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

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

Edited By Ramraj Saini | Updated on Nov 25, 2023 08:12 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4

NCERT Solutions for Exercise 6.4 Class 10 Maths Chapter 6 Triangles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 6.4 states the theorem “The square of the ratio of the corresponding sides of two comparable triangles is equal to the ratio of their areas”. Also discusses about criteria to find similarities between triangles and learn about simple geometry problems.

NCERT solutions for exercise 6.4 Class 10 Maths chapter 6 Triangles discusses mainly on the ratio of area and ratio of squared sides. 10th class Maths exercise 6.4 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths chapter 6 exercise 6.4

Triangles Class 10 Chapter 6 Exercise: 6.4

$\Delta ABC \sim \Delta DEF$ ( Given )

ar(ABC) = 64 $cm^2$ and ar(DEF)=121 $cm^2$ .

EF = 15.4 cm (Given )

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}$

$\frac{64}{121}=\frac{BC^2}{(15.4)^2}$

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

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

$\Rightarrow BC=11.2 cm$

Given: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.

AB = 2 CD ( Given )

In $\triangle AOB\, and\, \triangle COD,$

$\angle COD=\angle AOB$ (vertically opposite angles )

$\angle OCD=\angle OAB$ (Alternate angles)

$\angle ODC=\angle OBA$ (Alternate angles)

$\therefore \triangle AOB\, \sim \, \triangle COD$ (AAA similarity)

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{AB^2}{CD^2}$

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{(2CD)^2}{CD^2}$

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4.CD^2}{CD^2}$

$\Rightarrow \frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4}{1}$

$\Rightarrow ar(\triangle AOB)=ar(\triangle COD)=4:1$

Let DM and AP be perpendicular on BC.

$area\,\,of\,\,triangle=\frac{1}{2}\times base\times perpendicular$

$\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}$

In $\triangle APO\, and\, \triangle DMO,$

$\angle APO=\angle DMO$ (Each $90 \degree$ )

$\angle AOP=\angle MOD$ (Vertically opposite angles)

$\triangle APO\, \sim \, \triangle DMO,$ (AA similarity)

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

Since

$\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}$

$\Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{AP}{ MD}=\frac{AO}{DO}$

Let $\triangle ABC\, \sim \, \triangle DEF,$ , therefore,

$ar(\triangle ABC\,) = \,ar( \triangle DEF)$ (Given )

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}................................1$

$\therefore \frac{ar(\triangle ABC)}{ar(\triangle DEF)}=1$

$\Rightarrow \frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}=1$

$AB=DE$

$BC=EF$

$AC=DF$

$\triangle ABC\, \cong \, \triangle DEF$ (SSS )

D, E, and F are respectively the mid-points of sides AB, BC and CA of $\Delta ABC$ . ( Given )

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

In $\Delta BED \: \:and \: \: \Delta ABC$ ,

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

$\angle BDE=\angle BAC$ (corresponding angles )

$\Delta BED \: \:\sim \: \: \Delta ABC$ (By AA)

$\frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{DE^2}{AC^2}$

$\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{(\frac{1}{2}AC)^2}{AC^2}$

$\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{1}{4}$

$\Rightarrow ar(\triangle BED)=\frac{1}{4}\times ar(\triangle ABC)$

Let ${ar(\triangle ABC)$ be x.

$\Rightarrow ar(\triangle BED)=\frac{1}{4}\times x$

Similarly,

$\Rightarrow ar(\triangle CEF)=\frac{1}{4}\times x$ and $\Rightarrow ar(\triangle ADF)=\frac{1}{4}\times x$

$ar(\triangle ABC)=ar(\triangle ADF)+ar(\triangle BED)+ar(\triangle CEF)+ar(\triangle DEF)$

$\Rightarrow x=\frac{x}{4}+\frac{x}{4}+\frac{x}{4}+ar(\triangle DEF)$

$\Rightarrow x=\frac{3x}{4}+ar(\triangle DEF)$

$\Rightarrow x-\frac{3x}{4}=ar(\triangle DEF)$

$\Rightarrow \frac{4x-3x}{4}=ar(\triangle DEF)$

$\Rightarrow \frac{x}{4}=ar(\triangle DEF)$

$\frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{\frac{x}{4}}{x}$

$\Rightarrow \frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{1}{4}$

Let AD and PS be medians of both similar triangles.

$\triangle ABC\sim \triangle PQR$

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}............................1$

$\angle A=\angle P,\angle B=\angle Q,\angle \angle C=\angle R..................2$

$BD=CD=\frac{1}{2}BC\, \, and\, QS=SR=\frac{1}{2}QR$

Purring these value in 1,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}..........................3$

In $\triangle ABD\, and\, \triangle PQS,$

$\angle B=\angle Q$ (proved above)

$\frac{AB}{PQ}=\frac{BD}{QS}$ (proved above)

$\triangle ABD\, \sim \triangle PQS$ (SAS )

Therefore,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}................4$

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}=\frac{AC^2}{PR^2}$

From 1 and 4, we get

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

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AD^2}{PS^2}$

Let ABCD be a square of side units.

Therefore, diagonal = $\sqrt{2}a$

Triangles form on the side and diagonal are $\triangle$ ABE and $\triangle$ DEF, respectively.

Length of each side of triangle ABE = a units

Length of each side of triangle DEF = $\sqrt{2}a$ units

Both the triangles are equilateral triangles with each angle of $60 \degree$ .

$\triangle ABE\sim \triangle DBF$ ( By AAA)

Using area theorem,

$\frac{ar(\triangle ABC)}{ar(\triangle DBF)}=(\frac{a}{\sqrt{2}a})^2=\frac{1}{2}$

(A) 2: 1 (B) 1: 2 (C) 4 : 1 (D) 1: 4

Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

All angles of the triangle are $60 \degree$ .

$\triangle$ ABC $\sim \triangle$ BDE (By AAA)

Let AB=BC=CA = x

then EB=BD=ED= $\frac{x}{2}$

$\frac{ar(\triangle ABC)}{ar(\triangle BDE)}=(\frac{x}{\frac{x}{2}})^2=\frac{4}{1}$

Option C is correct.

(A) 2 : 3 (B) 4: 9 (C) 81: 16 (D) 16: 81

Sides of two similar triangles are in the ratio 4: 9.

Let triangles be ABC and DEF.

We know that

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{4^2}{9^2}=\frac{16}{81}$

Option D is correct.

## More About NCERT Solutions for Class 10 Maths Exercise 6.4

Class 10 Maths chapter 6 exercise 6.4: Most of the questions are based on the ratio of areas of similar triangles in exercise 6.4 Class 10 Maths, but we should be also aware of some basic geometry which helps to solve problems of NCERT solutions for Class 10 Maths exercise 6.4. We should know the properties of the trapezium and the criterion of similarity of triangles. Questions given in the Exercise 6.4 Class 10 Maths are very important for the board exam. Students can also access Triangles Class 10 Notes here and use them for quickly revision of the concepts related to Triangles.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 6.4

• Class 10 NCERT solutions Math is thought to be the best material for problem-solving Class 10 Maths chapter 6 exercise 6.4.
• NCERT Class 10 Maths chapter 6 exercise 6.4, All questions are updated from Class 10 Maths chapter 6 exercise 6.4 and contain all crucial questions from test pov.
• Exercise 6.4 Class 10 Maths, is founded on irrational numbers and the Fundamental Theorem of Arithmetic, both of which are key concepts in the chapter.

Also, See

## Subject Wise NCERT Exemplar Solutions

1. Do Similar triangles have same areas?

No, it is not always true but it is proved in NCERT solutions for Class 10 Maths 1 exercise 6.4 that ratio of areas of two similar triangles is equal to the square of ratio of corresponding sides of that triangles.

2. What is the area of equilateral triangle of side a?

No, it is not always true but it is proved in NCERT solutions for Class 10 Maths 1 exercise 6.4 that ratio of areas of two similar triangles is equal to the square of ratio of corresponding sides of that triangles.

3. Which criterion of similarity of triangle is used for right angled triangle?

RHS rule is used to prove similarity

4. If two similar triangles have same area, are they congruent?

Yes, by using theorem 6.6 given in NCERT solutions for Class 10 Maths 1 exercise 6.4, we can prove this.

5. What is the relation between the ratio of areas of similar triangles and their sides?

Ratio of areas of similar triangles is equal to the ratio of square of corresponding sides of that triangles.

Sides of similar triangles are in ratio 25:36, find the ratio of areas of triangle.

Ratio is 625:1296.

6. Are two triangles similar if they have two corresponding equal angles?

Yes, they are similar by AAA rule.

7. What do you understand by ASA rule?

Two triangles are congruent if two angles and the included side of one triangle are congruent with two angles and the included side of another triangle.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

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Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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