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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

December 26, 2018 by Dattu

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 6 Triangles Class 10 NCERT Solutions Ex 6.3.

  • Triangles Class 10 Ex 6.1
  • Triangles Class 10 Ex 6.2
  • Triangles Class 10 Ex 6.4
  • Triangles Class 10 Ex 6.5
  • Triangles Class 10 Ex 6.6

 

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 6
Chapter NameTriangles
ExerciseEx 6.3
Number of Questions Solved16
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3

NCERT Solutions for Class 10 Maths

Page No: 138

Question 1. State which pairs of triangles in Figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 37

Solution:
(i)    ∠A = ∠P = 600
∠B = ∠Q = 800
∠C = ∠R = 400
Therefore ∆ABC ~ ∆PQR     [by AAA rule]
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 38
(iii)   Triangles are not similar as the corresponding sides are not proportional.
(iv)    Triangles are not similar as the corresponding sides are not proportional.
(v)    Triangles are not similar as the corresponding sides are not proportional.
(vi)    In ∆DEF
∠D + ∠E + ∠F = 1800
(Sum of measures of angles of a triangle is 1800)
700 + 800 + ∠F = 1800
∠F = 300
Similarly in ∠PQR
∠P + ∠Q + ∠R = 1800
(Sum of measures of angles of a triangle is 1800)
∠P + 800 +300 = 1800
∠P = 700
Now In ∆DEF and ∆PQR
∠D = ∠P = 700
∠E = ∠Q = 800
∠F = ∠R = 300
Therefore ∆DEF ~ ∆PQR     [by AAA rule]

Page No: 139

Question 2. In the fig 6.35, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 39

Solution:
Since DOB is a straight line
Therefore ∠DOC + ∠COB = 1800
Therefore ∠DOC = 1800 – 1250
= 550
In ∠DOC,
∠DCO + ∠CDO + ∠DOC = 1800
∠DCO + 700 + 550 = 1800
∠DCO = 550
Since ∆ODC ~ ∆OBA,
Therefore ∠OCD = ∠OAB    [corresponding angles equal in similar triangles]
Therefore ∠ OAB = 550

Question 3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac { AO }{ OC } =\frac { OB }{ OD } \)

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 40
In ∆DOC and ∆BOA
AB || CD
Therefore ∠CDO = ∠ABO    [Alternate interior angles]
∠DCO = ∠BAO         [Alternate interior angles]
∠DOC = ∠BOA        [Vertically opposite angles]
Therefore ∆DOC ~ ∆BOA    [AAA rule)
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 41

Page No: 140

Question 4. In the figure, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 42
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 43

Solution:
In ∆PQR
∠PQR = ∠PRQ
Therefore PQ = PR    (i)
Given,
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 44

Question 5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 45
In ∆RPQ and ∆RST
∠RTS = ∠QPS              [given]
∠R = ∠R            [common angle]
∠RST = ∠RQP                      [ Remaining angles]
Therefore ∆RPQ ~ ∆RTS    [by AAA rule]

Question 6. In the figure, if ΔABE ≅ ΔACD, show that  ΔADE ~ ΔABC.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 46

Solution:
Since ∆ABE ~ ∆ACD
Therefore
AB = AC ——(1)
AD = AE—— (2)
Now, in ∆ADE and ∆ABC,
Dividing equation (2) by (1)
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 47

Question 7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 48

Solution:
(i)
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 49
In ∆AEP and ∆CDP
Since ∠CDP = ∠AEP = 900
∠CPD = ∠APE     (vertically opposite angles)
∠PCD = ∠PAE    (remaining angle)
Therefore by AAA rule,
∆AEP ~ ∆CDP
(ii)
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 50
In ∆ABD and ∆CBE
∠ADB = ∠CEB = 900
∠ABD = ∠CBE     (common angle)
∠DAB = ∠ECB    (remaining angle)
Therefore by AAA rule
∆ABD ~ ∆CBE
(iii)
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 51
In ∆AEP and ∆ADB
∠AEP = ∠ADB = 900
∠PAE = ∠DAB    (common angle)
∠APE = ∠ABD    (remaining angle)
Therefore by AAA rule
∆AEP ~ ∆ADB
(iv)
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 52
In ∆PDC and ∆BEC
∠PDC = ∠BEC = 900
∠PCD = ∠BCE    (common angle)
∠CPD = ∠CBE
Therefore by AAA rule
∆PDC ~ ∆BEC

Question 8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Solution:
∆ABE and ∆CFB
∠A = ∠C             (opposite angles of a parallelogram)
∠AEB = ∠CBF         (Alternate interior angles AE || BC)
∠ABE = ∠CFB         (remaining angle)
Therefore ∆ABE ~ ∆CFB    (by AAA rule)

Question 9. In figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 53

Solution:
In ∆ABC and ∆AMP
∠ABC = ∠AMP = 900
∠A = ∠A                 (common angle)
∠ACB = ∠APM            (remaining angle)
Therefore ∆ABC ~ ∆AMP        (by AAA rule)
Therefore \(\frac { CA }{ PA } =\frac { BC }{ MP } \)     (Corresponding sides are proportional)

Question 10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:
(i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 54
Since ∆ABC ~ ∆FEG
Therefore ∠A = ∠F
∠B = ∠E
As, ∠ACB = ∠FGE
Therefore ∠ACD = ∠FGH    (angle bisector)
And ∠DCB = ∠HGE        (angle bisector)
Therefore ∆ACD ~ ∆FGH    (by AAA rule)
And ∆DCB ~ ∆HGE        (by AAA rule)
For ∆ ACD and ∆FGH
\(\frac { CD }{ GH } =\frac { AC }{ FG } \)

Page No: 141

Question 11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 55

Solution:
In ∆ABD and ∆ECF,
Given that AB = AC        (isosceles triangles)
So, ∠ABD = ∠ECF
∠ADB = ∠EFC = 900
∠BAD = ∠CEF
Therefore ∆ABD ~ ∆ECF     (by AAA rule)

Question 12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 56
Median divides opposite side. 
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 57
Therefore ∆ABD ~ ∆PQM    (by SSS rule)
Therefore ∠ABD = ∠PQM    (corresponding angles of similar triangles)
Therefore ∆ABC ~ ∆PQR     (by SAS rule)

Question 13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2 = CB.CD

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 59
In ∆ADC and ∆BAC
Given that ∠ADC = ∠BAC
∠ACD = ∠BCA            (common angle)
∠CAD = ∠CBA            (remaining angle)
Hence,  ∆ADC ~ ∆BAC        [by AAA rule]
So, corresponding sides of similar triangles will be proportional to each other
\(\frac { CA }{ CB } =\frac { CD }{ CA } \)
Hence CA2  = CB × CD

Question 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 60

Question 15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 61
Let AB be a tower
CD be a pole
Shadow of AB is BE
Shadow of CD is DF
The  light rays from sun will fall on tower and pole at same angle and at the same time.
So, ∠DCF = ∠BAE
And ∠DFC = ∠BEA
∠CDF = ∠ABE        (tower and pole are vertical to ground)
Therefore ∆ABE ~ ∆CDF
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 62
So, height of tower will be 42 meters.

Question 16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 63
Since ∆ABC ~ ∆PQR
So, their respective sides will be in proportion
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 64
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R                     (2)
Since, AD and PM are medians so they will divide their opposite sides in equal halves.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 65
From equation (1) and (3)
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 66
So, we had observed that two respective sides are in same proportion in both triangles and also angle included between them is respectively equal
Hence, ∆ABD ~ ∆PQM             (by SAS rule)
NCERT Solutions for Class 10 Maths Chapter 6 Triangles 67

 

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.3, drop a comment below and we will get back to you at the earliest.

Filed Under: CBSE Tagged With: NCERT Solutions for Class 10 Maths

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