Question 37 (iii) (A)

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

Question 37 (iii) (A) - CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

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

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