 # ML Aggarwal Solutions Class 10 Mathematics Solutions for Similarity Exercise 13.3 in Chapter 13 - Similarity

In the figure (ii) given below, DE || BC and AD : DB = 1 : 2, find the ratio of the areas of ∆ADE and trapezium DBCE. Solution:-

The square of any ratio between any two corresponding sides of two similar triangles is equal to the ratio of their areas.

From the question it is given that, DE || BC and AD : DB = 1 : 2,

∠D = ∠B, ∠E = ∠C … [corresponding angles are equal]

∠A = ∠A … [common angles for both triangles]

Now, adding 1 for both side LHS and RHS,

(DB/AD) + 1 = (2/1) + 1

Area of ∆ADE/area of ∆ABC = (1/3)2

Area of ∆ADE/area of ∆ABC = 1/9

Area of ∆ABC = 9 area of ∆ADE

Area of trapezium DBCE

Area of ∆ABC – area of ∆ADE

Therefore, area of ∆ADE/area of trapezium = 1/8

Then area of ∆ADE : area of trapezium DBCE = 1: 8 Related Questions

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