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# RS Aggarwal Solutions: Exercise 6B - Triangles Class 10 Notes | EduRev

## Class 10 : RS Aggarwal Solutions: Exercise 6B - Triangles Class 10 Notes | EduRev

``` 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
```
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## Mathematics (Maths) Class 10

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