In this chapter, we provide KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4 pdf, free KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4 book pdf download. Now you will get step by step solution to each question.

**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.

Solution:

If ∆ABC ~ ∆DEF, then

ar(ΔABC)ar(ΔDEF)=BC2EF2

64121=BC2(15.4)2

BC2=64121=BC2(15.4)2

BC2=64121×237.161

∴ 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.

Solution:

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

To prove: Ar(ΔAOB)Ar(ΔCOD)=?

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

∆AOB ~ ∆COD

Ar(ΔAOB)Ar(ΔCOD)=AB2DC2

=(2DC)2DC2(⋅AB=2DC)

=4DC21DC2

Ar(ΔAOB)Ar(ΔCOD)=41

= 4 : 1

Question 3.

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

Ar(ΔABC)Ar(ΔDBC)=AODO

Solution:

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

To Prove: Ar(ΔABC)Ar(ΔDBC)=AODO

Construction: Draw AP ⊥ BC and DM ⊥ BC.

Area of ∆ = 12 × base × height

Ar(ΔABC)Ar(ΔDBC)=12×BC×AP12×BC×DM

=APDM

In ∆APO and ∆DMO,

∠APO = ∠DMO = 90° (construction)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ∆APO ~ ∆DMO

(∵ Similarity criterion is A.A.A.)

APDM=AODO

⇒Ar(ΔABC)Ar(ΔDBC)=AODO

Question 4.

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

Solution:

In ∆ABC ~ ∆PQR

Ar(ΔABC)Ar(ΔPQR)=(ABPQ)2=(BCQR)2=(ACPR)2 …………. (1)

Area of ∆ABC = Area of ∆PQR

⇒Ar(ΔABC)Ar(ΔPQR)=1 …………. (2)

Substituting the eqn (2) in eqn (1)

1=(ABPQ)2=(BCQR)2=(ACPR)2

⇒ 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.

Solution:

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

DE∥ACDE=12AC

EF∥ABEF=12AB

DF∥BCDF=12BC

Ar(ΔDEF)Ar(ΔABC)=DF2BC2(Theorem6)

Ar(ΔDEF)Ar(ΔABC)=(12BC)2BC2

Ar(ΔDEF)Ar(ΔABC)=14BC2BC1=14=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:

Let ∆ABC ~ ∆PQR

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

∆ABC ~ ∆PQR

ABPQ=BCQR=ACPR ………………… (1)

∠A = ∠P, ∠R = ∠Q, and ∠C = ∠R (2)

AD and PS are mid-point lines

BD=DC=BC2

and QS=SR=QR2

ABPQ=BDQS=ACPR …………… (3)

In ∆ABD and ∆PQS,

∠B = ∠Q ………. Eqn (2)

and ABPQ=BDQS ………. Eqn (3)

∴ ∆ABD ~ ∆PQS (SAS Postulate)

ABPQ=BDQS=ADPS ……………. (4)

Ar(ΔABC)Ar(ΔPQR)=(ABPQ)2=(BCQR)2=(ACPR)2

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

ABPQ=BCQR=ACPR=ADPS

⇒Ar(ΔBC)Ar(ΔPQR)=(ADPS)2

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.

Solution:

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

To Prove: (ΔABE)=12 Ar (ΔACF)

Ar(ΔABE)Ar(ΔACF)=AB2AC2Theorem6

=AB2AB2+BC2

∵ Pythagoras theorem

=AB22AB2

∴ AB = AC sides of square

=1×AB22×AB2

Ar(ΔABE)Ar(ΔACF)=12

**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

Solution:

C) 4 : 1

Ar(ΔABC)Ar(ΔBDE)=BC2BD2=BC2(12BC)2

=11/4=41

= 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

Solution:

D) 16 : 81

∴ (4)^{2} : (9)^{2} = 16 : 81.

**All Chapter KSEEB Solutions For Class 10 Maths**

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