CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard

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Question 13 - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at April 16, 2024 by

The perimeters of two similar triangles ∆ABC and ∆PQR are 35 cm & 45 cm respectively, then the ratio of the areas of the two triangles is______________

The perimeters of two similar triangles ∆ABC and ∆PQR are 35 cm and 45

Question 13 - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 2
Question 13 - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 3
Question 13 - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Part 4

Transcript

Question 13 The perimeters of two similar triangles ∆ABC and ∆PQR are 35 cm & 45 cm respectively, then the ratio of the areas of the two triangles is______________ Since ΔABC and ΔPQR are similar, the ratio of their corresponding sides is same 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐴𝐶/𝑃𝑅 Let 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐴𝐶/𝑃𝑅=𝑘 ∴ AB = kPQ BC = kQR AC = kPR Now, Perimeter of Δ ABC = AB + BC + AC = kPQ + kQR + kPR = k (PQ + QR + PR) = k × Perimeter of Δ PQR Therefore, (𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓" " ∆𝑃𝑄𝑅) = k So, we can write 𝐴𝐵/𝑃𝑄=𝐵𝐶/𝑄𝑅=𝐴𝐶/𝑃𝑅=(𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓" " ∆𝑃𝑄𝑅) " = k " We know that Ratio of area of similar triangles is equal to square of ratio of its corresponding sides Therefore, (𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝑃𝑄𝑅)=(𝐴𝐵/𝑃𝑄)^2 (𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝑃𝑄𝑅)=((𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓" " ∆𝑃𝑄𝑅))^2 Putting values (𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝑃𝑄𝑅)=(35/45)^2 (𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝑃𝑄𝑅)=(7/9)^2 (𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓" " ∆𝑃𝑄𝑅)=49/81 Hence, the required ratio is 49:81

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.