**Solving Practical Problems with Similar Triangles :**

In this section, we will learn how to solve practical problems with similar triangles.

**Example 1 :**

The government plans to develop a new industrial zone in an unused portion of land in a city.

The shaded portion of the map shown given below indicates the area of the new industrial zone. Find the area of the new industrial zone.

**Solution :**

By considering the lines AD and BC,the angles

∠AEB = ∠DEC (vertically opposite angles)

∠EAB = ∠EDC (alternate angles)

By using AA similarity criterion ∆ EAB ~ ∆ EDC

(AB/DC) = (EF/EG)

EF = (AB/DC) ⋅ EG

= (3/1) ⋅ 1.4

= 4.2 km

Area of new industrial zone = Area of ∆ EAB

= (1/2) ⋅ AB ⋅ EF

= (1/2) x 3 x 4.2

= 6.3 km²

So the area of new industrial zone is 6.3 km²

**Example 2 :**

A boy is designing a diamond shaped kite, as shown in the figure where AE = 16 cm, EC = 81 cm. He wants to use a straight cross bar BD. How long should it be?

**Solution :**

∆ EAD ~ ∆ EDC

So, EA/ED = ED/EC

ED² = EA ⋅ EC

= 16 ⋅ 81

= √16 ⋅ 81

= 4 ⋅ 9

= 36

ABD is the isosceles right triangle and the side AE is perpendicular to the side BD.

BE = ED

So,

BD = 2 ED

= 2(36) ==> 72 cm

**Example 3 :**

A student wants to determine the height of flag pole. He placed a small mirror on the ground so that he can see reflection of the top of the flagpole. The distance of the mirror from him is 0.5 m and the distance of the flagpole from the mirror is 3 m .If t he eyes are 1.5 m above the ground level, then find the height of the flagpole (The foot of student, mirror and the foot of flagpole lie along the a straight line)

**Solution :**

Let “C” be point of reflection

In triangles ABC and EDC

∠ABC = ∠EDC = 90 degree

∠BCA = ∠DCE

By using AA criterion ∆ABC ~ ∆EDC

AB/ED = BC/DC

1.5/ED = 0.5/3

4.5 = 0.5 ED

ED = 4.5/0.5

= 9 m

Therefore height of flag pole = 9 m

**Example 4 :**

A roof has a cross section as shown in the diagram

(i) Identify the similar triangles

(ii) Find the height h of the roof.

**Solution :**

Corresponding triangles from the given picture

(i) ∆ WZY ~ ∆ YZX

(ii) ∆ WYX ~ ∆ YZX

(iii) WZY ~ ∆ WYZ

From the (ii)

WY/YZ = XY/XZ

h/8 = 6/10

h = (6/10) ⋅ 8

h = 48/10

h = 4.8 m

After having gone through the stuff given above, we hope that the students would have understood, how to solve practical problems in similar triangles.

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