lesson on solid mensuration

7/27/2019 LESSON ON SOLID MENSURATION

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Elmar H. Mazon Jr. Solid Mensuration 7-8pm (TTh)

BS ME / 201210743 Prof. Mabutas

CYLINDERS AND PRISMS (V=Bh)

Introduction:

We examine cylinders and prisms, defining them, and measuring their surface

areas and volumes.

The Lesson:

We may think of a cylinder as a tin can, with two bases that are circles. Thesurface wrapping around the circles can be unwrapped and shown as a rectangle. A

diagram below shows a cylinder on the left. The rectangle at right is the side surface,

unwrapped from the cylinder. We know that its length is the circumference of the

circular base which is .

The surface area is the sum of the areas of the two circles that are the bases and

the area of the rectangle. Since the area of a circle is and the area of a

rectangle is lw, we can make a formula for the surface area of a

cylinder: .

The volume of a cylinder is found using the concept ofbase x height. The area of

the base is . The volume of the cylinder is .
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A prism is a (3-dimensional) polyhedron with bases that are

parallel, congruentpolygons. The sides are parallelograms. We show three examples

below. The important features are the areas of the bases and the height, which is

the perpendicular distance between the bases.

At the left is a rectangular solid. The middle diagram is of a prism with bases that

are hexagons. Notice that the slant height, s, is different than the height, h, since

the prism is tilted. The prism at right has bases that are pentagons and is not tilted.

We shall concentrate on prisms such as this which are called right prisms.

Volume and surface area can be found as follows:

Volume = base (area) x height

Surface Area = sum of areas of both bases and all sides

The area of the bases can be found if we know the dimensions and properties of

the polygons which form the bases. The area of all the sides (called the lateral area)can be found if we can measure the base and height(called the slant height and

labeled s) of each of the parallelograms which form the sides of the prism. Adding

the areas of the parallelograms which form the sides can be summarized by

multiplying the slantheightand the perimeter of a base. We have:Surface Area =

sum of areas of both bases and s x P where s is the slantheightand P is

the perimeter of a base.

If the bases are perpendicular to the parallelograms forming the sides (as in a right

prism), the slantheights = h.
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Let's Practice: We label the three prisms below A, B, and C. A is a rectangular solid. B

is a non-right prism. C is a right prism.

Suppose prism A has dimensions 5 x 4 x 7 feet. What are the surface area and volume?

We solve this using standard formulas for a rectangular solid.

The volume is 5 x 4 x 7 = 140 cubic feet. Two sides are 5 x 4, two are 4 x 7 and two are

5 x 7. The areas of these sides are 20, 28, and 35.

Thus the total surface area is twice the sum of these or 166 square feet.

Suppose prism B has bases which are regular hexagons ofside 2 meters, a heightof 3

meters, and a slantheightof 3.4 meters. What are the surface area and the volume?

The lateral area equals

s x P = 3.4 x 12 = 40.8 m2.

The area of a regular hexagon (base) is

m2.

This gives a total surface area of two bases plus the lateral area which equals

m2.

We now know the volume = area ofbase x heightequals

m3.

Suppose right prism C has bases which are regular pentagons ofradius 25
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centimeters. The heightis 40 centimeters. What are the surface area and the volume?

For the area of a base we use

.

This gives us area ofbase as

cm2.

We calculate volume = area ofbase x heightas

1486 x 40 = 59440 cm3 = 0.05944 m3.

The lateral area is s x P where P is the perimeter of the base of the regular

pentagon with radius 25 and s = h is the slant height.

A side can be found from

This gives us the length of a side of the base as

cm.

The lateral area equals

s x P = 40 x 5(29.39) = 5878 cm2.

The total surface area is two bases plus the lateral area or

8850 cm2 = 0.885 m2.

A tin can has a radius of 2 inches and a heightof 3 inches. What are the surface

area and volumes?

The volume is

in3.
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WORD PROBLEMS: VOLUME AND SURFACE AREA OF PRISMS

We will work with right prisms, in which the slantheightis the same as the

height. In order to solve problems which require application of the volume and

surface area for prisms, it is necessary to know how to use basic

formulas for volume (V) and surface area (SA) of a right prism:

V = area ofbase x height

SA = sum of both bases and the lateral area

know how to find the area of a base:

a) regular polygon: base area =b) regular hexagon: base area =

know how to find the length of a side of a regular polygon using

solve basic linear and quadratic equations.

A typical problem involving the volume or surface area of a prism gives us

one or more of the volume, lateral area, area of a base, heightand/or radius of

the prism. We will be required to calculate some of these quantities given

information about the others.

Suppose the heightof a right rectangular prism (rectangular solid) is 30 cm

and the volume is 480 cm3. If the base is a square, find the surface area of this

prism. A diagram is shown below.
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6/18

To get started, we sketch a diagram and label all of the given information to

determine the appropriate formula(s) we will be able to use.

Since we know the volume is 480 cm3, we will start with the formula for volume.

V = 480 cm3

V = lwh

The length land width ware both the same in a square base allowing us to use

the variable x for both land w. Because h is given as 30 cm we can write:

V = (x)(x)(30)

V = 30x2

30x2 = 480

x2 = 16x = 4 cm

We will now use x = 4 cm and h = 30 cm to calculate the area of all six

rectangular sides of this prism. Remember that the 2 bases are equal as are the

four sides making up the lateral area.

There are two square bases. Since the length of a side x = 4, there is a

total area of

2x2 = 2(4)2 = 32 cm2

The four sides are rectangles with dimensions of width 4 cm and height30 cm

for a total area of

4(4)(30) = 480 cm2

The total surface area is

32 cm2 + 480 cm2 = 512 cm2
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7/18

WORD PROBLEMS: VOLUME AND SURFACE AREA OF CYLINDERS

In order to solve problems which require application of the volume and surface

area for cylinders, it is necessary to

know how to use basic formulas for volume and surface area of a cylinder:

Volume: V = pr2h

Surface Area: S = 2pr2 + 2prh

Notice that these factored versions of the surface area formula are sometimes easier

to use:

S = 2pr2 + 2prh = 2p(r2 + rh) = 2pr(r + h)

Solve basic linear and quadratic equations.

A typical problem involving the volume or surface area of a cylinder gives us

the volume or heightand/or radius of the cylinder. We then need to calculate theunknown quantities based on the information given about the others.

Suppose that the height of a cylinder is 30 cm and its volume is 750p cm3. Find

its radius and surface area.
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8/18

To get started, we need to organize as much of the given information as possible

into a known formula. Since the volume (750p cm3) and height(30 cm) of

the cylinder are given, we will start with the equation for volume.

V = pr2h = 30pr2

V = 750p750p = 30pr2

r = 5 cm

Now we will find the surface area of the cylinder using our values for the radius and

height.

S = 2pr2 + 2prh

S = 2p(25) + 2p(5)(30)

S = 50p + 300p

S = 350p cm2
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9/18

Elmar H. Mazon Jr. Solid Mensuration 7-8pm (TTh)

BS ME / 201210743 Prof. Mabutas

PYRAMIDS AND CONES (V=1/3 Bh)

Introduction:

We examine 3-dimensional objects known as pyramids and cones, defining

them, and measuring their surface areas and volumes.

Definitions:

Pyramid: A 3-dimensional solid in which the base is a polygon and the sides are

triangles which meet in one pointcalled the vertex. We shall examine regular

pyramids in which the base is a regular polygon and the sides

are congruenttriangles.

A Right Circular Cone: A 3-dimensional solid in which the base is a circle.

The side of a cone is formed by straight lines which connect the circular base to a

vertex. The heightis the perpendicular distance from the vertex to the base and meets

the base in the center of the circle.

The Lesson:

The diagrams below show a pyramid and a cone. Both have

a heightofh and radius ofr. In the pyramid at left, r is the radius of theregularhexagon that is the base of the pyramid. In the cone at right, r is the radius of

the circular base. The slantheightis s in both diagrams.
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10/18

Pyramid:The perimeter of the base of the pyramid, which is a regular hexagon, is 6r since r is

the same length as the side of a regular hexagon.

The area of the base of the pyramid is given by

.

The area of the sides (lateral area), which are congruenttriangles, is given by

because in this case n = number of sides = 6, the base is equal to r, and the

slantheightof the triangles is s.

Notice that this can be rewritten as:

.

where P is the perimeter of the base.

Overview

In general, calculating the surface area of a pyramid requires finding the sum of the

areas of the base and the triangular sides.

The lateral area is given by

where s is the slantheightof a triangular side and P is the perimeter of the base.
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11/18

Remember that this is only true because we are assuming that the triangular

sides are congruent; that is, we are working with a regular pyramid. Note that in

a regular pyramid it is also true that (1) each triangle makes the same angle with

the base and (2) the heightconnects the vertex to the center of the regular

polygonal base.

Pyramid Summary:

Surface area of a regular

pyramid = area ofbase + .

Volume of a pyramid =

Cone Summary:

The area of the base of the cone is . The lateral area is given by .

The volume of a cone is given by .

Let's Practice:

i. In Egypt, the GreatPyramid of Giza is 145.75 m in heightand hasa square base of 229 m on a side. The triangular sides are congruentand form

anangle of 51 with the square base. What are the surface area and

the volume of this pyramid?

The base area is

2292 = 52441 m2.

The lateral area is

; where s is the slantheightof a triangular side.
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20Pyramid%27,%20500,%20500);


12/18

To calculate the slant height, we use the heightand the slantheight(as

a hypotenuse of a right triangle) as in the diagram shown above. Using the trig

function sine, we get

.

This gives us

.

The lateral area is

458s = 85875 m2.

The total surface area is

52441 + 85875 = 138,316 m2.

The volume is area ofbase x height=

x 2292 x 145.75 = x 7,643,275.75 m3 = 2,547,758.583 m3.

Despite the fact that well over 1,000,000 stone blocks weighing between 2 and 150

tons were manually installed, the exact measurements of the sides of the base and the

angles of the triangles forming the sides are off by less than 0.1% from that of a

perfectly regular pyramid.

ii. A cone has height5 feet and the radius of the base is 2 feet. What arethe surface area and the volume?

To find the volume we use

cubic feet.

To find the surface area we use

.

This means we must calculate the slantheights.

In order to do this, we note that the heightis perpendicular to the base of the cone at
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13/18

the center. We use the height, radius, and slantheightsto form a right triangle. A

diagram is shown below.

We now use the Pythagorean Theorem.

s2 = 52 + 22 = 25 + 4 = 29

Now solve for s:

s =

This gives us a surface area of

ft2.
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14/18

WORD PROBLEMS: VOLUME AND SURFACE AREA OF PYRAMIDS

We will work with regular pyramids, in which the triangular sides

are congruentand the base is a regular polygon. In order to solve problems

which require application of the volume and surface area for pyramids, it isnecessary to

know how to use basic formulas for volume and surface area of a regularpyramid:

i. Surface area of a regular pyramid = area ofbase +ii. Volume of a pyramid = .

know how to find the area of a base:

i. for a regular polygon: A =ii. for a hexagonal base: A =

know how to find the length of a side of a regular

polygon using .

use the Pythagorean Theorem in finding the radius, side of a base, or slantheight.

A typical problem involving the volume or surface area of a pyramid gives

us one or more of the volume, lateral area, area of a base, heightand/or radius of

the pyramid. We need to calculate some of these quantities given information

about the others.

Suppose that the heightof a regular square pyramid is 3 cm and

the length of one edge is 5 cm. What are the surface area and volume of this

pyramid? A diagram shown below includes the length of the radius CB.
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15/18

To get started, we must determine which quantities are known and look at

the formulas available for our use to to see which additional variables we need

to calculate.

For both surface area and volume, we need the area of the square base.

It is usually easier to find the area of the base if we first know the length ofone side. Since triangle ACB is a right triangle with hypotenuse 5 and one leg 3,

our square base has a radius of 4 . We can now use the formula

along with the knowledge that the radius r = 4 and the number of sides n = 4 to

find the length of one side of the square,

We now know the area of the square base is

A = s2 =

We can use this to find the volume which is

V =

To find the surface area, we need the perimeter of the base, P, and the slant

height, s

surface area = area ofbase +
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16/18

The perimeter of our square base is just four times the length of one side:

perimeter= 4( )

perimeter=

The slantheightis the length of AM in triangle ABD. Using the Pythagorean

Theorem and triangle ABM,

where MB = because it is half of one side

The surface area equals .
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17/18

WORD PROBLEMS: VOLUME AND SURFACE AREA OF CONES

We will work with right circular cones where the base is a circle and

the heightis the perpendicular distance from the vertex to the center of the base. In

order to solve problems which require application of the volume and surface area for

cones, it is necessary to

know how to use basic formulas for volume and surface area of a cone.1. The area of the base of a cone ofradius ris p r2.2. The volume of a cone is given by where h is the heightof the

cone.

3. The lateral area is given by p rs where s is the slant height.4. The surface area is given by S = p r2+ p rs

know how to solve basic linear and quadratic equations. know how to recognize right triangles and use the Pythagorean Theorem.

A typical problem involving the volume or surface area of a cone will give you

information about one or more of the following quantities and ask you to calculate the

others: volume, surface area, or heightand/or radius of the cone.

Suppose that the heightof a cone is 3 cm and its surface area is 24p cm2. Find

the heightand volume of this cone.

To get started, we need to write as much of the given information as possibleinto a known formula. We will start with the equation for surface area since we were

given the radius as 3 cm and the surface area as 24p.

S = 24p

S = p r2 + prs

S = p32 + p3s

S = 3p(3 + s)

Solving this equation for s we get

24p = 3p(3 + s)8 = 3 + s

s = 5 cm
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18/18

The information we have learned so far is summarized in the following diagram.

To calculate the volume we need to find the values ofrand h.

Since h, r, and s form a right triangle, we can use the Pythagorean Theorem to

calculate the value ofh

h2 + r2 = s2

h2 + 32 = 52

h2 = 25 - 9

h2 = 16

h = 4 cm

The triangle involving h, r, and s always allows us to use the Pythagorean

Theorem and s is always the hypotenuse.

We can now use r = 3 cm and h = 4 cm in the formula for volume:

Reference:Algebralab.org
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lesson on solid mensuration

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