Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.4 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions Chapter 6 Triangles Ex 6.4

Question 1.

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

Solution:

Question 2.

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.

Solution:

In ∆s AOB and COD, we have :

∠AOB = ∠COD [Vert.opp.∠s]

and ∠OAB = ∠OCD [Alternate ∠s]

∴ By AA criterion of similarity, we have:

Hence, area (∆ AOB) : area (∆ COD) = 4 : 1

Question 3.

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

= \(\frac { AO }{ DO }\).

Solution:

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

To prove :

= \(\frac { AO }{ DO }\).

Construction : Draw AE ⊥ BC and DF ⊥ BC.

Proof:

In ∆s AOE and DOF, we have :

∠AEO = ∠DFO = 90°

∠AOE = ∠DOF [Vertically opp. ∠s]

∴ By AA criterion of similarity, we have :

Question 4.

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

Solution:

Given : Two ∆s ABC and DEF such that ∆ ABC – ∆ DEF

and Area (∆ ABC) = Area (∆ DEF)

To prove : ∆ ABC ≅ ∆ DEF

Proof:

∆ ABC ~ ∆ DEF

So, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

and \(\frac { AB }{ DE }\) = \(\frac { BC }{ EF }\) = \(\frac { AC }{ DF }\)

To establish ∆ ABC ≅ ∆ DEF, it is sufficient to prove that

Hence ∆ ABC ≅ ∆ DEF [By SSS]

Question 5.

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

Solution:

Since D and E are respectively, the mid-points of the sides AB and BC of ∆ ABC, therefore

DE [| AC, or DE || FC … (1)

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

∆ ABC, therefore

DF || BC, or DF || EC … (2)

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

Similarly, ADEF is a parallelogram.

Now, in ∆s DEF and ABC, we have :

∠DEF = ∠A [Opp. Zs of ||gm ADEF]

and ∠EDF = ∠C [Opp. Zs of ||gm DECF]

∴ By AA criterion of similarity, we have :

∆ DE ≅ ∆ AC

Hence, Area (∆ DEF) : Area (∆ ABC) =1:4.

Question 6.

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

Solution:

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

To Prove : \(\frac{\text { Area }(\Delta \mathrm{ABC})}{\text { Area }(\Delta \mathrm{PQR})}=\frac{\mathrm{AD}^{2}}{\mathrm{PM}^{2}}\)

Proof : Since ∆ ABC ~ ∆ PQR, therefore

Question 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.

Solution:

Given : A square ABCD. Equilateral ∆s BCE and ACF have been drawn on side BC and the diagonal AC respectively.

To prove : Area (∆ BCE) = \(\frac { 1 }{ 2 }\)(Area ∆ ACF)

Proof: ∆ BCE ~ ∆ ACF [All equilateral triangles are similar]

Question 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

Solution:

Since ∆ ABC and ∆ BDE are equilateral triangles, they are equiangular and hence ∆ ABC – ∆ BDE

∴ (C) is the correct answer.

Question 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:

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 = (4)² : (9)² = 16 : 81

∴ (D) is the correct answer.