# Triangles : Exercise - 6.4 (Mathematics NCERT Class 10th)

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**Q.1Â Â Â Â Let and their areas be, respectively, 64 and 121 cm2. If EF = 15.4 cm, Find BC.**

**Sol.**

We have,

cm = 11.2 cm

**Q.2Â Â Â Â 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.**

**Sol.**

In AOB and COD, we have

[Vert. opp. ]

and, [Alernate ]

Therefore by AA-criterion of similarity, we have

Hence, area : area = 4 : 1

**Q.3Â Â Â Â In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that**

**Sol.**

Given : Two ABC and DBC which stand on the same base but on the opposite sides of BC.

To prove :

Construction : Draw and

Proof : In AOE and DOF, we have

[Vertically opp. ]

Therefore by AA-criterion of similarity, we have

........... (1)

Now,

[Using (1)]

Therefore

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

**Sol. **

Given : Two ABC and DEF such that

and, Area () = Area ()

To prove :

Proof :

, ,

and,

To establish , it is sufficient to prove that

AB = DE, BC = EF and AC = DF

Now, Area = Area

AB = DE, BC = EF, AC = DF

Hence, .

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

**Sol. **

Since, D and E are the mid-points of the sides AB and BC of respectively.

Therefore DE || AC

DE || FC ............. (1)

Since, D and F are the mid-points of the sides AB and AC of respectively.

Therefore DF || BC DF || EC ............ (2)

From (1) and (2), we can say that DECF is a parallelogram.

Similarly, ADEF isÂ a parallelogram

Now, in DEF and ABC, we have

[Opp. of ||gm ADEF]

and, [Opp. of ||gm DECF]

Therefore by AA-criterion of similarity, we have

[Because DE =======Â ]

Hence, Area : Area = 1 : 4.

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

Sol.

Given : , AD and PM are the medians of ABC and PQR respectively.

To prove :

Proof : Since

................ (1)

But, .............. (2)

From (1) and (2), we have

**Q.7Â Â 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 the diagonals.**

**Sol. **

Given : A square ABCD. Equilateral BCE and have been drawn on side and the diagonals AC respectively.

To prove : Area

Proof :

[Being equilateral so similar by AAA criterion of similarity]

[Because Diagonal = side AC = BC]

**Tick the correct answer and justify :**

**Q.8Â Â Â 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**

Â Â Â Â Â Â Â Â Â Â Â **(a) 2 : 1**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(b) 1 : 2**

Â Â Â Â Â Â Â Â Â Â Â **(c) 4 : 1**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(d) 1 : 4**

**Sol. **

Since and are equilateral, they are equiangular and hence

[Because D is the mid-point of BC therefore BC = 2BD]

Therefore (C) is the correct answer.

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

Â Â Â Â Â Â Â Â Â Â Â Â Â **(A) 2 : 3**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(B) 4 : 9**

Â Â Â Â Â Â Â Â Â Â Â Â Â **(C) 81 : 16**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **(D) 16 : 81**

**Sol. **

Since the ratio of the areas of two similar triangles equal to the ratio of the squares of any two corresponding sides. Therefore,

ratio of areas =Â

Therefore (D) is the correct answer