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NCERT Solutions for Class 10 Maths Exercise 6.4

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NCERT solutions for Maths Triangles

NCERT Solutions for Class 10 Maths Triangles

1. Let ABC DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.

Ans. We have,

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.

Ans. In s AOB and COD, we have,

AOB = COD[Vertically opposite angles]

OAB = OCD[Alternate angles]

By AA-criterion of similarity,

AOBCOD

Hence, Area (AOB) : Area (COD) = 4 : 1

NCERT Solutions for Class 10 Maths Exercise 6.4

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

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

To Prove:

Construction: Draw AEBC 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)]

NCERT Solutions for Class 10 Maths Exercise 6.4

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

Ans. Given: Two s ABC and DEF such that ABCDEF

And Area(ABC) = Area (DEF)

To Prove: ABCDEF

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

NCERT Solutions for Class 10 Maths Exercise 6.4

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

Ans. Since, D and E are the mid-points of the sides BC and CA of ABC respectively.

DEBA DEFA ……….(i)

Since, D and F are the mid-points 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

NCERT Solutions for Class 10 Maths Exercise 6.4

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

Ans. 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,

NCERT Solutions for Class 10 Maths Exercise 6.4

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.

Tick the correct answer and justify:

Ans. Given: A square ABCD,

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]

NCERT Solutions for Class 10 Maths Exercise 6.4

8. 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:

(A) 2: 1

(B) 1: 2

(C) 4: 1

(D) 1: 4

Ans. (C) Since ABC and BDE are equilateral, they are equiangular and hence,

ABCBDE

[D is the mid-point of BC]

NCERT Solutions for Class 10 Maths Exercise 6.4

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

Ans. (D) Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. Therefore,

Ratio of areas =

NCERT Solutions for Class 10 Maths Exercise 6.4

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11 thoughts on “NCERT Solutions for Class 10 Maths Exercise 6.4”

1. Very nice

2. Very useful……..

3. Hey, the fig. of Q 5 is wrong.Have a look at that.

4. Diagram of question 5 is wrong. Point F goes with AC not AB. Have a check guys. Do the correction or kids may suffer.

5. thaks team for ice crispy solutios

6. nyc

7. I LIKE IT SO MUCH

8. Good ?

9. fig.of ques.5 is wrong but allover nic

10. Good solve and very nice