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

Last updated date: 13th Jul 2024
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## NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Free PDF download of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4 (Ex 6.4) and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 10 Maths Chapter 6 Triangles Exercise 6.4 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise NCERT Solutions in your emails. You can also download Maths NCERT Solutions Class 10 to help you to revise the complete syllabus and score more marks in your examination. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution Class 10 Science, Maths solutions, and solutions of other subjects.

 Class: NCERT Solutions for Class 10 Subject: Class 10 Maths Chapter Name: Chapter 6 - Triangles Exercise: Exercise - 6.4 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
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## Access NCERT Solutions for Class 10 Maths Chapter - 6 - Triangles

EXERCISE NO: 6.4

1. Let $\Delta \mathbf{ABC\sim}\Delta \mathbf{DEF}$ and their areas be, respectively, $\mathbf{64}\ \mathbf{c}{{\mathbf{m}}^{\mathbf{2}}}$ and$\mathbf{121}\,\,\mathbf{c}{{\mathbf{m}}^{\mathbf{2}}}$. If $\mathbf{EF}=\mathbf{15}.\mathbf{4}$ cm, find $\mathbf{BC}$.

Ans: We are provided that, $\Delta ABC\text{ }\sim \Delta DEF$.

Therefore, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta DEF \right)}={{\left( \frac{AB}{DE} \right)}^{2}}={{\left( \frac{BC}{EF} \right)}^{2}}=\left( \frac{AC}{DF} \right)$

Also, it is provided that,

$EF=15.4$ cm,

$area\left( \Delta ABC \right)=64\ c{{m}^{2}}$,

$area\left( \Delta DEF \right)=121\,c{{m}^{2}}$.

Thus, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta DEF \right)}={{\left( \frac{BC}{EF} \right)}^{2}}$

Therefore, $\left( \frac{64}{121} \right)=\frac{B{{C}^{2}}}{15.4}$

$\Rightarrow \frac{BC}{15.4}=\frac{8}{11}$ cm

$\Rightarrow BC=\frac{8\times 15.4}{11}=11.2$ cm

2. Diagonals of a trapezium $\mathbf{ABCD}$ with $\mathbf{AB}||\mathbf{DC}$ intersect each other at the point $\mathbf{O}$. If $\mathbf{AB}=\mathbf{2CD}$, find the ratio of the areas of triangles $\mathbf{AOB}$ and $\mathbf{COD}$.

Ans:

It is provided that, $AB||CD$.

Therefore, $\angle OAB=\angle OCD$ and $\angle OBA=\angle ODC$ (since, these are alternate interior angles)

Now, for the triangles $\Delta AOB$ and $\Delta COD$, we have

$\angle AOB=\angle COD$ (Since, these are vertically opposite angles)

Similarly,

$\angle OAB=\angle OCD$

$\angle OBA=\angle ODC$

Hence, $\Delta AOB\sim \Delta COD$ (According to the AAA similarity property)

3. In the following figure, $\mathbf{ABC}$ and $\mathbf{DBC}$ are two triangles on the same base $\mathbf{BC}$. If $\mathbf{AD}$ intersects $\mathbf{BC}$ at $\mathbf{O}$, show that $\frac{\mathbf{area}\left( \mathbf{\Delta ABC} \right)}{\mathbf{area}\left( \mathbf{\Delta DBC} \right)}=\frac{\mathbf{AO}}{\mathbf{DO}}$.

Ans: On the line$BC$, sketch two perpendiculars, $AP$ and $DM$.

Recall that, area of a triangle $=\frac{1}{2}\times Base\times height$

Therefore, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta DBC \right)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times DM}=\frac{AP}{DM}$

For the triangle, $\Delta APO$ and $\Delta DMO$, we have

$\angle APO=\angle DMO$ (both the angles are of ${{90}^{\circ }}$)

$\angle AOP=\angle DOM$ (These angles are vertically opposite)

Thus, $\Delta APO\sim \Delta DMO$ (According to the $AA$ similarity property)

Therefore, $\frac{AP}{DM}=\frac{AO}{DO}$

Hence, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta DBC \right)}=\frac{AO}{DO}$

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

Ans:

Suppose that, $\Delta ABC\sim \Delta PQR$ are two similar triangles.

Then, we have

$\frac{area\left( \Delta ABC \right)}{area\left( \Delta PQR \right)}={{\left( \frac{AB}{PQ} \right)}^{2}}={{\left( \frac{BC}{QR} \right)}^{2}}={{\left( \frac{AC}{PR} \right)}^{2}}$                        …… (i)

Also, it is provided that, $area\left( \Delta ABC \right)=area\left( \Delta PQR \right)$.

This implies, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta PQR \right)}=1$                                           …… (ii)

Equating the equations (i) and (ii), it yields

$1={{\left( \frac{AB}{PQ} \right)}^{2}}={{\left( \frac{BC}{QR} \right)}^{2}}={{\left( \frac{AC}{PR} \right)}^{2}}$

Thus, $AB=PQ$ , $BC=QR$, and $AC=PR$.

Hence,$\Delta ABC\cong \Delta PQR$ (According to the $SSS$ congruence property)

5. $\mathbf{D},\,\,\mathbf{E}$ and $\mathbf{F}$ are respectively the mid-points of sides $\mathbf{AB},\,\,\mathbf{BC}$ and $\mathbf{CA}$ of $\mathbf{\Delta }\,\mathbf{ABC}$. Find the ratio of the area of $\mathbf{\Delta DEF}$ and $\mathbf{\Delta ABC}$.

Ans:

It is given that, $D$ is the mid-point of $AB$ and $E$ is the mid-point of $BC$ in the triangle $\Delta ABC$.

Therefore, $DE||AC$ and $DE=\frac{1}{2}AC$.

Again, for the triangles $\Delta BED$ and $\Delta BCA$, we have

$\angle BED=\angle BCA$ (corresponding angles)

Similarly,

$\angle BDE=\angle BAC$ and

$\angle EBA=\angle CBA$.

Therefore, $\Delta BED\sim \Delta BCA$ (According to the $AAA$ similarity property)

Thus, we have

$\frac{area\left( \Delta BED \right)}{area\left( \Delta BCA \right)}={{\left( \frac{DE}{AC} \right)}^{2}}$

$\Rightarrow \frac{area\left( \Delta BED \right)}{area\left( \Delta BCA \right)}=\frac{1}{4}$

$\Rightarrow area\left( \Delta BED \right)=\frac{1}{4}area\left( \Delta BCA \right)$

In a similar way,

$area\left( \Delta CFE \right)=\frac{1}{4}area\left( \Delta CBA \right)$ and

$area\left( \Delta ADF \right)=\frac{1}{4}area\left( \Delta ABA \right)$.

Again,

$area\left( \Delta DEF \right)=area\left( \Delta ABC \right)-\left[ area\left( \Delta BED \right)+area\left( \Delta CFE \right)+area\left( \Delta ADF \right) \right]$

$\Rightarrow area\left( \Delta DEF \right)=area\left( \Delta ABC \right)-\frac{3}{4}area\left( \Delta ABC \right)$

$\Rightarrow \frac{area\left( \Delta DEF \right)}{area\left( \Delta ABC \right)}=\frac{1}{4}$

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:

Suppose that, $\Delta ABC\sim \Delta PQR$ are two similar triangles. Also suppose that, $AD$, $PS$ are the medians of the triangle $\Delta ABC$ and $\Delta PQR$ respectively.

Therefore, $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$                        …… (i)

$\angle A=\angle P$, $\angle B=\angle Q$, and $\angle C=\angle R$         …… (ii)

Also, we have

$BD=DC=\frac{BC}{2}$,                                           …… (iii)

$QS=SR=\frac{QR}{2}$                                             ……. (iv)

From the equations (i), (ii), and (iii), we obtain

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}$                                          …… (v)

For the triangles $\Delta ABD$ and $\Delta PQS$, we have

$\angle B=\angle Q$ [by equation (ii)]

Also, $\frac{AB}{PQ}=\frac{BD}{QS}$ [by equation (v)]

Therefore, $\Delta ABD\sim \Delta PQS$ (According to the $SAS$ similarity property)

Thus, we can write that,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}$                                           …… (vi)

So, $\frac{area\left( \Delta ABC \right)}{area\left( \Delta PQR \right)}={{\left( \frac{AB}{PQ} \right)}^{2}}{{\left( \frac{BC}{QR} \right)}^{2}}={{\left( \frac{AC}{PR} \right)}^{2}}$.

Then, the equation (i) and (vi) together implies that,

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PS}$

Thus, we obtain

$\frac{area\left( \Delta ABC \right)}{area\left( \Delta PQR \right)}={{\left( \frac{AD}{PS} \right)}^{2}}$.

7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Ans:

In the above figure, $ABCD$ is a square of side $a$ units.

Then, its diagonal $=\sqrt{2a}$ units.

Suppose that, $\Delta ABE$ and $\Delta DBF$ are the two equilateral triangles.

Also, side of the equilateral triangle, $\Delta ABE$ described on one of its sides $=\sqrt{2a}$ units; and side of the equilateral triangle, $\Delta DBF$ described on one of its diagonals.

Now, recall that, each of angles of an equilateral triangle is ${{90}^{\circ }}$ and length of all of its sides are the same. So, all equilateral triangles are always similar to each other. Thus, the square of the ratio between the sides of these triangles will be equal to the ratio between their areas.

Hence, $\frac{area\text{ of }\Delta \text{ABE}}{area\text{ of }\Delta \text{DBF}}={{\left( \frac{a}{\sqrt{2}a} \right)}^{2}}=\frac{1}{2}$.

That is, area of $\Delta ABE=\frac{1}{2}\times$ area of $\Delta DBF$.

So, the required result is proved.

8. $\mathbf{ABC}$ and $\mathbf{BDE}$ are two equilateral triangles such that $\mathbf{D}$ is the mid-point of $\mathbf{BC}$. Ratio of the area of triangles $\mathbf{ABC}$ and $\mathbf{BDE}$ is

(a) $\mathbf{2}:\mathbf{1}$

(b) $\mathbf{1}:\mathbf{2}$

(c) $\mathbf{4}:\mathbf{1}$

(d) $\mathbf{1}:\mathbf{4}$

Ans:

Recall that, each of the angles of an equilateral is of ${{90}^{\circ }}$ and length all the sides of it are of equal. So, all equilateral triangles are similar to one another.

Thus, the square of the ratio between the sides of these triangles will be equal to the ratio between their areas.

Now, suppose that, sides of $\Delta ABC$ is of $x$ units.

Then, sides of $\Delta BDE$  is of $\frac{x}{2}$ units.

So, $\frac{area\text{ of }\Delta \text{ABC}}{area\text{ of }\Delta \text{BDE}}={{\left( \frac{x}{\frac{x}{2}} \right)}^{2}}=\frac{4}{1}$

Hence, the ratio between areas of triangles $ABC$ and $BDE$ is $4:1$.

That is, the correct answer is the option (c).

9. Sides of two similar triangles are in the ratio $\mathbf{4}:\mathbf{9}$. Areas of these triangles are in the ratio

(a) $\mathbf{2}:\mathbf{3}$

(b) $\mathbf{4}:\mathbf{9}$

(c) $\mathbf{81}:\mathbf{16}$

(d) $\mathbf{16}:\mathbf{81}$

Ans: Recall that, for similar triangles, the square of the ratio between the sides of these triangles will be equal to the ratio between their areas.

We are provided that the ratio of the sides of two similar triangles is $4:9$.

So, areas of these triangles will be in the following ratio

$={{\left( \frac{4}{9} \right)}^{2}}=\frac{16}{81}$.

That is, $16:81$.

Thus, option (d) is the correct answer.

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

Opting for the NCERT solutions for Ex 6.4 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.4 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Subject Triangles textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 6 Exercise 6.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 6 Exercise 6.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 6 Exercise 6.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 6: Triangles - Exercise 6.4

1. What is taught in the 6th chapter of Class 10th Science?

Ans: The 6th chapter of Class 10th Science deals with Triangle. It is one of the most interesting chapters of Geometry as it takes us through the different aspects and various in-depth concepts related to the geometrical figure triangle.

A triangle is basically a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles including the explanation of similar geometrical figures, different theorems related to the similarities of triangles, and the areas of similar triangles. The chapter concludes with a detailed explanation of the Pythagoras theorem and how to use it in solving problems. Go through the Chapter 6 of the NCERT textbook to learn more about Triangles and the concepts covered in it.

2. How many questions are there in Class 10 Maths Chapter 6 Exercise 6.4?

Ans: Exercise 6.4 of Class 10 Maths Chapter 6 consists of 14 sums. Among which, two are short answer type with Reasoning Questions, five are pure short answer questions, and the remaining two are the long answer type questions).

3. Why should I opt for Vedantu’s NCERT Solutions of Class 10 Maths Chapter 6 Exercise 6.4?

Ans: Vedantu’s NCERT Solutions of Class 10 Maths Chapter 6 Exercise 6.4 are delivered keeping in mind the types of students who would be using them. Let's take a look at the benefits of using this.

• The solutions are provided in layman's language and concentrate more on essential facts, terms, principles, theorems, and applications on different ideas.

• Complicated solutions that are difficult to understand are written in a simple and concise manner to save the students from going through all the unnecessary points which would cause strain on their minds.

• It provides a summary of the entire chapter along with the solutions of the exercises.

• The answers are treated systematically and created in an interesting manner.

• The content is kept compact, comprehensive and clear.

• Some answers are combined with necessary diagrams to make the concept easy to understand.

• The solutions are created as per the latest CBSE syllabus.

4. Why is the Triangles Chapter Important?

Ans: Class 10 maths chapter 6 maths Triangles is one of the important topics and can be looked at as a summarization of the concept of triangles and congruence of triangles which was discussed in class 9 Math. So, it will be easy to hold a grasp on this chapter. This chapter has a weightage of 14 marks in class 10 board exams. Nearly seven questions are being asked from this chapter in the final exam every year.

5. What do you understand by the Thales Theorem according to Chapter 6 of Class 10 Maths?

Ans: According to the Thales Theorem, the ratio of any two corresponding sides present in two equiangular triangles (triangles that have two equal corresponding angles) is always the same. This can also be referred to as the Basic Proportionality Theorem. Use Vedantu’s NCERT Solutions for Chapter 6 of Class 10 Maths to have a better understanding of how to solve the exercise. The whole exercise is solved in full detail. Using this NCERT Solutions will simplify the exercise for you.

6. What are some of the criteria to prove similarity between two triangles according to Chapter 6 of Class 10 Maths?

Ans: Some of the criteria to prove the similarity between two triangles are - AAA (angle-angle-angle), SSS (side - side - side), and SAS (side - angle - side). If you want to have a good grip over this exercise, you must download the NCERT Solutions for Chapter 6 of Class 10 Maths that is available on Vedantu free of cost. This NCERT Solution has tried to cover every aspect of the exercise - from formulas, theorems to step-by-step answers. Get the solution today!

7. Do I Need to Practice all Questions given in NCERT Exercise 6.4 of Chapter 6 of Class 10 Maths?

Ans: Yes, you should practice all the questions that are there in the NCERT Exercise 6.4 of Chapter 6 of Class 10 Maths to make sure that you do not miss any important questions that can be asked in the Class 10 board examination. Since this exercise is not optional, it is essential that you prioritize all questions. If you are facing difficulties in solving this exercise, then Vedantu’s NCERT Solutions of Chapter 6 of Class 10 Maths will help you out!

8. How to calculate the area of similar triangles as explained in Chapter 6 of Class 10 Maths?

Ans: To calculate the area of similar triangles, you need to keep in mind that the ratio of the areas of two similar triangles is equal to the ratio of the corresponding sides of these two triangles. By this logic, you will be able to calculate the area of similar triangles. If you are looking for more solutions from this exercise, download NCERT Solutions of Chapter 6 of Class 10 Maths4.

9. How can I ace Exercise 6.4 of Chapter 6 of Class 10 Maths?

Ans: Exercise 6.4 is based on the following theorem:

"The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides."

To master this exercise, learn and practice this theorem. Make sure that you practice the example questions of the NCERT textbook well. Also, practice the Exercise 6.4 questions diligently. You can refer to Vedantu's  NCERT solutions for the same. Additionally, you can refer to Vedantu's conceptual videos and free masterclasses for strengthening the concepts of this exercise.