SIMILAR TRIANGLES AND INDIRECT MEASUREMENT

In each figure, find x.

Problem 1 :

Solution:

8/6 = 104/x

Doing cross multiplication, we get

8x = 104(6)

x = 104(6) / 8

= 78

So, the required measure is 78 ft.

Problem 2 :

Solution:

1 ft = 12 inches

1 inch = 1/12

2 inch = 2/12

= 1/6 ft

= 0.16 ft

5 ft 2 inches = 5 + 0.16

= 5.16 ft

10/5.16 = 120/x

Doing cross multiplication, we get

10(x) = 120(5.16)

x = 120(5.16) / 10

= 61.92

So, the required height of the tree is 62 ft approximately.

Problem 3 :

Solution :

2/3 = x/20

Doing cross multiplication, we get

2(20) = 3x

3x = 40

x = 40/3

x = 13.3

So, the required length is 13.3 meter.

Problem 4 :

Solution :

x/55 = 50/40

Doing cross multiplication, we get

40x = 50(55)

x = 50(55)/40

= 68.75

So, the required length is 69 km.

Problem 5 :

Solution :

160/100 = (160 - x)/80

16/10 = (160 - x)/80

16(80) = 10(160 - x)

1280 = 1600 - 10x

10x = 1600 - 1280

10x = 320

x = 320/10

x = 32

So, the required value of x is 32 m.

Problem 6 :

Solution :

10/11.25 = x/9

10(9) = 11.25x

90 = 11.25x

x = 90/11.25

= 8

So, the value of x is 8 ft.

Problem 7 :

An office building 55 ft tall casts a shadow 30 ft long. How tall is a person standing nearby who casts a shadow 3 ft long?

Solution :

Let x be the height of the person.

55/30 = x/3

Doing cross multiplication, we get

3(55) = 30x

x = 3(55) / 30

x = 5.5

So, the required value of x is 5.5 ft.

Problem 8 :

A 20 ft pole casts a shadow 12 ft long. How tall is a nearby building that casts a shadow 20 ft long?

Solution:

Let x be the height of the building.

x/20 = 20/12

Doing cross multiplication, we get

12x = 20(20)

x = 20(20)/12

= 33.3

So, the required height of building is 33.3 ft.

Problem 9 :

A fire tower casts a shadow 30 ft long. A nearby tree casts a shadow 8 ft long. How tall is the fire tower if the tree is 20 ft tall?

Solution:

Let x be the tall of the fire tower.

x/30 = 20/8

8x = 20(30)

x = 600/8

= 75

So, the height of the fire tower is 75 ft.

Problem 10 :

A house casts a shadow 12 m long. A tree in the yard casts a shadow 8 m long. How tall is the tree if the house is 20 m tall?

Solution :

Let x be the tall of the tree.

20/12 = x/8

20(8) = 12x

x = 160/12

= 13.3

So, the height of the tree is 13.3 m.

Problem 11 :

A roof has a cross section that is a right triangle. The diagram shows the approximate dimensions of this cross section. Find the height h of the roof.

Solution :

Triangles YWX and ZYX are similar. By comparing the corresponding sides,

YX/ZX =YW/ZY

3.1/6.3 = h/5.5

h = 3.1(5.5) / 6.3

= 2.70

Approximately the height is 2.7 m.

Problem 12 :

To find the cost of installing a rock wall in your school gymnasium, you need to fi nd the height of the gym wall. You use a cardboard square to line up the top and bottom of the gym wall. Your friend measures the vertical distance from the ground to your eye and the horizontal distance from you to the gym wall. Approximate the height of the gym wall.

Solution :

Using geometric mean theorem (altitude)

8.5² = 5 ⋅ w

w = (8.5 ⋅ 8.5)/5

w = 14.45

Height of wall = 14.45 + 5

= 19.45 ft

Problem 13 :

Use the Geometric Mean Theorems to find AC and BD.

Solution :

BD² = AD ⋅ DC

BC² = DC ⋅ AC

AB² = AD ⋅ AC

Using Pythagorean theorem,

AC² = AB² + BC²

= 20² + 15²

= 400 + 225

= 625

AC = 25

From BC² = DC ⋅ AC, 

15² = DC ⋅ 25

DC = 225/25

= 9

From AB² = AD ⋅ AC,

20² = AD ⋅ 9

AD = 400/9

= 44.4

From BD² = AD ⋅ DC

= 44.4 (9)

= 399.6

BD = √399.6

BD = 19.98

Approximately 20.

Problem 14 :

Find the value(s) of the variable(s).

Solution :

12² = (a + 5) ⋅ 18

144 / 18 = a + 5

8 = a + 5

a = 8 - 5

a = 3

So, the value of a is 3.

Problem 14 :

Find the value(s) of the variable(s).

Solution :

6² = (b + 3) ⋅ 8

36/8 = b + 3

4.5 = b + 3

b = 4.5 - 3

= 1.5

So, the required value of b is 1.5