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.

Exercise 15(A) Solutions

Exercise 15(B) Solutions

Exercise 15(C) Solutions

Exercise 15(D) Solutions

Exercise 15(E) Solutions

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

Exercise 15(B) Page No: 218

1. In the following figure, point D divides AB in the ratio 3: 5. Find:

(i) AE/EC (ii) AD/AB (iii) AE/AC

Also if,

(iv) DE = 2.4 cm, find the length of BC.

(v) BC = 4.8 cm, find the length of DE.

Solution:

(i) Given, AD/DB = 3/5

And DE || BC.

So, by Basic Proportionality theorem, we have

AD/DB = AE/EC

AE/EC = 3/5

(ii) Given, AD/DB = 3/5

So, DB/AD = 5/3

Adding 1 both sides, we get

DB/AD + 1 = 5/3 + 1

(DB + AD)/ AD = (5 + 3)/ 3

AB/AD = 8/3

Therefore,

AD/AB = 3/8

(iii) In ∆ABC, as DE || BC

By BPT, we have

AD/DB = AE/ EC

So, AD/AB = AE/AC

From above, we have AD/AB = 3/8

Therefore,

AE/AC = 3/8

(iv) In ∆ADE and ∆ABC,

∠ADE = ∠ABC [As DE || BC, corresponding angles are equal.]

∠A = ∠A [Common angle]

Hence, ∆ADE ~ ∆ABC by AA criterion for similarity

So, we have

AD/AB = DE/BC

3/8 = 2.4/BC

BC = 6.4 cm

(v) Since, ∆ADE ~ ∆ABC by AA criterion for similarity

So, we have

AD/AB = DE/BC

3/8 = DE/4.8

DE = 1.8 cm

2. In the given figure, PQ ‖ AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find:

(i) CP/PA (ii) PQ (iii) If AP = x, then the value of AC in terms of x.

Solution:

(i) In ∆CPQ and ∆CAB,

∠PCQ = ∠APQ [As PQ || AB, corresponding angles are equal.]

∠C = ∠C [Common angle]

Hence, ∆CPQ ~ ∆CAB by AA criterion for similarity

So, we have

CP/CA = CQ/CB

CP/CA = 4.8/ 8.4 = 4/7

Thus, CP/PA = 4/3

(ii) As, ∆CPQ ~ ∆CAB by AA criterion for similarity

We have,

PQ/AB = CQ/CB

PQ/6.3 = 4.8/8.4

PQ = 3.6 cm

(iii) As, ∆CPQ ~ ∆CAB by AA criterion for similarity

We have,

CP/AC = CQ/CB

CP/AC = 4.8/8.4 = 4/7

So, if AC is 7 parts and CP is 4 parts, then PA is 3 parts.

Hence, AC = 7/3 x PA = (7/3)x

3. A line PQ is drawn parallel to the side BC of Δ ABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.

Solution:

In ∆ APQ and ∆ ABC,

∠APQ = ∠ABC [As PQ || BC, corresponding angles are equal.]

∠PAQ = ∠BAC [Common angle]

Hence, ∆APQ ~ ∆ABC by AA criterion for similarity

So, we have

AP/AB = AQ/AC

AP/9 = 4.2/6

Thus,

AP = 6.3 cm

4. In Δ ABC, D and E are the points on sides AB and AC respectively.

Find whether DE ‖ BC, if

(i) AB = 9cm, AD = 4cm, AE = 6cm and EC = 7.5cm. 

(ii) AB = 6.3 cm, EC = 11.0 cm, AD =0.8 cm and EA = 1.6 cm.

Solution:

(i) In ∆ ADE and ∆ ABC,

AE/EC = 6/7.5 = 4/5

AD/BD = 4/5 [BD = AB – AD = 9 – 4 = 5 cm]

So, AE/EC = AD/BD

Therefore, DE || BC by the converse of BPT.

(ii) In ∆ ADE and ∆ ABC,

AE/EC = 1.6/11 = 0.8/5.5

AD/BD = 0.8/5.5 [BD = AB – AD = 6.3 – 0.8 = 5.5 cm]

So, AE/EC = AD/BD

Therefore, DE || BC by the converse of BPT.

5. In the given figure, Δ ABC ~ Δ ADE. If AE: EC = 4: 7 and DE = 6.6 cm, find BC. If ‘x’ be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of ‘x’.

Solution:

Given,

Δ ABC ~ Δ ADE

So, we have

AE/AC = DE/BC

4/11 = 6.6/BC

BC = (11 x 6.6)/ 4 = 18.15 cm

And, also

As Δ ABC ~ Δ ADE, we have

∠ABC = ∠ADE and ∠ACB = ∠AED

So, DE || BC

And, AB/AD = AC/AE = 11/4 [Since, AE/EC = 4/7]

In ∆ ADP and ∆ ABQ,

∠ADP = ∠ABQ [As DP || BQ, corresponding angles are equal.]

∠APD = ∠AQB [As DP || BQ, corresponding angles are equal.]

Hence, ∆ADP ~ ∆ABQ by AA criterion for similarity

AD/AB = AP/AQ

4/11 = x/AQ

Thus,

AQ = (11/4)x

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Access other exercises of 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

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