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.
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?
∆ EAD ~ ∆ EDC
So, EA/ED = ED/EC
ED² = EA ⋅ EC
= 16 ⋅ 81
= √16 ⋅ 81
= 4 ⋅ 9
ABD is the isosceles right triangle and the side AE is perpendicular to the side BD.
BE = ED
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)
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.
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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