# NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Ex 6.4) Exercise 6.4

The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4, which are inclusive of solutions to one of the six exercises of Chapter 6 of the prescribed textbook of NCERT, require students to calculate the areas of triangles that are comparable. The answers to the questions in Exercise 6.4 are provided on the Extramarks website and mobile application in both online and downloadable PDF formats. The subject specialists in Mathematics at Extramarks have carefully prepared these solutions with step-by-step explanations and appropriate references.

It would be beneficial for students to know how to use the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 in order to solve the various questions that may be asked during the board examinations. Students’ self-confidence will rise as their problem-solving speed increases once they become skilled in using the techniques included in these answers.

Area of Similar Triangles is the subject of the fourth exercise in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4, Triangles. There are nine questions in this exercise, of which two require short answers with justification, five require short answers, and the final two require long answers. The Area of Similar Triangles and Theorem 6.6 are both topics covered in Exercise 6.4. The following is the theorem that serves as the foundation for Exercise 6.4:

• According to Theorem 6.6, the ratio of two identical triangles’ areas is equal to the square of the ratio of their corresponding sides.

There is no more efficient way to completely comprehend the principles taught in Class 10 than by completing the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4.

The NCERT Solutions for Class 10th Math Exercise 6.4 pertaining to the chapter on Triangles show that the ratio of two comparable triangles’ areas is equal to the square of their respective side ratios. All of the questions are built around this fundamental idea. It is best to read through this theorem derivation to comprehend it more comprehensively.These NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4 consist of nine questions, seven of which are word problems and three of which are multiple-choice questions.

To help them better understand the issue, teachers suggest that students create diagrams. Practising constructing diagrams will also improve a student’s capacity for visualisation. In order to obtain the correct values, it is also ideal to divide the problem into smaller components, solve each one separately, and then aggregate the results.

It is important that students read through the theorem based on identical triangles that is outlined in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4. Students are encouraged to practise answering questions, the theorem will be more easily understood. To solve the sums correctly, it is crucial to develop a clear conceptual understanding of the theme of the similarity criterion described in earlier chapters.

A set of solved examples is provided in the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 to help students better comprehend the subject and the methodology of writing well-structured solutions.

Students can download the PDF version of the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 from the Extramarks educational website. For the CBSE board examination, these NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 are inclusive of chapter-specific questions and answers, which are highly beneficial. The majority of the questions in CBSE examinations are taken from NCERT textbooks, which CBSE recommends. On the Extramarks website and mobile application, students may obtain the NCERT chapter-by-chapter solutions for the Mathematics textbook of Class 10 for all the chapters.

The Area of Identical Triangles is a theme that is introduced in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4. The ratio of two matching sides of comparable triangles is the same, as students have learned in earlier assignments from this chapter 6. There is a theorem encapsulated in the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 that establishes the relationship between the areas of comparable triangles and their corresponding sides.

The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 are indispensable learning resources which can aid students in writing well-coordinated solutions with logically sequenced steps. With numerous examples and exercise questions, important theorems have been clearly described and proven in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4. Students will learn about easy geometry problems and criteria to detect similarities between triangles with the aid of NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4. The ratio of area and ratio of squared sides are the key topics covered in the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4. The key topics covered in NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4 are the ratio of area to squared sides and the ratio of area to squared sides.Most of the issues in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 are concerned with topics like similarity in Triangles and Simple Geometry.

The Ratio of Area and Ratio of Squared Sides are the key topics covered in  NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4. Theorem 6.6 and the triangle similarity criterion are major themes in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4.

Solutions for the following themes are available on the Extramarks digital learning portal along with the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4:

• Triangles Exercise 6.1
• Triangles Exercise 6.2
• Triangles Exercise 6.3
• Triangles Exercise 6.4
• Triangles Exercise 6.5
• Triangles Exercise 6.6

Following are the benefits of NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4:

• When it comes to solving problems related to Class 10 Maths chapter 6 exercise 6.4, an imperative learning asset could be NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4.
• The solutions enclosed in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 have been engineered to include all of the important test-oriented problems.
• Irrational numbers and the Fundamental Theorem of Arithmetic, both essential ideas in the chapter, serve as the foundation of NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4.

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Ex 6.4) Exercise 6.4

When it comes to exam preparation, studying with the help of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4 can be extremely beneficial.There are numerous exercises in this chapter. Of which, the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4, pertaining to Exercise 6.4 have been made available in PDF format on Extramarks. Students can study this solution straight from the Extramarks website or mobile application, or they can download it as convenient for them.

The problems and questions from the exercise have been meticulously handled by a panel of highly proficient subject matter experts at Extramarks while adhering to all CBSE requirements. If a student is familiar with all the concepts from the topic of Triangles in the NCERT textbook and is fairly knowledgeable about the assessments provided alongside it, they can easily and efficiently earn the highest possible grade on the final examination. By using the NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4, students can easily comprehend the types of questions that may appear in the examination from this chapter, allowing them to adequately study for the final examination.

Triangles are the topic of Chapter 6 in the Mathematics textbook for Class 10. It is one of the most intriguing themes in geometry because it describes the various facets and in-depth concepts associated with the geometric figure triangle.

Essentially, a triangle is a flat figure with three sides and three angles. The explanation of related geometrical figures, several theorems relating to the similarities of triangles, and the areas of related triangles are just a few of the themes and sub topics covered in this chapter on Triangles. The Pythagoras theorem is thoroughly explained in the chapter’s conclusion, along with examples of how to apply it to issues. Students should read NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4 to learn more about triangles and the topics they cover.

There are 14 comprehensive questions in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4. Five of them are pure short answer questions, two more are long answer questions, and two more are short answer questions with reasoning questions.

One of the key concepts covered in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 is the significance of triangles, which may be seen as a summary of the triangles and congruence of triangles covered in Class 9 Mathematics. Consequently, it will be simple to understand this chapter. In the board examinations for the 10th grade, this chapter is worth 14 marks. Approximately seven examination questions are based on this chapter each year.

The needs and demands of students were taken into consideration while creatingNCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4. It would be useful to examine the advantages of using these learning assets:

• The solutions are given in simple terms and focus more on key terminologies, notions, theorems, and applications of various concepts.
• For difficult-to-understand questions, simple and direct writing is used  to make learning feasible for students. Along with the answers to the assessments included in the NCERT, these solutions offer a synopsis of the entire chapter.
• The answers are developed in an intriguing way and  handled methodically.
• The information is kept concise, thorough, and understandable.
• To make the understanding of concepts clearer, certain answers are paired with appropriate diagrams.
• The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 were developed in consideration of the most recent CBSE curriculum.

### Access NCERT Solutions for Class 10 Maths Chapter – 6 – Triangles

The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 include the technique for calculating the areas of similar triangles. The similarity of triangles, the criteria for the similarity of triangles, Pythagoras Theorem, and many other significant topics are covered in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4.Extramarks provides a collection of many practise questions covering each of the subtopics of the Triangles chapter’s subtopics. To excel in board examinations, students must practise all of the in-text questions. To access the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4, students are encouraged to register themselves on the Extramarks website and mobile application.

Triangles is an intriguing topic that would aid students in developing the conceptual foundations necessary to comprehend the much more challenging themes in the senior grades. Before moving on to the solutions, students can watch the interactive videos that Extramarks is providing. It would assist them in precisely resolving the issues.

Based on the most recent curriculum, Extramarks offers NCERT Solutions to every question in the Class 10 Mathematics Exercise 6.4 textbook. Practising all the questions provided by Extramarks will assist students in obtaining high grades. For the benefit of the students, Extramarks also offers interactive 3D Learning Videos, NCERT Exemplars, doubt-solving sessions, and other digital learning resources in addition to the NCERT solutions.

### NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

Following are some key details from NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4:

• Similar Shapes: Two figures are said to have similar shapes even if they are not identical in size.
• Comparability and Congruence: Comparable figures are all congruent figures, but the reverse is not true.
• When two polygons with the same number of sides have equal corresponding angles and proportional corresponding sides, they are said to be comparable (i.e., in the same ratio).
• Basic Proportionality Theorem: The other two sides of a triangle are divided in the same ratio if a line parallel to one side of a triangle intersects the other two sides at distinct spots.
• Contrary to the Basic Proportionality Theorem, a line is parallel to the third side of a triangle if it divides any two of the triangle’s sides in the same ratio.
• The internal bisector of a triangle’s angle divides the opposing side internally in the ratio of the sides containing the angle, according to the angle bisector theorem.
• Contrary to the Angle Bisector Theorem, the line across the vertex of a triangle bisects the angle at that vertex if it divides the opposite side in proportion to the other two sides.
• The exterior angle bisector theorem states that the ratio of the sides containing the angle to the angle’s external bisector divides the opposite side of a triangle.

Students can avail of the comprehensive NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 along with NCERT Solutions for all the chapters that constitute the Mathematics curriculum for Class 10, on the Extramarks online learning platform. The prescribed textbook of mathematics by the NCERT for Class 10 consists of the following chapters:

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progressions

Chapter 7: Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9: Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Areas Related to Circles

Chapter 13: Surface Areas and Volumes

Chapter 14: Statistics

Chapter 15: Probability

The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 are regarded as a very valuable tool for test preparation. Students who use Extramarks can have convenient access to NCERT questions and their answers. Since the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4 are compiled by subject specialists of Mathematics, they will undoubtedly help students get exemplary results. These NCERT Solutions for Class 10 Mathematics help students build a reliable academic foundation in the subject and equip them with the skills necessary to answer a variety of questions with ease.

Q.1

$\text{Let}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }{\text{DEF and their areas be, respectively, 64 cm}}^{\text{2}}{\text{and 121 cm}}^{\text{2}}\text{. If EF}=\text{15.4 cm, find BC.}$

Ans.

$\begin{array}{l}\text{It is given that}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{DEF.}\\ \therefore \frac{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)}{\text{ar}\left(\mathrm{\Delta }\text{DEF}\right)}=\frac{{\mathrm{BC}}^{2}}{{\mathrm{EF}}^{2}}\\ \mathrm{Given}\text{}\mathrm{that}\\ \mathrm{EF}=15.4\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{ar}\left(\mathrm{\Delta }\text{ABC}\right)=64{\text{cm}}^{2}\\ \text{ar}\left(\mathrm{\Delta DEF}\right)=121{\text{cm}}^{2}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)}{\text{ar}\left(\mathrm{\Delta }\text{DEF}\right)}=\frac{{\mathrm{BC}}^{2}}{{\mathrm{EF}}^{2}}\\ ⇒\frac{64}{121}=\frac{{\mathrm{BC}}^{2}}{{\left(15.4\right)}^{2}}\\ ⇒\frac{8}{11}=\frac{\mathrm{BC}}{15.4}\\ ⇒\mathrm{BC}=\frac{8×15.4}{11}\\ ⇒\mathrm{BC}=11.2\end{array}$

Q.2 Diagonals of a trapezium ABCD with AB ║ DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta C}\text{OD and}\mathrm{\Delta A}\text{OB,}\\ \angle \text{ODC}=\angle \text{OBA [Alternate interior angles as AB}\parallel \text{CD]}\\ \angle \text{DCO}=\angle \text{BAO [Alternate interior angles as AB}\parallel \text{CD]}\\ \angle \mathrm{C}\text{OD}=\angle \mathrm{A}\text{OB [Vertically opposite angles]}\\ \therefore \mathrm{\Delta C}\text{OD}~\mathrm{\Delta A}\text{OB [AAA similarity critarion]}\\ ⇒\frac{\mathrm{ar}\left(\mathrm{\Delta C}\text{OD}\right)}{\mathrm{ar}\left(\mathrm{\Delta A}\text{OB}\right)}=\frac{{\mathrm{CD}}^{2}}{{\mathrm{AB}}^{2}}\\ ⇒\frac{\mathrm{ar}\left(\mathrm{\Delta C}\text{OD}\right)}{\mathrm{ar}\left(\mathrm{\Delta A}\text{OB}\right)}=\frac{{\mathrm{CD}}^{2}}{{\left(2\mathrm{CD}\right)}^{2}}=\frac{1}{4}\text{[Given AB}=2\mathrm{CD}\right]\\ ⇒\mathrm{ar}\left(\mathrm{\Delta A}\text{OB}\right):\mathrm{ar}\left(\mathrm{\Delta C}\text{OD}\right)=4:1\end{array}$

Q.3 In the following figure, ABC and DBC are two triangles on the same base BC.
If AD intersects BC at O, show that ar(ΔABC) (ΔDBC) = AO DO.

Ans.

$\text{We draw perpendiculars AP and DM on line BC.}$

$\begin{array}{l}\text{We know that area of a triangle}=\frac{1}{2}×\mathrm{Base}×\mathrm{Height}\\ \therefore \frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta DBC}\right)}=\frac{\frac{1}{2}\mathrm{BC}×\mathrm{AP}}{\frac{1}{2}\mathrm{BC}×\mathrm{DM}}=\frac{\mathrm{AP}}{\mathrm{DM}}\\ \mathrm{In}\text{}\mathrm{\Delta }\text{APO and}\mathrm{\Delta DM}\text{O,}\\ \angle \mathrm{APO}=\angle \mathrm{DMO}=90\mathrm{°}\text{}\\ \angle \mathrm{A}\text{OP}=\angle \mathrm{D}\text{OM [Vertically opposite angles]}\\ \therefore \mathrm{\Delta }\text{APO}~\mathrm{\Delta DM}\text{O [By AA similarity critarion]}\\ ⇒\frac{\mathrm{AP}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}\\ ⇒\frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta DBC}\right)}=\frac{\mathrm{AO}}{\mathrm{DO}}\end{array}$

Q.4 If the areas of two similar triangles are equal, prove that they are congruent.

Ans.

$\begin{array}{l}\mathrm{Let}\text{}\mathrm{\Delta ABC}~\mathrm{\Delta PQR},\text{then we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^{2}={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^{2}={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^{2}\\ ⇒\frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^{2}={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^{2}={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^{2}\text{[Given}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta PQR}\right)\right]\\ ⇒1={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^{2}={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^{2}={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^{2}\\ ⇒1=\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\\ ⇒\mathrm{AB}=\mathrm{PQ},\text{}\mathrm{BC}=\mathrm{QR}\text{and}\mathrm{AC}=\mathrm{PR}\\ \therefore \mathrm{\Delta ABC}\cong \mathrm{\Delta PQR}\text{[By SSS congruence criterion]}\end{array}$

Q.5

$\begin{array}{l}\text{D, E and F are respectively the mid-points of sides AB, BC and CA of}\mathrm{\Delta }\text{ABC. Find the ratio of the areas}\\ \text{of}\mathrm{\Delta }\text{DEF and}\mathrm{\Delta }\text{ABC.}\end{array}$

Ans.

$\begin{array}{l}\text{D and E are mid-points of}\mathrm{\Delta }\text{ABC.}\\ \therefore \mathrm{DE}\parallel \mathrm{AC}\text{and DE}=\frac{1}{2}\mathrm{AC}\\ \mathrm{In}\text{}\mathrm{\Delta }\text{BED and}\mathrm{\Delta }\text{BCA,}\\ \angle \text{BED}=\angle \text{BCA [Corresponding angles]}\\ \angle \text{BDE}=\angle \text{BAC [Corresponding angles]}\\ \angle \text{EBD}=\angle \text{CBA [Common angles]}\\ \therefore \mathrm{\Delta }\text{BED}~\mathrm{\Delta }\text{BCA [AAA similarity criterion]}\\ ⇒\frac{\mathrm{ar}\left(\mathrm{\Delta }\text{BED}\right)}{\mathrm{ar}\left(\mathrm{\Delta }\text{BCA}\right)}={\left(\frac{\mathrm{DE}}{\mathrm{AC}}\right)}^{2}={\left(\frac{\frac{1}{2}\mathrm{AC}}{\mathrm{AC}}\right)}^{2}=\frac{1}{4}\\ ⇒\mathrm{ar}\left(\mathrm{\Delta }\text{BED}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{BCA}\right)\\ \mathrm{Similarly},\\ \mathrm{ar}\left(\mathrm{\Delta CF}\text{E}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{CBA}\right)\text{and}\mathrm{ar}\left(\mathrm{\Delta ADF}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)\\ \text{Also,}\\ \text{}\mathrm{ar}\left(\mathrm{\Delta D}\text{EF}\right)=\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)-\left[\mathrm{ar}\left(\mathrm{\Delta }\text{BED}\right)+\mathrm{ar}\left(\mathrm{\Delta CF}\text{E}\right)+\mathrm{ar}\left(\mathrm{\Delta ADF}\right)\right]\\ ⇒\mathrm{ar}\left(\mathrm{\Delta D}\text{EF}\right)=\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)-\frac{3}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)\\ ⇒\frac{\mathrm{ar}\left(\mathrm{\Delta D}\text{EF}\right)}{\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)}=\frac{1}{4}\end{array}$

Q.6 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Ans.

$\begin{array}{l}\mathrm{Let}\text{}\mathrm{us}\text{}\mathrm{take}\text{two triangles ABC and PQR such that}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta PQR}.\\ \mathrm{Let}\text{AD and PS be the medians of}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta PQR}\text{respectively.}\\ \text{Now,}\\ \text{}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta PQR}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\text{and}\angle \text{A}=\angle \mathrm{P}\text{,}\angle \mathrm{B}=\angle \mathrm{Q},\text{}\angle \mathrm{C}=\angle \mathrm{R}.\\ \mathrm{Since}\text{AD and PS are medians, so we have}\\ \text{BD}=\mathrm{DC}=\frac{\mathrm{BC}}{2}\text{and QS}=\mathrm{SR}=\frac{\mathrm{QR}}{2}\\ \text{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta PQS},\\ \angle \mathrm{B}=\angle \mathrm{Q}\text{and}\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QS}}\\ \therefore \mathrm{\Delta }\text{ABD}~\mathrm{\Delta PQS}\text{[SAS similarity criterion]}\\ ⇒\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QS}}=\frac{\mathrm{AD}}{\mathrm{PS}}\\ \text{Now,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta PQR}\\ ⇒\frac{\mathrm{ar}\left(\mathrm{\Delta A}\text{BC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^{2}\\ ⇒\frac{\mathrm{ar}\left(\mathrm{\Delta A}\text{BC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^{2}\end{array}$

Q.7

$\begin{array}{l}\text{Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the}\\ \text{equilateral triangle described on one of its diagonals.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{ABCD is a square and length of its one side is}\mathrm{a}\text{unit.}\\ \text{Therefore, length of its diagonal}=\sqrt{2}\mathrm{a}\\ \text{Equilateral triangles as per question are formed here as}\\ \mathrm{\Delta }\text{ABE and}\mathrm{\Delta DBF}.\\ \text{Length of a s}\mathrm{ide}\text{}\mathrm{of}\text{equilateral}\mathrm{\Delta }\text{ABE}=\mathrm{a}\\ \text{and}\\ \text{length of a s}\mathrm{ide}\text{}\mathrm{of}\text{equilateral}\mathrm{\Delta DBF}=\sqrt{2}\mathrm{a}\\ \mathrm{We}\text{know that equilateral triangles are similar to each other.}\\ \text{Therefore, ratio of their areas is equal to ratio of squares of}\\ \text{their sides.}\\ \mathrm{i}.\mathrm{e}.,\text{}\frac{\mathrm{ar}\left(\mathrm{\Delta A}\text{BE}\right)}{\mathrm{ar}\left(\mathrm{\Delta DBF}\right)}={\left(\frac{\mathrm{a}}{\sqrt{2}\mathrm{a}}\right)}^{2}=\frac{1}{2}\end{array}$

Q.8

$\begin{array}{l}\text{ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles}\\ \text{ABC and BDE is}\\ \text{(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{BDE are equilateral and D is the}\\ \text{mid point of BC. Therefore, BD}=\frac{\mathrm{BC}}{2}\\ \mathrm{We}\text{know that equilateral triangles are similar to each other}\\ \text{and so ratio of their areas is equal to square of the ratio of}\\ \text{their sides.}\\ \therefore \text{}\frac{\mathrm{ar}\left(\mathrm{\Delta A}\text{BC}\right)}{\mathrm{ar}\left(\mathrm{\Delta BDE}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{BD}}\right)}^{2}={\left(\frac{\mathrm{BC}}{\mathrm{BD}}\right)}^{2}\text{[}\mathrm{\Delta A}\text{BC is equilateral]}\\ \text{}={\left(\frac{\mathrm{BC}}{\frac{\mathrm{BC}}{2}}\right)}^{2}=\frac{4}{1}\text{}\\ ⇒\mathrm{ar}\left(\mathrm{\Delta A}\text{BC}\right):\mathrm{ar}\left(\mathrm{\Delta BDE}\right)=4:1\\ \text{Hence, the correct answer is (C).}\end{array}$

Q.9

$\begin{array}{l}\text{Sides of two similar triangles are in the ratio 4}:\text{9}.\text{\hspace{0.17em}Areas of these triangles are in the ratio}\\ \left(\text{A}\right)\text{2}:\text{3}\left(\text{B}\right)\text{4}:\text{9}\left(\text{C}\right)\text{81}:\text{16}\left(\text{D}\right)\text{16}:\text{81}\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\text{is given that sides of two similar triangles are in the}\\ \text{ratio 4}:\text{9.}\\ \text{Therefore,}\\ \text{Ratio of areas of these triangles}=\frac{{4}^{2}}{{9}^{2}}=\frac{16}{81}\\ \text{Hence, the correct answer is (D).}\end{array}$