Solving Word Problems in Similar Triangles :
In this section, let us learn how to solve word problems in similar triangles.
Example 1 :
lengths of the three sides of triangle ABC are 6 cm, 4 cm and 9 cm. Triangle
PQR ad BC are congruent. One of the lengths of the sides of triangle PQR is 35
cm. What is the greatest perimeter possible for triangle PQR
From the given information let us draw a rough diagram.
∆ PQR ~ ∆ ABC
PQ/AB = QR/BC = PR/AC = perimeter of ∆ PQR/Perimeter of ∆ BC
Let QR = 35
The corresponding sides must be QR and BC.
Perimeter of ∆ PQR/ Perimeter of ∆ ABC = QR/BC= 35/4
Perimeter of triangle PQR = (35/4) ⋅ 19
So, perimeter of triangle PQR is 166.25 cm².
Example 2 :
In the figure given below, the sides DE and BC are parallel and (AD/B) = 3/5, calculate the value of
(i) area of triangle ADE/are of triangle ABC
(ii) area of trapezium BCED/area of triangle ABC
In triangle ABC, the sides DE and BC are parallel
Area of ∆ ADE/ Area of ∆ ABC = AD²/AB²
(ii) Area of ∆ ADE = 9 k
Area of ∆ ADE = 64 k
Area of trapezium BCDE = area of ∆ ABC – area of ∆ ADE
= 64 k – 9 k
= 55 k
Area of trapezium BCDE/Area of ∆ ABC = 55 k/64 k
Example 3 :
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) x EG
= (3/1) x 1.4
= 4.2 km
Area of new industrial zone = Area of ∆ EAB
= (1/2) ⋅ AB ⋅ EF
= (1/2) ⋅ 3 ⋅ 4.2
= 6.3 km²
So the area of new industrial zone is 6.3 km²
After having gone through the stuff given above, we hope that the students would have understood, how to solve word problems in similar triangles.
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