# CBSE Solutions for Class 10 Maths

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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.

Answer :

In  ABC, it is given that DEBC.
Applying Thales' theorem, we get:
AD/DB = AE/EC
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.

Answer :

In ABC, it is given that DE BC.

Applying Thales' Theorem, we get:

AD/DB = AE/EC

Adding 1 to both sides, we get:

AD/DB +1 = AE/EC+1

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.

Answer :

In ABC, it is given that DEBC.

Applying Thales' theorem, we get:

AD/DB = AE/EC

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.

Answer :

In ABC, it is given that DEBC.

Applying Thales' theorem, we get:

AD/AB=AE/AC

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.

Answer :

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have:

AD/DB = AE/EC

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.

Answer :

In ABC, it is given that DEBC.

Applying Thales' theorem, we have:

AD/DB  = AE/EC

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 AD=(7x−4) cm, AE=(5x−2) cm,

DB=(3x+4) cm and EC=3x cm.

Answer :

In △ABC, it is given that DE∥BC.

Applying Thales' theorem, we have:

AD/DB = AE/EC

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.

Answer :

We have:
AD/DB = 5.7/9.5 = 0.6 cm

AE/EC= 4.8/8 = 0.6 cm

Hence,AD/DB=AE/EC

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.

Answer :

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,AD/DB = 5.2/6.5=4/5

AE/EC = 4.2/7

Thus, AD/DB≠AE/EC

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.

Answer :

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,
AD/DB=6.3/4.5=7/5
AE/EC=5.6/4=7/5
⇒AD/DB=AE/EC
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.

Answer :

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,
AD/DB = 7.2/4.8=3/2
AE/EC = 6.4/3.6= 16/9
Thus, AD/DB≠AE/EC
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.

Answer :

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.

Answer :

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.

Answer :

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.

Answer :

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 Answer :

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.

Answer :

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

Answer :

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

Answer :

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

Answer :

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

Answer :

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

Answer :

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

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### Chapter 6 : Triangles

Triangle is one of the most interesting and exciting chapters of the unit Geometry as it takes us through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles including the detailed explanation of similar figure, different theorems related to the similarities of triangles with proof, and the areas of similar triangles. The chapter concludes by explaining the Pythagoras theorem and the ways to use it in solving problems.

### Browse & Download CBSE Books For Class 10 - All Subjects

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.