This exercise contains problems based on all the concepts discussed in this Chapter. Moreover, this exercise concludes the Chapter with different models of problems. The Selina Solutions for Class 10 Maths is the right resource for every student who is looking to improve his/her ability of solving problems and to get a strong grip over the concepts of ICSE Class 10 Maths. For a better understanding of the concepts, students can make use of Concise Selina Solutions for Class 10 Maths Chapter 15 Similarity (With Applications to Maps and Models) Exercise 15(E) PDF, which is provided in the links given below.
Selina Solutions Concise Maths Class 10 Chapter 15 Similarity (With Applications to Maps and Models) Exercise 15(E) Download PDF
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Exercise 15(E) Page No: 229
1. In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5 cm and BC = 18 cm.
Find:
(i) AY/YC (ii) YC/AC (iii) XY
Solution:
Given, XY || BC.
So, In Δ AXY and Δ ABC
∠AXY = ∠ABC [Corresponding angles]
∠AYX = ∠ACB [Corresponding angles]
Hence, ∆AXY ~ ∆ABC by AA criterion for similarity.
As corresponding sides of similar triangles are proportional, we have
(i) AX/AB = AY/AC
9/13.5 = AY/AC
AY/YC = 9 / 4.5
AY/YC = 2
AY/YC = 2/1
(ii) We have,
AX/AB = AY/AC
9/13.5 = AY/ AC
YC/AC = 4.5/13.5 = 1/3
(iii) As, ∆AXY ~ ∆ABC
AX/AB = XY/BC
9/13.5 = XY/18
XY = (9 x 18)/ 13.5 = 12 cm
2. In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE = 15 cm,
Calculate:
(i) EC (ii) AF (iii) PE
Solution:
(i) In Δ AEB and Δ FEC,
∠AEB = ∠FEC [Vertically opposite angles]
∠BAE = ∠CFE [Since, AB||DC]
Hence, ∆AEB ~ ∆FEC by AA criterion for similarity.
So, we have
AE/FE = BE/EC = AB/FC
15/EC = 9/13.5
EC = 22.5 cm
(ii) In Δ APB and Δ FPD,
∠APB = ∠FPD [Vertically opposite angles]
∠BAP = ∠DFP [Since, AB||DF]
Hence, ∆APB ~ ∆FPD by AA criterion for similarity.
So, we have
AP/FP = AB/FD
6/FP = 9/31.5
FP = 21 cm
So, AF = AP + PF = 6 + 21 = 27 cm
(iii) We already have, ∆AEB ~ ∆FEC
So,
AE/FE = BE/CE = AB/FC
AE/FE = 9/13.5
(AF – EF)/ FE = 9/13.5
AF/EF – 1 = 9/13.5
27/EF = 9/13.5 + 1 = 22.5/ 13.5
EF = (27 x 13.5)/22.5 = 16.2 cm
Now, PE = PF – EF = 21 – 16.2 = 4.8 cm
3. In the following figure, AB, CD and EF are perpendicular to the straight line BDF.
If AB = x and; CD = z unit and EF = y unit, prove that:
1/x + 1/y = 1/z
Solution:
In Δ FDC and Δ FBA,
∠FDC = ∠FBA [As DC || AB]
∠DFC = ∠BFA [common angle]
Hence, ∆FDC ~ ∆FBA by AA criterion for similarity.
So, we have
DC/AB = DF/BF
z/x = DF/BF …. (1)
In Δ BDC and Δ BFE,
∠BDC = ∠BFE [As DC || FE]
∠DBC = ∠FBE [Common angle]
Hence, ∆BDC ~ ∆BFE by AA criterion for similarity.
So, we have
BD/BF = z/y ….. (2)
Now, adding (1) and (2), we get
BD/BF + DF/BF = z/y + z/x
1 = z/y + z/x
Thus,
1/z = 1/x + 1/y
– Hence Proved
4. Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: AB/PQ = AD/PM.
Solution:
Given, ∆ABC ~ ∆PQR
AD and PM are the medians, so BD = DC and QM = MR
So, we have
AB/PQ = BC/QR [Corresponding sides of similar triangles are proportional.]
Then,
AB/PQ = (BC/2)/ (QR/2) = BD/QM
And, ∠ABC = ∠PQR i.e. ∠ABD = ∠PQM
Hence, ∆ABD ~ ∆PQM by SAS criterion for similarity.
Thus,
AB/PQ = AD/PM.
5. Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that: AB/PQ = AD/PM.
Solution:
Given, ∆ABC ~ ∆PQR
So,
∠ABC = ∠PQR i.e. ∠ABD = ∠PQM
Also, ∠ADB = ∠PMQ [Both are right angles]
Hence, ∆ABD ~ ∆PQM by AA criterion for similarity.
Thus,
AB/PQ = AD/PM
6. Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: AB/PQ = AD/PM
Solution:
Given, ∆ABC ~ ∆PQR
And, AD and PM are the angle bisectors.
So,
∠BAD = ∠QPM
Also, ∠ABC = ∠PQR i.e. ∠ABD = ∠PQM
Hence, ∆ABD ~ ∆PQM by AA criterion for similarity.
Thus,
AB/PQ = AD/PM
7. In the following figure, ∠AXY = ∠AYX. If BX/AX = CY/AY, show that triangle ABC is isosceles.
Solution:
Given,
∠AXY = ∠AYX
So, AX = AY [Sides opposite to equal angles are equal.]
Also, from BPT we have
BX/AX = CY/AY
Thus,
AX + BX = AY + CY
So, AB = AC
Therefore, ∆ABC is an isosceles triangle.
8. In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown.Â
Prove that: AB/BC = PQ/QR
Solution:
Let join AR such that it intersects BQ at point X.
In ∆ACR, BX || CR. By BPT, we have
AB/BC = AX/XR … (1)
In ∆APR, XQ || AP. By BPT, we have
PQ/QR = AX/XR … (2)
From (1) and (2),
AB/BC = PQ/QR
– Hence Proved
9. In the following figure, DE || AC and DC || AP. Prove that: BE/EC = BC/CP.
Solution:
Given, DE || AC
So,
BE/EC = BD/DA [By BPT]
And, DC || AP
So,
BC/CP = BD/DA [By BPT]
Therefore,
BE/EC = BC/CP
10. In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.
Calculate: (i) EF (ii) AC
Solution:
(i) In ∆PCD and ∆PEF,
∠CPD = ∠EPF [Vertically opposite angles]
∠DCE = ∠FEP [As DC || EF, alternate angles.]
Hence, ∆PCD ~ ∆PEF by AA criterion for similarity.
So, we have
27/EF = 15/7.5
Thus,
EF = 13.5
(ii) And, as EF || AB
∆CEF ~ ∆CAB by AA criterion for similarity.
EC/AC = EF/AB
22.5/AC = 13.5/22.5
Thus, AC = 37.5 cm
11. In ΔABC, ∠ABC = ∠DAC, AB = 8 cm, AC = 4 cm and AD = 5 cm.
(i) Prove that ΔACD is similar to ΔBCA.
(ii) Find BC and CD
(iii) Find the area of ΔACD: area of ΔABC
Solution:
(i) In ∆ACD and ∆BCA,
∠DAC = ∠ABC [Given]
∠ACD = ∠BCA [Common angles]
Hence, ∆ACD ~ ∆BCA by AA criterion for similarity.
(ii) Since, ∆ACD ~ ∆BCA
We have,
AC/BC = CD/CA = AD/AB
4/BC = CD/4 = 5/8
4/BC = 5/8
So, BC = 32/5 = 6.4 cm
And,
CD/4 = 5/8
Thus, CD = 20/8 = 2.5 cm
(iii) As, ∆ACD ~ ∆BCA
We have,
Ar(∆ACD)/ Ar(∆BCA) = AD2/ AB2 = 52/82
Ar(∆ACD)/ Ar(∆BCA) = 25/64
12. In the given triangle P, Q and R are mid-points of sides AB, BC and AC respectively. Prove that triangle QRP is similar to triangle ABC.
Solution:
In ∆ABC, as PR || BC by BPT we have
AP/PB = AR/RC
And, in ∆PAR and ∆BAC,
∠PAR = ∠BAC [Common]
∠APR = ∠ABC [Corresponding angles]
Hence, ∆PAR ~ ∆BAC by AA criterion for similarity
So, we have
PR/BC = AP/AB
PR/BC = ½ [Since, P is the mid-point of AB]
PR = ½ BC
Similarly,
PQ = ½ AC
RQ = ½ AB
So,
PR/BC = PQ/AC = RQ/AB
Therefore,
∆QRP ~ ∆ABC by SSS similarity.
13. In the following figure, AD and CE are medians of ∆ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB,
(ii) AG: GD = 2: 1
Solution:
(i) In ∆BFD and ∆BEC,
∠BFD = ∠BEC [Corresponding angles]
∠FBD = ∠EBC [Common]
Hence, ∆BFD ~ ∆BEC by AA criterion for similarity.
So,
BF/BE = BD/BC
BF/BE = ½ [Since, D is the mid-point of BC]
BE = 2BF
BF = FE = 2BF
Thus,
EF = FB
(ii) In ∆AFD, EG || FD and using BPT we have
AE/EF = AG/GD …. (1)
Now, AE = EB [Since, E is the mid-point of AB]
AE = 2EF [As, EF = FB, by (1)]
So, from (1) we have
AG/GD = 2/1
Therefore, AG: GD = 2: 1
14. Two similar triangles are equal in area. Prove that the triangles are congruent.
Solution:
Let’s consider two similar triangles as ∆ABC ~ ∆PQR
So,
Ar(∆ABC)/ Ar(∆PQR) = (AB/PQ)2 = (BC/QR)2 = (AC/PR)2
Since,
Area of ∆ABC = Area of ∆PQR [Given]
Hence,
AB = PQ
BC = QR
AC = PR
So, as the respective sides of two similar triangles are all of same length.
We can conclude that,
∆ABC ≅ ∆PQR [By SSS rule]
– Hence Proved
15. The ratio between the altitudes of two similar triangles is 3: 5; write the ratio between their:
(i) medians. (ii) perimeters. (iii) areas.
Solution:
The ratio between the altitudes of two similar triangles is same as the ratio between their sides.
So,
(i) The ratio between the medians of two similar triangles is same as the ratio between their sides.
Thus, the required ratio = 3: 5
(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides.
Thus, the required ratio = 3: 5
(iii) The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.
Thus, the required ratio = (3)2: (5)2 = 9: 25
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