NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.4

Chapter 6 Triangles NCERT Solutions Exercise 6.4 for Class 10 Maths assists students in broadening their problem-solving capabilities in the realm of similar triangles. In the exercise, we will be explaining the area of a triangle in terms of another triangle and its respective sides. The idea creates a strong foundation for solving different questions and provides a significant relationship between the areas of two similar triangles, ultimately leading to solving many intricate problems. So, let's begin discovering this significant concept of the exercise.

1.0Download NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.4 : Free PDF

2.0Introduction to Similar Triangles

Let's first learn about similar triangles before directly moving into the ratio of areas of similar triangles. Two triangles are referred to as similar when their corresponding sides are in proportion & corresponding angles are equal. This implies that although the form of the triangles is the same, their dimensions are different. Similar triangles form the basis of geometry, especially in topics involving proportion and scaling. The relationship between the sides of any two similar triangles forms the foundation for understanding how their areas are related.

3.0NCERT Solutions Exercise 6.4: Key Concepts

The entire exercise is on the concept of comparing the two triangles' areas, with the proportionality or the ratio of the sides of the triangles. Let's see some of the important concepts utilised in the exercise to answer the questions that are part of it:

Area of Similar Triangles

Most of the questions in the exercise are done using the similarity relation between the areas of two similar triangles and their corresponding sides. The relation is converted to a theorem, which says that the ratio of the areas of any two similar triangles equals the square of the ratio of their corresponding sides. To grasp this better, consider two similar triangles, ABC and PQR. As per the theorem:

$\frac{ar\Delta ABC}{ar\Delta PQR}=(\frac{AB}{PQ}{)}^2=(\frac{BC}{QR}{)}^2=(\frac{CA}{RP}{)}^2$

The theorem is established by the use of the classical formula for the area of a triangle, as well as the method of similarity of triangles.

Exercise Overview

As discussed above, the questions of Exercise 6.4 are actually based on finding the ratio of the areas of two similar triangles using the properties of similar triangles. To master these questions, one needs to follow a step-by-step method. The following are step-by-step methods for solving these types of questions:

• Identifying corresponding sides: For similar triangle questions, be careful to identify corresponding sides because a small error can change the result entirely.
• Use Geometric Properties: The exercise includes various questions based on the various geometrical shapes, such as trapeziums, parallelograms, etc. These questions tend to ask for the identification of areas where the theorem of areas can be applied to solve them based on the need of the question.
• Using the formula: Use the formula of area ratio with caution, particularly in composite shapes, where the triangles are combined with some other geometric shapes.
• Practice: The more you practice similar triangle problems and areas, the more proficient you will be in using them.

By following these structured approaches, you will be able to easily solve the exercises in NCERT Solutions Class 10 Maths Chapter 6 Triangles - Exercise 6.4. Key concepts of proportionality, similarity, and properties of shapes in geometry are applied to calculate the ratio of areas, solve unknowns, and understand triangle proportions.

4.0NCERT Solutions Class 10 Maths Chapter 6 Triangles Exercise 6.4 : Detailed Solutions

1. Let ΔABC ~ ΔDEF and their areas be 64 cm² and 121 cm² respectively. If EF = 15.4 cm, find BC.

Sol. ΔABC ~ ΔDEF (Given)

=> ar(ABC)/ar(DEF) = BC²/EF² (By theorem 6.7)

=> 64/121 = BC²/EF² => (BC/EF)² = (8/11)²

=> BC/EF = 8/11 => BC = (8/11) × EF

=> BC = (8/11) × 15.4 cm = 11.2 cm

2. Diagonals of 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.

Sol.

In △AOB and △COD,

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∴ By AA similarity

△AOB ∼ △COD

So, ar(△AOB) / ar(△COD) = (AB/CD)²

= (2/1)² {AB = 2CD}

= 4 : 1

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

Solution:

Draw AL perpendicular to BC and DM perpendicular to BC (see figure).

ΔOLA ~ ΔOMD (AA similarity criterion)

=> AL/DM = AO/DO

Now, ar(ΔABC)/ar(ΔDBC) = (1/2 × BC × AL) / (1/2 × BC × DM)

= AL/DM = AO/DO

Hence, ar(ΔABC)/ar(ΔDBC) = AO/DO

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

Sol. Let ΔABC ~ ΔPQR and

area(ΔABC) = area(ΔPQR) (Given)

i.e., area(ΔABC)/area(ΔPQR) = 1

=> AB²/PQ² = BC²/QR² = CA²/PR² = 1

=> AB = PQ, BC = QR and CA = PR

=> ΔABC ≅ ΔPQR (By SSS)

5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC.

Sol.

DF = (1/2) BC, DE = (1/2) AC, EF = (1/2) AB [By midpoint theorem]

So, DF/BC = DE/AC = EF/AB = 1/2

Therefore, ΔDEF ~ ΔCAB (By SSS)

So, (ar ΔDEF) / (ar ΔABC) = (DE/AC)² = (1/2)² = 1/4 or 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.

Sol. In the figure, AD is a median of ΔABC and PM is a median of ΔPQR. Here, D is the midpoint of BC and M is the midpoint of QR.

Now, we have ΔABC ~ ΔPQR.

=> ∠B = ∠Q (Corresponding angles are equal)

Also AB/PQ = BC/QR (Ratio of corresponding sides are equal)

=> AB/PQ = 2BD/2QM (∵ D is mid-point of BC and M is mid-point of QR)

=> AB/PQ = BD/QM

In ΔABD and ΔPQM,

∠ABD = ∠PQM

and AB/PQ = BD/QM

=> ΔABD ~ ΔPQM (SAS similarity)

=> AB/PQ = AD/PM

Now, ar(ΔABC)/ar(ΔPQR) = AB²/PQ² (By theorem 6.7)

=> ar(ΔABC)/ar(ΔPQR) = AD²/PM² (∵ AB/PQ = AD/PM)

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.

Sol. ABCD is a square having sides of length = a.

Then the diagonal BD = a√2.

We construct equilateral triangles PAB and QBD.

=> ΔPAB ~ ΔQBD (Equilateral triangles are similar)

=> ar(ΔPAB) / ar(ΔQBD) = AB² / BD² = a² / (a√2)² = 1/2

=> ar(ΔPAB) = (1/2) ar(ΔQBD).

8. Tick the correct answer and justify:

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

(1) 2:1

(2) 1:2

(3) 4:1

(4) 1:4

Sol.

(BC = 2BD)

Since both are equilateral triangles.

Triangle ABC ~ Triangle EBD

(ar Triangle ABC) / (ar Triangle BDE) = (BC/BD)² = (2/1)² = 4:1

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

(1) 2:3

(2) 4:9

(3) 81:16

(4) 16:81

Sol. (area of 1st Triangle) / (area of 2nd Triangle) = (4/9)² = 16/81

5.0Benefits of Class 10 Maths Chapter 6 Triangles Exercise 6.4

• Improves accuracy in solving proofs and theorem-based questions.
• Encourages logical thinking through application-based problems.
• Helps in understanding the relationship between sides and angles in triangles.
• Boosts confidence in handling real-life geometry problems.