# KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4

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

Question 1.
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm2 and 121 cm2, 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)
Ar(ΔABC)Ar(ΔPQR)=(ABPQ)2=(BCQR)2=(ACPR)2
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. 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.

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