Similarity Of Triangles Class 10th Mathematics Gujarat Board Solution

Class 10th Mathematics Gujarat Board Solution
Exercise 6.1
  1. According to the definition of similarity of triangles, which are the conditions…
  2. For ΔPQR and ΔXYZ, the correspondence PQR ↔ YZX is a similarity. m∠P = 2m∠Q and…
  3. The correspondence ABC ↔ PQR between Δ ABC and Δ PQR is a similarity. AB : PQ =…
  4. Δ PQR ~ ΔDEF for the correspondence PQR ↔ EDF. If PQ + QR = 15, DE + DF = 10 and…
  5. In ΔABC and ΔPQR, ABC ↔ QPR is a similarity. The perimeter of AABC is 15 and…
  6. In ΔXYZ ~ ΔDEF consider the correspondence XYZ ↔ EDF. If deltaxyz/deltadef = 3/4…
  7. Using the definition of similarity prove that all the isosceles right angled…
  8. For ΔABC and ΔXYZ, ABC ↔ XYZ is a similarity. If ab/4 = bc/6 = ac/3 , AC = 3 and…
  9. State whether the following statements are true or false. Give reasons for your…
  10. ΔABC ~ ΔPQR for the correspondence ABC ↔ QRP. If m∠A = 50, m∠C = 30, then m∠R…
  11. ΔLMN ~ ΔXYZ for the correspondence LMN ↔ ZYX. If m∠Z = 50, m∠X = 40, then m∠L…
  12. If the correspondence ABC ↔ EFD is a similarity in AABC and ADEF, then of the…
  13. ABC ~ PQR is a similarity in ΔABC and ΔPQR. If the perimeter of ΔABC is 12 and…
Exercise 6.2
  1. In ΔABC, a line parallel to bar bc intersects bar ab and bar ac in D and E…
  2. In ΔABC, the bisector of ∠C intersects bar ab in F. If 2AF = 3FB and AC = 7.2…
  3. In ΔXYZ, the bisector of ∠Y intersects bar zx in P. (1) If XP : PZ = 4 : 5 and…
  4. In Δ ABC, the bisector of ∠A intersects bar bc in D. Prove that bd = bc x…
  5. □ ABCD is a trapezium such that vector ab || bar cd . m in bar ad and n in bar…
  6. In ΔABC, D and E are the mid-points of bar bc and bar ac respectively. bar ad…
  7. In ΔPQR, x in bar qr , such that Q-X-R. A line parallel to bar pr and passing…
  8. In ΔABC, x in bar bc and B-X-C. A line passing through X and parallel to vector…
  9. In ΔABC, the bisector of ∠A intersects bar bc in D and the bisector of ∠ADC…
  10. In ΔABC, D is the mid-point of bar bc and P is the mid-point of bar ad . vector…
Exercise 6.3
  1. In figure 6.27, Prove that BR = CS.
  2. □ ABCD is a parallelogram. Prove that ABD and BDC are similar.
  3. In □ABCD, M and N are the mid-points of bar ad and bar bc . If vector ab ∥ bar…
  4. In □ ABCD, A—P—D, B—Q—C. If AB ∥ PQ and PQ || DC, prove that AP X QC = PD X BQ.…
  5. In Δ ABC, the bisector of ∠ A intersects BC in D. The bisector of ∠ADB…
  6. In Δ ABC, Δ PQR and ΔXYZ correspondences ABC ↔ PQR, PQR ↔ XYZ are similarity.…
  7. State giving reasons, whether the following statements are true or false: In all…
Exercise 6.4
  1. ∠B is a right angle in ΔABC and bar bd is an altitude to hypotenuse. AB = 8, BC…
  2. In □m ABCD, T ∈ bar bc and vector dt intersects bar bd in M and vector dc in O.…
  3. In □m ABCD, M is the mid-point of bar bc . vector dm and vector ab intersect in…
  4. P and Q are the mid-points of vector ab and bar ac in ΔABC. If the area of ΔAPQ…
  5. □ABCD is a rhombus. vector ac integrate erdection bar bd = {0}. Prove that the…
  6. In □PQRS, bar pr integrate erdection bar qs = {T}, PS = QR, bar pq | bar rs .…
  7. In ΔABC, a line parallel to bar bc , passes through the mid-point of vector ab .…
  8. P, Q, R are the mid-points of the sides of ΔABC. X, Y, Z are the mid-points of…
  9. In Δ ABC, x in bar bc , y in bar c_a , z in bar ab such that bar xy | vector ab…
  10. Two triangles are similar. Prove that if sides in one pair of corresponding…
Exercise 6
  1. In ΔABC, p in bar ab such that ap/pb = m/n , m , n are positive real numbers. A…
  2. D, E and F are the mid-points of bar bc , bar c_sa and vector ab respectively in…
  3. Can two similar triangles have same area? If yes, in which case they have the…
  4. The correspondence ABC ↔ DEF is similarity in ΔABC and ΔDEF. bar am is an…
  5. Explain with reasons, whether the following statements are true or false: (1)…
  6. bar ad and bar be are the altitudes of ΔABC. If AB = 12, AC = 9.9, AD = 8.1 and…
  7. D, E, F are respectively the mid-points of bar pq , bar qr , bar pr of ΔPQR.…
  8. In ΔABC, bar am and bar cn are altitudes. If AB = 12, BC = 15 and AM = 9.6,…
  9. Areas of two similar triangles are 25 and 16. The ratio of the perimeters of…
  10. Area of ΔABC = 36 and area of ΔPQR = 64. The correspondence ABC ↔ PQR is a…
  11. In ΔABC, A—M—B, A—N—C and bar mn || bar bc . If AM = 8.4, AN = 6.4, CN = 19.2,…
  12. The correspondence ABC ↔ PQR is a similarity in Δ ABC and ΔPQR. AB = 16, AC =…
  13. The correspondence ABC ↔ XYZ is a similarity in ΔABC and ΔXYZ. ABC = 72, BC =…
  14. □ ABCD is trapezium in which bar ad || bar bc . The diagonals intersect in P.…
  15. In ΔABC, m∠B = 90 and bar bd is an altitude. The correspondence BDA ↔ _____…
  16. The correspondence ABC ↔ ZXY is a similarity in ΔABC and ΔXYZ. If AB = 12, BC =…
  17. ABC ↔ DEF is a similarity in ΔABC and ΔADEF, m∠A = 40, m∠E + m∠F =A. 40 B. 80…
  18. In ΔABC, M ∈ bar ab , N ∈ bar ac such that bar mn || bar bc ,……….is not true.A.…
  19. In ΔABC, B—M—C and A—N—C, bar mn || bar ab . If NC : NA = 1 : 3 and CM = 4,…
  20. In ΔXYZ and ΔPQR, XYZ ↔ PQR is similarity. XY = 12, YZ = 8, ZX = 16, PR = 8.…
  21. The bisector of ∠P in ΔPQR intersects bar qr in D. If QD: RD = 4 : 7 and PR =…
  22. The lengths of the sides bar bc , bar ca , bar ab of ΔABC are in the ratio 3 :…
  23. The correspondence ABC ↔ YZX in ΔABC and ΔXYZ is similarity. m∠B + m∠C = 120.…
  24. Correspondence ABC ↔ DEF of bar ac ABC and ΔDEF is similarity. If AB + BC = 10…
  25. The lengths of the sides of ΔDEF are 4, 6, 8. ΔDEF ~ ΔPQR for correspondence…
  26. The bisector of ∠B intersects bar ac in D. If BA = 12 and BC = 16, AD = 9, then…
  27. In Δ ABC, bar pq || bar bc , P ∈ bar ab , Q ∈ bar ac . If 4AP = AB and AQ = 4,…
  28. In ΔABC, the correspondence ABC ↔ BAC is similarity …….. of the following is…
  29. In ΔABC, A—P—B, A—Q—C and bar pq || bar bc . If PQ = 5, AP = 4, AB = 12, then…
  30. In Δ PQR, P—M—Q and P—N—R. If PQ = 18, PM = 12, PR = 9 and NR = …, then bar mn…

Exercise 6.1
Question 1.

According to the definition of similarity of triangles, which are the conditions for correspondence DEF ↔ ZXY between ΔDEF and ΔXYZ to be a similarity?


Answer:

Given, ∆DEF ↔ ∆ZXY,


There are mainly two conditions for correspondence DEF ZXY between DEF and XYZ to be a similarity.


⟹The corresponding angles are congruent.


i.e. m∠D = m∠X ∠D ≅ ∠X


m∠E = m∠Y ∠E ≅ ∠Y


m∠F = m∠Z ∠F ≅ ∠Z


⟹The lengths of the corresponding sides are in proportion.


i.e. 



Question 2.

For ΔPQR and ΔXYZ, the correspondence PQR ↔ YZX is a similarity. m∠P = 2m∠Q and m∠X = 120. Find m∠Y.


Answer:

Given, ΔPQR and ΔXYZ

P = 2Q and X = 120


If the correspondence PQRYZX is a similarity then


m∠P = m∠X ... (i)


m∠Q = m∠Y ... (ii)


m∠R = m∠Z ... (iii)


From eq. (i) and (ii)


∠P = 120


Because mP = 2mQ


Hence, 2Q = 120


Q = 60


And ∠Q = ∠Y


Hence, ∠Y = 60.



Question 3.

The correspondence ABC ↔ PQR between Δ ABC and Δ PQR is a similarity. AB : PQ = 4 : 5. If AC = 6, then find PR. If QR = 15, then find BC.


Answer:

Given, ΔABC and ΔPQR

And AB : PQ = 4 : 5


AC = 6 and QR = 15


The correspondence ABC  PQR between ABC and PQR is a similarity.


Hence, 



PR = 


PR = 7.5


And 



BC = 


PR = 12



Question 4.

Δ PQR ~ ΔDEF for the correspondence PQR ↔ EDF. If PQ + QR = 15, DE + DF = 10 and PR = 6, find EF.


Answer:

Given, PQR and DEF are similar for the correspondence PQR  EDF.

PQ + QR = 15 and DE + DF = 10


PR = 6


If the correspondence PQRDEF is a similarity.


Hence, 


And 


From these above two equations




EF 


EF = 4



Question 5.

In ΔABC and ΔPQR, ABC ↔ QPR is a similarity. The perimeter of AABC is 15 and perimeter of ΔPQR is 27. If BC = 7 and QR = 9, find PR and AC.


Answer:

Given, In ABC and PQR, ABCQPR is a similarity

The perimeter of ABC is 15 and perimeter of PQR is 27


And BC = 7 and QR = 9


AB + BC + CA = 15


AB + 7 + CA = 15


AB + CA = 8 ...... (i)


PQ + QR + RP = 27


PQ + 9 + RP = 27


PQ + RP = 18 ...... (ii)


Due to similarity 



Hence, 




PR = 


PR = 14.4


Hence, 



AC = 


AC = 5



Question 6.

In ΔXYZ ~ ΔDEF consider the correspondence XYZ ↔ EDF.

If  find  and .


Answer:

Given, XYZ and DEF are similar for the correspondence XYZEDF

And 


By similarity law,









Question 7.

Using the definition of similarity prove that all the isosceles right angled triangles are similar.


Answer:

Prove that all the isosceles right angled triangles are similar


According to these figures,


In ∆ABC, AB = BC and m∠B = 90


In ∆XYZ, XY = YZ and m∠Y = 90 i.e. ∆ABC and ∆XYZ are isosceles right angled triangles.


To prove: ∆ABC and ∆XYZ are similar triangles.


Proof: ∆ABC and ∆XYZ are both isosceles right angled triangles in which m∠B = m∠Y = 90


Also AB = BC and XY = YZ, then



Question 8.

For ΔABC and ΔXYZ, ABC ↔ XYZ is a similarity. If , AC = 3 and XY = 5, find YZ and XZ.


Answer:

Given, For ΔABC and ΔXYZ, ABC ↔ XYZ is a similarity


And AC = 3 and XY = 5


Hence, 


AB = 4 and BC = 6


Due to similarity of both the triangles,




 ⟹ YZ =  = 7.5


And,


⟹ XZ =  = 3.75



Question 9.

State whether the following statements are true or false. Give reasons for your answer:

(1) If ΔPQR and ΔABC are similar and none of them is equilateral, then all the six correspondences between ΔPQR and ΔABC are similarities.

(2) All congruent triangles are similar.

(3) All similar triangles are congruent.

(4) If the correspondences ABC ↔ BAC is similarity, then AABC is an isosceles triangle.

(5) The correspondence PQR ↔ YZX between ΔPQR and ΔYZX is a similarity. If m∠P = 60, m∠R = 40, then m∠Z = 80.


Answer:

(1) False


For equilateral triangles, all the six correspondences are similarity. But in triangles other than equilateral, the measures of all the angles are not same.


(2) True


Congruent triangles are equal with respect to size and shape, while for triangles to be similar, it is sufficient that their shapes are same.


(3) False


Similar triangles are equal in shape but not in size. For triangles to be congruent, they must be equal in both, shape as well as size. Hence, all similar triangles are not congruent.


(4) True


For ∆ABC, the correspondence ABC↔BAC is a similarity.


∴ ∠A≅∠B


∴BC≅AC


Hence, two sides of ABC are congruent and therefore ∆ABC is an isosceles triangle.


(5) Given statement is true.



Question 10.

ΔABC ~ ΔPQR for the correspondence ABC ↔ QRP. If m∠A = 50, m∠C = 30, then m∠R =
A. 80

B. 50

C. 30

D. 100


Answer:

The triangle ABC and triangle PQR are similar in correspondence ABC↔QRP


∴ mQ = mA = 50


And mP = mC = 30


In ∆PQR,


mP + mQ + mR = 180


30 + 50 + mR = 180


mR = 100


Question 11.

ΔLMN ~ ΔXYZ for the correspondence LMN ↔ ZYX. If m∠Z = 50, m∠X = 40, then m∠L + m∠N =
A. 10

B. 90

C. 110

D. 80


Answer:

In ∆LMN and ∆XYZ, the correspondence LMN  ZYX is a similarity.


∴ mL = mZ = 50


And mN = mX = 40


Hence, mL + mN = 50 + 40 = 90


Question 12.

If the correspondence ABC ↔ EFD is a similarity in AABC and ADEF, then of the following is not true.
A. 

B. 

C. 

D. 


Answer:

the correspondence ABC  EFD is a similarity in ∆ABC and ∆DEF


∴ 


Question 13.

ABC ~ PQR is a similarity in ΔABC and ΔPQR. If the perimeter of ΔABC is 12 and perimeter of a ΔPQR is 20, then AB : PQ =
A. 3 : 5

B. 5 : 3

C. 4 : 3

D. 3 : 4


Answer:

Given, ABC PQR is a similarity in ABC and PQR and the perimeter of ABC is 12 and perimeter of a PQR is 20




Hence, AB : PQ = 3 : 5



Exercise 6.2
Question 1.

In ΔABC, a line parallel to  intersects  and in D and E respectively. Fill in the blanks shown in the table :



Answer:

Given, In ΔABC, a line parallel to  intersects and in D and E respectively as shown in fig.


In ABC, a line parallel to BC intersects AB and AC in D and E respectively.


∴  ...... (i)


 ...... (ii)


 …… (iii)


Also, AB = AD + DB and AC = AE + EC


1) AB = AD + DB


AB = 3.6 + 2.4


∴AB = 6


Now from eq (i)




AE = 


AE = 2.7


But, AC = AE + EC


AC = 2.7 + 1.8 = 4.5


AC = 4.5


2) AC = AE + EC


4.2 = AE + 3.15


AE = 1.05


From eq (ii)




AD = 


AD = 1.55


Then AB = AD + DB


∴ 6.2 = 1.55 + DB


DB = 4.65


3) AC = AE + EC


∴ AC = 6.4 + 8.0


AC = 14.4


∴  From eq (i)



AD = 


AD = 9.6


AB = AD + DB


AB = 9.6 + 12


AB = 21.6


4) From eq (ii)




AC = 


AC = 13.8


AB = AD + DB


18.4 = 7.2 + DB


DB = 18.4 – 7.2


DB = 11.2


Then, AC = AE + EC


∴ 13.8 = 5.4 + EC


EC = 8.4


5) From eq (iii)




AB = 


AB = 6.8


AB = AD + DB


6.8 = AD + 3.4


AD = 3.4


AC = AE + EC


5.1 = AE + 2.55


AE = 2.55




Question 2.

In ΔABC, the bisector of ∠C intersects  in F. If 2AF = 3FB and AC = 7.2 find BC.


Answer:

Given,


Given, 2AF = 3FB



The bisector of ∠C intersects AB in F.




BC = 


BC = 4.8



Question 3.

In ΔXYZ, the bisector of ∠Y intersects  in P.

(1) If XP : PZ = 4 : 5 and YZ = 6.5, find XY.

(2) if XY : YZ = 2 : 3 and XP = 3.8, find PZ and ZX.


Answer:

Given, XP : PZ = 4 : 5

XY : YZ = 2 : 3



In ∆XYZ, the bisector of angle Y intersects ZX in P.


(1) ∴ 



XY = 


XY = 5.2


(2) 



PZ = 


PZ = 5.7


ZX = PZ + XP


ZX = 5.7 + 3.8


ZX = 9.5



Question 4.

In Δ ABC, the bisector of ∠A intersects  in D. Prove that

and 


Answer:

Given, In ΔABC, the bisector of ∠A intersects  in D

TO PROVE :


1) 


2) 



Proof: In ABC, the bisector of A intersects  in D


∴ 


By applying componendo


∴ 



By applying invertendo



BD  ...... (i)


Similarly,






DC =  PROVED.



Question 5.

□ ABCD is a trapezium such that  ||  and  such that  || . Prove that  .


Answer:

Given, ABCD is a trapezium such that  || and  such that  || 


⟹ First draw DB to intersect MN at P.


Proof : AB || CD and MN || AB


∴ MN || CD


Now, MN || AB and M-P-N


Hence, MP || AP


Similarly, as MN || CD and M-P-N


PN || CD


In ∆ABD, MP || AB


 ... (i)


In ∆BCD, PN || CD


 ... (ii)


From eq (i) and (ii)


 PROVED.



Question 6.

In ΔABC, D and E are the mid-points of  and  respectively.  and  intersect in G. A line m passing through D and parallel to  intersects  in K. Prove that AC = 4CK.


Answer:

Given, In ABC, D and E are the mid-points of  and  respectively.  and intersect in G. A line m passing through D and parallel to intersects in K


To prove : AC = 4CK


Proof : In ∆ABC, E is the midpoint of AC.


∴ CE = AC


∴ AC = 2CE ...... (i)


In ∆ABC, D is the midpoint of BC.


∴CD = CB


 -...... (ii)


In ∆CBE, C-D-B, C-K-E



From eq (ii)



Hence, CE = 2CK


Substituting this into eq (i)


AC = 2 (2CK)


AC = 4CK



Question 7.

In ΔPQR, , such that Q-X-R. A line parallel to  and passing through X intersects in Y. A line parallel to px and passing through Y intersects  in Z. Prove that 


Answer:

Given, In PQR, , such that Q-X-R. A line parallel to and passing through X intersects in Y. A line parallel to px and passing through Y intersects in Z


To prove : 


Proof : In PQR, XY || PR


∴  ...... (i)


In PQX, YZ || PX


∴  ...... (ii)


From eq (i) and (ii),




Question 8.

In ΔABC,  and B-X-C. A line passing through X and parallel to  intersects  in Y. A line passing through X and parallel to  intersects in Z. Prove that CY2 = AC • CZ.


Answer:

Given, In ABC, and B-X-C. A line passing through X and parallel to  intersects  in Y. A line passing through X and parallel to intersects in Z as shown in figure.


To prove : CY2 = AC × CZ


Proof : In ∆ABC, XY || AB


∴  ...... (i)


In ∆YBC, XZ || BY


∴  ...... (ii)


From eq (i) and (ii)



∴ CY2 = AC × CZ



Question 9.

In ΔABC, the bisector of ∠A intersects  in D and the bisector of ∠ADC intersects in E. Prove that AB × AD × EC = AC × BD × AE.


Answer:

Given, In ABC, the bisector of ∠A intersects  in D and the bisector of ∠ADC intersects in E as shown in figure.


To prove: AB × AD × EC = AC × BD × AE


Proof: In ∆ABC, the bisector of ∠A intersects BC in D.


∴  ...... (i)


In ∆ADC, the bisector of ∠ADC intersects AC in E.


∴  ...... (i)


By multiplying eq (i) and (ii)



∴ 


∴ AB × AD × EC = AC × BD × AE



Question 10.

In ΔABC, D is the mid-point of and P is the mid-point of intersects  in Q. Prove that (i) CQ = 2AQ (ii) BP = 3PQ


Answer:

Given: In ΔABC,


BD = DC [∵ D is mid-point of BC]


AP = PD [∵ P is mid-point of AD]


Construction: Draw a line l from D parallel to AC and let it intersect BQ at S and AB at R.


Draw line m from D parallel to AC intersecting AC at T.


Draw a line from C parallel to BQ.


Extend AD and let it intersect the parallel drawn from C at point E.



Proof:


In ΔADT,


PQ||DT [∵ by construction]


AP = PD [∵ Given]


Hence, by mid-point theorem,


AQ = QT …(1)


In ΔBQC,


BD = DC [∵ given: d is mid-point of BC]


DS|| QC [∵ by construction]


Hence, by mid-point theorem,


QT = TC …(2)


By eq. (1), (2) and (3) we get that,


AQ = QT =TC …(4)


⇒ AQ = 2QC [As QC = QT + TC]


In ΔPDB and ΔCDE,


∠CDE = ∠BDP [vertically opposite angles]


BD = DC [Given]


∠PBD = ∠ECD [alternate interior angles between BP||CE]


Therefore, ΔCDE ≅ ΔPDB.


CE = BP [By CPCT] …(5)


In ΔAPQ and ΔAEC,


∠A = ∠A [common angle]


∠APQ = ∠AEC [Corresponding angles when PQ||CE]


∠AQP = ∠ACE [Corresponding angles when PQ||CE]


By AAA similarity,


ΔAPQ ∼ ΔAEC


And,



 [from (4)]


 [From (5)]


PQ = 3BP


Hence proved.




Exercise 6.3
Question 1.

In figure 6.27, 

Prove that BR = CS.



Answer:

In Δ ABC

Given, A-P- B, B- R-S-C and A-Q-C


and


To prove – BR = CS


Proof : In Δ ABC given 


∴ ………..eq(1)


(∵ If a line to one side of a Δ intersects the other two sides of the Δ in distinct points, the segments of the other sides of the Δ in the same half plane of the line are proportional to the corresponding sides of the Δ)


Similarly, 


∴  ………………..2


From 1 and 2, we get


 ……..eq(3)


Similarly, 


∴ ……………4


From 3 and 4



⇒ BR = CS hence proved



Question 2.

□ ABCD is a parallelogram. Prove that ABD and BDC are similar.


Answer:


Given ABCD is a ∥gram


To prove : - Δ ABD ∼ Δ BDC


Proof :- For the correspondence ABD ↔ CDB between ΔABD and Δ BDC


∠ BAD ≅ ∠ DCB (opposite ∠s of a ∥gram


∠ABD ≅ ∠CDB (alternate ∠s)


∠ADB ≅ ∠CDB (alternate ∠s)


AB = CD (opposite sides of a ∥ gram ABCD )


∴ 


Also AD = CB (opposite sides of a ∥ gram ABCD)


∴  = 1


And BD = DB (common)


 = 1


Hence, for the correspondence, ABD ↔ CDB between ΔABD and ΔBDC


The corresponding angles are congruent and the lengths of its corresponding sides are in proportion.


∴for the correspondence ABD ↔ CDB, ΔABD and ΔBDC are similar.



Question 3.

In □ABCD, M and N are the mid-points of  and  . If ∥  prove that 


Answer:


Given: In □ ABCD, M and N are the mid points ofand  and  ∥ 


To Prove:  ∥ 


Construction:- extend  towards D and  towards C to intersect each other at P ( ∵ AB > CD)


Proof:- M is the mid-point of 


∴ DA = 2 DM …………1


Similarly, N is the mid-point of 


∴ CB = 2 CN…………..eq(2)


In Δ PAB ,  ∥ 


∴ 


( ∵ If a line ∥ to one side of a Δ intersects the other two sides of the Δ in distinct points, the segments of the other sides of the Δ in the same half plane of the line are proportional to the corresponding sides of the Δ)


∴  (from 1 and 2)



In Δ PMN, P-D-M , P-C-N and



And 


 ∥ 



Question 4.

In □ ABCD, A—P—D, B—Q—C. If AB ∥ PQ and PQ || DC, prove that

AP X QC = PD X BQ.



Answer:

Given: In □ ABCD, A—P—D, B—Q—C,  ∥  and 


To prove: AP×QC = PD×BQ


Proof:  ∥  and 


∴  ∥  ∥ 


So,  ,  ,  are three ∥ linesand  and  are their transversal


∴ 


(∵ if three or more than three ∥ lines are intercepted by two transversal, the segments cut off on the transversal between the same parallel lines are proportional)


∴ AP × QC = PD × BQ hence proved



Question 5.

In Δ ABC, the bisector of ∠ A intersects BC in D. The bisector of ∠ADB intersects AB in F and the bisector of ∠ ADC intersects in E. Prove that AF X AB X CE = AE × AC × BF.


Answer:


Given : In Δ ABC, the bisector of ∠ A intersects  in D. The bisector of ∠ADB intersects  in F and the bisector of ∠ ADC intersects in E


To prove : AF × AB × CE = AE × AC × BF


Proof : in Δ ABC, the bisector of ∠A intersects  at D


∴ 1


(∵ in a Δ the bisector of an angle divides the side opposite to the angle in the segments whose lengths are in the ratio of their corresponding sides)


Similarly, In Δ ADB, the bisector of ∠D intersects  at F


 ………………2


And in Δ ADC , the bisector of ∠D intersects  at E


 …………..eq(3)


Multiplying 1, 2 and 3, we get




And
(∵  )


∴ 


∴ AF × BD × CE =AE × CD × BF



Question 6.

In Δ ABC, Δ PQR and ΔXYZ correspondences ABC ↔ PQR, PQR ↔ XYZ are similarity. Prove that ABC ↔ XYZ is similarity.


Answer:

Given:- In Δ ABC, Δ PQR and ΔXYZ correspondences ABC ↔ PQR, PQR ↔ XYZ are similarity

To Prove:- ABC ↔ XYZ is similarity


Proof:- In ΔABC and ΔPQR , the correspondence ABC ↔ PQR is a similarity


∴ ∠A ≅ ∠P, ∠ B ≅ ∠Q , ∠C ≅ ∠R ………..1


…………………..eq(2)


In Δ PQR and ΔXYZ , the correspondence PQR ↔ XYZ is a similarity


∴ ∠ P ≅∠X , ∠ Q ≅ ∠Y , ∠R≅ ∠Z………………3


……………………………………..4


From 1 and 3 , we get


∠A ≅ ∠x, ∠ B ≅ ∠Y , ∠C ≅ ∠Z


Multiplying 2 and 4, we get




Thus for the correspondence ABC ↔ XYZ between ΔABC and ΔXYZ, the corresponding angles are congruent and the lengths of corresponding sides are in proportion.


∴ the correspondence ABC ↔ XYZ between ΔABC and ΔXYZ is a similarity.



Question 7.

State giving reasons, whether the following statements are true or false: In all the following questions the line does not contain a side of the triangle.

(1) A line can be drawn in the plane of a triangle not intersecting any of the sides of a triangle.

(2) A line can be drawn in the plane of a triangle which is not passing through any of the three vertices and intersecting all the three sides of the triangle.

(3) If a line drawn in the plane of a triangle intersects the triangle at only one point, the line passes through a vertex of the triangle.

(4) If a line intersects two of the three sides of a triangle in two distinct points and does not intersect the third side, then the line is parallel to the third side.

(5) In the plane of Δ ABC, a line / can be drawn such that  = {P}, = {Q} and = ϕ.


Answer:

(1) The given statement is true because for a line in the plane of a Δ, there are 3 possibilities


i) The line does not intersect the Δ


ii) The line intersects the Δ at one point


iii) The line intersects the Δ at two points


The first possibility is satisfied; hence the given statement is true.



(2) The given statement is false because according to the theorem, if a line lying in the plane of a Δ and not passing through any vertex intersects one side, then it does intersect one more side but does not intersect the third side. Thus, a line not passing through a vertex of a Δ cannot intersect all the three sides.


(3) The given statement is true because a line intersecting a Δ and not passing through any vertex will intersect the Δ at two points.



(4) The given statement is false because if a line intersects two sides of a Δ and does not intersect the third side, it can intersect the line containing the third side.



(5) The given statement is true because in the plane of ΔABC, if line l intersects  and  at distinct points P and Q respectively, then according to the theorem, if a line lying in the plane of a Δ and not passing through any vertex intersects one side, then it does intersect one more side but does not intersect the third side, it cannot intersect.




Exercise 6.4
Question 1.

∠B is a right angle in ΔABC and  is an altitude to hypotenuse. AB = 8, BC = 6. Find the area of ΔBDC.


Answer:


In ∆ABC, ∠B is a right angle, AB = 8 and BC = 6




= 24 …(1)


In ∆ABC, ∠B is a right angle and BD is an altitude,


In ∆ABC,


∠A + ∠C = 90⁰ and,


In ∆BDC,


∠DBC + ∠C = 90⁰


So, ∠A = ∠DBC …(2)


In ∆ABC and ∆BDC,


∠DAB ≅ ∠DBC [from (2)]


∠ADB ≅ ∠BDC [by right angles]


The correspondence ADB ↔ BDC is a similarity by AA corollary.


Areas of similar triangles are proportional to the squares of their corresponding sides.




 …(3)


Now, ADB + BDC = ABC





BDC = 8.64


The area of ∆BDC is 8.64.



Question 2.

In □m ABCD, T ∈  and  intersects  in M and in O. Prove that AM2 = MT • MO.


Answer:

In  ABCD, T and  intersects  in M and in O.



To prove:


AM2 = MT •MO


In ∆AMB and ∆OMD


∠AMD ≅ ∠OMD (Vertically opposite angles)


∠MBA ≅ ∠MDO (Alternate angles)


The correspondence AMD↔OMB is a similarity


 …(1)


In ∆AMD and ∆TMB


∠AMD ≅ ∠TMB (Vertically opposite angles)


∠MAD ≅ ∠MTB (Alternate angles)


The correspondence AMD↔TMB is a similarity


 …(2)


Multiplying (1) and (2),




Hence, AM2 = MT •MO



Question 3.

In □m ABCD, M is the mid-point of  and  intersect in N. Prove that DN = 2MN.


Answer:

In ABCD, M is the mid-point of and  intersect in N.



To prove : DN = 2MN


Proof: M is the mid-point of BC


MB = MC


 …(1)


In ∆MBN and ∆MCD


∠BMN ≅ ∠CMD (Vertically opposite angles)


∠MNB ≅ ∠MDC (Alternate angles)


The correspondence MBN↔MCD is a similarity



 (from 1)


MN = MD


Now, D-M-N


DN = DM + MN


DN = DM + MN


DN = MN + MN (MD = MN)


DN = 2MN


Hence proved.



Question 4.

P and Q are the mid-points of  and in ΔABC. If the area of ΔAPQ = 12√3, find the area of ΔABC.


Answer:


In ∆ABC, P and Q are mid-points of AB and AC respectively.



The correspondence ∆APQ↔∆ABC is a similarity by SSS theorem.


Now,


Areas of similar triangles are proportional to the squares of their corresponding sides.





ABC = 12√3 × 4


ABC = 48√3


The area of ∆ABC is 48√3.



Question 5.

□ABCD is a rhombus.  = {0}. Prove that the area of ΔOAB = (area of □ ABCD).


Answer:

Given: ABCD is a rhombus.  = {0}.



To prove: Area of OAB = (area of □ ABCD)


Proof:


O is the mid-point of AC as well as BD.


Further, in rhombus ABCD,


AB ≅ BC ≅ CD ≅ DA


In ∆OAB and ∆OBC,


OA ≅ OC


OB ≅ OB


AB ≅ CB


∆OAB and ∆OBC are congruent by SSS theorem for congruence.


Thus, their areas are equal.


Similarly, ∆OAB, ∆OBC, ∆OCD and ∆ODA are all congruent triangles having equal areas.


∆OAB = ∆OBC = ∆OCD = ∆ODA


Now, ABCD = ∆OAB + ∆OBC + ∆OCD + ∆ODA


ABCD = ∆OAB + ∆OAB + ∆OAB + ∆OAB


ABCD = 4∆OAB



Thus,




Question 6.

In □PQRS,  = {T}, PS = QR,. Prove that ΔTPS is similar to ΔQTR.


Answer:


Given: In PQRS, PR ∩ QS = {T}, PS = QR and PQ||RS.


To prove: ∆TPS and ∆QTR are similar.


Construction: Through R, draw a line parallel to PS to intersect PQ at M.


Proof:


In PMRS, PM||RS (P-M-Q) and PS||MR (construction).


Thus, PMRS is a parallelogram.


PS = MR


Further, PS = QR (given)


In ∆RMQ, ∠RMQ ≅ ∠RQM …(i)


Further, the corresponding angles formed by transversal PQ to PS||MR are congruent.


∠RMQ ≅ ∠SPM (corresponding angles)


∠RQM ≅ ∠SPM (by (i))


∠RQP ≅ ∠SPQ (P-M-Q)


In ∆SPQ and ∆RQP,


∠SPQ ≅ ∠RQP


PQ ≅ QP (same line segment)


SP ≅ RQ (given)


The correspondence ∆SPQ↔∆RQP is a congruence by SAS theorem of congruence.


∠PSQ ≅ ∠QRP


∠PST ≅ ∠QRT (S-T-Q and R-T-P)


In ∆TPS and ∆QTR


∠STP ≅ ∠RTQ (Vertically opposite angles)


∠PST ≅ ∠QRT


∆TPS and ∆QTR are similar by AA corollary.


Thus, by AA corollary the correspondence ∆TSP↔∆TRQ is a similarity.



Question 7.

In ΔABC, a line parallel to , passes through the mid-point of . Prove that the line bisects  .


Answer:


Given: In ∆ABC, a line parallel to BC, passes through the mid-point of AB.


To prove: Line m bisects AC.


Proof: In the plane of ∆ABC, line m is parallel to BC and intersects AB at a point other than a vertex of the triangle.


Thus, m intersects AC.


Let m ∩ AC = {E}


In ∆ABC, D is the mid-point of AB.


AD = DB



In ∆ABC, A-D-B, A-E-C and DE||BC.




AE = EC and A-E-C.


E is the mid-point of AC.


Thus, line m bisects AC.



Question 8.

P, Q, R are the mid-points of the sides of ΔABC. X, Y, Z are the mid-points of the sides of ΔPQR. If the area of ΔXYZ is 10, find the area of APQR and the area of ΔABC.


Answer:


In ∆ABC, P, Q, R are the mid-points of the sides AB, BC and CA respectively.



The correspondence ∆ABC↔∆QRP is a similarity.


Areas of similar triangles are proportional to the squares of their corresponding sides.



∆ABC = ∆POR


Similarly, X,Y and Z are the mid-points of the sides of ∆PQR, we get


∆PQR = 4∆XYZ


∆PQR = 4×10


∆PQR = 40


Thus, the area of ∆PQR is 40 sq. units.


∆ABC = 4∆PQR


∆ABC = 4×40


∆ABC = 160


Hence, the area of ∆ABC is 160 sq. units.



Question 9.

In Δ ABC,  such that  . Prove that X, Y, Z are the mid-points of  , respectively.


Answer:


Given: In ∆ABC, X∊BC, Y∊CA, Z∊AB.


Also, XY||AB, YZ||BC and ZX||AC.


To prove: X,Y and Z are the mid-points of BC, CA and AB respectively.


Proof: In ∆ABC, YZ||BC.


 …(1)


In ∆ABC, XY||AB.


 …(2)


From (1) and (2),


 …(3)


In ∆ABC, ZX||AC.


 …(4)


From (3) and (4),



AZ2 = BZ2


AZ = BZ


Hence, A-Z-B and AZ = BZ.


Thus, Z is the mid-point of AB.


Similarly, Y is the mid-point of AC and X is the midpoint of BC.


Therefore, X, Y and Z are the mid-points of BC, CA and AB respectively.



Question 10.

Two triangles are similar. Prove that if sides in one pair of corresponding sides are congruent, then the triangles are congruent.


Answer:

Given: The correspondence ∆ABC↔∆PQR is a similarity and AB ≅ PQ.


To prove: ∆ABC and ∆PQR are congruent.


Proof:


The correspondence ∆ABC↔∆PQR is a similarity.



Further, AB ≅ PQ


AB = PQ




AB = PQ, BC = QR, AC = PR


AB ≅ PQ, BC ≅ QR, AC ≅ PR


The correspondence ∆ABC↔∆PQR is a congruence by SSS theorem of congruence.


Hence, ∆ABC and ∆PQR are congruent.




Exercise 6
Question 1.

In ΔABC,  such that  are positive real numbers. A line passing through P and parallel to  intersects in Q. Prove that (m + n)2 (area of ΔAPB) = m2(area of ΔABC)


Answer:

We have


Given: In ∆ABC,  such that , where m and n are positive real numbers.


QP ∥ BC


Prove that: (m + n)2 (area of ∆APB) = m2 (area of ∆ABC)


Proof: Since 


 [by invertendo componendo invertendo rule]


⇒ 


In ∆APQ and ∆ABC,


∠APQ ≅ ∠ABC [∵, corresponding angles]


∠AQP ≅ ∠ACB [∵, corresponding angles]


⇒ The correspondence APQ ↔ ABC is a similarity.


Remember the property, areas of similar triangles are proportional to the squares of the corresponding sides.


∴ 


⇒ 


⇒ 


⇒ 


By cross-multiplication, we get


(m + n)2 (Area of ∆APB) = m2 (Area of ∆ABC)


Thus, proved.



Question 2.

D, E and F are the mid-points ofand  respectively in ΔABC. Prove that the area of □ BDEF = 1/2 area of ΔABC.


Answer:

We have


Given: D is the midpoint of , E is the midpoint of  & F is the midpoint of .


⇒ AF = FB, AE = EC & BD = DC


To Prove: Area of □BDEF = 1/2 Area of ∆ABC


Proof: Since, given is that D, E and F are midpoints of  respectively.


We know, AE = EC.


⇒ AE = FD [∵, EC = FD]


⇒  [∵, ] …(i)


Also, we know that, AF = FB.


⇒ AF = ED [∵, FB = ED]


⇒  [∵, ] …(ii)


Now, in ∆AFE and ∆DEF, we have


AF ≅ ED, AE ≅ FD & FE ≅ EF


⇒ By SSS theorem of congruence, we can say that the congruency AFE ↔ DEF is a similarity.


Similarly, in ∆AFE and ∆FBD, we have


AF ≅ FB, AE ≅ FD & FE ≅ BD


⇒ By SSS theorem of congruence, we can say that the congruency AFE ↔ FBD is a similarity.


Similarly, the correspondence AEF ↔ EDC is a similarity.


So, ∆AFE, ∆DEF, ∆FBD and ∆EDC are all congruent triangles.


⇒ ∆AFE = ∆DEF = ∆FBD = ∆EDC …(iii)


Now,


BDEF = ∆DEF + ∆FBD


⇒ BDEF = ∆AFE + ∆AFE [from equation (iii)]


⇒ BDEF = 2 ∆AFE …(iv)


And


∆ABC = ∆AFE + ∆DEF + ∆FBD + ∆EDC


⇒ ∆ABC = ∆AFE + ∆AFE + ∆AFE + ∆AFE [from equation (iii)]


⇒ ∆ABC = 4 ∆AFE


⇒ ∆ABC = 2 (2 ∆AFE) …(v)


Comparing equations (iv) and (v), we get


∆ABC = 2 (BDEF)


⇒ BDEF = 1/2 ∆ABC


⇒ Area of □BDEF = 1/2 (Area of ∆ABC)


Thus, proved.



Question 3.

Can two similar triangles have same area? If yes, in which case they have the same area?


Answer:

Yes, two similar triangles can have same area.

Note, when two triangles are congruent then all corresponding sides as well as corresponding angles of one triangle are equal to those of the other triangle.


⇒ If two triangles are congruent, they are similar too.


And congruent triangles have same area too.


Hence, the similar triangles need to be congruent to have similar areas.



Question 4.

The correspondence ABC ↔ DEF is similarity in ΔABC and ΔDEF.  is an altitude of ΔABC and  is an altitude of ΔDEF. Prove that AB X DN = AM X DE.


Answer:

We have


Given: The correspondence ABC ↔ DEF is similarity in ∆ABC and ∆DEF.


 is an altitude of ∆ABC and  is an altitude of ∆DEF.


To Prove: AB × DN = AM × DE


Proof: Since, the correspondence ABC ↔ DEF is a similarity.


By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


So, we can write ∠B ≅ ∠E. …(i)


In ∆ABM and ∆DEN,


From result (i),


∠B ≅ ∠E


Also, ∠M ≅ ∠N [since, they are right angles of ∆ABM and ∆DEN respectively]


⇒ By AA-corollary, the correspondence ABM ↔ DEN is a similarity in ∆ABC and ∆DEF]


Then, again by definition of similarity of correspondences in triangles we can say that,



By cross-multiplication, we get


⇒ AB × DN = AM × DE


Hence, proved.



Question 5.

Explain with reasons, whether the following statements are true or false:

(1) AAA criterion of similarity of triangles cannot be the criterion for congruence of triangles.

(2) SAS criterion for congruence of triangles cannot be a criterion for similarity of triangles.

(3) Two congruent triangles have the same area.

(4) Two similar triangles always have the same area.

(5) Area of similar triangles are proportional to the squares of measures of their corresponding angles.


Answer:

(1). Similarity of triangles and congruence of triangles are closely related, but has a fine line of difference.


Similar triangles are the same shape, but not necessarily the same size. The triangles are congruent if, in addition to this, their corresponding sides are of equal length.


By AAA criterion of similarity of triangles, we get the shapes of triangles equal but, for congruence of triangles their sizes must be equal too.


∴ AAA criterion of similarity of triangles cannot be the criterion for congruence of triangles.


Hence, the statement is true.


(2). For congruence of triangles, the corresponding sides are of equal length while for similarity of triangles, measure of the corresponding sides of two triangles must be proportional.


SAS criterion for congruence of triangles not only makes sure that the triangles have equal length of corresponding sides but also makes sure that the triangles have same shape.


∴ SAS criterion for congruence of triangles can be a criterion for similarity of triangles.


Hence, the statement is false.


(3). Congruency of triangles ensure that the triangles have same shape as well as equal lengths of corresponding sides.


⇒ The shape and size of two congruent triangles are equal.


∴ Two congruent triangles have the same area.


Hence, the statement is true.


(4). Similarity of triangles indicate that they have the same shape but their sizes are not same.


Area of triangles are equal when their size are equal, shape of triangle cannot determine its area.


∴ Two similar triangles doesn’t always have the same area.


Hence, the statement is false.


(5). Area of similar triangles are proportional to the squares of their corresponding sides.


Since, the measure of corresponding angles of similar triangles are always equal.


∴ Area of similar triangles are not proportional to the squares of their corresponding angles.


Hence, the statement is false.



Question 6.

Fill in the blanks so that the following statements are true:

and  are the altitudes of ΔABC. If AB = 12, AC = 9.9, AD = 8.1 and BE = 7.2, perimeter of ΔABC = _____.


Answer:

We have



Given that: AD and BE are altitudes of ∆ABC. ⇒ ∠ADC = ∠


In ∆ADC and ∆BEC,


∠ADC = ∠BEC [∵, ∠ADC = ∠BEC = 90° as AD and BE ]


∠ACD = ∠BCE [∵, ∠ACD and ∠BCE are same angles of the same triangles, so they obviously are equal]


⇒ By AA-corollary, we can say that ∆ADC ∼ ∆BEC for the correspondence of ADC ↔ BEC.


By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ 


⇒ 


⇒ 


⇒ BC = 8.8


So, perimeter of ∆ABC is given by


Perimeter of ∆ABC = AB + BC + AC


⇒ Perimeter of ∆ABC = 12 + 8.8 + 9.9


⇒ Perimeter of ∆ABC = 30.7


Thus, perimeter of ∆ABC = 30.7.



Question 7.

Fill in the blanks so that the following statements are true:

D, E, F are respectively the mid-points of  , of ΔPQR. The correspondence DEF ↔ _____ is similarity.


Answer:

We have



Given: D is the midpoint of PQ ⇒ PD = DQ = 1/2 PQ


But PD = DQ = FE [∵, F and E are also midpoints of PR and QR respectively]


So, EF = 1/2 PQ


Or  …(i)


E is the midpoint of QR ⇒ QE = ER = 1/2 RQ


But QE = ER = DF [∵, D and F are also midpoints of PQ and PR respectively]


So, DF = 1/2 RQ


Or  …(ii)


F is the midpoint of PR ⇒ PF = FR = 1/2 RP


But PF = FR = DE [∵, D and E are midpoints of PQ and QR respectively]


So, DE = 1/2 RP


Or  …(iii)


Comparing equations (i), (ii) & (iii), we get



We can also arrange it as,



∴ The correspondence DEF ↔ RPQ is a similarity by SSS theorem.


Thus, answer is RPQ.



Question 8.

Fill in the blanks so that the following statements are true:

In ΔABC,  and are altitudes. If AB = 12, BC = 15 and AM = 9.6, then CN = _____.


Answer:

We have



Given: AM ⊥ BC and CN ⊥ AB.


AB = 12, BC = 15 & AM = 9.6


To find: CN = ?


For this,


Take ∆AMB and ∆CNB,


∠AMB = ∠CNB [∵, ∠AMB = ∠CNB = 90°]


∠ABM = ∠CBN [∵, ∠AMB and ∠CBN are same angles of the same triangle ABC]


⇒ By AA corollary, ∆AMB ∼ ∆CNB for the correspondence AMB ↔ CNB.


By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ 


Or 


⇒ 


⇒ 


⇒ 


⇒ CN = 12


Thus, answer is 12.



Question 9.

Fill in the blanks so that the following statements are true:

Areas of two similar triangles are 25 and 16. The ratio of the perimeters of the triangles is _______.


Answer:

Given: Area of two similar triangles.


Area of ∆1 = 25


Area of ∆2 = 16


Recall two properties:


First,


Ratio of areas of two similar triangles = Ratio of the squares of the corresponding sides


Second,


Ratio of perimeter of two similar triangles = Ratio of corresponding sides


Using the first property, we can write that



⇒ 


Taking under root of both sides, we get



⇒ 


⇒ 


Now, using second property, we can say



Thus, answer is 5/4.



Question 10.

Fill in the blanks so that the following statements are true:

Area of ΔABC = 36 and area of ΔPQR = 64. The correspondence ABC ↔ PQR is a similarity. If AB = 12, then PQ = ______.


Answer:

Given: Area of ∆ABC = 36,


Area of ∆PQR = 64


AB = 12


The correspondence ABC ↔ PQR is a similarity.


To find: PQ = ?


For this,


Recall the property,


Ratio of areas of two similar triangles = Ratio of squares of the corresponding sides


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ PQ = 2 × 8


⇒ PQ = 16


Thus, answer is 16.



Question 11.

Fill in the blanks so that the following statements are true:

In ΔABC, A—M—B, A—N—C and  ||. If AM = 8.4, AN = 6.4, CN = 19.2, then AB = ______.


Answer:

We have



Given: A – M – B, A – N – C


MN ∥ BC


AM = 8.4


AN = 6.4


CN = 19.2


To find: AB = ?


Take ∆ABC,


We know that, MN ∥ BC. So, by property we can write as:



⇒ 


⇒ 


⇒ 


⇒ BM = 25.2


We know AM = 8.4 and BM = 25.2.


Now, AB = AM + BM [∵, from the diagram]


⇒ AB = 8.4 + 25.2


⇒ AB = 33.6


Thus, the answer is 33.6.



Question 12.

Fill in the blanks so that the following statements are true:

The correspondence ABC ↔ PQR is a similarity in Δ ABC and ΔPQR. AB = 16, AC = 8, PQ = 24, BC = 12, then QR + PR = ____.


Answer:

Given: The correspondence ABC ↔ PQR is a similarity in ∆ABC and ∆PQR.


AB = 16,


AC = 8,


PQ = 24 &


BC = 12


To find: QR + PR = ?


By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ 


⇒ 


⇒ 


To find (QR + PR), cross multiply it.


⇒ (QR + PR) × AB = (BC + AC) × PQ


Substituting values,


⇒ (QR + PR) × 16 = (12 + 8) × 24


⇒ (QR + PR) × 16 = 20 × 24


⇒ 


⇒ QR + PR = 30


Thus, the answer is 30.



Question 13.

Fill in the blanks so that the following statements are true:

The correspondence ABC ↔ XYZ is a similarity in ΔABC and ΔXYZ. ABC = 72, BC = 6, YZ = 10. Then XYZ = ______.


Answer:

Given: The correspondence ABC ↔ XYZ is a similarity in ∆ABC and ∆XYZ.


ABC = 72,


BC = 6 &


YZ = 10


To find: XYZ = ?


So, by property we can say that,


Ratio of areas of two similar triangles = Ratio of squares of the corresponding sides


⇒ 


⇒ 


Taking reciprocal of both sides, we get



Substituting values in the above equation, we get


⇒ 


⇒ 


⇒ 


⇒ XYZ = 200


Thus, the answer is 200.



Question 14.

Fill in the blanks so that the following statements are true:

□ ABCD is trapezium in which  ||. The diagonals intersect in P. If PD = 9, AP = 5, PB = 7.2, then AC = ______.


Answer: 

We have



Given: ABCD is a trapezium.


AD ∥ BC


PD = 9,


AP = 5 &


PB = 7.2


To find: AC = ?


We have got the trapezium ABCD, in which


AD ∥ BC & P is the intersection point of AC and BD.


Take ∆PAD and ∆PCB,


∠APD = ∠CPB [∵, Vertically opposite angles are equal]


∠PAD = ∠PCB [∵, alternate angles]


∠PDA = ∠PBC [∵, alternate angles]


⇒ By AAA similarity theorem, ∆PAD ∼ ∆PCB for the correspondence PAD ↔ PCB.


Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ 


Reciprocating it, we get



⇒ 


⇒ 


⇒ PC = 4


From the diagram, note that


AC = AP + PC


⇒ AC = 5 + 4


⇒ AC = 9


Thus, the answer is 9.



Question 15.

Fill in the blanks so that the following statements are true:

In ΔABC, m∠B = 90 and  is an altitude. The correspondence BDA ↔ _____ between ΔBDA and ΔBDC is a similarity.


Answer: 

We have



Given: In ∆ABC,


∠ABC = ∠ADB = ∠BDC = 90°


In ∆BDA and ∆CDB,


∠BDA = ∠CDB = 90°


∠ABD = ∠BCD [measure of both being equal, i.e., (90° - m∠A)]


This can be explained as,


In ∆BDA, by angle sum property of triangle,


∠BDA + ∠DAB + ∠ABD = 180°


⇒ 90° + ∠A + ∠ABD = 180° [∵, ∠BDA = 90° & ∠DAB = ∠A]


⇒ ∠ABD = 180° - 90° - ∠A


⇒ ∠ABD = 90° - ∠A …(i)


Similarly, in ∆ABC by angle sum property of triangle,


∠ABC + ∠BAC + ∠BCA = 180°


⇒ 90° + ∠A + ∠BCD = 180° [∵, ∠ABC = 90°, ∠BAC = ∠A & ∠BCA = ∠BCD (from the figure)]


⇒ ∠BCD = 180° - 90° - ∠A


⇒ ∠BCD = 90° - ∠A …(ii)


By equations (i) and (ii), we get


∠ABD = ∠BCD


Hence, by AA corollary the correspondence BDA ↔ CDB is similarity between ∆BDA and ∆BDC.


Thus, the answer is CDB.



Question 16.

Fill in the blanks so that the following statements are true:

The correspondence ABC ↔ ZXY is a similarity in ΔABC and ΔXYZ. If AB = 12, BC = 9, CA = 7.5 and ZX = 10, then YZ + XY = ____.


Answer:

Given: The correspondence ABC ↔ ZXY is a similarity in ∆ABC and ∆XYZ.


AB = 12,


BC = 9,


CA = 7.5 &


ZX = 10


To find: YZ + XY = ?


Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ 


⇒ 


⇒ 


Reciprocating both sides,


⇒ 


⇒ 


⇒ 


⇒ 


Thus, the answer is 13.75.



Question 17.

ABC ↔ DEF is a similarity in ΔABC and ΔADEF, m∠A = 40, m∠E + m∠F =
A. 40

B. 80

C. 140

D. 180


Answer:

Given: m ∠A = 40°


The correspondence ABC ↔ DEF is a similarity between ∆ABC and ∆DEF.


To find: m ∠E + m ∠F = ?


Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ m ∠A = m ∠D


⇒ m ∠A = m ∠D = 40°


⇒ m ∠D = 40°


In ∆DEF,


m ∠D + m ∠E + m ∠F = 180°


⇒ 40° + m ∠E + m ∠F = 180°


⇒ m ∠E + m ∠F = 180° - 40°


⇒ m ∠E + m ∠F = 140°


∴ Answer is 140.


Thus, option (c) is correct.


Question 18.

In ΔABC, M ∈  , N ∈  such that ||  ,……….is not true.
A. AN ⋅ MB = AM ⋅ NC

B. AM ⋅ MB = AN ⋅ NC

C. AB ⋅ AN = AM ⋅ AC

D. AB ⋅ NC = AC ⋅ MB


Answer:

We have the diagram as



Given: In ∆ABC, MN ∥ BC.


By property of triangles, we can say that


,


 &



Cross multiplication of the above equations,


⇒ AM × NC = AN × MB


Or AM . NC = AN . MB is true.


AM × AC = AN × AB


Or AM . AC = AN . AB is true.


MB × AC = NC × AB


Or MB . AC = NC . AB is true.


So, AM . MB = AN . NC is not true.


Thus, option (b) is correct.


Question 19.

In ΔABC, B—M—C and A—N—C, ||. If NC : NA = 1 : 3 and CM = 4, then BC = ………..
A. 12

B. 16

C. 8

D. 1/2


Answer:

We have the diagram as,



Given: In ∆ABC,


MN ∥ AB


NC:NA = 1:3


⇒ 


CM = 4


To find: BC = ?


In ∆ABC,


Since, MN ∥ AB


⇒ 


⇒ 


⇒ MB = 4 × 3 [∵,  ⇒ ]


⇒ MB = 12


In B – M – C,


BC = MB + CM


⇒ BC = 12 + 4


⇒ BC = 16


Thus, option (b) is correct.


Question 20.

In ΔXYZ and ΔPQR, XYZ ↔ PQR is similarity. XY = 12, YZ = 8, ZX = 16, PR = 8. So, PQ + QR = ………….
A. 20

B. 10

C. 15

D. 9


Answer:

For ∆XYZ and ∆PQR, the correspondence XYZ ↔ PQR has similarity.


Also, given that


XY = 12,


YZ = 8,


ZX = 16 &


PR = 8


To find: PQ + QR = ?


Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ 


⇒ 


Or 


⇒ 


⇒ 


⇒ 


⇒ 


⇒ PQ + QR = 10


Thus, option (b) is correct.


Question 21.

The bisector of ∠P in ΔPQR intersects  in D. If QD: RD = 4 : 7 and PR = 14, then PQ = ………..
A. 8

B. 4

C. 12

D. 16


Answer:

We have



Given: PD is the bisector of ∠P.


QD:RD = 4:7


PR = 14


To find: PQ = ?


In ∆PQR, the bisector of P intersects the line QR at D. By property of triangles,


⇒ 


⇒ 


⇒ 


⇒ PQ = 2 × 4


⇒ PQ = 8


Thus, option (a) is correct.


Question 22.

The lengths of the sidesof ΔABC are in the ratio 3 : 4 : 5. Correspondence ABC ↔ PQR is similarity. If PR = 12, the perimeter of ΔPQR is ……….
A. 12

B. 36

C. 24

D. 27


Answer:

Given: In ∆ABC,


BC:CA:AB = 3:4:5


From ∆ABC and ∆PQR, the correspondence ABC ↔ PQR is similarity.


In ∆PQR,


PR = 12


To find: Perimeter of ∆PQR = ?


In ∆ABC, since we have BC:CA:AB = 3:4:5


Let the lengths of BC, CA and AB be 3t, 4t and 5t respectively. (where, t > 0)


Perimeter of ∆ABC = 3t + 4t + 5t


⇒ Perimeter of ∆ABC = 12t, t > 0


Also, from given we have,


In ∆ABC and ∆PQR, the correspondence ABC ↔ PQR is similarity.


So, from property we can say that,


Ratio of perimeters of ∆ABC and ∆PQR = Ratio of their corresponding sides


⇒ 


Substituting the given values, we get



⇒ 


⇒ Perimeter of ∆PQR = 36


Thus, option (b) is correct.


Question 23.

The correspondence ABC ↔ YZX in ΔABC and ΔXYZ is similarity. m∠B + m∠C = 120. So, m∠Y = ………
A. 70

B. 55

C. 110

D. 60


Answer:

Given that,


From ∆ABC and ∆XYZ, the correspondence ABC ↔ YZX is similarity.


Also, m ∠B + m ∠C = 120°


To find: m ∠Y = ?


In ∆ABC, by angle sum property of triangles we can say that,


m ∠A + m ∠B + m ∠C = 180°


⇒ m ∠A + (m ∠B + m ∠C) = 180°


⇒ m ∠A + 120° = 180°


⇒ m ∠A = 180° - 120°


⇒ m ∠A = 60° …(i)


Now, from ∆ABC and ∆XYZ, the correspondence ABC ↔ YZX is similarity.


Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ ∠A ≅ ∠Y


∴ m ∠A = m ∠Y


From equation (i), we get


m ∠A = m ∠Y = 60°


∴ m ∠Y = 60°


Thus, option (d) is correct.


Question 24.

Correspondence ABC ↔ DEF of  ABC and ΔDEF is similarity. If AB + BC = 10 and DE + EF = 12 and AC = 6, then DF………..
A. 6

B. 5

C. 7.2

D. 16


Answer:

Given: AB + BC = 10,


DE + EF = 12 &


AC = 6


The correspondence ABC ↔ DEF is similarity between ∆ABC and ∆DEF.


To find: DF = ?


From ∆ABC and ∆DEF, the correspondence ABC ↔ DEF is a similarity.


Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ 


⇒ 


⇒ 


By reciprocating on both sides,


⇒ 


⇒ 


⇒ 


⇒ DF = 7.2


Thus, option (c) is correct.


Question 25.

The lengths of the sides of ΔDEF are 4, 6, 8. ΔDEF ~ ΔPQR for correspondence DEF ↔ QPR. If the perimeter of ΔPQR = 36, then the length of the smallest side of ΔPQR is ………..
A. 6

B. 2

C. 4

D. 8


Answer:

Given: Lengths of sides of ∆DEF are 4, 6 and 8.


∆DEF ∼ ∆PQR for the correspondence DEF ↔ PQR.


Perimeter of ∆PQR = 36


To find: length of the smallest side of ∆PQR = ?


Since, sides of ∆DEF are 4, 6 and 8.


⇒ Perimeter of ∆DEF = 4 + 6 + 8


⇒ Perimeter of ∆DEF = 18


Since, ∆DEF ∼ ∆PQR for the correspondence DEF ↔ PQR.


∴ 


⇒ 


⇒ 


⇒ 


⇒ Smallest side of ∆PQR = 4 × 2


⇒ Smallest side of ∆PQR = 8


Thus, option (d) is correct.


Question 26.

The bisector of ∠B intersects  in D. If BA = 12 and BC = 16, AD = 9, then AC = ……
A. 15

B. 21

C. 18

D. 8


Answer:

We have



Given: ∠ABD = ∠CBD


BA = 12,


BC = 16 &


AD = 9


To find: AC = ?


∵, In ∆ABC bisector of ∠B intersects AC at D.


By theorem of proportionality, we have



By reciprocating on both sides,


⇒ 


Substituting the values, we get


⇒ 


⇒ 


⇒ CD = 12


We can express AC as,


AC = AD + CD


⇒ AC = 9 + 12


⇒ AC = 21


Thus, option (b) is correct.


Question 27.

In Δ ABC,  ||, P ∈ , Q ∈ . If 4AP = AB and AQ = 4, then AC =
A. 12

B. 4

C. 8

D. 16


Answer:

We have,



Given: In ∆ABC,


PQ ∥ BC,


4 AP = AB &


AQ = 4


To find: AC = ?


We know, 4 AP = AB


⇒  …(i)


In ∆ABC, we know that PQ ∥ BC.



⇒ 


⇒ 


⇒ 


⇒ AC = 4 × 4


⇒ AC = 16


Thus, the option (d) is correct.


Question 28.

In ΔABC, the correspondence ABC ↔ BAC is similarity …….. of the following is true.
A. ∠B ≅ ∠C

B. ∠C ≅ ∠A

C. ∠A ≅ ∠ B

D. ∠A ≅ ∠B ≅ ∠C


Answer:

Given: In ∆ABC, the correspondence ABC ↔ BAC is similarity.


Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


∠A ≅ ∠B, ∠B ≅ ∠A and ∠C ≅ ∠C are all true.


Thus, option (c) is true.


Question 29.

In ΔABC, A—P—B, A—Q—C and  ||. If PQ = 5, AP = 4, AB = 12, then BC = …….
A. 9.6

B. 20

C. 15

D. 5


Answer:

We have



In ∆ABC,


PQ ∥ BC


PQ = 5,


AP = 4 &


AB = 12


In ∆ABC and ∆APQ, we have


∠A = ∠A [∵, ∠BAC and ∠PAQ are equal angles as they are same angles]


∠ABC = ∠APQ [∵, Alternate angles are equal]


∠ACB = ∠AQP [∵, Alternate angles are equal]


⇒ By AAA-corollary of similarity, ∆ABC ∼ ∆APQ for the correspondence ABC ↔ APQ.


Now by definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.


⇒ 


⇒ 


⇒ BC = 5 × 3


⇒ BC = 15


Thus, option (c) is correct.


Question 30.

In Δ PQR, P—M—Q and P—N—R. If PQ = 18, PM = 12, PR = 9 and NR = …, then  || 
A. 

B. 3

C. 24

D. 6


Answer:

We have



Given: P – M – Q and P – N – R


In ∆PQR, MN ∥ QR.


PQ = 18,


PM = 12 &


PR = 9


To find: NR = ?


In ∆PQR, when MN ∥ QR.


⇒ 


Substituting the given values in the above equation,



⇒ 


⇒ PN = 6


Observe the diagram,


PR = PN + NR


⇒ NR = PR – PN


⇒ NR = 9 – 6


⇒ NR = 3


Thus, option (b) is correct.


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