Understanding Changes in Dimensions, Perimeters, Areas, and Volumes

Area of $K L M N$	Area of $P Q R S$
$A_{1} = K L \cdot L M$	$A_{2} = P Q \cdot Q R$

Area of

K L M N

Area of

P Q R S

A_{1} = K L \cdot L M

A_{2} = P Q \cdot Q R

Surface Area of Solid $A$	Surface Area of Solid $B$
$S A_{1} = 2 (a_{1} \cdot a_{2} + a_{1} \cdot a_{3} + a_{2} \cdot a_{3})$	$S A_{2} = 2 (b_{1} \cdot b_{2} + b_{1} \cdot b_{3} + b_{2} \cdot b_{3})$

Surface Area of Solid

A

Surface Area of Solid

B

S A_{1} = 2 (a_{1} \cdot a_{2} + a_{1} \cdot a_{3} + a_{2} \cdot a_{3})

S A_{2} = 2 (b_{1} \cdot b_{2} + b_{1} \cdot b_{3} + b_{2} \cdot b_{3})

Volume of Solid $A$	Volume of Solid $B$
$V_{1} = a_{1} \cdot a_{2} \cdot a_{3}$	$V_{2} = b_{1} \cdot b_{2} \cdot b_{3}$

Volume of Solid

A

Volume of Solid

B

V_{1} = a_{1} \cdot a_{2} \cdot a_{3}

V_{2} = b_{1} \cdot b_{2} \cdot b_{3}

The surface areas of two similar pyramids are $54$ square meters and $24$ square meters. The volume of the smaller pyramid is $8$ cubic meters.

Find the volume of the larger pyramid.

Let's go over the information we have. The smaller pyramid has a surface area of 24 square meters and a volume of 8 cubic meters. We want to find the volume V_2 of a similar pyramid with a surface area of 54 square meters.

	Surface Area (m^2)	Volume (m^3)
Smaller Pyramid	24	8
Larger Pyramid	54	V_2

First we will find the scale factor between the pyramids using what we know about the surface areas of similar solids.

Surface Areas of Similar Solids |- If the scale factor of two similar solids is ab or a:b, then the ratio of their surface areas is a^2b^2 or a^2:b^2.

Since the ratio of the areas of similar figures is the square of the scale factor, we need to find the square root of the ratio of the areas to find the scale factor. A_1/A_2 = (a/b)^2 ⇔ a/b = sqrt(A_1/A_2) Let's substitute the surface areas and find the scale factor.

This means that the larger pyramid is 32 times the size of the smaller pyramid. Using this scale factor and what we know about the volumes of similar solids, we can find the volume of the larger pyramid V_2.

Volumes of Similar Solids |- If the scale factor of two similar solids is ab or a:b, then the ratio of their volumes is a^3b^3 or a^3:b^3.

This means that the ratio of the volumes of the pyramids is equal to the cube of the scale factor. Volumes of Pyramids V_2/V_1 = (3/2)^3 Finally we substitute V_1=8 and solve for V_2.

The volume of the larger pyramid is 27 cubic meters.

A cone is cut by a plane parallel to its base. The small cone on top is similar to the large cone.

Cone cut by a plane parallel to its base

If the volume of the cone above the plane is

\frac{1}{8}

of the volume of the larger cone

C,

what is the height of cone

C ?

We are given that a plane is parallel to the base of cone C. The volume V_1 of the cone above the plane is 18 of the volume V_2 of the larger cone. V_1/V_2 = 1/8 To find the height of the larger cone, we need to find the length scale factor of the cones. Recall that the ratio of the volumes of similar solids is equal to the cube of the ratio of their corresponding side lengths. Volumes of Similar Solids V_1/V_2=(a/b)^3 ⇔ a/b = sqrt(V_1/V_2) With this in mind, let's calculate the cube root of the given volume scale factor, 18.

The length scale factor between the smaller cone on top and larger cone is 12. cc Volume Scale Factor & Length Scale Factor 1:8 & 1:2 Now that we have the length scale factor, we can write a proportion to find the height of larger cone. We know that the height of the smaller cone is 12 feet. Let's use h to represent the height of the larger cone. 1/2=12/h Finally, let's solve for h!

The height of the larger cone is 24 feet.

The following nesting dolls are similar.

If the smallest doll is

1

inch tall and each subsequent doll is

1

inch taller than previous one, which statement is correct about their surface areas?

We know the smallest nesting doll is 1 inch tall and that each doll after that is 1 inch taller than the previous one. Let's find the length of each doll.

Since the dolls are similar solids, the ratio of their surface areas is equal to the square of the ratio of their heights. Let's express the surface area of each doll in terms of the surface area of the first one. If we let the height of the dth doll in the set be d and its surface area S_d, we can write a general relation. Surface Areas of The Dolls S_d/S_1 = (d/1)^2 Let's solve the equation for S_d.

This equation tells us that the surface area of each doll is equal to the square of its height multiplied by the surface area of the first doll. Let's make a table that shows the surface area of each doll in terms of S_1.

	Height in Inches (d)	Surface Area (S_d)
Doll 1	1	S_1
Doll 2	2	S_2 = 2^2* S_1= 4 S_1
Doll 3	3	S_3 =3^2* S_1 = 9 S_1
Doll 4	4	S_4 =4^2* S_1= 16 S_1
Doll 5	5	S_5 =5^2* S_1=25 S_1
Doll 6	6	S_6 = 6^2* S_1=36 S_1

Notice that the relationship between the surface areas of the first and fourth dolls is among the given options. Therefore, it is the correct answer. S_4=4^2 * S_1 ✓

Let's verify that the other statements are incorrect. S_2 & = 2 * S_1 S_3 & = 3^2 * S_2 S_6 & = 6* S_5 From the table, we know that the surface area of the second doll is 4 times the surface area of the first doll. This means that the option showing the surface area of the second doll is incorrect. S_2 = 2 * S_1 & * S_2 = 4 * S_1 & ✓ Let's now check if S_3=3^3 * S_2 is correct. To do so, we will substitute the surface areas in the expression with their equivalent values expressed in terms of S_1 from our table.

The expression on the left is not equal to the expression on the right, so this option is also incorrect. S_3 = 3 * S_2 & * Finally, let's consider the last equation. We will follow the same procedure as before.

We ended with different expressions on the right and left sides of the equation, which means that this is also incorrect. S_6 = 6 * S_5 & *

Changes in Dimensions

Catch-Up and Review

How Do the Dimensions of Similar Figures Change After a Scale?

Relations Between the Similar Polygons

Perimeters of Similar Polygons Theorem

Proof

Making a Miniature Model of a Basketball Court

Hint

Solution

Areas of Similar Figures

Proof

Making a Miniature Model of a Spectator Stand

Hint

Solution

Finding the Length Scale, Perimeter, or Area of Similar Polygons

Surface Areas of Similar Solids

Proof

Designing a Miniature Basketball

Hint

Solution

Volumes of Similar Solids

Proof

Designing a Miniature Model of a Stadium

Hint

Solution

Miniature Basketball Players

Hint

Solution

Changes in Dimensions

Recommended exercises

	12 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions