Question 37 (iii) (A)

If 2AB = 5DE and △ ABC ∼ △ DEF then show that (perimeter of △ABD)/(perimeter of △DEF) is constant.

 

 

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Question 37 (iii) (A) If 2AB = 5DE and △ ABC ∼ △ DEF then show that (𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 △𝐴𝐵𝐷)/(𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 △𝐷𝐸𝐹) is constant.Given 2AB = 5DE 𝐴𝐵/𝐷𝐸=5/2 Also, given that △ ABC ∼ △ DEF And, Sides of similar triangles are proportional So, 𝐴𝐵/𝐷𝐸=𝐴𝐶/𝐷𝐹=𝐵𝐶/𝐸𝐹 And since 𝐴𝐵/𝐷𝐸=5/2 Therefore, 𝑨𝑩/𝑫𝑬=𝑨𝑪/𝑫𝑭=𝑩𝑪/𝑬𝑭=𝟓/𝟐 So, we can write AB = 𝟓/𝟐 𝑫𝑬 , AC = 𝟓/𝟐 𝑫𝑭, BC = 𝟓/𝟐 𝑬𝑭 Now, (𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 △𝑨𝑩𝑫)/(𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 △𝑫𝑬𝑭)=(𝐴𝐵 + 𝐴𝐶 + 𝐵𝐶)/(𝐷𝐸 + 𝐷𝐹 + 𝐸𝐹) =(5/2 𝐷𝐸 + 5/2 𝐷𝐹 + 5/2 𝐸𝐹)/(𝐷𝐸 + 𝐷𝐹 + 𝐸𝐹) =(5/2(𝐷𝐸 + 𝐷𝐹 + 𝐸𝐹))/(𝐷𝐸 + 𝐷𝐹 + 𝐸𝐹) =5/2 Thus, (𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 △𝑨𝑩𝑫)/(𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 △𝑫𝑬𝑭)=5/2 is constant Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo