- GST
- Banking
- Shares and Dividends
- Quadratic Equations in One Variable
- Factorization
- Ratio and Proportion
- Matrices
- Arithmetic and Geometric Progression
- Reflection
- Section Formula
- Equation of Straight Line
- Similarity
- Locus
- Circles
- Constructions
- Mensuration
- Trigonometric Identities
- Trigonometric Tables
- Heights and Distances
ML Aggarwal Solutions Class 10 Mathematics Solutions for Similarity Exercise 13.3 in Chapter 13 - Similarity
Prev
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]
Consider the ∆ADE and ∆ABC,
∠A = ∠A … [common angles for both triangles]
Therefore, ∆ADE ~ ∆ABC
But, AD/DB = ½
Then, DB/AD = 2/1
Now, adding 1 for both side LHS and RHS,
(DB/AD) + 1 = (2/1) + 1
(DB + AD)/AD = (2 + 1)
Therefore, ∆ADE ~ ∆ABC
Then, area of ∆ADE/area of ∆ABC = AD2/AB2
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
9 area of ∆ADE – area of ∆ADE
8 area of ∆ADE
Therefore, area of ∆ADE/area of trapezium = 1/8
Then area of ∆ADE : area of trapezium DBCE = 1: 8 
Facebook
Whatsapp
Copy Link
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
