Open App

Class 10 Exam > Class 10 Notes > Mathematics Class 10 (Maharashtra SSC Board) > Textbook Solutions: Similarity

Textbook Solutions: Similarity | Mathematics Class 10 (Maharashtra SSC Board) PDF Download

Q: 3. What are the properties of similar triangles?

Ans. The properties of similar triangles include:1. Corresponding angles are equal.2. The lengths of corresponding sides are in proportion.3. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Download, print and study this document offline

Please wait while the PDF view is loading

 Page 1


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Page 2


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Answer : Here, ?ABC and ?ADB has common Base. 
?  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
=  
=  
Q. 3. In adjoining figure 1.14 seg PS ? seg RQ, seg QT ? seg PR. If RQ = 6, PS = 6 
and PR = 12, then find QT. 
 
Answer : 
 
Considering, Area of (?PQR) with base QR 
? PS will be the Height 
 
 
Page 3


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Answer : Here, ?ABC and ?ADB has common Base. 
?  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
=  
=  
Q. 3. In adjoining figure 1.14 seg PS ? seg RQ, seg QT ? seg PR. If RQ = 6, PS = 6 
and PR = 12, then find QT. 
 
Answer : 
 
Considering, Area of (?PQR) with base QR 
? PS will be the Height 
 
 
Now, consider the Area of (?PQR) with base PR 
? QT will be the Height 
? The triangle is the same 
? The area will be the same irrespective of the base taken. 
And we know that area of triangle =  × Base× Height 
? ×QR×PS  
=  × PR × QT 
?  × 6 × 6   
=  × 12 × QT 
? QT = 3 
Q. 4. In adjoining figure, AP ? BC, AD || BC, then find A(?ABC) : A (?BCD) 
 
 
Answer : 
 
Page 4


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Answer : Here, ?ABC and ?ADB has common Base. 
?  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
=  
=  
Q. 3. In adjoining figure 1.14 seg PS ? seg RQ, seg QT ? seg PR. If RQ = 6, PS = 6 
and PR = 12, then find QT. 
 
Answer : 
 
Considering, Area of (?PQR) with base QR 
? PS will be the Height 
 
 
Now, consider the Area of (?PQR) with base PR 
? QT will be the Height 
? The triangle is the same 
? The area will be the same irrespective of the base taken. 
And we know that area of triangle =  × Base× Height 
? ×QR×PS  
=  × PR × QT 
?  × 6 × 6   
=  × 12 × QT 
? QT = 3 
Q. 4. In adjoining figure, AP ? BC, AD || BC, then find A(?ABC) : A (?BCD) 
 
 
Answer : 
 
We can re-draw the fig. 1.15 (as shown above) where we add DO 
Which will be height of ?BCD. 
Now,  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
?  
(? the distance between the two parallel lines is always equal ? AP = DO) 
?  = 1:1 
Q. 5. In adjoining figure PQ ? BC, AD ? BC then find following ratios. 
 
 
Page 5


Similarity 
 
Practice Set 1.1 
Q. 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and 
height is 6. Find the ratio of areas of these triangles. 
Answer : We know that area of triangle =  × Base× Height 
? Area (triangle 1) =  ×9× 5 
=  
? Area (triangle 2) =  ×10× 6 
= 30 
? The ratio of areas of these triangles will be =  
=  
=  
=  
Q. 2. If figure 1.13 BC ? AB, AD ? AB, BC = 4, AD = 8, then find 
?? (??????? )
?? (??????? )
. 
 
 
Answer : Here, ?ABC and ?ADB has common Base. 
?  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
=  
=  
Q. 3. In adjoining figure 1.14 seg PS ? seg RQ, seg QT ? seg PR. If RQ = 6, PS = 6 
and PR = 12, then find QT. 
 
Answer : 
 
Considering, Area of (?PQR) with base QR 
? PS will be the Height 
 
 
Now, consider the Area of (?PQR) with base PR 
? QT will be the Height 
? The triangle is the same 
? The area will be the same irrespective of the base taken. 
And we know that area of triangle =  × Base× Height 
? ×QR×PS  
=  × PR × QT 
?  × 6 × 6   
=  × 12 × QT 
? QT = 3 
Q. 4. In adjoining figure, AP ? BC, AD || BC, then find A(?ABC) : A (?BCD) 
 
 
Answer : 
 
We can re-draw the fig. 1.15 (as shown above) where we add DO 
Which will be height of ?BCD. 
Now,  
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
?  
?  
(? the distance between the two parallel lines is always equal ? AP = DO) 
?  = 1:1 
Q. 5. In adjoining figure PQ ? BC, AD ? BC then find following ratios. 
 
 
Answer : We know that area of triangle =  × Base × Height 
(i)  
(PROPERTY: Areas of triangles with equal heights are proportional to their 
corresponding bases.) 
(ii) 
(PROPERTY: Areas of triangles with equal bases are proportional to their 
corresponding heights.) 
(iii)  
(PROPERTY: Areas of triangles with equal heights are proportional to their 
corresponding bases.) 
(iv)  
=  
 
Practice Set 1.2 
Q. 1. Given below are some triangles and lengths of line segments. Identify in 
which figures, ray PM is the bisector of ?OPR. 
 
 
Answer :

	Mathematics Class 10 (Maharashtra SSC Board) 26 videos\|208 docs\|38 tests

Mathematics Class 10 (Maharashtra SSC Board)

26 videos|208 docs|38 tests

Join Course for Free

FAQs on Textbook Solutions: Similarity - Mathematics Class 10 (Maharashtra SSC Board)

1. What is similarity in geometry?

Ans. Similarity in geometry refers to a relationship between two shapes where they have the same shape but may differ in size. This means that corresponding angles are equal, and the lengths of corresponding sides are in proportion.

2. How can we determine if two triangles are similar?

Ans. Two triangles are similar if they satisfy any of the following criteria: 1. Angle-Angle (AA) criterion: If two angles of one triangle are equal to two angles of another triangle. 2. Side-Angle-Side (SAS) criterion: If one angle of a triangle is equal to one angle of another triangle, and the sides including these angles are in proportion. 3. Side-Side-Side (SSS) criterion: If the corresponding sides of the two triangles are in proportion.

3. What are the properties of similar triangles?

Ans. The properties of similar triangles include: 1. Corresponding angles are equal. 2. The lengths of corresponding sides are in proportion. 3. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

4. How do we find the scale factor of similar figures?

Ans. The scale factor of similar figures can be found by measuring the lengths of corresponding sides from the two figures. The scale factor is the ratio of the lengths of one side of a figure to the length of the corresponding side of the other figure. For example, if one side of the first figure is 4 units and the corresponding side of the second figure is 8 units, the scale factor is 4:8 or simplified to 1:2.

5. Can similarity be applied to real-life situations?

Ans. Yes, similarity can be applied in various real-life situations such as in architecture, engineering, and design. For instance, architects use similar shapes when creating scale models of buildings, ensuring that the proportions are maintained while the size is reduced. Similarly, maps represent larger areas in a smaller, similar format.

About this Document

4.97/5 Rating

Oct 03, 2025 Last updated

Related Exams

Class 10

Document Description: Textbook Solutions: Similarity for Class 10 2025 is part of Mathematics Class 10 (Maharashtra SSC Board) preparation. The notes and questions for Textbook Solutions: Similarity have been prepared according to the Class 10 exam syllabus. Information about Textbook Solutions: Similarity covers topics like and Textbook Solutions: Similarity Example, for Class 10 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Textbook Solutions: Similarity.

Introduction of Textbook Solutions: Similarity in English is available as part of our Mathematics Class 10 (Maharashtra SSC Board) for Class 10 & Textbook Solutions: Similarity in Hindi for Mathematics Class 10 (Maharashtra SSC Board) course. Download more important topics related with notes, lectures and mock test series for Class 10 Exam by signing up for free. Class 10: Textbook Solutions: Similarity | Mathematics Class 10 (Maharashtra SSC Board)

Description

Full syllabus notes, lecture & questions for Textbook Solutions: Similarity | Mathematics Class 10 (Maharashtra SSC Board) - Class 10 | Plus excerises question with solution to help you revise complete syllabus for Mathematics Class 10 (Maharashtra SSC Board) | Best notes, free PDF download

Information about Textbook Solutions: Similarity

In this doc you can find the meaning of Textbook Solutions: Similarity defined & explained in the simplest way possible. Besides explaining types of Textbook Solutions: Similarity theory, EduRev gives you an ample number of questions to practice Textbook Solutions: Similarity tests, examples and also practice Class 10 tests

	Mathematics Class 10 (Maharashtra SSC Board) 26 videos\|208 docs\|38 tests

Mathematics Class 10 (Maharashtra SSC Board)

26 videos|208 docs|38 tests

Join Course for Free

Download as PDF

Explore Courses for Class 10 exam

Summary

Textbook Solutions: Similarity | Mathematics Class 10 (Maharashtra SSC Board)

Free

MCQs

Exam

ppt

Textbook Solutions: Similarity | Mathematics Class 10 (Maharashtra SSC Board)

Objective type Questions

mock tests for examination

shortcuts and tricks

Previous Year Questions with Solutions

past year papers

practice quizzes

Semester Notes

study material

Extra Questions

Viva Questions

video lectures

Important questions

Textbook Solutions: Similarity | Mathematics Class 10 (Maharashtra SSC Board)

Sample Paper

pdf

;

Additional Information about Textbook Solutions: Similarity for Class 10 Preparation

Textbook Solutions: Similarity Free PDF Download

The Textbook Solutions: Similarity is an invaluable resource that delves deep into the core of the Class 10 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the Textbook Solutions: Similarity now and kickstart your journey towards success in the Class 10 exam.

Importance of Textbook Solutions: Similarity

The importance of Textbook Solutions: Similarity cannot be overstated, especially for Class 10 aspirants. This document holds the key to success in the Class 10 exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

Textbook Solutions: Similarity Notes

Textbook Solutions: Similarity Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to Textbook Solutions: Similarity. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, Textbook Solutions: Similarity Notes on EduRev are your ultimate resource for success.

Textbook Solutions: Similarity Class 10 Questions

The "Textbook Solutions: Similarity Class 10 Questions" guide is a valuable resource for all aspiring students preparing for the Class 10 exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study Textbook Solutions: Similarity on the App

Students of Class 10 can study Textbook Solutions: Similarity alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the Textbook Solutions: Similarity, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of Textbook Solutions: Similarity is prepared as per the latest Class 10 syllabus.

Education Revolution