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
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm2 and 121 cm2, If EF = 15.4 cm., find BC.
If ∆ABC ~ ∆DEF, then
∴ BC = 8 × 1.4
∴ BC = 11.2 cm
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.
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
= 4 : 1
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
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
In ∆APO and ∆DMO,
∠APO = ∠DMO = 90° (construction)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ∆APO ~ ∆DMO
(∵ Similarity criterion is A.A.A.)
If the areas of two similar triangles are equal, prove that they are congruent.
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)
⇒ AB = PQ, BC = QR, AC = PR
∴ ∆ABC ≅ ∆PQR (∵ S.S.S postulate)
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.
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
Prove that the ratio of the areas , of two similar triangles is equal to the square of the ratio of their corresponding medians.
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
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)
From Eqn. (1) and Eqn. (4), we have
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.
Data: Equilateral triangle ABE is on AB of square ABCD. Equilateral triangle ACF is on diagonal AC.
To Prove: (ΔABE)=12 Ar (ΔACF)
∵ Pythagoras theorem
∴ AB = AC sides of square
Tick the correct answer and justify:
∆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
C) 4 : 1
= 4 : 1
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
D) 16 : 81
∴ (4)2 : (9)2 = 16 : 81.
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