NCERT Class 10 Maths Solutions Chapter - 6 Triangles, Ex - 6.3

Ex - 6.3

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:


Solution

(i)    A = P = 60�
B = Q = 80�
C = R = 40�
Therefore ABC ~ PQR     [by AAA rule]

(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 = 180�
(Sum of measures of angles of a triangle is 180�)
70� + 80� + F = 180�
F = 30� 
Similarly in PQR
P + Q + R = 180�
(Sum of measures of angles of a triangle is 180�)
P + 80� +30� = 180�
P = 70� 
Now In DEF and PQR
D = P = 70�
E = Q = 80�
F = R = 30�
Therefore DEF ~ PQR     [by AAA rule]

Question 2.  In figure , ODC ~ OBA, BOC = 125� and CDO = 70�. Find DOC, DCO and OAB


Solution

Since DOB is a straight line
Therefore DOC + COB = 180�
Therefore DOC = 180� - 125�
                 = 55�
In DOC,
DCO + CDO + DOC = 180�
DCO + 70� + 55� = 180�
DCO = 55�
Since ODC ~ OBA,
Therefore  OCD = OAB    [corresponding angles equal in similar triangles]

Therefore  OAB = 55�

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 

Solution

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)

Question 4.  

Solution

In PQR 
PQR = PRQ
Therefore PQ = PR    (i)
Given,

Question 5.  S and T are point on sides PR and QR of PQR such that P = RTS. Show that RPQ ~ RTS.

Solution

In RPQ and RST
RTS = QPS              [given]
R = R            [common angle]
RST = RQP                      [ Remaining angles]
Therefore RPQ ~ RTS    [by AAA rule]

Question 6.  In figure, If ABE ~ ACD, show that ADE ~ ABC.

Solution

Since ABE ~ ACD
Therefore     AB = AC               (1)
AD = AE                (2)
Now, in ADE and ABC,
Dividing equation (2) by (1)

Question 7.  In figure , altitudes AD and CE of ABC intersect each other at the point P. Show that:

Solution

(i)

In AEP and CDP
Since CDP = AEP = 90�
CPD = APE     (vertically opposite angles)
PCD = PAE    (remaining angle)
Therefore by AAA rule,
AEP ~ CDP 

(ii)   

In ABD and CBE
ADB = CEB = 90�
ABD = CBE     (common angle)
DAB = ECB    (remaining angle)
Therefore by AAA rule
ABD ~ CBE 

(iii)

In AEP and ADB
AEP = ADB = 90�
PAE = DAB    (common angle)
APE = ABD    (remaining angle)
Therefore by AAA rule
AEP ~ ADB 

(iv)

In PDC and BEC
PDC = BEC = 90�
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 ABE ~ CFB

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:

Solution

In ABC and AMP
ABC = AMP = 90�
A = A                 (common angle)
ACB = APM            (remaining angle)
Therefore ABC ~ AMP        (by AAA rule)

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  ABC and  EFG respectively. If  ABC ~ FEG, Show that:

Solution

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)

Question 11.  In 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 ABD ~ ECF

Solution

In ABD and ECF,
Given that AB = AC        (isosceles triangles)
So, ABD = ECF    
ADB = EFC = 90�
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 PQR. Show that ABC ~ PQR.

Solution

Median divides opposite side. 
 
 
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

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


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

ABC ~ PQR

Solution

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

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

So, height of tower will be 42 meters.

Question 16.  If AD and PM are medians of triangles ABC and PQR, respectively where ABC ~ PQR prove that

Solution

Since ABC ~ PQR
So, their respective sides will be in proportion
 
Also, A = P, B = Q, C = R                     (2)
Since, AD and PM are medians so they will divide their opposite sides in equal halves.
 
From equation (1) and (3) 
 
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)

Ex-6.4