NCERT Solutions for Class 10 Maths Exercise 6.4 Chapter 6 Triangles – FREE PDF Download
NCERT Class 10 Maths Ch 6 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
6 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Triangles solutions will help you understand the chapter thoroughly.
NCERT Solutions for Class 10 Maths Chapter 6 – Triangles
1. Let 
ABC 
DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
BC = 
cm = 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.
s AOB and COD, we have, 
AOB = COD[Vertically opposite angles]
OAB = OCD[Alternate angles]
By AA-criterion of similarity,
AOB
COD
Hence, Area (AOB) : Area (COD) = 4 : 1
3. In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that 
s ABC and DBC which stand on the same base but on the opposite sides of BC. 
To Prove: 
Construction: Draw AE
BC and DFBC.
Proof: In s AOE and DOF, we have, AEO = DFO = 
and AOE = DOF[Vertically opposite)
AOEDOF[By AA-criterion]
……….(i)
Now, 
[using eq. (i)]
4. If the areas of two similar triangles are equal, prove that they are congruent.
s ABC and DEF such that ABCDEF And Area(ABC) = Area (DEF)
To Prove: ABC
DEF
Proof: ABCDEF
A = D, B = E, C = F
And 
To establish ABCDEF, it is sufficient to prove that, AB = DE, BC = EF and AC = DF
Now, Area(ABC) = Area (DEF)
= 1
= 1
AB = DE, BC = EF, AC = DF
Hence,ABCDEF
5. D, E and F are respectively the midpoints of sides AB, BC and CA of ABC. Find the ratio of the areas of DEF and ABC.
ABC respectively. DEBA DEFA ……….(i)
Since D and F are the midpoints of the sides BC and AB of ABC respectively.
DFCA DEAE ……….(ii)
From (i) and (ii), we can say that AFDE is a parallelogram.
Similarly, BDEF is a parallelogram.
Now, in s DEF and ABC, we have
FDE = A[opposite angles of gm AFDE]
And DEF = B[opposite angles of gm BDEF]
By AA-criterion of similarity, we have DEF ABC
[
DE = 
AB]
Hence, Area (DEF): Area (ABC) = 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 medi
Given:
ABCPQR, AD and PM are the medians of s ABC and PQR respectively. 
To Prove: 
Proof: Since ABCPQR
……….(1)
But, 
……….(2)
From eq. (1) and (2), we have,
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 the diagonals.
Equilateral s BCE and ACF have been drawn on side BC and the diagonal AC respectively.
To Prove: Area (BCE) = Area (ACF)
Proof: BCEACF
[Being equilateral so similar by AAA criterion of
similarity]
[ Diagonal = 
side AC = BC]
Tick the correct answer and justify:
8. ABC and BDE are two equilateral triangles such that D is the midpoint 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
ABC and BDE are equilateral, they are equiangular and hence, ABCBDE
[D is the midpoint of BC]
(C) is the correct answer.
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
Ratio of areas = 
(D) is the correct answer.