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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.

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  1. NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.4
  2. Download Free PDF of NCERT Solutions for Class 10 Maths chapter 6 exercise 6.4
  3. Assess NCERT Solutions for Class 10 Maths chapter 6 exercise 6.4
  4. More About NCERT Solutions for Class 10 Maths Exercise 6.4
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 6.4
  6. NCERT Solutions of Class 10 Subject Wise
  7. Subject Wise NCERT Exemplar Solutions
NCERT Solutions for Exercise 6.4 Class 10 Maths Chapter 6 - Triangles
NCERT Solutions for Exercise 6.4 Class 10 Maths Chapter 6 - Triangles

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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Triangles Class 10 Chapter 6 Exercise: 6.4

Q1 Let \Delta ABC \sim \Delta DEF and their areas be, respectively, 64 cm^2 and 121 cm^2 . If EF = 15.4 cm, find BC.

Answer:

\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

Q2 Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Answer:

1635931770237

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

Q3 In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \frac{ar (ABC)}{ar ( DBC )} = \frac{AO}{DO}

1635931784662

Answer:

1635931796737

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}

Q5 D, E, and F are respectively the mid-points of sides AB, BC and CA of \Delta ABC . Find the ratio of the areas of \Delta DEF \: \:and \: \: \Delta ABC

Answer:

1635932236410

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}

Q6 Proves that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer:

1635932258592

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}

Q7 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.

Answer:

1635932273637

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}

Q8 Tick the correct answer and justify : ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is

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

Answer:

1635932291018

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.

Q9 Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

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

Answer:

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

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Question (FAQs)

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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