Selina Solutions Concise Maths Class 10 Chapter 15 Similarity

 

Selina Solutions Concise Maths Class 10 Chapter 15 Similarity 

When two shapes are the same size but there is a difference in size, they are called the same size. This chapter deals with the similarity of triangles. Basics of understanding the corresponding sides and corresponding angles of congruent triangles, different terms for the symmetry of two triangles, the original proportionality theorem, the relationship between the areas of two triangles, the similarity in shape changes and finally maps and models. Applications are key concepts. Under this chapter. Students who find it difficult to solve problems in this chapter or other may refer to Selina Solutions for Class 10 Mathematics prepared by experienced faculty members at maths Pdf Solution . It is designed to build confidence among students who are preparing for their ICSE exam. Selina Solutions Concise Maths Class 10 Chapter 15 Parity (with Maps and Models Application) PDF is available in the link below.

Selina Solutions Concise Maths Class 10 Chapter 15 Similarity (With Applications to Maps and Models)

Exercise 15(A) Solutions

Exercise 15(B) Solutions

Exercise 15(C) Solutions

Exercise 15(D) Solutions

Exercise 15(E) Solutions

Access Selina Solutions Concise Maths Class 10 Chapter 15 Similarity (With Applications to Maps and Models)

Exercise 15(D) Page No: 229

1. A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A’ B’ C’ Calculate:

(i) the length of AB, if A’ B’ = 6 cm. 

(ii) the length of C’ A’ if CA = 4 cm. 

Solution:

Given that, Δ ABC has been enlarged by scale factor m of 2.5 to Δ A’B’C’.

(i) A’B’ = 6 cm

So,

AB(2.5) = A’B’ = 6 cm

AB = 2.4 cm

(ii) CA = 4 cm

We know that,

CA(2.5) = C’A’

C’A’ = 4 x 2.5 = 10 cm

2. A triangle LMN has been reduced by scale factor 0.8 to the triangle L’ M’ N’. Calculate:

(i) the length of M’ N’, if MN = 8 cm.

(ii) the length of LM, if L’ M’ = 5.4 cm.

Solution:

Given, Δ LMN has been reduced by a scale factor m = 0.8 to Δ L’M’N’.

(i) MN = 8 cm

So, MN (0.8) = M’N’

(8)(0.8) = M’N’

M’N’ = 6.4 cm

(ii) L’M’ = 5.4 cm

So, LM (0.8) = L’M’

LM (0.8) = 5.4

LM = 6.75 cm

3. A triangle ABC is enlarged, about the point 0 as centre of enlargement, and the scale factor is 3. Find:

(i) A’B’, if AB = 4 cm.

(ii) BC, if B’C’ = 15 cm.

(iii) OA, if OA’ = 6 cm

(iv) OC’, if OC = 21 cm

Also, state the value of: 

(a) OB’/OB (b) C’A’/CA

Solution:

Given that, Δ ABC is enlarged and the scale factor m = 3 to the Δ A’B’C’.

(i) AB = 4 cm

So, AB(3) = A’B’

(4)(3) = A’B’

A’B’ = 12 cm

(ii) B’C’ = 15 cm

So, BC(3) = B’C’

BC(3) = 15

BC = 5 cm

(iii) OA’ = 6 cm

So, OA (3) = OA’

OA (3) = 6

OA = 2 cm

(iv) OC = 21 cm

So, OC(3) = OC’

21 x 3 = OC’

OC’ = 63 cm

The ratio of the lengths of the two corresponding sides of two triangles.

Δ ABC is enlarged and the scale factor m = 3 to the Δ A’B’C’

Hence,

(a) OB’/OB = 3

(b) C’A’/CA = 3

Exercise 15(A) Solutions

Exercise 15(B) Solutions

Exercise 15(C) Solutions

Exercise 15(D) Solutions

Exercise 15(E) Solutions

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