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Ex 6.4, 2 - Diagonals of a trapezium ABCD with AB || DC - Area of similar triangles

Ex 6.4, 2 - Chapter 6 Class 10 Triangles - Part 2
Ex 6.4, 2 - Chapter 6 Class 10 Triangles - Part 3

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.