Question 30 (Choice - 1) - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Last updated at April 16, 2024 by Teachoo

The perimeters of two similar triangles are 25cm and 15cm respectively. If one side of the first triangle is 9cm, find the length of the corresponding side of the second triangle.

Note
: This
is similar
to Question 13 –
Maths
Standard Sample Paper 2020

Check the answer here https://
www.teachoo.com
/10979/3130/Question-13/category/CBSE-Class-10-Sample-Paper-for-2020-Boards---
Maths
-Standard/

Question 30 (Choice - 1) The perimeters of two similar triangles are 25cm and 15cm respectively. If one side of the first triangle is 9cm, find the length of the corresponding side of the second triangle.Let two triangles be ΔABC and ΔPQR
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
First, finding the Relation between
Ratio of sides of similar Triangles
And
Ratio of Perimeters of similar Triangles
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 "
Given that
The perimeters of two similar triangles are 25cm and 15cm respectively. If one side of the first triangle is 9cm
Therefore,
𝑨𝑩/𝑷𝑸=(𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝒐𝒇" " ∆𝑨𝑩𝑪)/(𝑷𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓 𝒐𝒇" " ∆𝑷𝑸𝑹) " "
Putting values
9/𝑃𝑄=25/15
9/𝑃𝑄=5/3
PQ = 3/5 × 9
PQ = 𝟐𝟕/𝟓 = 5.4 cm

Made by

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

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