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 given pairs of figures are similar. Express the length scale factor between the figures as a fraction in simplest form.

A pair of similar triangles, one with a side length of 15 units and the other with the corresponding side length of 9 units

A pair of similar quadrilaterals, one with a side length of 2.7 units and the other with the corresponding side length of 3.6 units

We want to find the length scale factor between the given triangles.

Since the triangles are similar, we can express the scale factor as a ratio of the corresponding side lengths. 15/9 Notice that the numerator and denominator of the fraction have common factors. Let's simplify it!

The scale factor 53 refers to the ratio of the side length of the larger triangle to the corresponding side length of the smaller triangle. Alternatively, we can express the scale factor as the ratio of the side length of the smaller triangle to the corresponding side length of the larger triangle. 5/3 or 3/5

We want to find the scale factor between the corresponding side lengths of the given similar quadrilaterals.

We can calculate the scale factor by dividing the side length of the larger quadrilateral by the corresponding side length of the smaller quadrilateral. 2.7/3.6 Let's simplify this fraction.

We can also express this scale factor as its multiplicative inverse to show the scale factor of the bigger figure to the smaller figure. 3/4 or 4/3

The figure will be enlarged by a scale factor of $2.4 .$

A figure with side lengths of 2, 2, 2, 4, 5, and 6 units, with one side length unspecified

What is the perimeter of the enlarged figure?

Let's start by finding the perimeter of the given figure.

First, we need to find the missing side length. To do so, we can create a right triangle where the missing side length is its hypotenuse.

Next, let's apply the Pythagorean theorem.

The missing diagonal length is 5 units. Now that we know all the side lengths of the figure, we can add up all seven side lengths to find the perimeter of the figure. 5+2+6+2+2+4+5= 26 The perimeter of the figure is 26 units. We know that the figure will be enlarged by a length scale factor of 2.4. This means that each side length will be scaled by a factor of 2.4. Since the figures will be similar, we can consider the Perimeters of Similar Polygons Theorem.

If two polygons are similar, then the ratio of their perimeters is equal to the ratio of their corresponding side lengths.

If we let P_1 represent the perimeter of the given figure and P_2 represent the perimeter of the enlarged figure, then we can write the following proportion using the theorem. P_2/P_1 = k ⇒ P_2 = P_1 * k Let's multiply the perimeter of the figure by the given scale factor to find the perimeter of the enlarged figure. 26 * 2.4= 62.4 The perimeter of the enlarged figure will be 62.4 units.

A smaller copy of a school's

2

foot by

3

foot rectangular flag is being made for the front of the school's homework planner. The dimensions of the copy will be

\frac{1}{8}

of the school flag. What is the area of the copy? Round the answer to one decimal place.

We know that the dimensions of the smaller copy of the flag are 18 those of the original. We want to find the area of the smaller flag. Let's first calculate the area of the original flag. Since the flag is a rectangle, its area is the product of its width and length. Area of The Original Flag 2 * 3=6 The area of the original flag is 6 square feet. Since the flag and its smaller copy are similar figures, the ratio of their areas is equal to square of the ratio of the their corresponding side lengths, or the length scale factor. We are already given this value. Area of the smaller copy/Area of the original flag = ( 1/8)^2 Let's substitute the area of the original flag into the equation and calculate the area of the copy.

The area of the smaller flag is about 0.1 square feet.

Consider the following two similar cans.

The length scale factor between the cans is

2 : 7 .

If the surface area of the smaller can is

150

square centimeters, calculate the surface area of the larger can.

We can find the surface area of the larger can by using the surface area of the smaller can and the length scale factor because the cans are similar. The ratio between the surface areas of the figures is equal to square of the ratio of their corresponding side lengths. SA_1/SA_2=(l_1/l_2 )^2 The ratio of the corresponding side lengths is the same as the length scale factor, which for the cans is given as 2:7, or 27. SA_1/SA_2=( 2/7)^2 The surface area of the smaller can is 150 square centimeters. Since the scale factor is less than 1, we will represent the surface area of the smaller can as SA_1 and the larger one as SA_2.

Let's substitute the scale factor and the surface area of the smaller can into the equation and solve it for SA_2.

The surface area of the larger can is 1837.5 square centimeters.

The model of a new apartment building is shown. The architect plans for the actual building to be $150$ times the size of the model.

If the volume of the model is

2.1

cubic meters, what will the volume of the new apartment building be when it is completed?

We want to find the volume of the apartment building given that its dimensions will be 150 times those of the model. In other words, the ratio between a length l_m in the model and its corresponding length l_r in the residence is 1150. l_m/l_r = 1/150 Since the model and the actual building will be similar figures, the ratio of the volumes must be the cube of the ratio of their corresponding side measures, or the scale factor between the figures. V_m/V_a=( 1/150)^3 Let's substitute the volume of the model into the above equation and solve it for V_a.

The volume of the new apartment building will be 7 087 500 cubic meters.

Determine whether the given cylinders are similar.

A cylinder with a radius of 4 cm and a surface area of 13 square inches, and another cylinder with a radius of 6 cm and a surface area of 37 square inches

We want to know if the given cylinders are similar. We are given their radii and surface areas. Recall that if two figures are similar, then the ratio of their surface areas is equal to square of the ratio of their corresponding length measures. SA_1/SA_2= (l_1/l_2 )^2 We know the radii and surface areas of the given cylinders.

	Radius (cm)	Surface Area (in.)^2
Cylinder 1	r_1 = 4	SA_1 = 13
Cylinder 2	r_2 = 6	SA_2 = 37

Let's substitute these values into the equation and see if the equation holds.

As we can see, the ratio of the surface areas is not equal to the square of the ratio of the corresponding side lengths. Therefore, the cylinders are not similar solids.

Assume that two snails have similar shells. The volume of the older snail's shell is $12$ cubic centimeters, while the volume of the younger snail's shell is $3$ cubic centimeters.

If the younger snail's shell is

2.1

centimeters high, what is the height of the older snail's shell? Round the result to one decimal place.

We want to find the height of the older snail's shell. Since the shells are similar, the ratio of their volumes is equal to the cube of the ratio of their corresponding lengths. Let's represent the height and volume of the younger snail as h_y and V_y, and those of the older snail as h_o and V_o, respectively. This gives us the following equation. V_o/V_y=(h_o/h_y)^3 We know that the height of the young snail's shell is 2.1 centimeters and its volume is 3 cubic centimeters. The volume of the older snail's shell is 12 cubic centimeters.

	Height (cm)	Volume (cm^3)
Young Snail	2.1	3
Old Snail	h_o	12

We can find the height of the shell of the older snail by substituting the given values into our equation. We can ignore the units because all the measures are given in centimeters. Let's solve for h_o!

The height of the shell of the older snail is about 3.3 centimeters.

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