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 of the questions present in the exercise 6.4 is given below, in both pdf and scrollable image format. These 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 first and second term examination. Once the students get proficient in these solutions, their problem solving speed will increase, in turn boosting their self confidence.
Download PDF of NCERT Solutions for Class 10 Maths Chapter 6- Triangles Exercise 6.4
Access Answers of Maths NCERT class 10 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 â€“ 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, ratio of their areas are 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 are 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.
Since, 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:
Given, D, E and F are respectively the mid-points of sides AB, BC and CA of Î”ABC.
In Î”ABC,
F is the mid-point of AB (Already given)
E is the mid-point of AC (Already given)
So, by the mid-point theorem, we have,
FE || BC and FE = 1/2BC
â‡’ FE || BC and FE || BD [BD = 1/2BC]
Since, opposite sides of parallelogram are equal and parallel
âˆ´ BDEF is parallelogram.
Similarly, in Î”FBD and Î”DEF, we have
FB = DE (Opposite sides of parallelogram BDEF)
FD = FD (Common sides)
BD = FE (Opposite sides of parallelogram BDEF)
âˆ´ Î”FBDÂ â‰… Î”DEF
Similarly, we can prove that
Î”AFE â‰… Î”DEF
Î”EDC â‰… Î”DEF
As we know, if triangles are congruent, then they are equal in area.
So,
Area(Î”FBD) = Area(Î”DEF) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(i)
Area(Î”AFE) = Area(Î”DEF) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(ii)
and,
Area(Î”EDC) = Area(Î”DEF) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iii)
Now,
Area(Î”ABC) = Area(Î”FBD)Â + Area(Î”DEF)Â + Area(Î”AFE)Â +Â Area(Î”EDC) â€¦â€¦â€¦(iv)
Area(Î”ABC) = Area(Î”DEF)Â +Â Area(Î”DEF)Â +Â Area(Î”DEF)Â +Â Area(Î”DEF)
From equation (i),Â (ii)Â andÂ (iii),
â‡’ Area(Î”DEF) = (1/4)Area(Î”ABC)
â‡’ Area(Î”DEF)/Area(Î”ABC) = 1/4
Hence, Area(Î”DEF): Area(Î”ABC) = 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€¦â€¦(i)
and, AB/DE = BC/EF = CA/FD â€¦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 equationÂ (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
As 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)
Since, Î”APC and Î”BQC are both equilateral triangles, as per given,
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. 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 triangle. D is the midpoint of BC.
Â
âˆ´ BD = DC = 1/2BC
Let each side of triangle is 2a.
As, Î”ABC ~ Î”BDE
âˆ´ Area(Î”ABC)/Area(Î”BDE) = AB^{2}/BD^{2}Â = (2a)^{2}/(a)^{2}Â = 4a^{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, Sides of two similar triangles are in the ratio 4 : 9.
Let ABC and DEF are two similar triangles, such that,
Î”ABC ~ Î”DEF
And AB/DE = AC/DF = BC/EF = 4/9
As, 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. The Exercise 6.4 deals with the Area of Similar Triangles and one theorem, Theorem 6.6. The theorem that forms the base of the 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 properly, there is no other effective method than solving the NCERT Solutions. Solving these solutions not only helps you in understanding the concepts, but also helps in getting acquainted with different types of questions that could be asked in the CBSE Class 10 first and second term exam.