triangle. This meansthat we cannot use the formula that we have been using to find the area of this triangle.We do have the length of all three sides of the triangle. This means that we can use HeronsFormula to find the area of this triangle.

For this formula, it does not matter which side we label a, b, or c. For our purposes, we willlet a be 6, b be 7, and c be 8. Now that we have a, b, and c, we need to calculate s so thatwe can plug a, b, c , and s into the formula. We get s = 1

2(6 + 7 + 8) = 1

2(21) = 10.5. Now

we can plug everything in to Herons formula to find the area of this triangle to be

A =√s(s− a)(s− b)(s− c) =

√10.5(10.5− 6)(10.5− 7)(10.5− 8) ≈ 20.33316 square units.

Another formula that we are interested in is the Pythagorean Theorem. This applies to onlyright triangles. The Pythagorean Theorem relates the lengths of the sides of a right triangle.

Pythagorean Theorem

a2 + b2 = c2

a = leg (one side of the triangle that makes up the right angle)

b = leg (another side of the triangle that makes up the right angle

c = hypotenuse (side opposite the right angle)

When using the Pythagorean Theorem, it is important to make sure that we always use thelegs of the triangle for a and b and the hypotenuse for c.

Example 8. Find the length of the third side of the triangle below.

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Solution: The figure is a right triangle (as indicated by the box in one of the angles ofthe triangle). We need to decide what the side we are looking for is in terms of a leg orthe hypotenuse of the triangle. The hypotenuse is the side of the triangle opposite the rightangle. That would be the side that has length 26 in our picture. Thus c will be 26 in ourformula. This means that the other two sides of the triangle are legs a and b. We will let abe 10, and we will thus be looking for b. When we plug all this into the formula, we get

8 Geometry

a2 + b2 = c2

102 + b2 = 262

100 + b2 = 676

We now need to solve this equation for b.

100 + b2 = 676

b2 = 576

b =√

576 = 24

Since we are asked for a length, we have that the third side is 24 units. (Notice that this isnot square units since we are not finding an area).

We are now going to move on to circles. We already mentioned the perimeter (circumference)of a circle in the perimeter sections. We also need a formula for finding the area of a circle.

Area of a Circle

A = πr2

r = radius of the circle

π = the number that is approximated by 3.141593

For circle problems we need to remember that the circumference (perimeter) of a circle isonly the curved part of the circle. It does not include either the radius or diameter of thecircle. In order to find the perimeter and area, we will need the radius of the circle. Recallthat the diameter of a circle is a segment from on point on the circle to another pointon the circle that passes through the center of the circle. A radius of a circle is half of adiameter (i.e. a segment that from one point on the circle to the center of the circle).

Example 9. Find the perimeter and area of the circle below.

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Solution: The number 11 in the figure above is the length of the diameter of this circle.We need the radius to be able to use the formulas. The radius is half of the diameter.Thus r = 1

2(diameter) = 1

2(11) = 5.5 . We are now ready to find both the perimeter

(circumference) and area by plugging 5.5 into each formula for r. Perimeter:

C = 2πr = 2(π)(5.5) = 11π ≈ 34.557519

MAT 142 - Module 9

The answer 11π units is an exact answer for the perimeter. The answer 34.557519 units isan approximation.

Area:

A = πr2 = π(5.5)2 = 30.25π ≈ 95.0331778

The answer 30.25π square units is an exact answer for the area. The answer 95.0331778square units is an approximation.

Example 10. Find the perimeter and area of the semicircle below.

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*Solution: In this problem, we have a semicircle (a half of a circle). The diameter of thishalf circle is 120. Thus the radius is r == 1

2(120) = 60 .

Perimeter: We now can find the perimeter. We will find the circumference of the whole circleand then divide it by 2 since we only have half of a circle. This will give us C = 2πr = 120πas the circumference for the whole circle and 60π units as the circumference of half of thecircle.

Unfortunately this is not the answer for the perimeter of the figure given. The circumferenceof a circle is the curved part. The straight line segment in our figure would not have beenpart of the circumference of the whole circle, and thus it is not included as part of half ofthe circumference. We only have the red part shown in the picture below.

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We still need to include the straight segment that is 120 units long. Thus the whole perimeterof the figure is

curved part + straight part = 60π + 120 ≈ 308.495559units.

Area: We can also find the area. Here we will plug 60 in for r in the area formula for awhole circle and then divide by 2 for the half circle. This will give us A = π(60)2 = 3600πsquare units for the area of the whole circle and 1800π square units for the area of the halfcircle. Now are we done with finding the area or is there more that we need to do like wedid in finding the perimeter? If we think back to the definition of the area, (it is the numberof square units needed to cover the figure) we should see that there is nothing further to do.The inside of the figure was covered already. We just have half as much inside as we didbefore.

Example 11. Walk 10 yards south, then 12 west and then 3 yards south. How far from theoriginal starting point are you?

Solution: To solve this problem, it would help to have a picture.

10 Geometry

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Above we can see a description of how we walked in the problem. What we are asked tofind is how far it is from where we started to where we ended. That distance would berepresented by a straight line from the start to the finish. This will be shown in the picturebelow by a black dashed line.

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We need to figure out how long this line is. We might be tempted to say that we havetwo right triangles and that the line we are looking for is just the sum of each triangleshypotenuse. The only problem with that is that we do not know the lengths of the legs ofeach of the right triangles. Thus we are not able to apply the Pythagorean Theorem in thisway.

We are on the right track though. This is a Pythagorean Theorem problem. If we add acouple of line segments, we can create a right triangle whose hypotenuse is the line we arelooking for.

MAT 142 - Module 11

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As you can see above, the orange segments along with the red segment that was 10 yardsand the dashed segment make up a right triangle whose hypotenuse is the dashed segment.We can figure out how long each of the legs of this triangle are fairly easily. The orangesegment that is an extension of the red segment is another 3 yards. Thus the vertical leg ofthis triangle is 10 + 3 = 13 yards long. The orange segment which is horizontal is the samelength as the green segment. This means that the horizontal leg of this triangle is 12 yardslong. We can now use the Pythagorean Theorem to calculate how far away from the originalstarting point we are.

132 + 122 = c2

169 + 144 = c2

313 = c2

√313 = c ≈ 17.6918

Here an exact answer does not really make sense. It is enough to approximate that we endup 17.6918 yards from where we started.

Volume and Surface Area

We will continue our study of geometry by studying three-dimensional figures. We will lookat the two-dimensional aspect of the outside covering of the figure and also look at thethree-dimensional space that the figure encompasses.

The surface area of a figure is defined as the sum of the areas of the exposed sides of anobject. A good way to think about this would be as the area of the paper that it would taketo cover the outside of an object without any overlap. In most of our examples, the exposedsides of our objects will polygons whose areas we learned how to find in the previous section.When we talk about the surface area of a sphere, we will need a completely new formula.

The volume of an object is the amount of three-dimensional space an object takes up. Itcan be thought of as the number of cubes that are one unit by one unit by one unit that ittakes to fill up an object. Hopefully this idea of cubes will help you remember that the unitsfor volume are cubic units.

12 Geometry

Surface Area of a Rectangular Solid

SA = 2(lw + lh+ wh)

l = length of the base of the solid

w = width of the base of the solid

h = height of the solid

Volume of a Solid with a Matching Base and Top

V = Ah

A = area of the base of the solid


Volume of a Rectangular Solid

(specific type of solid with matching base and top)

V = lwh

l = length of the base of the solid

w = width of the base of the solid


Example 12. Find the volume and the surface area of the figure below

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Solution: This figure is a box (officially called a rectangular prism). We are given thelengths of each of the length, width, and height of the box, thus we only need to plug intothe formula. Based on the way our box is sitting, we can say that the length of the baseis 4.2 m; the width of the base is 3.8 m; and the height of the solid is 2.7 m. Thus we canquickly find the volume of the box to be V = lwh = (4.2)(3.8)(2.7) = 43.092 cubic meters.

Although there is a formula that we can use to find the surface area of this box, you shouldnotice that each of the six faces (outside surfaces) of the box is a rectangle. Thus, thesurface area is the sum of the areas of each of these surfaces, and each of these areas isfairly straight-forward to calculate. We will use the formula in the problem. It will give usSA = 2(lw + lh+ wh) = 2((4.2× 3.8) + (4.2× 2.7) + (3.8× 2.7)) = 75.12 square meters.

A cylinder is an object with straight sides and circular ends of the same size. The volumeof a cylinder can be found in the same way you find the volume of a solid with a matching

MAT 142 - Module 13

base and top. The surface area of a cylinder can be easily found when you realize that youhave to find the area of the circular base and top and add that to the area of the sides. If youslice the side of the cylinder in a straight line from top to bottom and open it up, you willsee that it makes a rectangle. The base of the rectangle is the circumference of the circularbase, and the height of the rectangle is the height of the cylinder.

Volume of a Cylinder

V = Ah

A = area of the base of the cylinder

h = height of the cylinder

Surface Area of a Cylinder

SA = 2(πr2) + 2πrh

r = the radius of the circular base of the cylinder

h = height of the cylinder

π = the number approximated by 3.141593

Example 13. Find the volume and surface area of the figure below

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=356$7(!E5($*!!

!!!!!!! ;F);)<G)B;<BCAC)BAC))BC) )) =+=+"=+= 505=E 4H3$5(!

3-,/4@!!>0,4!,4!$-!(#$7/!$-4:(5@!!E-!$&&52#,%$/(!$-4:(5!,4!A<I@;JKGF!

4H3$5(!3-,/4@!

L2'3%(!26!$!M?',-.(5!

E0L =

4(;(%<"(.&".(#2(%<"(=.*"(#2(%<"(3',>/0"&(<(;(%<"(<">6<%(#2(%<"(3',>/0"&(

=356$7(!E5($!26!$!M?',-.(5!

505=E !! )BC) ) +=

&(;(%<"(&.0>-*(#2(%<"(3>&3-,.&(=.*"(#2(%<"(3',>/0"&(<(;(%<"(<">6<%(#2(%<"(3',>/0"&(?(;(%<"(/-$="&(%<.%(>*(.@@&#A>$.%"0(='(BC8D89EB(

(

Solution: This figure is a cylinder. The diameter of its circular base is 12 inches. Thismeans that the radius of the circular base is r = 1

2d = 1

2(12) = 6 inches. The height of the

cylinderi s 10 inches. To calculate the volume and surface area, we simply need to plug intothe formulas.Surface Area:SA = 2(πr2) + 2πrh = 2(π × 62) + 2π(6)(10) = 72π + 120π = 192π square units. This is anexact answer. An approximate answer is 603.18579 square units.Volume:In order to plug into the formula, we need to recall how to find the area of a circle (the baseof the cylinder is a circle). We will the replace A in the formula with the formula for thearea of a circle. V = Ah = πr2h = π(62)(10) = 360π cubic inches. An approximation of thisexact answer would be 1130.97336 cubic inches.

Our next set of formulas is going to be for spheres. A sphere is most easily thought of as aball. The official definition of a sphere is a three-dimensional surface, all points of which areequidistant from a fixed point called the center of the sphere. A circle that runs along thesurface of a sphere to that it cuts the sphere into two equal halves is called a great circle of

14 Geometry

that sphere. A great circle of a sphere would have a diameter that is equal to the diameterof the sphere.

Surface Area of a Sphere

SA = 4πr2



Volume of a Sphere

V = 43πr3



Example 14. Find the volume and surface area of the figure below

!"#$"%&'()#%"*(+#,-$"(./0(1-&2.3"(4&".( 5.6"(78(#2(9:(

!"#$%&'()*(

+,-.(/0'(12&3$'(#-.(4356#7'(#5'#(26(/0'(6,835'(9'&2:(

;

<) ,-

((

=2&3/,2-*(

>0,4(,4(#(4%0'5'?((@'(#5'(8,1'-(/0#/(/0'(.,#$'/'5(26(/0'(4%0'5'(,4(

;

<) (,-70'4?((@'(-''.(/2(7#&73&#/'(/0'(5#.,34(26(/0'(4%0'5'(/2(

7#&73&#/'(/0'(12&3$'(#-.(4356#7'(#5'#?((>0'(5#.,34(26(#(4%0'5'(,4(

0#&6(26(,/4(.,#$'/'5?((>0,4($'#-4(/0#/(/0'(5#.,34(,4(

;AB<?AAC

BD

;

BD

B

A

;

<)

B

A

B

A==!

"

#$%

&=!

"

#$%

&== .5 (,-70'4?((@'(7#-(-2:(E34/(

%&38(/0,4(-3$9'5(,-(/2(/0'(625$3&#4(/2(7#&73&#/'(/0'(12&3$'(#-.(

4356#7'(#5'#?(

F2&3$'*(( ( ) DGA<H<?BG;AB<?A)

G

)

G )) !== ""5F (739,7(,-70'4?(

=356#7'(I5'#*(( ( ) B;BAD?GA;AB<?AGG BB !== ""5=I (4J3#5'(,-70'4?(

(

(

!"#$%&'(G*(

+,-.(/0'(12&3$'(26(/0'(6,835'(( ;(

A)(

<(

A<(

(

F2&3$'(26(#(

=2&,.(:,/0(#(

K#/70,-8(L#4'(

#-.(>2%(

I0F =

4;(.&".(#2(%<"(=.*"(#2(%<"(*#,>0(

<(;(<">6<%(#2(%<"(*#,>0(

Solution:This is a sphere. We are given that the diameter of the sphere is 358

inches. Weneed to calculate the radius of the sphere to calculate the volume and surface area. Theradius of a sphere is half of its diameter. This means that the radius is r = 1

2d = 1

2(35

8) =

12(298

) = 2916

= 1.8125 inches. We can now just plug this number in to the formulas to calculatethe volume and surface area.Volume: V = 4

3πr3 = 4

3π(1.81253) ≈ 24.941505 cubic inches.

Surface Area: SA = 4πr2 = 4π(1.81252) ≈ 41.28219 square inches.

Example 15. Find the volume of the figure

!"#$"%&'()#%"*(+#,-$"(./0(1-&2.3"(4&".( 5.6"(78(#2(9:(

!"#$%&'()*(

+,-.(/0'(12&3$'(#-.(4356#7'(#5'#(26(/0'(6,835'(9'&2:(

;

<) ,-

((

=2&3/,2-*(

>0,4(,4(#(4%0'5'?((@'(#5'(8,1'-(/0#/(/0'(.,#$'/'5(26(/0'(4%0'5'(,4(

;

<) (,-70'4?((@'(-''.(/2(7#&73&#/'(/0'(5#.,34(26(/0'(4%0'5'(/2(

7#&73&#/'(/0'(12&3$'(#-.(4356#7'(#5'#?((>0'(5#.,34(26(#(4%0'5'(,4(

0#&6(26(,/4(.,#$'/'5?((>0,4($'#-4(/0#/(/0'(5#.,34(,4(

;AB<?AAC

BD

;

BD

B

A

;

<)

B

A

B

A==!

"

#$%

&=!

"

#$%

&== .5 (,-70'4?((@'(7#-(-2:(E34/(

%&38(/0,4(-3$9'5(,-(/2(/0'(625$3&#4(/2(7#&73&#/'(/0'(12&3$'(#-.(

4356#7'(#5'#?(

F2&3$'*(( ( ) DGA<H<?BG;AB<?A)

G

)

G )) !== ""5F (739,7(,-70'4?(

=356#7'(I5'#*(( ( ) B;BAD?GA;AB<?AGG BB !== ""5=I (4J3#5'(,-70'4?(

(

(

!"#$%&'(G*(

+,-.(/0'(12&3$'(26(/0'(6,835'(( ;(

A)(

<(

A<(

(

F2&3$'(26(#(

=2&,.(:,/0(#(

K#/70,-8(L#4'(

#-.(>2%(

I0F =

4;(.&".(#2(%<"(=.*"(#2(%<"(*#,>0(

<(;(<">6<%(#2(%<"(*#,>0(Solution:This figure is a solid with the same shape base and top. The shape of the base and

top is a trapezoid. Thus we will need to remember the formula for the area of a trapezoid.

MAT 142 - Module 15

For this trapezoid, the lengths of the bases are 13 and 8 units. It does not matter which ofthese we say is b1 and which is b2. The height of the trapezoid is 5 units. The height of thesolid is 15 units. We will start by plugging the information about the trapezoidal base intothe formula for the area of a trapezoid. Once we have this area, we will plug that and theheight of the solid into the volume formula.Area of the trapezoidal base:A = 1

2(b1 + b2)h = 1

2(13 + 8)5 = 52.5 square units.

Volume of trapezoidal solid:V = Ah = (52.5)(15) = 787.5 cubic units.

Our next formulas will be for finding the volume of a cone or a pyramid. These twoformulas are grouped together since they are very similar. Each is basically 1

3times the area

of the base of the solid times the height of the solid. In the case of the cone, the base is acircle. In the case of the pyramid, we will have a base that is a rectangle. The height in bothcases is the perpendicular distance from the apex to the plane which contains the base.

A pyramid is a solid figure with a polygonal base (in our case a rectangle) and triangularfaces that meet at a common point (the apex). A cone is the surface of a conic solid whosebase is a circle. This is more easily thought of as a pointed ice-cream cone whose top iscircular and level.

Volume of a Rectangular Pyramid

V = 13lwh

l = the length of the base of the pyramid

w = the width of the base of the pyramid

h = the perpendicular height of the pyramid

Volume of a Cone

V = 13πr2h

r = the radius of the circular base of the cone

π = the number that is apporximated by 3.141593

h = the perpendicular height of the cone

Example 16. Find the volume of the figure.

!"#$"%&'()#%"*(+#,-$"(./0(1-&2.3"(4&".( 5.6"(78(#2(89(

!

!

!

!

!

!

!

!

!

!

"#$%&'(!)*!

+,-.!/0(!12'3%(!24!/0(!4,536(7!

!!

82'3/,2-*!

8,-9(!/0(!4,536(!0$:!$!9,693'$6!;$:(!$-.!'22<:!',<(!$-!,9(!96($%!

92-(=!/0,:!%3:/!;(!$!92-(7!!>-!26.(6!/2!4,-.!/0(!12'3%(!24!$!92-(=!

?(!-((.!/0(!6$.,3:!24!/0(!9,693'$6!;$:(!$-.!/0(!0(,50/!

@&(6&(-.,93'$6!0(,50/A!24!/0(!92-(7!!B0(!0(,50/!,:!5,1(-!$:!CD!

9(-/,%(/(6:7!!B0(!2/0(6!%($:36(%(-/!24!CE!9(-/,%(/(6:!,:!/0(!

.,$%(/(6!24!/0(!9,693'$6!;$:(7!!F(!/03:!%3:/!9$'93'$/(!/0(!6$.,3:!

/2!5(/!!!!!6 =

C

D. =

C

D@CEA = )!9(-/,%(/(6:7!!F(!$6(!-2?!6($.G!/2!&'35!

,-/2!/0(!12'3%(!24!$!92-(!426%3'$7!!!!!!H =

C

I!6D0 =

C

I! )D( )@CDA = CEE! !

93;,9!9(-/,%(/(6:!,:!/0(!(#$9/!12'3%(7!!J-!$&&62#,%$/,2-!24!/0,:!

12'3%(!?23'.!;(!ICK7C)LDMM!93;,9!9(-/,%(/(6:7!

!

H2'3%(!24!$!N2-(!!

!

06H D

I

C!=

&(:(%;"(&.0<-*(#2(%;"(3<&3-,.&(=.*"(#2(%;"(3#/"(>(:(%;"(/-$="&(%;.%(<*(.??&#@<$.%"0(='(ABCDC8EA(

;(:(%;"(?"&?"/0<3-,.&(;"<6;%(#2(%;"(3#/"((

CD!9%!

CE!9%!

16 Geometry

Solution: Since the figure has a circular base and looks like an ice cream cone, this must bea cone. In order to find the volume of a cone, we need the radius of the circular base and theheight (perpendicular height) of the cone. The height is given as 12 centimeters. The othermeasurement of 10 centimeters is the diameter of the circular base. We thus must calculatethe radius to get r = 1

2d = 1

2(10) = 5 centimeters. We are now ready to plug into the volume

of a cone formula. V = 13πr2h = 1

3π(52)(12) = 100π cubic centimeters is the exact volume.

An approximation of this volume would be 314.159266 cubic centimeters.


!"#$"%&'()#%"*(+#,-$"(./0(1-&2.3"(4&".( 5.6"(78(#2(9:(

!"#$%&'()*(

+,-.(/0'(12&3$'(24(/0'(4,536'7(

(

(

(

(

(

(

(

(

(

(

82&3/,2-*(

90'(:#;'(24(/0,;(4,536'(,;(#(6'</#-5&'(#-.(/0'(;,.';(24(/0'(4,536'(

#6'(/6,#-5&';=(/03;(/0,;(4,536'(,;(#(6'</#-53&#6(%>6#$,.7((90'(0',50/(

?%'6%'-.,<3&#6(0',50/@(,;(AB(,-<0';7((90'(&'-5/0(24(/0'(:#;'(,;(C(

,-<0';=(#-.(/0'(D,./0(24(/0'(:#;'(,;(E(,-<0';7((8,-<'(D'(0#1'(#&&(24(

/0'(%#6/;(426(/0'(12&3$'(426$3&#=(D'(<#-(F3;/(%&35(,-/2(/0'(12&3$'(

24(#(6</#-53&#6(%>6#$,.(426$3&#(/2(5'/(

((((G =

A

H&D0 =

A

H?C@?E@?AB@ =

HEB

I(<3:,<(,-<0';7((J-(#%%62",$#/,2-(24(

/0,;(12&3$'(D23&.(:'(AA)7))))C(<3:,<(,-<0';7(

(

K3;/(&,L'(D,/0(#6'#;=(D'(<#-(#..(#-.(;3:/6#</(12&3$';(24(.,44'6'-/(;2&,.;(/2(

4,-.(/0'(12&3$'(24(#(;2&,.(/0#/(,;(#(<2$:,-#/,2-(24($26'(/0#-(2-'(;2&,.(26(

/0#/(0#1'(2-'(;2&,.(6'$21'.(462$(#-2/0'67(

(

!"#$%&'(C*(

+,-.(/0'(12&3$'(24(/0'(4,536'7(

E(

I(

A)((

(

82&3/,2-*(

E(,-(

C(,-(

AB(,-(

Solution: The base of this figure is a rectangle and the sides of the figure are triangles, thusthis figure is a rectangular pyramid. The height (perpendicular height) is 10 inches. Thelength of the base is 7 inches, and the width of the base is 5 inches. Since we have all ofthe parts for the volume formula, we can just plug into the volume of a rctangular pyramidformula to get V = 1

3lwh = 1

3(7)(5)(10) = 350

3cubic inches. An approximation of this volume

would be 116.66667 cubic inches.


!"#$"%&'()#%"*(+#,-$"(./0(1-&2.3"(4&".( 5.6"(78(#2(9:(

!"#$%&'()*(

+,-.(/0'(12&3$'(24(/0'(4,536'7(

(

(

(

(

(

(

(

(

(

(

82&3/,2-*(

90'(:#;'(24(/0,;(4,536'(,;(#(6'</#-5&'(#-.(/0'(;,.';(24(/0'(4,536'(

#6'(/6,#-5&';=(/03;(/0,;(4,536'(,;(#(6'</#-53&#6(%>6#$,.7((90'(0',50/(

?%'6%'-.,<3&#6(0',50/@(,;(AB(,-<0';7((90'(&'-5/0(24(/0'(:#;'(,;(C(

,-<0';=(#-.(/0'(D,./0(24(/0'(:#;'(,;(E(,-<0';7((8,-<'(D'(0#1'(#&&(24(

/0'(%#6/;(426(/0'(12&3$'(426$3&#=(D'(<#-(F3;/(%&35(,-/2(/0'(12&3$'(

24(#(6</#-53&#6(%>6#$,.(426$3&#(/2(5'/(

((((G =

A

H&D0 =

A

H?C@?E@?AB@ =

HEB

I(<3:,<(,-<0';7((J-(#%%62",$#/,2-(24(

/0,;(12&3$'(D23&.(:'(AA)7))))C(<3:,<(,-<0';7(

(

K3;/(&,L'(D,/0(#6'#;=(D'(<#-(#..(#-.(;3:/6#</(12&3$';(24(.,44'6'-/(;2&,.;(/2(

4,-.(/0'(12&3$'(24(#(;2&,.(/0#/(,;(#(<2$:,-#/,2-(24($26'(/0#-(2-'(;2&,.(26(

/0#/(0#1'(2-'(;2&,.(6'$21'.(462$(#-2/0'67(

(

!"#$%&'(C*(

+,-.(/0'(12&3$'(24(/0'(4,536'7(

E(

I(

A)((

(

82&3/,2-*(

E(,-(

C(,-(

AB(,-(

Solution: The first thing that we need to do is figure out what type of figure this is. If werotate the solid 90 degrees to the right, we get a figure that looks like this

!"#$"%&'()#%"*(+#,-$"(./0(1-&2.3"(4&".( 5.6"(78(#2(98(

!"#$%&'()$)"&*+$)",)$-#$*##.$)/$./$&($%&+0'#$/0)$-",)$)12#$/%$%&+0'#$

)"&($&(3$$4%$-#$'/),)#$)"#$(/5&.$67$.#+'##($)/$)"#$'&+")8$-#$+#)$,$%&+0'#$

)",)$5//9($5&9#$)"&($

$

$

$

$

$

$

$

$

$

!"&($5//9($5&9#$,$:15&*.#'$-&)"$)"#$;&..5#$;&((&*+3$$<$+//.$-,1$)/$

)"&*9$,=/0)$)"&($%&+0'#$&($,$'/55$/%$2,2#'$)/-#5(3$$>#$,'#$)'1&*+$)/$

%&*.$/0)$)"#$?/50;#$/%$)"#$2,2#'$)/-#5(3$$!"#$=#()$-,1$)/$./$)"&($&($

)/$%&+0'#$/0)$-",)$)"#$?/50;#$/%$)"#$5,'+#'$:15&*.#'$&($-&)"/0)$)"#$

;&((&*+$2,')3$$>#$:,*$)"#*$%&*.$)"#$?/50;#$/%$)"#$(;,55#'$:15&*.#'$

/'$;&((&*+$2,')3$$@&*,5518$-#$-&55$(0=)',:)$)"#$?/50;#$/%$)"#$(;,55#'$

:15&*.#'$%'/;$)"#$?/50;#$/%$)"#$5,'+#'$:15&*.#'$)/$+#)$)"#$?/50;#$

/%$/0'$:0''#*)$(/5&.3$

$

>#$-&55$(),')$-&)"$)"#$5,'+#'$:15&*.#'3$$!"#$',.&0($/%$)"#$:&':05,'$

=,(#$&($+&?#*$,($A$0*&)(3$$!"#$"#&+")$/%$)"#$5,'+#'$:15&*.#'$&($BC$

0*&)(3$$>#$:,*$)"#*$:,5:05,)#$)"#$?/50;#$/%$)"#$5,'+#'$:15&*.#'$)/$

=#$$$$$D = <" = !'E" = ! FAEGFBCG = H77! cubic units.

I#J)$-#$-&55$:,5:05,)#$)"#$?/50;#$/%$)"#$(;,55#'$:15&*.#'3$$!"#$

',.&0($/%$)"#$:&':05,'$=,(#$/%$)"#$(;,55#'$:15&*.#'$&($E$0*&)(3$$!"#$

"#&+")$/%$)"#$(;,55#'$:15&*.#'$&($BC$0*&)(3$$>#$:,*$:,5:05,)#$)"#$

?/50;#$/%$)"#$(;,55#'$:15&*.#'$-/05.$=#$

$$$$D = <" = !'E" = ! FEEGFBCG = CH! $:0=&:$0*&)(3$

$

>#$*/-$(0=)',:)$)"#$?/50;#$/%$)"#$(;,55#'$:15&*.#'$%'/;$)"#$

?/50;#$/%$)"#$5,'+#'$:15&*.#'$)/$+#)$)"#$?/50;#$/%$/0'$(/5&.3$

D/50;#$/%$5,'+#'$:15&*.#'$K$?/50;#$/%$(;,55#'$:15&*.#'$L$

$$H77! " CH! = MMC! $:0=&:$0*&)(3$$!"&($&($,22'/J&;,)#51$B7AA3ANABM$

:0=&:$0*&)(3$

$

D/50;#$/%$,$

O15&*.#'$

<"D =

4(:(%;"(.&".(#2(%;"(<.*"(#2(%;"(

3',=/0"&(

;(:(%;"(;"=6;%(#2(%;"(3',=/0"&(

This looks like a cylinder with the middle missing. A good way to think about this figure isa roll of paper towels. We are trying to find out the volume of the paper towels. The bestway to do this is to figure out what the volume of the larger cylinder is without the missing

MAT 142 - Module 17

part. We can then find the volume of the smaller cylinder or missing part. Finally, we willsubtract the volume of the smaller cylinder from the volume of the larger cylinder to get thevolume of our current solid.

We will start with the larger cylinder. The radius of the circular base is given as 5 units.The height of the larger cylinder is 16 units. We can then calculate the volume of the largercylinder to be V = Ah = πr2h = π(52)(16) = 400π cubic units.

Next we will calculate the volume of the smaller cylinder. The radius of the circular baseof the smaller cylinder is 2 units. The height of the smaller cylinder is 16 units. We cancalculate the volume of the smaller cylinder would be V = Ah = πr2h = π(22)(16) = 64πcubic units.

We now subtract the volume of the smaller cylinder from the volume of the larger cylinderto get the volume of our solid. Volume of larger cylinder − volume of smaller cylinder= 400π − 64π = 336π cubic units. This is approximately 1055.57513 cubic units.

Example 19. From an 8.5-inch by 11-inch piece of cardboard, 2-inch square corners are cutout and the resulting flaps are folded up to form an open box. Find the volume and surfacearea of the box.

Solution: For this problem, it will be really helpful to make the box described above. Youstart with a standard piece of paper. You the cut out the dashed square indicated belowfrom each corner. You make sure each side of the square is 2 inches in length.

!"#$"%&'()#%"*(+#,-$"(./0(1-&2.3"(4&".( 5.6"(78(#2(9:(

!"#$%&'()*(

+,-$(#.()/012.34(56(7712.34(%2'3'(-8(3#,95-#,9:(;12.34(<=>#,'(3-,.',<(#,'(

3>?(->?(#.9(?4'(,'<>&?2.@(8&#%<(#,'(8-&9'9(>%(?-(8-,$(#.(-%'.(5-"/((+2.9(

?4'(A-&>$'(#.9(<>,8#3'(#,'#(-8(?4'(5-"/(

(

(

B-&>?2-.*(

+-,(?42<(%,-5&'$:(2?(C2&&(5'(,'#&&6(4'&%8>&(?-($#D'(?4'(5-"(

9'<3,25'9(#5-A'/((E->(<?#,?(C2?4(#(<?#.9#,9(%2'3'(-8(%#%',/((E->(

?4'(3>?(->?(?4'(9#<4'9(<=>#,'(2.923#?'9(5'&-C(8,-$('#34(3-,.',/((

E->($#D'(<>,'('#34(<29'(-8(?4'(<=>#,'(2<(;(2.34'<(2.(&'.@?4/(

(

(

(

(

(

(

(

(

F4#?(6->(#,'(&'8?(C2?4(2<(?4'(<4#%'(5'&-C/(

(

(

(

(

(

(

(

(

E->(.-C(8-&9(>%(#&-.@(?4'(9#<4'9(&2.'<(?-(3,'#?'(#(5-"/((G4'(5-"(

?4#?(C'(4#A'(3,'#?'9(2<(#(,'3?#.@>&#,(<-&29/((G42<(5-"(4#<(.-(?-%/((

H-?(4#A2.@(#(?-%(C2&&(.-?(#88'3?(?4'(A-&>$'(-8(?4'(5-"/(((

(

F'(-.&6(.''9(?-(9'?',$2.'(?4'(&'.@?4(-8(?4'(5#<'(-8(?4'(5-":(?4'(

C29?4(-8(?4'(5#<'(-8(?4'(5-":(#.9(?4'(4'2@4?(-8(?4'(5-"/((G4'(,'9(

9#<4'9(&2.'<(,'%,'<'.?<(?4'(&'.@?4(-8(?4'(5#<'(-8(?4'(5-"/((G4'(

-,2@2.#&(&'.@?4(-8(?4'(%#%',(C#<(77(2.34'</((F'(,'$-A'9(;(2.34'<(

8,-$(?4'(?-%(-8(?4'(%#@'(#.9(C'(#&<-(,'$-A'9(;(2.34'<(8,-$(?4'(

What you are left with is the shape below.

!"#$"%&'()#%"*(+#,-$"(./0(1-&2.3"(4&".( 5.6"(78(#2(9:(

!"#$%&'()*(

+,-$(#.()/012.34(56(7712.34(%2'3'(-8(3#,95-#,9:(;12.34(<=>#,'(3-,.',<(#,'(

3>?(->?(#.9(?4'(,'<>&?2.@(8&#%<(#,'(8-&9'9(>%(?-(8-,$(#.(-%'.(5-"/((+2.9(

?4'(A-&>$'(#.9(<>,8#3'(#,'#(-8(?4'(5-"/(

(

(

B-&>?2-.*(

+-,(?42<(%,-5&'$:(2?(C2&&(5'(,'#&&6(4'&%8>&(?-($#D'(?4'(5-"(

9'<3,25'9(#5-A'/((E->(<?#,?(C2?4(#(<?#.9#,9(%2'3'(-8(%#%',/((E->(

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You now fold up along the dashed lines to create a box. The box that we have created is arectangular solid. This box has no top. Not having a top will not affect the volume of thebox.

We only need to determine the length of the base of the box, the width of the base of thebox, and the height of the box. The red dashed lines represent the length of the base of thebox. The original length of the paper was 11 inches. We removed 2 inches from the top ofthe page and we also removed 2 inches from the bottom of the page. Thus each red dashedline is 11− 2− 2 = 7 inches.

18 Geometry

The green dashed lines represent the width of the base of the box. The original width of thepaper was 8.5 inches. We removed 2 inches from the left side of the page and also removed2 inches from the right side of the page. Thus each green dashed line is 8.5 − 2 − 2 = 4.5inches.

We now need to think about the height of the box. Since we have folded up the sides forform the height of the box, we just need to determine how tall those sides are. Since theywere made by cutting out 2-inch square from each corner, these sides must be 2 inches high.

Now we are ready of calculate the volumes V = lwh = (7)(4.5)(2) = 63 cubic inches.

Now we need to calculate the surface area of our box. Since there is no top to this box, wecan start formula for the surface area of a box. We will then need to subtract off the area ofthe top of the box. This will give us SA = 2(lw+wh+lh) = 2((7×4.5)+(4.5×2)+(7)) = 109square inches for the box with the top included. The top would have the same area as thebase of the box. This would be A = lw = (7)(4.5) = 31.5 square inches. Thus the surfacearea of our figure is total surface area − area of the top = 109− 31.5 = 77.5 square inches.

There is another way to calculate the surface area of this box. The surface area is the amountof paper it would take to cover the box without overlap. You should notice that this is thesame as the amount of paper we used to make the box. Thus, it is enough to calculate thearea of the paper as shown here.

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Example 20. A propane gas tank consists of a cylinder with a hemisphere at each end. Findthe volume of the tank if the overall length is 20 feet and the diameter of the cylinder is 6 feet.

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Solution: We are told this tank consists of a cylinder (one its side) with a hemisphere ateach end. A hemisphere is half of a sphere. To find the volume, we need to find the volumeof the cylinder and the volumes of each hemisphere and then adding them together.

Lets start with the hemisphere sections. The diameter of the circular base of the cylinderis indicated to be 6 feet. This would also be the diameter of the hemispheres at each endof the cylinder. We need the radius of the sphere to find its volume. Once we calculate thevolume of the whole sphere, we multiply it by 1

2to find the volume of the hemisphere (half

of a sphere). We calculate the radius to be r = 12d = 1

2(6) = 3 feet. Now we can calculate

the volume of the two hemispheres at the ends of the tank.

MAT 142 - Module 19

Left Hemisphere:The volume of a whole sphere is V = 4

3πr3 = 4

3π(33) = 36π cubic feet. We now multiply this

volume by 12

to find the volume of the hemisphere to get 12V = 1

2(36π) = 18π cubic feet.

Right Hemisphere:The volume of a whole sphere is V = 4

3πr3 = 4

3π(33) = 36π cubic feet. We now multiply this

volume by 12

to find the volume of the hemisphere to get 12V = 1

2(36π) = 18π cubic feet.

You might notice that we went through exactly the same process with exactly the samenumber for each hemisphere. We could have shortened this process by realizing that puttingtogether the two hemispheres on each end, which were of the same diameter, would create awhole sphere. We could just have calculated the volume of this whole sphere.

Now we need to calculate the volume of the cylinder. We need the radius of the cylinderand the height of the cylinder to find its volume. The radius of the cylinder is the same asthe radius of the hemispheres at each end. Thus the radius of the cylinder is 3 feet. It maylook like the height of the cylinder is 20 feet. It turns out that this is not the case. The 20feet includes the hemispheres at each end. We need to subtract the part of the 20 feet thatrepresents the hemispheres.

!

20 ft

6 ft

radius of hemisphere = 3 ft radius of

hemisphere = 3 ft

height of cylinder

Looking at the figure above, we see that the distance from the end of the left hemisphere tothe left end of the cylinder is the radius of the hemisphere. The radius of the hemisphereis 3 feet. Similarly, the distance from the right end of the cylinder to the end of the rightcylinder is also the radius of the hemisphere. The radius of the hemisphere is 3 feet. If wenow subtract these two distances from the overall length of the tanks, we will have the heightof the cylinder to be h = 20−3−3 = 14 feet. We can now calculate the volume of the cylinder.

Cylinder:V = Ah = πr2h = π(32)(14) = 126π cubic feet.

We can now find the volume of the tank by adding together the volumes of the cylinder, theright hemisphere, and the left hemisphere. We get the volume to be V = 126π+18π+18π =162π cubic feet. This is approximated by 508.938 cubic feet.

Example 21. A regulation baseball (hardball) has a great circle circumference of 9 inches;a regulation softball has a great circle circumference of 12 inches.

a. Find the volumes of the two types of balls.b. Find the surface areas of the two types of balls.

20 Geometry

Solution:Part a:In order to find the volume of a sphere, we need the radius of the sphere. In this problem,we are not given the radius. Instead we are given the circumference of a great circle of thesphere. From this information, we can calculate the radius of the great circle, which is theradius of the sphere.

Radius of the baseball:We calculate the radius of the baseball by plugging in the circumference of the great circle ofthe baseball into the formula for the circumference of the circle and solve for r (the radius).

C = 2πr

9 = 2πr

9

2π= r

Now that we have the radius of the baseball, we can calculate the volume, by plugging theradius into the formula for the volume of a sphere.

V =4

3πr3

V =4

3π

(9

2π

)3

V =4× 93 × π3× 23 × π3

V =2916π

24π3

V =243

2π2cubic inches

This is approximately 12.3105 cubic inches.

Radius of the softball:We will go through the same process with the softball to calculate the radius of the softball.

C = 2πr

12 = 2πr

12

2π= r

6

π= r

Now that we have the radius of the softball, we can calculate the volume, by plugging theradius into the formula for the volume of a sphere.

MAT 142 - Module 21

V =4

3πr3

V =4

3π

(6

π

)3

V =4× 63 × π

3× π3

V =864π

3π3

V =288

π2cubic inches

This is approximately 29.1805 cubic inches.

Part b:In order to find the surface area of a sphere, we need the radius of the sphere. We calculatedthe radius for each type of ball in part a. We only need to plug this information in to theformula for surface area of a sphere.

Surface area of a baseball:

SA = 4π

(9

2π

)2

SA =4π(9)2

22π2

SA =324π

4π2

SA =81

πsquare inches

This is approximately 25.7831 square inches.

Surface area of a softball:

SA = 4π

(6

π

)2

SA =4π(6)2

π2

SA =144π

π2

SA =144

πsquare inches

This is approximately 45.83662 square inches.

Our final example is an application problem. We will need to be able to use dimensionalanalysis, volumes, and common sense in order to be able to answer the question.

22 Geometry

Example 22. Mike Jones bought an older house and wants to put in a new concrete drive-way. The driveway will be 30 feet long, 10 feet wide, and 9 inches thick. Concrete (a mixtureof sand, gravel, and cement) is measured by the cubic yard. One sack of dry cement mixcosts $7.30, and it takes four sacks to mix up 1 cubic yard of concrete. How much will itcost Mike to buy the cement?

Solution: The driveway that is being poured will be a rectangular solid or box. Thusin order to answer this question, we will first need to find the volume of this box (theamount of cubic units it will take to fill this box). The problem tells us that concrete ismeasured by the cubic yard. This lets us know that those are the units we will want tocalculate with. None of the dimensions of the driveway are given in yards. We will need touse our dimensional analysis (unit conversion) to convert all of the measurements to yards.This could be done at a later point, but this is the easiest place to take care of the conversion.

Convert length:30 feet

1× 1 yard

3 feet= 10yards

Convert width:10 feet

1× 1 yard

3 feet=

10

3yards. We are not going to estimate this value since the approximation

with introduce error. As in the finance section, we dont want to round until the end of theproblem or in a place where it is absolutely necessary.

Convert height:9 inches

1× 1 yard

36 inches=

1

4yards = .25yards. Since the decimal representation of this number

is a terminating decimal, we can use this representation in our calculations.

Volume of driveway:We are now ready to calculate the volume of the driveway. V = lwh = (10)(10

3)(.25) = 25

3cubic yards.Now, we are told that it takes 4 bags of cement to make one cubic yard of concrete. Sowe will now calculate how many bags of cement to buy. Mikes driveway is 25

3cubic yards

and each of those cubic yards requires 4 bags of cement. Thus Mike will need 253× 4 = 100

3bags or approximately 33.333 bags. Here is where common sense needs to come in. There isno store that will sell .333 bags of cement mix. Stores only sell whole bags of cement mix.Thus we will need to round up to the next whole bag (we cant round down or we will nothave enough cement mix to complete the driveway). This means that Mike will need to buy34 bags of cement mix. Each of these bags will cost $7.30. This means that Mike will pay7.30× 34 = $248.20 for the cement mix for this driveway.

Similar Triangles

We will continue our study of geometry by looking at similar triangles. Two triangles aresimilar if their corresponding angles are congruent (or have the same measurement).

MAT 142 - Module 23

!

In the picture above, the corresponding angles are indicated in the two triangles by the samenumber of hash marks. In other words, the angle with one hash mark in the smaller trianglecorresponds to the angle with one hash mark in the larger triangle. These angles have thesame measure or are congruent.

There are also corresponding sides in similar triangles. In the triangles above side a corre-sponds to side d (for example). These sides do not necessarily have the same measure. Theydo, however, form a ratio that is the same no matter which pair of corresponding sides theratio is made from. Thus we can write the following equation

a

d=b

e=c

f

Notice that the sides of one particular triangle are always written on top of the fractionsand the sides of the other triangle are always written on the bottom of the fractions. It doesnot matter which triangle is put in which part of the fraction as longs as we are consistentwithin a problem.

Similar Triangles Ratios

ad

= be

= cf

!

It should be noted that although our triangles are in the same relative position, this is notneeded for triangles to be similar. One of the triangles can be rotated or reflected.

Please note that pictures below are not drawn to scale.

Example 23. Given that the triangles are similar, find the lengths of the missing sides.

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Solution: There is one side missing in the triangle on the left. This side is labeled x. Thereis also one side missing from the triangle on the right. This side is labeled y. The side x inthe triangle on the left corresponds to the side labeled 10 in the triangle on the right. We

24 Geometry

know this because these sides connect the angle with one hash mark to the angle with threehash marks in each of the triangles. The side 100 in the triangle on the left corresponds tothe side labeled 8 in the triangle on the right. Finally the side 90 in the triangle on the leftcorresponds to the side labeled y in the triangle on the right. We will need this informationto find values for x and y.

Find x :We will start by finding the value for x. We will need to form a ratio with the side labeledx and is corresponding side in the triangle on the right. We will also need a pair of corre-sponding side both of whose measurements we have and make a ratio out of those. Oncewe make an equation using these ratios, we will just need to solve the equation for x. I amgoing to choose to place the sides of the triangle on the left on the top of each of the ratios.

x

10=

100

88x = 10(100)

8x = 1000

x = 125units

Find y :We will continue by finding the value for y. We will need to form a ratio with the sidelabeled y and is corresponding side in the triangle on the right. We will also need a pairof corresponding side both of whose measurements we have and make a ratio out of those.Although we know the measurements of all the other sides of the triangle, it is best to avoidusing a side we just calculate to make further calculations if possible. The reason for thisis that if we have made a mistake in our previous calculation, we would end up with anincorrect answer here as well. Once we make an equation using these ratios, we will justneed to solve the equation for y. I am going to choose to place the sides of the triangle onthe left on the top of each of the ratios.

90

y=

100

8

8(90) = 100y

720 = 100y

7.2units = y

Example 24. Given that the triangles are similar, find the lengths of the missing sides.

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MAT 142 - Module 25

Solution: Our first job here is to determine which sides in the triangle on the left correspondto which sides in the triangle on the right. The side in the triangle on the left labeled yjoins the angle with one hash mark and the angle with two has marks. The side in the tri-angle on the right that does this is the side labeled 37.25. Thus these are corresponding sides.

In a similar manner, we can see that the side labeled 3.2 in the triangle on the left corre-sponds to the side labeled x on the triangle on the right. Finally the side labeled 2.8 in thetriangle on the left corresponds to the side labeled 45 in the triangle on the right.

Find x :We will start by finding the value for x. We will need to form a ratio with the side labeledx and is corresponding side in the triangle on the right. We will also need a pair of corre-sponding side both of whose measurements we have and make a ratio out of those. Oncewe make an equation using these ratios, we will just need to solve the equation for x. I amgoing to choose to place the sides of the triangle on the left on the top of each of the ratios.

3.2

x=

2.8

453.2(45) = x(2.8)

144 = 2.8x

144

2.8units = x

This is approximately 51.42857 units.

Find y :We will continue by finding the value for y. We will need to form a ratio with the sidelabeled y and is corresponding side in the triangle on the right. We will also need a pairof corresponding side both of whose measurements we have and make a ratio out of those.Although we know the measurements of all the other sides of the triangle, it is best to avoidusing a side we just calculate to make further calculations if possible. The reason for thisis that if we have made a mistake in our previous calculation, we would end up with anincorrect answer here as well. Once we make an equation using these ratios, we will justneed to solve the equation for y. I am going to choose to place the sides of the triangle onthe left on the top of each of the ratios.

y

37.25=

2.8

45y(45) = 37.25(2.8)

45y = 104.3

y =104.3

45units

This is approximately 3.21778 units.

Our final example is an application of similar triangles.

26 Geometry

Example 25. A 3.6-foot tall child casts a shadow of 4.7 feet at the same instant that atelephone pole casts a shadow of 15 feet. How tall is the telephone pole?

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When we look at the picture above, we see that we have two triangles that are similar. Weare looking for the height of the pole. The pole side of the triangle on the right correspondsto the side of the child side of the triangle on the left. The two shadow sides of the trianglesalso correspond to each other. Thus we can write an equation of ratios to find the height ofthe pole.

3.6

pole=

4.7

15

3.6(15) = pole(4.7)

54 = 4.7pole

54

4.7= pole

Thus the pole is approximately 11.48936 feet high.

Right Triangle Trigonometry

We will complete our study with a further study of right triangles. We will look at trigono-metric value as defined by ratios of the sides of a right triangle.

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571/'%38*%!In the figure above, you can see the sides of a right triangle labeled. The side labeledhypotenuse is always opposite the right angle of the right triangle. The names of the othertwo sides of the right triangle are determined by the angle that is being discusses. In ourcase, we will be discusing the sides in terms of the angle labeled A. The angle A is form bythe hypotenuse of the right triangle and the side of the right triangle that is called adjacent.The adjacent side will always make up part of the angle that is being discussed and not bethe hypotenuse. The side of the right triangle that does not form part of angle A is calledthe opposite side. The opposite side will never form part of the angle being discussed.

MAT 142 - Module 27

The trigonometric function values of a particular value can be as the ratio of a particularpair of sides of a right triangle containing an angle of that measure. We will look at threeparticular trigonometric ratios.

Trigonometric Ratios

sin(A) = oppositehypotenuse

cos(A) = adjacenthypotenuse

tan(A) = oppositeadjacent

sin = shortened form of the sine function

cos = shortened form of the cosine function

tan = shortened from of the tangent function

A = the angle value

We will use these ratios to answer questions about triangles below and then we will gothrough a couple of application problems.

Example 26. Use Trigonometric ratios to find the unknown sides and angles in the righttriangles below:

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Solution: We are being asked to find values for x, y, and B. We will do the angle B first.

Angle B :We can find the measure of angle B without using any trigonometric ratios. What we needto remember to find this value is that the sum of the three angles of a triangle will alwaysadd up to 180 degrees. It does not matter the size or shape of the triangle. The sum ofthe three angles will always be 180 degrees. We know that one angle is a right angle. Itsmeasure is 90 degrees. The measure of the other angle is given to be 60 degrees. Thus wejust need to calculate

90 + 60 +B = 180

150 +B = 180

B = 30

Thus the measure of angle B is 30 degrees.

28 Geometry

Side x :We will now work to find the length of side x. We need to start by determining which anglewe are going to use for our problem. As in the past, it is best to use an angle that is given.Thus we will be using the angle labeled 60 degrees. Our next step is to determine which sidex is relative to the angle labeled 60 degrees. Since the side labeled x is opposite the rightangle, it is the hypotenuse. There are two trigonometric ratios that include the hypotenuse.Thus, we need to determine which one to use. This will be determined by the other side ofthe triangle whose measure we know. This is the side labeled 10. Since this side is not one ofthe sides of the triangle that makes up the angle labeled 60 degrees, it is the opposite side.This means that we need to use the trigonometric ratio that has both the hypotenuse andthe opposite side. That ratio is the sine ratio. We will plug into that equation and solve forx.

sin(A) =opposite

hypotenuse

sin(60) =10

xsin(60)

1=

10

xx sin(60) = 1(10)

x =10

sin(60)

Through out this solution, we have left sin(60) in this form. This saves us from needing toround until the end of the problem. sin(60) is just a number that at the end of the problemcan be calculated by our calculator. Our answer is approximately x ≈ 11.54701.

Side y :We will finish by finding the length of side y. As in the previous par of the problem, it is bestto use the angle labeled 60 degrees. Our next step is to determine which side y is relative tothe angle labeled 60 degrees. Since the side labeled y is forms part of the angle labeled 60degrees, it is the adjacent side. There are two trigonometric ratios that include the adjacentside. Since the other side that is given is the side labeled 10 and this side is the oppositeside (see explanation above), we will need to use the trigonometric ratio that has both theopposite side and the adjacent side. That ratio is the tangent ratio. We will plug into thatequation and solve for y.

MAT 142 - Module 29

tan(A) =opposite

adjacent

tan(60) =10

y

tan(60)

1=

10

y

y tan(60) = 1(10)

y =10

tan(60)

Through out this solution, we have left tan(60) in this form. This saves us from needing toround until the end of the problem. tan(60) is just a number that at the end of the problemcan be calculated by our calculator. Our answer is approximately y ≈ 5.773503.


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Solution: As we solve this problem, we will leave out the explanations of how we determinethe names of the sides of the triangle.

Angle A:As with the previous problem, the sum of the angles of a triangle is 180 degrees. Thus wecalculate

90 + 45 + A = 180

135 + A = 180

A = 45

Thus the measure of angle A is 45 degrees.

Side x :Side x forms part of the angle that is labeled to be 45 degrees, thus this is the adjacent side.We are also given the measure of the side opposite the angle to be 3. Thus we will want touse the tangent ratio.

30 Geometry

tan(45) =3

xtan(45)

1=

3

xx tan(45) = 1(3)

x =3

tan(45)

Our answer is x = 3.

We did not need to use trigonometric ratios to find x. We could have used the fact that ourtriangle has two angles that are equal. Such a triangle is an isosceles triangle. We shouldrecall that the sides of an isosceles triangle opposite the equal angles are equal in length.Thus since one of the side was length 3, the side labeled x is also of length 3.

Side y :Side y is opposite the right angle of the triangle and thus is the hypotenuse. We also havegiven opposite side to be 3. Thus to find y, we will need to use the sine ratio.

sin(45) =3

y

sin(45)

1=

3

y

y sin(45) = 1(3)

y =3

sin(45)

Our answer is approximately y ≈ 4.24264.


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Solution:Angle B :As with the previous problem, the sum of the angles of a triangle is 180 degrees. Thus wecalculate

90 + 53.4 +B = 180

143.4 +B = 180

B = 36.6

MAT 142 - Module 31

Thus the measure of angle B is 36.6 degrees.

Side c:We will use the given angle labeled 53.4 degrees as our angle for solving this problem. Sidec forms part of the triangle that is opposite the right angle. Thus it is the hypotenuse. Weare also given the measure of the side adjacent to the angle we are using to be 5.6. Thus wewill want to use the cosine ratio.

cos(53.4) =5.6

ccos(53.4)

1=

5.6

cc cos(53.4) = 1(5.6)

c =5.6

cos(53.4)

Our answer is approximately c ≈ 9.39243.

Side a:We will use the given angle labeled 53.4 degrees as our angle for solving this problem. Sidea does not form the angle we are using. Thus it is the opposite side. We are also given themeasure of the side adjacent to the angle we are using to be 5.6. Thus we will want to usethe tangent ratio.

tan(53.4) =a

5.6tan(53.4)

1=

a

5.65.6 tan(53.4) = 1(a)

5.6 tan(53.4) = a

Our answer is approximately a ≈ 7.54041.


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Solution: In this problem, we are not given any angle to use. Instead we will need to changethe labels of our sides as we solve each of the angles in turn. We will start by finding side c.Since this is a right triangle, we can use the Pythagorean theorem to find the length of c.

Side c:The legs (a and b) are given to be 7.0 and 8.0. It does not matter which we label a andwhich we label b.

The approximate length of side c is 10.63015.

32 Geometry

Angle A:As we solve for angle A, we need to label the sides whose measures are given relative toangle A. The side labeled 8.0 forms part of the angle A. Thus it is the adjacent side. Theside labeled 7.0 does not form any part of the angle A. Thus it is the opposite side. Thetrigonometric ratio that includes both the adjacent and opposite sides is the tangent ratio.We will fill in the information to that equation and solve for A.

We need to know how to solve for A in this equation. As in our section on exponentialfunctions and their inverses, there is an inverse function (a functions that undoes) for thetangent function. On the calculator it is labeled tan−1. Thus we can finally solve for A bycalculating

Thus the measure of angle A is approximately A ≈ 41.18593 degrees.

Angle B :As we solve for angle B, we need to relabel the sides whose measures are given relative toangle B. The side labeled 7.0 forms part of the angle B. Thus it is the adjacent side. Theside labeled 8.0 does not form any part of the angle B. Thus it is the opposite side. Thetrigonometric ratio that includes both the adjacent and opposite sides is the tangent ratio.We will fill in the information to that equation and solve for A.

We will once again use the inverse function of the tangent function. On the calculator it islabeled tan−1. Thus we can finally solve for B by calculating

Thus the measure of angle B is approximately B ≈ 48.81407 degrees.


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Solution:Angle A:As with the previous problem, the sum of the angles of a triangle is 180 degrees. Thus wecalculate

90 + 43.9 + A = 180

133.9 + A = 180

A = 46.1

Thus the measure of angle A is 46.1 degrees.

Side a:We will use the given angle labeled 43.9 degrees as our angle for solving this problem. Side aforms part of the angle labeled 43.9 degrees. Thus it is the adjacent side. We are also giventhe measure of the hypotenuse to be .86. Thus we will want to use the cosine ratio.

MAT 142 - Module 33

cos(43.9) =a

.86cos(43.9)

1=

a

.86.86 cos(43.9) = 1(a)

.86 cos(43.9) = a

Our answer is approximately a ≈ 0.619674.

Side b:We will use the given angle labeled 43.9 degrees as our angle for solving this problem. Sideb does not form the angle we are using. Thus it is the opposite side. We are also given themeasure of the hypotenuse to be .86. Thus we will want to use the sine ratio.

sin(43.9) =b

.86sin(43.9)

1=

b

.86.86 sin(43.9) = 1(b)

.86 sin(43.9) = b

Our answer is approximately b ≈ 0.596326.

We will finish by looking at some application problems for our right triangle trigonometricratios.

Example 31. A support cable runs from the top of the telephone pole to a point on theground 47.2 feet from its base. If the cable makes an angle of 28.7◦ with the ground, find(rounding to the nearest tenth of a foot)

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a. the height of the poleb. the length of the cable

Solution:The picture above shows that we have a right triangle situation.Part a:The pole is opposite the angle of 28.7 degrees that is given. The other side of the trianglethat we know to be 47.2 feet forms part of the angle of 28.7 degrees. Thus it is the adjacentside. We will be able to use the tangent ratio to solve this problem since it includes boththe opposite and adjacent sides.

34 Geometry

tan(A) =opposite

adjacent

tan(28.7) =pole

47.2tan(28.7)

1=

pole

47.247.2 tan(28.7) = 1(pole)

47.2 tan(28.7) = pole

Thus the pole is approximately 25.8 feet tall.

Part b:The cable is opposite the right angle of triangle and thus is the hypotenuse. We still knowthat the adjacent side is 47.2 feet. We will be able to use the cosine ratio to solve thisproblem since it includes both the hypotenuse and adjacent sides.

tan(A) =adjacent

hypotenuse

cos(28.7) =47.2

cablecos(28.7)

1=

47.2

cable(cable) cos(28.7) = 1(47.2)

cable =47.2

cos(28.7)

Thus the cable is approximately 53.8 feet tall.

Example 32. You are hiking along a river and see a tall tree on the opposite bank. Youmeasure the angle of elevation of the top of the tree and find it to be 62.0◦. You thenwalk 45 feet directly away from the tree and measure the angle of elevation. If the secondmeasurement is 48.5◦, how tall is the tree? Round your answer to the nearest foot.

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Solution:

This problem will require a little more algebra than the previous problems. We will startby looking at the two different triangles that we have and writing trigonometric ratios that

MAT 142 - Module 35

include the tree for each of our triangles. We will start by looking at the bigger triangles(shown in red below).

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In this red triangle, the tree is opposite the angle that is given to be 48.5 degrees. We arealso given the length of 45 feet as part of the side that is adjacent to the angle given to be48.5 degrees. Since we have part of this adjacent side, we are going to label the other partof the side to be x feet. Thus the whole adjacent side is 45 + x feet long. Since we have theopposite and the adjacent sides, we can use the tangent ratio. For the red triangle, we have

tan(x) =opposite

adjacent

tan(48.5) =tree

45 + x

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In this blue triangle, the tree is opposite the angle that is given to be 62 degrees. Based onwhat we did for the red triangle, we know that the length of the side adjacent to the anglegiven to be 62 degrees is x feet long. Since we have the opposite and the adjacent sides, wecan use the tangent ratio. For the blue triangle, we have

tan(x) =opposite

adjacent

tan(62) =tree

x

Now we are ready for the algebra. We have two equations tan(48.5) = tree45+x

and tan(62) =treex

with two unknowns (variable). We can use the substitution method (solve one of theequations for one of the variable and then plug that in to the other equation) to determinethe height of the tree. One way to go here is to solve the equation tan(62) = tree

xfor x. This

will give us

36 Geometry

tan(62) =tree

xtan(62)

1=

tree

xx tan(62) = 1(tree)

x =tree

tan(62)

We can now plug this expression for x into the equation tan(48.5) = tree45+x

and solve for theheight of the tree.

tan(48.5) =tree

45 + x

tan(48.5)

1=

tree

45 + xWrite both sides as fractions.

(45 + x) tan(48.5) = tree Cross multiply.

45 tan(48.5) + x tan(48.5) = tree Apply distributive property.

45 tan(48.5) +

(tree

tan(62)

)tan(48.5) = tree Substitute in expression for x.

45 tan(48.5) = tree−(

tree

tan(62)

)tan(48.5) Collect variable on the same side.

45 tan(48.5) = tree

(1−

(1

tan(62)

)tan(48.5)

)Factor out the variable tree.

45 tan(48.5)(1−

(1

tan(62)

)tan(48.5)

) Divide both sides by value in parentheses.

We now just need to plug this expression into our calculator to find out the height of thetree. The tree is approximately 127 feet tall.

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