Basic Proportionality theorem and its converse are widely discussed in this exercise. The problems of this exercise are well framed in order to give a deep understanding of this theorem. Selina Solutions for Class 10 Maths can be utilised by students to clarify any doubts in solving problems. All the solutions are developed by subject matter experts having vast academic experience. Avail Selina Solutions Concise Maths Class 10 Chapter 15 Similarity (With Applications to Maps and Models) Exercise 15(B) PDF, from the links provided below.
Selina Solutions Concise Maths Class 10 Chapter 15 Similarity (With Applications to Maps and Models) Exercise 15(B) Download PDF
Access 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.Â
(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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