Students can Download Class 10 Maths Chapter 2 Triangles Ex 2.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Ex 2.4

Question 1.

Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm^{2} and 121 cm^{2}, If EF = 15.4 cm., find BC.

Answer:

If ∆ABC ~ ∆DEF, that

∴ BC = 8 × 1.4

∴ BC = 11.2 cm

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.

Answer:

Data: Diagonals of a trapezium ABCD with AB||DC intersect each other at the point O. We have AB = 2 CD.

To prove:

As per similarity criterion, ∆AOB and ∆COD are,

∆AOB ~ ∆COD

= 4 : 1

Question 3.

In the given figure. 2.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Answer:

Data: ABC and DBC are two triangles on the same base BC. If AD intersects BC at O.

To Prove:

Construction: Draw AP ⊥ BC and DM ⊥ BC.

Area of ∆ = \(\frac{1}{2}\) × base × height

In ∆APO and ∆DMO,

∠APO = ∠DMO = 90° (construction)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ∆APO ~ ∆DMO

(∵ Similarity criterion is A.A.A.)

Question 4.

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

Answer:

In ∆ABC ~ ∆PQR

Area of ∆ABC = Area of ∆PQR )

Substituting the eqn (2) in eqn (1)

⇒ AB = PQ, BC = QR, AC = PR

∴ ∆ABC ≅ ∆PQR (∵ S.S.S postulate)

Question 5.

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

Answer:

Data: D. E and F are respectively midpoints of sides AB, BC and CA of ∆ABC.

To Prove: Ratio of the areas of ∆DEF and ∆ABC.

If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side and it is half of that side.

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.

Answer:

Let ∆ABC ~ ∆PQR

AD and PS are mid-point lines of ∆ABC and ∆PQR.

∆ABC ~ ∆PQR

∠A = ∠P, ∠R = ∠Q, and ∠C = ∠R (2)

AD and PS are mid-point lines

In ∆ABD and ∆PQS,

∠B = ∠Q ………. Eqn (2)

∴ ∆ABD ~ ∆PQS (SAS Postulate)

From Eqn. (1) and Eqn. (4), we have

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 its diagonals.

Answer:

Data: Equilateral triangle ABE is on AB of square ABCD. Equilateral triangle ACF is on diagonal AC.

To Prove:

∵ Pythagoras theorem

∴ AB = AC sides of square

Tick the correct answer and justify:

Question 8.

∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point 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

Answer:

C) 4 : 1

= 4 : 1

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

Answer:

D) 16 : 81

∴ (4)^{2} : (9)^{2} = 16 : 81.