Out of the 6 exercises present in the CBSE Class 10 Chapter 6 Triangles, Exercise 6.4 deals with the calculation of areas of triangles that are similar. The solutions to the questions present in Exercise 6.4 are given below in both PDF and scrollable image format. These NCERT Class 10 solutions are prepared with proper reference, giving step-by-step explanations, by the Maths experts at BYJUâ€™S.

Understanding the proper methods to solve the NCERT Solutions for Class 10 Maths will help the students in solving the different types of questions that are likely to be asked in the board examination. Once the students get proficient in these NCERT Class 10 Maths solutions, their problem-solving speed will increase, boosting their self-confidence.

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

### Access other exercise solutions of Class 10 Maths Chapter 6 – Triangles

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)

Exercise 6.3 Solutions 16 Questions (1 Main Question with 6 Sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)

Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)

### Access Answers of Maths NCERT Class 10 Chapter 6 – Triangles Exercise 6.4

**1. Let Î”ABC ~ Î”DEF and their areas be, respectively, 64 cm ^{2}Â and 121 cm^{2}. If EF = 15.4 cm, find BC.**

**Solution: **Given, Î”ABC ~ Î”DEF,

Area of Î”ABC = 64 cm^{2}

Area of Î”DEF =Â 121 cm^{2}

EF = 15.4 cm

As we know, if two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

= AC^{2}/DF^{2}Â = BC^{2}/EF^{2}

âˆ´ 64/121 = BC^{2}/EF^{2}

â‡’ (8/11)^{2}Â = (BC/15.4)^{2}

â‡’ 8/11 = BC/15.4

â‡’ BC = 8Ã—15.4/11

â‡’ BC = 8 Ã— 1.4

â‡’ BC = 11.2 cm

**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.**

**Solution: **

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In Î”AOB and Î”COD, we have

âˆ 1 = âˆ 2 (Alternate angles)

âˆ 3 = âˆ 4 (Alternate angles)

âˆ 5 = âˆ 6 (Vertically opposite angle)

âˆ´ Î”AOB ~ Î”COD [AAA similarity criterion]

As we know, if two triangles are similar, then the ratio of their areas is equal to the square of the ratio of their corresponding sides. Therefore,

Area of (Î”AOB)/Area of (Î”COD) = AB^{2}/CD^{2}

= (2CD)^{2}/CD^{2}Â [âˆ´ AB = 2CD]

âˆ´ Area of (Î”AOB)/Area of (Î”COD)

= 4CD^{2}/CD^{2} = 4/1

Hence, the required ratio of the area of Î”AOB and Î”COD = 4:1

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

**Solution:**

Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.

We have to prove: Area (Î”ABC)/Area (Î”DBC) = AO/DO

Let us draw two perpendiculars, AP and DM, on line BC.

We know that area of a triangle = 1/2Â Ã— BaseÂ Ã— Height

In Î”APO and Î”DMO,

âˆ APO = âˆ DMO (Each 90Â°)

âˆ AOP = âˆ DOM (Vertically opposite angles)

âˆ´ Î”APO ~ Î”DMO (AA similarity criterion)

âˆ´ AP/DM = AO/DO

â‡’ Area (Î”ABC)/Area (Î”DBC) = AO/DO

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

**Solution: **

Say, Î”ABC and Î”PQR are two similar triangles and equal in area.

Now, let us prove Î”ABC â‰… Î”PQR

Î”ABC ~ Î”PQR

âˆ´ Area of (Î”ABC)/Area of (Î”PQR) = BC^{2}/QR^{2}

â‡’ BC^{2}/QR^{2}Â =1 [Since, Area(Î”ABC) = (Î”PQR)

â‡’ BC^{2}/QR^{2}

â‡’ BC = QR

Similarly, we can prove that

AB = PQ and AC = PR

Thus, Î”ABC â‰… Î”PQR [SSS criterion of congruence]

**5. D, E and F are, respectively, the mid-points of sides AB, BC and CA of Î”ABC. Find the ratio of the area of Î”DEF and Î”ABC.**

**Solution: **

âˆ´Â DE || AC and

DE = (1/2) AC (Mid-point theorem) …. (1)

âˆ BEDÂ =Â âˆ BCAÂ (Corresponding angles)

âˆ BDEÂ =Â âˆ BACÂ (Corresponding angles)

âˆ EBDÂ =Â âˆ CBA (Common angles)

âˆ´Î”BEDâˆ¼Î”BCAÂ (AAA similarity criterion)

^{2}

â‡’ar (Î”BED) / ar (Î”BCA)Â = (1/4)Â [From (1)]

â‡’ar (Î”BED)Â = (1/4) ar (Î”BCA)

Similarly,

Also,

â‡’arÂ (Î”DEF)Â = arÂ (Î”ABC)Â âˆ’ (3/4) arÂ (Î”ABC)Â = (1/4)Â ar (Î”ABC)

â‡’ar (Î”DEF) /Â ar (Î”ABC)Â = (1/4)

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

**Solution: **

Given: AM and DN are the medians of triangles ABC and DEF, respectively and Î”ABC ~ Î”DEF.

We have to prove: Area(Î”ABC)/Area(Î”DEF) = AM^{2}/DN^{2}

Since, Î”ABC ~ Î”DEF (Given)

âˆ´ Area(Î”ABC)/Area(Î”DEF) = (AB^{2}/DE^{2}) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦…**(i)**

and, AB/DE = BC/EF = CA/FD â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦…**(ii)**

In Î”ABM and Î”DEN,

Since Î”ABC ~ Î”DEF

âˆ´ âˆ B = âˆ E

AB/DE = BM/EN [Already proved in equation **(i)**]

âˆ´ Î”ABC ~ Î”DEF [SAS similarity criterion]

â‡’ AB/DE = AM/DN â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..**(iii)**

âˆ´ Î”ABM ~ Î”DEN

The areas of two similar triangles are proportional to the squares of the corresponding sides.

âˆ´ area(Î”ABC)/area(Î”DEF) = AB^{2}/DE^{2}Â = AM^{2}/DN^{2}

Hence, proved.

^{
}**7.Â 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.**

**Solution:**

Given, ABCD is a square whose one diagonal is AC. Î”APC and Î”BQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Area(Î”BQC) = Â½ Area(Î”APC)

Î”APC and Î”BQC are both equilateral triangles,

âˆ´ Î”APC ~ Î”BQC [AAA similarity criterion]

âˆ´ area(Î”APC)/area(Î”BQC) = (AC^{2}/BC^{2}) = AC^{2}/BC^{2}

Since, Diagonal = âˆš2 side = âˆš2 BC = AC

â‡’ area(Î”APC) = 2Â Ã— area(Î”BQC)

â‡’ area(Î”BQC) = 1/2area(Î”APC)

Hence, proved.

**Tick the correct answer and justify.**

**8. ABC and BDE are two equilateral triangles, such that D is the mid-point of BC. The ratio of the area of triangles ABC and BDE is
(A) 2:1
(B) 1:2
(C) 4:1
(D) 1:4
**

**Solution: **

Given**, **Î”ABC and Î”BDE are two equilateral triangles. D is the midpoint of BC.

âˆ´ BD = DC = 1/2BC

Let each side of the triangle be 2*a*.

Î”ABC ~ Î”BDE

âˆ´ Area(Î”ABC)/Area(Î”BDE) = AB^{2}/BD^{2}Â = (2*a*)^{2}/(*a*)^{2}Â = 4*a*^{2}/*a*^{2}Â = 4/1 = 4:1

Hence, the correct answer is (C).

**9. 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**

**Solution: **

Given, the sides of two similar triangles are in the ratio 4:9.

Let ABC and DEF be two similar triangles, such that,

Î”ABC ~ Î”DEF

And AB/DE = AC/DF = BC/EF = 4/9

The ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides.

âˆ´ Area(Î”ABC)/Area(Î”DEF) = AB^{2}/DE^{2Â }

âˆ´ Area(Î”ABC)/Area(Î”DEF) =Â (4/9)^{2Â }= 16/81 = 16:81

Hence, the correct answer is (D).

The fourth exercise of Class 10 NCERT Maths Chapter 6 Triangles deals with the topic Area of Similar Triangles. There are 9 questions in this exercise, out of which 2 are short answers with reasoning type of questions, 5 are short answer type questions, and the remaining 2 are long answer type of questions. Exercise 6.4 deals with the Area of Similar Triangles and one theorem, Theorem 6.6. The theorem that forms the base of Exercise 6.4 is given below:

**Theorem 6.6**: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

To understand the concepts that are taught in Class 10 well, there is no other effective method than solving NCERT Solutions. Solving these solutions not only helps you understand the concepts but also helps in getting acquainted with different types of questions that could be asked in the CBSE Class 10 board exam.

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