Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  RD Sharma Solutions: Triangles (Exercise 6B)

Triangles (Exercise 6B) RD Sharma Solutions | Mathematics (Maths) Class 10 PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 Page 1


 
Exercise 4B
1. In each of the given pairs of triangles, find which pair of triangles are similar. State the
similarity criterion and write the similarity relation in symbolic form:
Sol:
(i)
We have:
BAC = PQR = 
ABC = QPR = 
ACB = PRQ = 
QPR
(ii)
We have:
But , EDF (Included angles are not equal)
Thus, this  triangles are not similar.
(iii)
We have:
Also, ACB = PQR = 
-
(iv)
We have
Page 2


 
Exercise 4B
1. In each of the given pairs of triangles, find which pair of triangles are similar. State the
similarity criterion and write the similarity relation in symbolic form:
Sol:
(i)
We have:
BAC = PQR = 
ABC = QPR = 
ACB = PRQ = 
QPR
(ii)
We have:
But , EDF (Included angles are not equal)
Thus, this  triangles are not similar.
(iii)
We have:
Also, ACB = PQR = 
-
(iv)
We have
 
-
(v)
A + (Angle Sum Property)
2. BOC = 115
0
and CDO = 70
0
.
Find (i) DCO (ii) DCO (iii) OAB (iv) OBA.
Sol:
(i)
It is given that DB is a straight line.
Therefore,
(ii)
Therefore,
(iii)
-
Therefore,
(iv)
-
Therefore,
Page 3


 
Exercise 4B
1. In each of the given pairs of triangles, find which pair of triangles are similar. State the
similarity criterion and write the similarity relation in symbolic form:
Sol:
(i)
We have:
BAC = PQR = 
ABC = QPR = 
ACB = PRQ = 
QPR
(ii)
We have:
But , EDF (Included angles are not equal)
Thus, this  triangles are not similar.
(iii)
We have:
Also, ACB = PQR = 
-
(iv)
We have
 
-
(v)
A + (Angle Sum Property)
2. BOC = 115
0
and CDO = 70
0
.
Find (i) DCO (ii) DCO (iii) OAB (iv) OBA.
Sol:
(i)
It is given that DB is a straight line.
Therefore,
(ii)
Therefore,
(iii)
-
Therefore,
(iv)
-
Therefore,
 
3. If AB = 8cm, BO = 6.4cm, OC = 3.5cm and
CD = 5cm, find (i) OA (ii) DO.
Sol:
(i) Let OA be X cm.
-
Hence, OA = 5.6 cm
(ii) Let OD be Y cm
-
Hence, DO = 4 cm
4. In the given figure, if ADE = B, show that  
BE = 2.1cm and BC = 4.2cm, find DE.
Sol:
Given :
Let DE be X cm
-
5. The perimeter of two similar triangles ABC and PQR are 32cm and 24cm respectively. If PQ
= 12cm, find AB.
Sol:
It is given that triangles ABC and PQR are similar.
Therefore,
Page 4


 
Exercise 4B
1. In each of the given pairs of triangles, find which pair of triangles are similar. State the
similarity criterion and write the similarity relation in symbolic form:
Sol:
(i)
We have:
BAC = PQR = 
ABC = QPR = 
ACB = PRQ = 
QPR
(ii)
We have:
But , EDF (Included angles are not equal)
Thus, this  triangles are not similar.
(iii)
We have:
Also, ACB = PQR = 
-
(iv)
We have
 
-
(v)
A + (Angle Sum Property)
2. BOC = 115
0
and CDO = 70
0
.
Find (i) DCO (ii) DCO (iii) OAB (iv) OBA.
Sol:
(i)
It is given that DB is a straight line.
Therefore,
(ii)
Therefore,
(iii)
-
Therefore,
(iv)
-
Therefore,
 
3. If AB = 8cm, BO = 6.4cm, OC = 3.5cm and
CD = 5cm, find (i) OA (ii) DO.
Sol:
(i) Let OA be X cm.
-
Hence, OA = 5.6 cm
(ii) Let OD be Y cm
-
Hence, DO = 4 cm
4. In the given figure, if ADE = B, show that  
BE = 2.1cm and BC = 4.2cm, find DE.
Sol:
Given :
Let DE be X cm
-
5. The perimeter of two similar triangles ABC and PQR are 32cm and 24cm respectively. If PQ
= 12cm, find AB.
Sol:
It is given that triangles ABC and PQR are similar.
Therefore,
 
6. The corresponding sides of two similar triangles ABC and DEF are BC = 9.1cm and EF =
Sol:
-
Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their
corresponding sides.
m
Therefore,
7. In the given figure, CAB = 90
0
and AD BC. 
AB = 1m and BC = 1.25m, find AD.
Sol:
-
AD = 
= 0.6 m or 60 cm
8. In the given figure, ABC = 90
0
and BD AC.
If AB = 5.7cm, BD = 3.8cm and CD = 5.4cm,
find BC.
Sol:
It is given that ABC  is a right angled triangle and BD is the altitude drawn from the right
angle to the hypotenuse.
Page 5


 
Exercise 4B
1. In each of the given pairs of triangles, find which pair of triangles are similar. State the
similarity criterion and write the similarity relation in symbolic form:
Sol:
(i)
We have:
BAC = PQR = 
ABC = QPR = 
ACB = PRQ = 
QPR
(ii)
We have:
But , EDF (Included angles are not equal)
Thus, this  triangles are not similar.
(iii)
We have:
Also, ACB = PQR = 
-
(iv)
We have
 
-
(v)
A + (Angle Sum Property)
2. BOC = 115
0
and CDO = 70
0
.
Find (i) DCO (ii) DCO (iii) OAB (iv) OBA.
Sol:
(i)
It is given that DB is a straight line.
Therefore,
(ii)
Therefore,
(iii)
-
Therefore,
(iv)
-
Therefore,
 
3. If AB = 8cm, BO = 6.4cm, OC = 3.5cm and
CD = 5cm, find (i) OA (ii) DO.
Sol:
(i) Let OA be X cm.
-
Hence, OA = 5.6 cm
(ii) Let OD be Y cm
-
Hence, DO = 4 cm
4. In the given figure, if ADE = B, show that  
BE = 2.1cm and BC = 4.2cm, find DE.
Sol:
Given :
Let DE be X cm
-
5. The perimeter of two similar triangles ABC and PQR are 32cm and 24cm respectively. If PQ
= 12cm, find AB.
Sol:
It is given that triangles ABC and PQR are similar.
Therefore,
 
6. The corresponding sides of two similar triangles ABC and DEF are BC = 9.1cm and EF =
Sol:
-
Therefore, their corresponding sides will be proportional.
Also, the ratio of the perimeters of similar triangles is same as the ratio of their
corresponding sides.
m
Therefore,
7. In the given figure, CAB = 90
0
and AD BC. 
AB = 1m and BC = 1.25m, find AD.
Sol:
-
AD = 
= 0.6 m or 60 cm
8. In the given figure, ABC = 90
0
and BD AC.
If AB = 5.7cm, BD = 3.8cm and CD = 5.4cm,
find BC.
Sol:
It is given that ABC  is a right angled triangle and BD is the altitude drawn from the right
angle to the hypotenuse.
 
By AA similarity theorem, we get :
-
Hence, BC = 8.1 cm
9. In the given figure, ABC = 90
0
and BD AC.
If BD = 8cm, AD = 4cm, find CD.
Sol:
It is given that ABC is a right angled triangle
and BD is the altitude drawn from the right angle to the hypotenuse.
Therefore, by AA similarity theorem, we get :
-
CD = 
10. P and Q are
AQ = 3cm and QC = 6cm, show that BC = 3PQ.
Sol:
We have :
Therefore, by AA similarity theorem, we get:
-
Hence, 
BC = 3PQ
Read More
124 videos|457 docs|77 tests

Top Courses for Class 10

124 videos|457 docs|77 tests
Download as PDF
Explore Courses for Class 10 exam

Top Courses for Class 10

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Summary

,

Sample Paper

,

Previous Year Questions with Solutions

,

Objective type Questions

,

pdf

,

Important questions

,

MCQs

,

past year papers

,

Triangles (Exercise 6B) RD Sharma Solutions | Mathematics (Maths) Class 10

,

Semester Notes

,

shortcuts and tricks

,

practice quizzes

,

video lectures

,

Free

,

Triangles (Exercise 6B) RD Sharma Solutions | Mathematics (Maths) Class 10

,

study material

,

Viva Questions

,

ppt

,

Exam

,

mock tests for examination

,

Extra Questions

,

Triangles (Exercise 6B) RD Sharma Solutions | Mathematics (Maths) Class 10

;