Ex 6.4, 2
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Given: ABCD is trapezium where AB II DC
and diagonals intersect at O
& AB = 2 CD
To find: (ar ∆ AOB)/(ar ∆ COD)
Solution:
Since we need to find ratio of area of ∆ 𝐴𝑂𝐵 and ∆ 𝐶𝑂𝐷.
Lets first prove ∆ 𝐴𝑂𝐵 and ∆ 𝐶𝑂𝐷 are similar
In ∆ AOB and ∆ COD
∠ AOB = ∠ COD
∠ OAB = ∠ OCD
Hence ,Δ AOB ~ Δ COD
We know that if two triangle are similar ,
Ratio of areas is equal to square of ratio of its corresponding sides
Hence (𝑎𝑟 ∆ 𝐴𝑂𝐵)/(𝑎𝑟 ∆ 𝐶𝑂𝐷)=( 𝐴𝐵/𝐶𝐷)^2
(𝑎𝑟 ∆ 𝐴𝑂𝐵)/(𝑎𝑟 ∆ 𝐶𝑂𝐷)=((2 𝐶𝐷)/𝐶𝐷)^2
(𝑎𝑟 ∆ 𝐴𝑂𝐵)/(𝑎𝑟 ∆ 𝐶𝑂𝐷)=((2 )/1)^2
(𝑎𝑟 ∆ 𝐴𝑂𝐵)/(𝑎𝑟 ∆ 𝐶𝑂𝐷)=4/1
∴ ar Δ AOB : ar Δ COD =4:1
Hence ratio of areas is 4 :1

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.