HOW TO FIND AREA AND PERIMETER OF SIMILAR FIGURES

Problem 1 :

Triangle HIJ is similar to triangle STR, what is the perimeter of triangle STR

Solution :

Since the given sides are similar,

• HI and ST are corresponding sides.
• IJ and TR are corresponding sides.
• HJ and SR are corresponding sides.

HI : ST = 4 : 10

HI : ST = 2 : 5

Perimeter of STR :

= ST + TR + SR

= 10 + 15 + 12

= 37

Problem 2 :

If the two triangles are similar

Find

i) Scale factor

ii) Ratio of perimeter

iii)  Ratio of area

Solution :

The sides which are having side lengths 21 and 7, they are corresponding sides.

The sides lengths 15 and 5 are corresponding, then 18 and 6 are corresponding.

Ratio = 21 : 7 ==> 3 : 1

i) Scale factor = 3 : 1

ii)  Ratio of perimeter = 3 : 1

iii)  Ratio of area = 3² : 1²

Problem 3 :

Two triangles have a scale factor of 2/3. The area of the larger triangle is 12 cm². What is the area of smaller triangle.

Solution :

Area of smaller triangle / Area of larger triangle = (2/3)²

Area of larger triangle = 12 cm²

Area of smaller triangle / 12 = (2/3)²

Area of smaller triangle = 12 (4/9)

Area of smaller triangle = 5.3 cm²

Problem 4 :

If the length of each side of triangle is cut to 1/3 of its original size, what happens to the area of the triangle ?

The new area is _______________ of the original area.

Solution :

From the given information, every side is being divided into 1/3 of the original size. So, the ratio is 1 : 3.

Relationship between scale factor and area :

Area of smaller triangle : Area of larger triangle = (1 : 3)²

Area of smaller triangle : Area of larger triangle = 1 : 9

Area of smaller triangle = (1/9) of area of larger triangle.

Problem 5 :

For the two triangles below to be similar, which of the following be true ?

a)  x = 2y/3    b)  c = 3y/2     c)  x = 3y     d)  x = y

Solution :

Since the given shapes are similar,

EF : KJ ==> 6 : 9 ==> 2 : 3

FG : KL ==> x : y

x : y = 2 : 3

x/y = 2/3

x = 2y/3

Problem 6 :

An architect is building a model of a tennis court for a new client. On the model, the court is 6 inches wide and 13 inches long. An official tennis court is 36 feet wide. What is the length of a tennis court ?

Solution :

Let x be the length of tennis court.

Since the shapes are similar, the ratio between model to official tennis court is 

6 : 13 = 36 : x

6/13 = 36/x

6x = 36(13)

x = 36(13)/6

x = 6 (13)

x = 78

So, length of official tennis court is 78 feet.

Problem 7 :

Your family has decided to put a rectangular patio in your backyard, similar to the shape of your backyard. Your backyard has a length of 45 feet and a width of 20 feet. The length of your new patio is 18 feet. Find the perimeters of your backyard and of the patio.

Solution :

Length of backyard = 45 ft

Width = 20 ft

Length of new patio = 18 ft

Since these two shapes are similar, corresponding sides will be in the same ratio as well the ratio of the perimeters will also be equal to the ratio of the corresponding sides.

Perimeter of backyard = 2(45 + 20)

= 2(65)

= 130 ft

Perimeter of backyard : perimeter of patio = 45 : 18

130 : Perimeter of patio = 45 : 18

Perimeter of patio (45) = 18(130)

Perimeter of patio = 18(130)/45

= 52 ft

Problem 8 :

Rectangle A is similar to rectangle B. Rectangle A has side lengths of 6 and 12. Rectangle B has a side length of 18. What are the possible values for the length of the other side of rectangle B? Select all that apply.

a) 6       b) 9    c) 24     d) 36

Solution :

Rectangle A is similar to rectangle B.

Scale factor for A to B is = 6 : 18

= 1 : 3

Let w be the width of the rectangle.

1 : 3 = 12 : w

1/3 = 12/w

w = 3(12)

w = 36

So, option d is correct.

Problem 9 :

What is the perimeter of the rectangle?

b) What is the area of the rectangle?

c) If the width of the rectangle is doubled, what do you think happens to the perimeter?

d) If the width of the rectangle is doubled, what do you think happens to the area?

Solution :

a) Perimeter of the rectangle = 2(length + width)

= 2(7 + 5)

= 2(12)

= 24 m

b) Area of rectangle = length x width

= 7 x 5

= 35 square meter

c) When width is doubled, new width = 2(5)

= 10 m

Perimeter of the rectangle = 2(7 + 10)

= 2 (17)

= 34 m

d) Area of rectangle = 7 x 10

= 70 square meter