**Perimeters and Areas of Similar Figures Worksheet :**

Worksheet given in this section is much useful to the students who would like to practice problems on perimeters and areas similar figures.

Before look at the worksheet, if you would like to know the stuff related to perimeters and areas similar figures,

**Problem 1 : **

Pentagons ABCDE and LMNPQ shown below are similar.

(a) Find the ratio (red to blue) of the perimeters of the pentagons.

(a) Find the ratio (red to blue) of the areas of the pentagons.

**Problem 2 : **

The red and blue figures shown below are similar. Find the ratio (red to blue) of their perimeters and of their areas.

**Problem 3 :**

The red and blue figures shown below are similar. Find the ratio (red to blue) of their areas using the theorem and justify your answer.

**Problem 4 :**

A trading pit at the Chicago Board of Trade is in the shape of a series of regular octagons. One octagon has a side length of about 14.25 feet and an area of about 980.4 square feet. Find the area of a smaller octagon that has a perimeter of about 76 feet.

**Problem 5 :**

Mr. Alex is buying photographic paper to print a photo in different sizes. An 8 inch by 10 inch sheet of the paper costs $0.42. What is a reasonable cost for a 16 inch by 20 inch sheet ?

**Problem 1 : **

Pentagons ABCDE and LMNPQ shown below are similar.

(a) Find the ratio (red to blue) of the perimeters of the pentagons.

(a) Find the ratio (red to blue) of the areas of the pentagons.

**Solution :**

The ratio of the lengths of corresponding sides in the pentagons is

= 5 / 10

= 1 / 2

= 1 : 2

**Solution (a) : **

Because the ratio of the lengths of the corresponding sides is 1 : 2, the ratio of the perimeters is also 1 : 2.

So, the perimeter of pentagon ABCDE is half the perimeter of pentagon LMNPQ.

**Solution (b) : **

The ratio of the lengths of the corresponding sides in the pentagons is 1 : 2.

Using the Theorem, the ratio of the areas is

= 1^{2} : 2^{2}

= 1 : 4

So, the area of pentagon ABCDE is one fourth the area of pentagon LMNPQ.

**Problem 2 : **

The red and blue figures shown below are similar. Find the ratio (red to blue) of their perimeters and of their areas.

**Solution : **

**Ratio of the perimeters :**

The ratio of the lengths of corresponding sides in the hexagon is

= 3 / 9

= 1 / 3

= 1 : 3

Hence, the ratio of the perimeters is also 1 : 3.

**Ratio of the areas :**

The ratio of the lengths of the corresponding sides in the pentagons is 1 : 3.

Using the Theorem, the ratio of the areas is

= 1^{2} : 3^{2}

= 1 : 9

**Problem 3 :**

The red and blue figures shown below are similar. Find the ratio (red to blue) of their areas using the theorem and justify your answer.

**Solution : **

The ratio of the lengths of corresponding sides in the parallelograms is

= 6 / 4

= 3 / 2

= 3 : 2

The ratio of the lengths of the corresponding sides in the parallelograms is 3 : 2.

Using the Theorem, the ratio of the areas is

= 3^{2} : 2^{2}

= 9 : 4

**Justification : **

Area of **red** parallelogram is

= base ⋅ height

= 6 ⋅ 5

= 30 square units

Area of **blue** parallelogram is

= base ⋅ height

= 4 ⋅ 3 1/3

= 4 ⋅ 10/3

= 40/3 square units.

Ratio of the areas of parallelograms is

= 30 : 40/3

= 90 : 40

= 9 : 4

Hence, the ratio of the areas found 9 : 4 using Theorem is reasonable.

**Problem 4 :**

A trading pit at the Chicago Board of Trade is in the shape of a series of regular octagons. One octagon has a side length of about 14.25 feet and an area of about 980.4 square feet. Find the area of a smaller octagon that has a perimeter of about 76 feet.

**Solution : **

All regular octagons are similar because all corresponding angles are congruent and the corresponding side lengths are proportional.

Draw and label a sketch.

Find the ratio of the side lengths of the two octagons, which is the same as the ratio of their perimeters.

The ratio of the areas of the smaller octagon to the larger is

= a^{2} : b^{2}

= 2^{2} : 3^{2}

= 4 : 9

= 4 / 9

Calculate the area of the smaller octagon. Let A represent the area of the smaller octagon.

Then, we have

A / 980.4 = 4 / 9

Use cross product property.

9A = 4(980.4)

Simplify.

9A = 3921.6

Divide each side by 9.

9A/9 = 3921.6/9

A ≈ 435.7

The area of the smaller octagon is about 435.7 square feet.

**Problem 5 :**

Mr. Alex is buying photographic paper to print a photo in different sizes. An 8 inch by 10 inch sheet of the paper costs $0.42. What is a reasonable cost for a 16 inch by 20 inch sheet ?

**Solution : **

The ratio of the lengths of the corresponding sides of the two rectangular pieces of paper is

= 8/16 or 10/20

= 1/2

= 1 : 2

The ratio of the areas of the pieces of paper is

= 1^{2} : 2^{2}

= 1 : 4

So, the area of larger piece of paper is four times the area of smaller piece of paper.

Because the cost of the paper should be a function of its area, the larger piece of paper should cost about four times as much as the cost of smaller piece of paper.

Hence, the cost of larger piece of paper is

= 4(0.42)

= $1.68

After having gone through the stuff given above, we hope that the students would have understood, "Perimeters and Areas of Similar Figures Worksheet".

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

HTML Comment Box is loading comments...

You can also visit our following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Trigonometry word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**