MA120S - Mr. S. Koslowsky

is a 3-dimensional

1.4 - Surface Areas of Right Pyramids and RiqhfCones

Pyramids

A r\Q)Y^ ^r^^\^object that has an 0^-^tX directly above the center of thebase. It also has triangular faces and a base that is a^0 \v) 0)0^ The shape of the ^^^^ tells you

the name of the pyramid. The V^'|O)\\Y of thepyramid is the distance from the center of the base to theapex. When the base is a f -^.Oj ^A\ o< r polygon, thetriangular faces are congruent (same shape and size).The ^\ o.n.'V ^-^ 'i o^ ^:\ is the height of a triangular face.

he^ht Slcnt height

Finding Surface Areas of Right Pyramids

Example #1 - Regular Tetrahedron

Calculate the surface area of a regular tetrahedron, with a side length of 5.0 m, to thenearest square meter

^ ^^^u-\ ^^lO-M^\^^

S v<\

^ \ ^..^ -')-s,.^

v

<.^

^ v ^.v^ '- \^

^j^T'i-^ ^ ^.^ ^

r.<M^ ^\!< (7\v\ ^',^3

A'-^^ ~ ^ ^ ^ ,-v^^ -5.

^,^^ n \

/\ - \^.^^ T(\\

MA120S - Mr. S. Koslowsky\o^A -s^ : <^^^ ^> Q-r^sk^

.S-\\ V ^.^1 V 'i,'\,\'(>'- [[ol. ^T^Example #2 - Right Rectangular Pyramid ^-^-A right rectangular pyramid has base dimensions 4 m by6 m, and a height of 8 m.Calculate the surface area of the pyramid to the nearest square meter.

6

^V^V. ^^^

J^-_ ^ ^ ^^ ^ \^\ ^

3 \ ^'s%^ ^Aw\ Yt'i^^"^\^v^

^ J^~^s^^.^ w

Vr^V V^'^\^\

^\K v t~- ^,"cFiv.^.vSY'-^,^ ^

('i^^ ^f 1^A'. ^ ^Bl^-^ . ^.^^ ^ \f^tr^^, <'\r\^ui\^ ^r^< ^

-^ "^---~- - d--*,. . ^ \(^^ __- ^ V-'^ I,^ _, .\ ^^r~"

\ ^.^\ ^-. Uv. -\ (^. 1^ ^(^V ^ \ ^^\ ^.V,^ Y(P.U^^^,

n<m

;n.o<Now let's figure out the surface area of a regular right pyramid in general (no numbers)We'll do a right pyramid with a ?c\^g\<^ base.The base has length L and the slant height is 3

.^OIV<- ^r1.^^ ^ ^f,^\^

\':i ^

LA ^ ^ L.^

-^ ~-^-^

-y

s^^\^ '-^ ^f,^^r o{ \^^

<^^^ ^)(r^ ^^ 0\

Here's the general formula for the surface area of a right pyramid with a regular polygonbase and slant height s:

^»./\. ^-^S^^r'.^t^r Q^- \^^^ .\ (^WK 0<T^u<

MA120S - Mr. S. Koslowsky

Cones

AJ^_O^_ ^\rc^\ ^r cofv^ ,533.dimensional object that has a ^ \f 0^\ fr^r base and a^ wr ^ ^ lateral area.The^^X_ is directly

above the center of the base. Similar to a pyramid, the heightand slant height are as shown in this diagram.

The general formula for the surface area of a right cone with slant height s and baseradius r is:

^,/\.^ ^<"5 V ^r'X

\^^F<A\ \^^^ ^^^ (t>ri\Y<M ^*Y

Example #3 - Surface Area of a Right ConeA right cone has a base radius of 4 m and a height of 10 m. Calculate the surface areaof this cone to the nearest square meter

A~^^^ ^S ^T^u^^ ^Y^^^-V

VS^

YU|' ^ \0^ ^5'v

\\<0

\0,~l-?=s\^

^.^\^ ^(^(\OT1) ^ ^^^. /\ ^ w.^^\ ^ ^^.^'1

v

MA120S - Mr. S. Koslowsky

Example #4 - Finding an Unknown MeasurementThe lateral area of a cone is 100 cm2. The diameter of the cone is 6 cm. Determine theheight of the cone to the nearest tenth of a centimeter.

-s,A^ ^^3^ ^ < ^ ^.ro<^\^^ ^^> ^T*\

\fJ\^\^Tt<^

^>^\

\^^ ^^\0(^ -^^^^\^& ^ ^\^v^

-g^vY ~^o^Y

^'- ^o.s.\ ^

Vs

AO.^\\

^

^^vr ,U~\^^ °\~-\\\.^

J^ ^.^\<~: \0,\^>w

\^/v^ ^\^w

Example #5 - Finding an Unknown Measurement (part 2)A model of the Great Pyramid of Giza is constructed for a museum display. The surfacearea of the triangular faces is 3000 in2. The side length of the square base is 50 in.Determine the height of the model to a tenth of an inch.

'5 /^ ^ ~^ ^ ^v' ^w ^ W^^)\ (\^y^ ^r^o^.^\ ^^ p\

.1,&

^0(j6 -:1 \s^^^ ^ ^^^&~. _ , ^66 -;\^ (^

v^v^--v \ ^et^ \6^V ^ ^^^t6 \ uo

^^ 5

lt\i

J^^i[^^_^v^

MA120S - Mr. S. Koslowsky

1.5 - Volumes of Right Pyramids and Right Cones

Pyramids

When thinking about the volume of a right pyramid, it's helpful to first consider ar .<^vv't ^f<^YV Right prisms can have a base that is any shape.

The Volume of a right prism is:^= \s^n ^<-^<N\^-L ^(.'l^VV^ ^v

The volume of a r \ °tVV ^ ^| rtf\ ^\', ^ yvith the same base and the

Vs

same

y^

height is 1/3 that of the prism. So the formula is:

V ^ ^ AY or V~- ^s

Example #1 - Calculating the Volume of a Right Square PyramidCalculate the volume of a right square pyramid with a base length of 2 ft. and a slantheight of 7 ft.

v^ ^v^v7

^

^ ^ \^^J^v.^^

Vv'- ^.v

^)^^ ifV< ^y\

|^r <^\^ : ^ ^V

v\^ f\^

V. ^^.^ v^ °\.^^ ^\^

MA120S - Mr. S. Koslowsky

We could write a formula for the volume of a right rectangular pyramid:

V'- ^p\v ^frv< ^^^ Lw\ ^V^^jY V^ iv/v^^ %r

Example #2 - Calculating the Volume of a Right Rectangular PyramidDetermine the volume of a right rectangular pyramid with base dimensions 3.6 m by4.7 m and height 6.9 m. Answer to the nearest tenth of a cubic meter.

^^

^.1

vLV^

^>

V-. ^3,0^.^'^)

V/\\^.^^

'^>

^7s-T:nz~Cones

Right cylinders and ^'I^VV CQXY^^ have the same relationship betweentheir VpV^Y^tS The volume of a right cone is 1/3 the volume of a rightcylinder with the same ^o o-^^ and ^ ^ > ^^1TVolume of right cylinder:

V - ^r^Y\

Volume of right cone:

MA120S - Mr. S. Koslowsky

Example #3 - Calculating the Volume of a ConeDetermine the volume of a right cone with a diameter of 8 mm and a height of 13 mm.Answer to the nearest cubic millimeter.

/, _-^. ^K^^ -^n\\<\

'A<^'(«V--

^

^'1^, ;^~^)

,^\ w^^^

Example #4 - Finding an Unknown MeasurementA right cone has a height of 8 m and a volume of 300 m3. Determine the radius of thebase of the cone to the nearest meter.

v^

^^ ^)0^

^^v.~^

^ <^(^^

.>^

^ <v (^'T^^rv

"^^A^

^,^^ tv\

v^.^1^f

MA120S - Mr. S. Koslowsky

1.6 - Surface Area and Volume of a Sphere

A sphere is an infinite set of ^0\AT5 in space that are the same distance awayfrom a fixed point, which is the ^^^'\tr A line segment that joins the centerto any of these points is called a fo^^iV^^ .

Surface Area of a SphereTo think about a sphere's surface area, relate it to the curved surface of a

^ ' ^^ ^v)l^j(lr (not including the top and bottom circles). Draw acylinder and a sphere that have the same diameter. The height of the cylinder shouldbe the same as the diameter. These 2 surfaces have the same surface area!!!

(^^r

^Vv^ ^.\5 ^f^^ ," cyt,^^^^

The surface area of the curved part of a cylinder is:

^,/\.^^^^V -) ^l\-^r(\f^/\^, V\^^-r

Since the height = diameter, we can call the height 2r (2 times radius).

4 i^r \

5 (\-- ^^ f(^^)<$f\ ^ 1-^ <v rK

So this is the formula for the surface area of a sphere with radius r:

^'?\JL(t 1^- :-^^fv

MA120S - Mr. S. Koslowsky

Example #1 - Determine the Surface Area of a SphereThe diameter of a softball is approximately 4 in. Determine the surface area of asoftball to the nearest square inch.

<f^^< iA.^

.5P.,C ^-TX^T

^(\^ ^T\[^)5 A--^ 0.11 ^ \

Example #2 - Determine the Diameter of a SphereThe surface area of a soccer ball is approximately 250 square inches. What is thediameter of a soccer ball to the nearest tenth of an inch?

^A^^-T^f^

Z^& ^^.^ri"^-n -^^

Y\J 1<^<^^ ^1(-

'\(^.^^^Y,^"\Y

^,-T^^Jf

^.^\=rw

MA120S - Mr. S. Koslowsky

Volume of a SphereImagine a sphere covered with very small ^ ^/^^ F ^-3Each square has lines from it's corners to the center, whichforms a r ' of^'V _ ^\jro\y^\^ The volume ofthe sphere would be the ^OV\A^^ of all these pyramids.

Let's derive the formula!^^^ ^ ^3V

<\ \^<-S:V|CA '.

O\\A^^. Q{ S^Vtr^ ^3'^^ 0+ ^ruvw'^ vol^^^\/ ^ ^^.^V 0-^- ^\\[-^[V^^ ^r^(\\^, o)VV)

v^ ::- ^ ^^w ^ ^< ^c^^\r

Vs'- ^(^r^,\^>

Vs'- ^^T^ ^

r

\f

1^ ^^,^*i 5 fo^^\\r^5^^ ':>^r<l

3^h. t^ra< ^\r^^

13 ^V^i^^./<.

\^r^}

Example #3 - Determine the Volume of a Sphere

The moon approximates a sphere with diameter 2160 mi. What is the approximatevolume of the moon? Express your answer infscientific notation with 2 decimals.

^f^ \^0 ^

v--^c'V^ ^-i\ (\QiQ) V^ ^.^t Y\O\^

v^ ^^^ ^\ \^ ^

MA120S - Mr. S. Koslowsky

Example #4 - Determine the Surface Area and Volume of a HemisphereA hemisphere has radius 5.0 cm. Calculate the surface area and volume of thehemisphere to the nearest tenth each.

^ k^r^»<^ P\< ty^?:)f»\

^- ^ Vs, c\rc\v o«,^ Vo^V(- ^^V^i

C\r<\ '< ^^ ^ ^^c^.V/\.^r-

/\ ^ ^ (^Yf\--1<i ^u, t^

^,_ ^^ ^ ^^"^

A^

^»V-\ 3.A,

A = ^^ (^Yf\^Y.^ ^

^^VT^.0^ "^X^~^T

1\ ^^T^

v^

v^

v

i ^^^ ^ ^^ ^ ^^ 'lvr^- -^

- -^ r\ r

V ~- ^ r\ (^V'- ^ ^ (\^V'- ^^\.l^ ^

MA120S - Mr. S. Koslowsky

JL.7 - Solving Problems Involving Composite Objects

A composite object consists of ^W^ or more distinct objects.To find the volume of a composite object, find the volume of each part, then

c^ ^ ^ the volumes.Example #1 - Volume

Consider a right cylinder with a hemisphere placed on top. Both objects have a radiusof 18.0 cm. The cylinder has a height of 32.0 cm. Determine the volume of this objectto the nearest tenth of a cubic centimeter.

Y?^', ^^r^_

/^\

-^^

.r^ \\ c^v

1>Y^T^v=

^dV\^\ ^^

- ^^ t\r .^

v^ ^'"^ <\ 0^

^/^ -^\\ ^^°\>oY ^;

cvjVw ^r

v ^ ^ ^^v^^^ (^

^/r-r\(^^-V^^^~a.,^

^ ^^, ^^vf^ ^ \-\ ^\^\^\ ^^v^_ii^^- ~ " """" 11^~1^~y\)

To calculate the surface area of a composite object, only include the surfaces that areon the M \^^'S,^<L of the object. Find these areas, then 0^ ^ ^ them.Example #2 - Surface Area

Consider a cube with a right square pyramid on top of it. The cube has side length 5 m.The right pyramid has a slant height of 4 m. Find the surface area of this compositeobject.

t^fl

(Krl^ 0^ ^ ^^<<

h-- {^ ^ ^S °^^ ^ A^-3

\

\^ ^y^\^ ncv^

^ ^vf

^^VO^ ^

^ 6\^\ \^ \^

.^\

MA120S - Mr. S. Koslowsky

Example #3

Consider a right rectangular prism with a right cylinder sitting on top. The prism hasdimensions 5 cm by 6 cm by 2 cm. The cylinder has radius 2 cm and height 10 cm.Find the volume of this composite object.

.r^Y ^^)\'^ ^<r- ^oV

V^^^Yv'.^^ ^^^1.^^\{\^v^\^.^ ^

\A'(^ Y ^.'|^V ^(iVv.^^

v^v^v^v'-^^v^ ^^ ^

^ u V^ , \Y^^<» ^ ^ \^^.^ ^

Find the surface area of this composite object.

Uorr^^V^ ^lo^

oV \v^\^ -\v.

W6^V^ <5vA^V(MV J\\^^<\. ^

^\)T,c ^<\

<J\L^V^

u

^\r^\<t ^o^ ^V\ ^^ m^o^\<tC(K}^ AVj^ ^^ dr^^ ^('\8\

s<v< ^.

^..^TV '5,A.

'S^^^'Y^- ^ ^^'^ ^ v^ v ~^ ^ ^^

\

^^fw-V^ ^yvr^ of- ^A)V^^^f\^. ^^rV\5 A _ ^^(t)^

TA

^^^Y ^ ^-^ C^ \ ^ 1^ ~- vyci . ^ ^

\^ VA v^^\ ^^\,

^.Y «^^ ^0 YO\\|-:S.^-(\.^ t^L