# CBSE Solutions for Class 10 Maths

#### Select CBSE Solutions for class 10 Subject & Chapters Wise :

D and E are points on the sides AB and AC, respectively, of a ΔABC, such that DEBC. If AD = 3.6 cm, AB = 10 cm and AE = 4.5 cm, find EC and AC.

In  ABC, it is given that DEBC.
Applying Thales' theorem, we get:
AD = 3.6 cm , AB = 10 cm, AE = 4.5 cm​
DB = 10 −- 3.6 = 6.4 cm
or, 3.6/6.4= 4.5EC

or, EC = 6.4×4.5/3.6

or, EC =8 cm

D and E are points on the sides AB and AC, respectively, of a ΔABC, such that DEBC. If AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm, find AD.

In ABC, it is given that DE BC.

Applying Thales' Theorem, we get:

Adding 1 to both sides, we get:

AB/DB= AC/EC

13.3/DB = 11.9/5.1

DB = 13.3×5.1/11.9 = 5.7 cm

Therefore, AD = AB − DB = 13.5 − 5.7 = 7.6 cm

D and E are points on the sides AB and AC, respectively, of a ΔABC, such that DEBC. If AD/DB=4/7 and AC=6.6 cm, find AE.

In ABC, it is given that DEBC.

Applying Thales' theorem, we get:

4/7= AE/EC

Adding 1 to both the sides, we get:

11/7= AC/EC

EC = 6.6×7/11 = 4.2 cm

Therefore, AE = AC −EC= 6.6−4.2 = 2.4 cm

D and E are points on the sides AB and AC, respectively, of a ΔABC, such that DEBC. If AD/AB=8/15 and EC=3.5 cm, find AE.

In ABC, it is given that DEBC.

Applying Thales' theorem, we get:

8/15= AE/AE + EC

8/15 = AE/(AE+3.5)

8AE + 28 = 15AE

7AE = 28

AE = 4 cm

D and E are points on the sides AB and AC respectively of a ΔABc such that DEBC. Find the value of x, when AD=x cm, DB=(x−2)cm,AE=(x+2) cm and EC=(x−1) cm.

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have:

x/x−2=x+2/x−1

x(x−1) = (x−2)(x+2)

x2−x = x2−4

x=4 cm

D and E are points on the sides AB and AC respectively of a ΔABc such that DEBC. Find the value of x, when AD=4 cm, DB=(x−4) cm, AE=8 cmand EC=(3x−19) cm.

In ABC, it is given that DEBC.

Applying Thales' theorem, we have:

4x−4 = 83x−19

4(3x−19) = 8(x−4)

12x −76 = 8x – 32

4x = 44

x = 11 cm

D and E are points on the sides AB and AC respectively of a ΔABc such that DEBC. Find the value of x, when DB=(3x+4) cm and EC=3x cm.

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have:

7x−43x+4 = 5x−23x

3x(7x−4) =(5x−2)(3x+4)

21x2 − 12x = 15x2 +14 x−8

6x2−26x+8 = 0

(x−4)(6x−2) = 0

x = 4, 13∵ x≠13     (as if x=13 then AE will become negative)

x =4 cm

and E are points on the sides AB and AC respectively of a ΔABC. In each of the following cases, determine whether DEBC or not. AD=5.7 cm, DB=9.5 cm, BD=4.8 cm and EC=8 cm.AD=5.7 cm, DB=9.5 cm, BD=4.8 cm and EC=8 cm.

We have:
AD/DB = 5.7/9.5 = 0.6 cm

AE/EC= 4.8/8 = 0.6 cm

Applying the converse of Thales' theorem, we conclude that DEBC

and E are points on the sides AB and AC respectively of a ΔABC. In each of the following cases, determine whether DEBC or not. AB=11.7 cm, AC=11.2 cm, BD=6.5 cm and AE=4.2 cm.AB=11.7 cm, AC=11.2 cm, BD=6.5 cm and AE=4.2 cm.

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

AE/EC = 4.2/7

Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.

and E are points on the sides AB and AC respectively of a ΔABC. In each of the following cases, determine whether DEBC or not. AB=10.8 cm, AD=6.3 cm, AC=9.6 cm and EC=4 cm.AB=10.8 cm, AD=6.3 cm, AC=9.6 cm and EC=4 cm.

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 = 4 cm
Therefore,
AE = 9.6 −- 4 = 5.6 cm
Now,
AE/EC=5.6/4=7/5
Applying the converse of Thales' theorem,
we conclude that DE∥BC.

and E are points on the sides AB and AC respectively of a ΔABC. In each of the following cases, determine whether DEBC or not. AD=7.2 cm, AE=6.4 cm, AB=12 cm and AC=10 cm.AD=7.2 cm, AE=6.4 cm, AB=12 cm and AC=10 cm.

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,
AE/EC = 6.4/3.6= 16/9
Applying the converse of Thales' theorem,
we conclude that DE is not parallel to BC.

In a ΔABC,  AD is the bisector or A. If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.

It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC=AB/AC
⇒5.6/DC=6.4/8
⇒DC = 8×5.6/6.4 = 7 cm

In a ΔABC,  AD is the bisector or A. If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC =AB/AC
Let BD be x cm.
Therefore, DC = (6−x) cm
⇒x6−x = 10/14
⇒14x = 60−10x
⇒24x = 60
⇒x = 2.5 cm
Thus, BD = 2.5 cm
DC = 6−2.5 = 3.5 cm

In a ΔABC,  AD is the bisector or A. If AB = 5.6 cm, BD = 3.8 cm and BC = 6 cm, find AC.

It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC=AB/AC
BD = 3.2 cm, BC = 6 cm
Therefore, DC = 6−3.2 = 2.8 cm
⇒3.2/2.8=5.6/AC

⇒AC = 5.6×2.8/3.2=4.9 cm

In a ΔABC,  AD is the bisector or A. If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

It is given that AD bisects ∠A.
Applying angle−bisector theorem in △ABC, we get:
BD/DC = AB/AC
⇒BD/3 = 5.6/4
⇒BD = 5.6×3/4
⇒BD = 4.2 cm
Hence,  BC = 3 + 4.2 = 7.2 cm

M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that DM/MN=DC/BN Given: ABCD is a parallelogram
To prove: DM/MN=DC/BN
Proof: In △DMC and △NMB
∠DMC =∠NMB      (Vertically opposite angle)
∠DCM =∠NBM       (Alternate angles)
By AAA- similarity
△DMC ~ △NMB
∴DM/MN=DC/BN

A 13 m long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.

Let AB and AC be the ladder and height of the building.
It is given that:
AB = 13 m and AC = 12 m
We need to find the distance of the foot of the ladder from the building, i.e, BC.
In right-angled triangle ABC, we have: AB2 = AC2 + BC2

BC = √ (132 − 122) = √ (169 – 144) = √25 = 5m
Hence, the distance of the foot of the ladder from the building is 5 m 1    ∠BAC A    ∠ADE 2    ∠ABC B    ∠AED 3    ∠ACB C    ∠CAD 4    ∠CED D    180ᵒ - ∠BCE

1-C, 2-A, 3-B, 4-D

ABCD is a parallelogram. 1    ∠MCD A    ∠BMN 2    ∠CDM B    ∠MBN 3    ∠CMD C    ∠ADC 4    ∠ABC D    ∠MNB

1-B, 2-D, 3-A, 4-C

AB ӀӀ EF ӀӀ CD , ∠B = ∠A = 60ᵒ 1    ∠PDC A    60ᵒ 2    ∠DCF B    ∠DEF 3    AP C    ∠BFE 4    ∠P D    BP

1-B, 2-C, 3-D, 4-A 1    ∠B A    ∠ANM 2    BM B    180ᵒ - ∠B 3    ∠M C    CN 4    ∠C D    ∠C

1-D, 2-C, 3-B, 4-A

BC ӀӀ EF 1    ∠EFC A    ∠BOC 2    ∠FEB B    ∠EOC 3    ∠FOE C    ∠EBC 4    ∠BOF D    ∠BCF

1-D, 2-C, 3-A, 4-B