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

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

``` Page 1

1.
(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
(ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD.
(iii) If = and AC = 6.6cm, find AE.
(iv) If = and EC = 3.5cm, find AE.
Sol:
(i) BC.
AD = 3.6 cm , AB = 10 cm, AE = 4.5cm
DB = 10  3.6 = 6.4cm
Or,
Or, EC =
Or, EC= 8 cm
Thus, AC = AE + EC
= 4.5 + 8 = 12.5 cm
(ii) t is given that DE || BC.
Adding 1 to both sides, we get :
+1
(iii) BC.
Adding 1 to both the sides, we get :
EC = = 4.2 cm
Therefore,
Page 2

1.
(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
(ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD.
(iii) If = and AC = 6.6cm, find AE.
(iv) If = and EC = 3.5cm, find AE.
Sol:
(i) BC.
AD = 3.6 cm , AB = 10 cm, AE = 4.5cm
DB = 10  3.6 = 6.4cm
Or,
Or, EC =
Or, EC= 8 cm
Thus, AC = AE + EC
= 4.5 + 8 = 12.5 cm
(ii) t is given that DE || BC.
Adding 1 to both sides, we get :
+1
(iii) BC.
Adding 1 to both the sides, we get :
EC = = 4.2 cm
Therefore,

AE = AC EC = 6.6 4.2 = 2.4 cm
(iv)
8AE + 28 = 15AE
7AE = 28
AE = 4cm
2. D and E are
the value of x, when
(i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm.
(ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm.
(iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm.
Sol:
(i) || BC.
:
(ii)
:
=
4 (3x-19) = 8 (x-4)
12x 76 = 8x 32
4x = 44
x = 11 cm
(iii) || BC.
3x (7x-4) = (5x-2) (3x+4)
Page 3

1.
(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
(ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD.
(iii) If = and AC = 6.6cm, find AE.
(iv) If = and EC = 3.5cm, find AE.
Sol:
(i) BC.
AD = 3.6 cm , AB = 10 cm, AE = 4.5cm
DB = 10  3.6 = 6.4cm
Or,
Or, EC =
Or, EC= 8 cm
Thus, AC = AE + EC
= 4.5 + 8 = 12.5 cm
(ii) t is given that DE || BC.
Adding 1 to both sides, we get :
+1
(iii) BC.
Adding 1 to both the sides, we get :
EC = = 4.2 cm
Therefore,

AE = AC EC = 6.6 4.2 = 2.4 cm
(iv)
8AE + 28 = 15AE
7AE = 28
AE = 4cm
2. D and E are
the value of x, when
(i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm.
(ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm.
(iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm.
Sol:
(i) || BC.
:
(ii)
:
=
4 (3x-19) = 8 (x-4)
12x 76 = 8x 32
4x = 44
x = 11 cm
(iii) || BC.
3x (7x-4) = (5x-2) (3x+4)

21 12x = 15 + 14x-8
6 26x + 8 = 0
(x-4) (6x-2) =0
x = 4,
(as if x =
x = 4 cm
3. D and E are points on the sides AB and AC
(i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm.
(ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm.
(iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm.
(iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm.
Sol:
(i) We have:
Hence,
We conclude that DE || BC.
(ii) We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 -6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 4.2 = 7 cm
Now,
Thus,
We conclude that DE is not parallel to BC.
(iii) We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,
Page 4

1.
(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
(ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD.
(iii) If = and AC = 6.6cm, find AE.
(iv) If = and EC = 3.5cm, find AE.
Sol:
(i) BC.
AD = 3.6 cm , AB = 10 cm, AE = 4.5cm
DB = 10  3.6 = 6.4cm
Or,
Or, EC =
Or, EC= 8 cm
Thus, AC = AE + EC
= 4.5 + 8 = 12.5 cm
(ii) t is given that DE || BC.
Adding 1 to both sides, we get :
+1
(iii) BC.
Adding 1 to both the sides, we get :
EC = = 4.2 cm
Therefore,

AE = AC EC = 6.6 4.2 = 2.4 cm
(iv)
8AE + 28 = 15AE
7AE = 28
AE = 4cm
2. D and E are
the value of x, when
(i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm.
(ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm.
(iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm.
Sol:
(i) || BC.
:
(ii)
:
=
4 (3x-19) = 8 (x-4)
12x 76 = 8x 32
4x = 44
x = 11 cm
(iii) || BC.
3x (7x-4) = (5x-2) (3x+4)

21 12x = 15 + 14x-8
6 26x + 8 = 0
(x-4) (6x-2) =0
x = 4,
(as if x =
x = 4 cm
3. D and E are points on the sides AB and AC
(i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm.
(ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm.
(iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm.
(iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm.
Sol:
(i) We have:
Hence,
We conclude that DE || BC.
(ii) We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 -6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 4.2 = 7 cm
Now,
Thus,
We conclude that DE is not parallel to BC.
(iii) We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,

DB = 10.8 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4cm
Therefore,
AE = 9.6 4 = 5.6 cm
Now,
(iv) We have :
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 7.2 = 4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 6.4 = 3.6 cm
Now,
This,
heorem,
We conclude that DE is not parallel to BC.
4. In a ABC, AD is the bisector of A.
(i) If AB = 6.4cm, AC = 8cm and BD = 5.6cm, find DC.
(ii) If AB = 10cm, AC = 14cm and BC = 6cm, find BD and DC.
(iii) If AB = 5.6cm, BD = 3.2cm and BC = 6cm, find AC.
(iv) If AB = 5.6cm, AC = 4cm and DC = 3cm, find BC.
Sol:
(i) It is give that AD bisects  A.
Applying angle
Page 5

1.
(i) If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
(ii) If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD.
(iii) If = and AC = 6.6cm, find AE.
(iv) If = and EC = 3.5cm, find AE.
Sol:
(i) BC.
AD = 3.6 cm , AB = 10 cm, AE = 4.5cm
DB = 10  3.6 = 6.4cm
Or,
Or, EC =
Or, EC= 8 cm
Thus, AC = AE + EC
= 4.5 + 8 = 12.5 cm
(ii) t is given that DE || BC.
Adding 1 to both sides, we get :
+1
(iii) BC.
Adding 1 to both the sides, we get :
EC = = 4.2 cm
Therefore,

AE = AC EC = 6.6 4.2 = 2.4 cm
(iv)
8AE + 28 = 15AE
7AE = 28
AE = 4cm
2. D and E are
the value of x, when
(i) AD = x cm, DB = (x 2) cm, AE = (x + 2) cm and EC = (x 1) cm.
(ii) AD = 4cm, DB = (x 4) cm, AE = 8cm and EC = (3x 19) cm.
(iii) AD = (7x 4) cm, AE = (5x 2) cm, DB = (3x + 4) cm and EC = 3x cm.
Sol:
(i) || BC.
:
(ii)
:
=
4 (3x-19) = 8 (x-4)
12x 76 = 8x 32
4x = 44
x = 11 cm
(iii) || BC.
3x (7x-4) = (5x-2) (3x+4)

21 12x = 15 + 14x-8
6 26x + 8 = 0
(x-4) (6x-2) =0
x = 4,
(as if x =
x = 4 cm
3. D and E are points on the sides AB and AC
(i) AD = 5.7cm, DB = 9.5cm, AE = 4.8cm and EC = 8cm.
(ii) AB = 11.7cm, AC = 11.2cm, BD = 6.5cm and AE = 4.2cm.
(iii) AB = 10.8cm, AD = 6.3cm, AC = 9.6cm and EC = 4cm.
(iv) AD = 7.2cm, AE = 6.4cm, AB = 12cm and AC = 10cm.
Sol:
(i) We have:
Hence,
We conclude that DE || BC.
(ii) We have:
AB = 11.7 cm, DB = 6.5 cm
Therefore,
AD = 11.7 -6.5 = 5.2 cm
Similarly,
AC = 11.2 cm, AE = 4.2 cm
Therefore,
EC = 11.2 4.2 = 7 cm
Now,
Thus,
We conclude that DE is not parallel to BC.
(iii) We have:
AB = 10.8 cm, AD = 6.3 cm
Therefore,

DB = 10.8 6.3 = 4.5 cm
Similarly,
AC = 9.6 cm, EC = 4cm
Therefore,
AE = 9.6 4 = 5.6 cm
Now,
(iv) We have :
AD = 7.2 cm, AB = 12 cm
Therefore,
DB = 12 7.2 = 4.8 cm
Similarly,
AE = 6.4 cm, AC = 10 cm
Therefore,
EC = 10 6.4 = 3.6 cm
Now,
This,
heorem,
We conclude that DE is not parallel to BC.
4. In a ABC, AD is the bisector of A.
(i) If AB = 6.4cm, AC = 8cm and BD = 5.6cm, find DC.
(ii) If AB = 10cm, AC = 14cm and BC = 6cm, find BD and DC.
(iii) If AB = 5.6cm, BD = 3.2cm and BC = 6cm, find AC.
(iv) If AB = 5.6cm, AC = 4cm and DC = 3cm, find BC.
Sol:
(i) It is give that AD bisects  A.
Applying angle

(ii) It is given that AD bisects .
Applying angle bisector
Let BD be x cm.
Therefore, DC = (6- x) cm
(iii) It is given that AD bisector
Applying angle get:
BD = 3.2 cm, BC = 6 cm
Therefore, DC = 6- 3.2 = 2.8 cm
(iv) It is given that AD bisects
Applying angle
5. M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB
produced at N. Prove that
(i)
D M D C
M N B N
(ii)
D N A N
D M D C
Sol:
(i) Given: ABCD is a parallelogram
```
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## Mathematics (Maths) Class 10

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