KC Sinha Mathematics Solution Class 10 Chapter 5 Triangles Exercise 5.2

Exercise 5.1
Exercise 5.2
Exercise 5.3
Exercise 5.4
Exercise 5.5

Exercise 5.2

Question 1

In ΔABC, P and Q are two points on AB and AC respectively such that PQ || BC and $\frac{A P}{P B}=\frac{2}{3}$, then find$\frac{A Q}{Q C}$.

Sol :

Given: PQ || BC
and $\frac{A P}{P B}=\frac{2}{3}$
By Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.

$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$
$\Rightarrow \frac{A P}{P B}=\frac{A Q}{Q C}=\frac{2}{3}$

Question 2

In figures (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Sol :
(i)

Given: DE || BC
$\therefore \frac{A D}{D B}=\frac{A E}{E C}$
[by basic proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.]
$\Rightarrow \frac{1.5}{3}=\frac{1}{\mathrm{EC}}$ [given: AD= 1.5cm, DB =3cm & AE =1cm]
$\Rightarrow \mathrm{EC}=\frac{3}{1.5}$
$\Rightarrow \mathrm{EC}=\frac{3 \times 10}{15}$
$\Rightarrow \mathrm{EC}=\frac{30}{15}$
⇒ EC = 2cm

(ii)

Given: DB = 7.2 cm, AE = 1.8 cm and EC = 5.4 cm
and DE || BC
$\therefore \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
[by basic proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.]
$\Rightarrow \frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}$
$\Rightarrow A D=\frac{7.2 \times 1.8}{5.4}$
$\Rightarrow A D=\frac{72 \times 18}{54 \times 10}$
$\Rightarrow A D=\frac{24}{10}$
⇒ AD = 2.4cm

Question 3

In a ΔABC, DE || BC, where D is a point on AB and E is a point on AC, then

(i) $\frac{E C}{D B}$ =………

(ii)$\frac{A D}{A E}$ =………
(iii) $\frac{A B}{D B}$ =………

(iv) $\frac{E C}{D B}$=………

Sol :

(i) Given: DE || BC
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
$\therefore \frac{A D}{D B}=\frac{A E}{E C}$ [by basic proportionality theorem]

(ii) Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
By basic proportionality theorem, we know that
$\frac{A D}{D B}=\frac{A E}{E C}$
$\Rightarrow \frac{A D}{A B-A D}=\frac{A E}{A C-A E}$
$\Rightarrow \frac{A D}{A D\left(\frac{A B}{A D}-1\right)}=\frac{A E}{A E\left(\frac{A C}{A E}-1\right)}$
$\Rightarrow \frac{A B}{A D}-1=\frac{A C}{A E}-1$
$\Rightarrow \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}$
$\Rightarrow \frac{A D}{A E}=\frac{A B}{A C}$

(iii) From part (i), we know that $\frac{A D}{D B}=\frac{A E}{E C}$
On adding 1 to both the sides, we get
$\frac{A D}{D B}+1=\frac{A E}{E C}+1$
$\Rightarrow \frac{A D+D B}{D B}=\frac{A E+E C}{E C}$
$\Rightarrow \frac{A B}{D B}=\frac{A C}{E C}$

(iv) From part (iii), we have $\frac{A B}{D B}=\frac{A C}{E C}$
$\Rightarrow \frac{E C}{D B}=\frac{A C}{A B}$

Question 4

If in Δ ABC, DE || BC and DE cuts sides AB and AC at D and E respectively such that AD: DB = 4: 5, then find AE: EC.

Sol :

Given: DE || BC
and $\frac{A D}{D B}=\frac{4}{5}$
To find: AE: EC
Given: DE || BC
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
$\therefore \frac{A D}{D B}=\frac{A E}{E C}$ [by basic proportionality theorem]
$\Rightarrow \frac{A D}{D B}=\frac{A E}{E C}=\frac{4}{5}$

Question 5

In the adjoining figure, DE || BC. Find x.

Sol :
Given: AD = x
DB =16, AE = 34 and EC = 17
Given: DE || BC
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
So, by basic proportionality theorem
$\therefore \frac{A D}{D B}=\frac{A E}{E C}$
$\Rightarrow \frac{x}{16}=\frac{34}{17}$
$\Rightarrow \frac{\mathrm{x}}{16}=2$
⇒ x = 32

Question 6

In the adjoining figure, AD = 2 cm, DB = 3 cm, AE = 5 cm and DE || BC, then find EC.

Sol :
Given: AD = 2cm, DB =3cm, AE = 5cm
and DE || BC
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
So, by basic proportionality theorem
$\therefore \frac{A D}{D B}=\frac{A E}{E C}$
$\Rightarrow \frac{2}{3}=\frac{5}{\mathrm{EC}}$
$\Rightarrow \mathrm{EC}=\frac{5 \times 3}{2}$
⇒ EC = 7.5 cm

Question 7

In the adjoining figure, DE || BC, AD = 2.4 cm, AE = 3.2 cm, CE = 4.8 cm, find BD.

Sol :
Given: AD = 2.4cm, AE = 3.2cm and EC = 4.8cm
and DE || BC
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
So, by basic proportionality theorem
$\therefore \frac{A D}{D B}=\frac{A E}{E C}$
$\Rightarrow \frac{2.4}{\mathrm{DB}}=\frac{3.2}{4.8}$

$\Rightarrow \frac{2.4}{\mathrm{DB}}=\frac{2}{3}$

$\Rightarrow \mathrm{DB}=\frac{2.4 \times 3}{2}$

⇒ DB = 3.6 cm
or BD = 3.6 cm

Question 8

If DE has been drawn parallel to side BC of ΔABC cutting AB and AC at points D and E respectively, such that $\frac{A D}{D B}=\frac{3}{4}$, then find the value of $\frac{A E}{E C}$.

Sol :

Given: DE || BC
and $\frac{A D}{D B}=\frac{3}{4}$
To find : $\frac{A \mathrm{E}}{\mathrm{EC}}$
Given: DE || BC
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
$\therefore \frac{A D}{D B}=\frac{A E}{E C}$ [by basic proportionality theorem]
$\Rightarrow \frac{A D}{D B}=\frac{A E}{E C}=\frac{3}{4}$

Question 9

In the adjoining figure, P and Q are points on sides AB and AC respectively of mix such that PQ || BC and AP= 8 cm, AB =12 cm, AQ = 3x cm, QC = (x + 2) cm. Find x.

Sol :
Given: AP = 8cm , AB =12cm, AQ = (3x)cm and QC = (x+2)cm
and PQ || BC
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$ [by basic proportionality theorem]
$\Rightarrow \frac{A P}{A B-A P}=\frac{A Q}{Q C}$
$\Rightarrow \frac{8}{12-8}=\frac{3 x}{x+2}$
$\Rightarrow \frac{8}{4}=\frac{3 x}{x+2}$
$\Rightarrow 2=\frac{3 x}{x+2}$
⇒ 2(x+2) = 3x
⇒ 2x + 4 = 3x
⇒ 2x – 3x = - 4
⇒ x = 4

Question 10

In the adjoining figure, DE || BC, find x.

Sol :
Given: AD = 4, DB =x – 4, AE = x – 3 and EC = 3x – 19
and DE || BC
Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
So, by basic proportionality theorem
$\therefore \frac{A D}{D B}=\frac{A E}{E C}$
$\Rightarrow \frac{4}{x-4}=\frac{x-3}{3 x-19}$
⇒ 4(3x – 19) = (x – 4)(x – 3)
⇒ 12x – 76 = x² – 3x – 4x + 12
⇒ 12x – 76 = x² –7x + 12
⇒ x² –7x + 12 – 12x + 76 = 0
⇒ x² –19x + 88 = 0
Solving the Quadratic equation by splitting themiddle term, we get,
⇒ x² –11x – 8x + 88 = 0
⇒ x(x – 11) – 8(x – 11) = 0
⇒ (x – 8)(x – 11) =0
⇒ x = 8 and 11

Question 11

If D and E are points on sides AB and AC respectively of ΔABC and AB = 12 cm, AD = 8 cm, AE = 12 cm, AC =18 cm, then prove that DE || BC.

Sol :

Given: AB = 12cm, AD =8cm, AE = 12cm and AC = 18cm

To Prove: DE || BC
In ΔABC,

$\frac{A D}{D B}=\frac{A D}{A B-A D}$

$=\frac{8}{12-8}=\frac{8}{4}=2$

and $\frac{A E}{E C}=\frac{A E}{A C-E C}$
$=\frac{12}{18-12}=\frac{12}{6}=2$

Thus, $\frac{A D}{D B}=\frac{A E}{E C}$

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
Hence, DE || BC [by converse of basic proportionality theorem]
Hence, Proved.

Question 12

P and Q are points on sides AB and AC respectively of Δ ABC. For each of the following cases, state whether PQ || BC.

(i) AP= 8 cm, PB = 3 cm, AC = 22 cm and AQ =16 cm.
(ii) AB= 1.28 cm, AC = 2.56 cm, AP= 0.16 cm and AQ = 0.32 cm
(iii) AB = 5 cm, AC =10 cm, AP= 4 cm, AQ = 8 cm.
(iv) AP= 4 cm, PB= 4.5 cm, AQ = 4 cm, QC = 5 cm.
Sol :

(i) Given: AP= 8 cm, PB = 3 cm, AC = 22 cm and AQ =16 cm
To find: PQ || BC
In ΔABC,
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{8}{3}$
and $\frac{A Q}{Q C}=\frac{16}{A C-A Q}=\frac{16}{22-16}=\frac{16}{6}=\frac{8}{3}$

Thus ,

$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
Hence, PQ || BC [by converse of basic proportionality theorem]
Hence, Proved.

(ii) Given: AB= 1.28 cm, AC = 2.56 cm, AP= 0.16 cm and AQ = 0.32 cm
To find: PQ || BC
In ΔABC,
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{0.16}{\mathrm{AB}-\mathrm{AP}}=\frac{0.16}{1.28-0.16}=\frac{0.16}{1.12}=\frac{1}{7}$

and $\frac{A Q}{Q C}=\frac{16}{A C-A Q}=\frac{0.32}{2.56-0.32}=\frac{0.32}{2.24}=\frac{1}{7}$

Thus, $\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
Hence, PQ || BC [by converse of basic proportionality theorem]
Hence, Proved.

(iii) Given: AB = 5 cm, AC =10 cm, AP= 4 cm, AQ = 8 cm
To find: PQ || BC
In ΔABC,
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{4}{\mathrm{AB}-\mathrm{AP}}=\frac{4}{5-4}=\frac{4}{1}=4$
and $\frac{A Q}{Q C}=\frac{8}{A C-A Q}=\frac{8}{10-8}=\frac{8}{2}=4$

Thus,

$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
Hence, PQ || BC [by converse of basic proportionality theorem]
Hence, Proved.

(iv) Given: AP= 4 cm, PB= 4.5 cm, AQ = 4 cm, QC = 5 cm
To find: PQ || BC
In ΔABC,
$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{4}{4.5}=\frac{4 \times 10}{45}=\frac{8}{9}$
and $\frac{A Q}{Q C}=\frac{4}{5}$

Thus, $\frac{\mathrm{AP}}{\mathrm{PB}} \neq \frac{\mathrm{AQ}}{\mathrm{QC}}$

Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.
⇒ PQ is not parallel to BC

Question 13

In the adjoining figure, AD is the bisector of ∠BAC. If BC = 10 cm, BD = 6 cm AC = 6 cm, then find AB.

Sol :

Given: AD is the bisector of ∠BAC
and by Angle-Bisector theorem which states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides.
$\therefore \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$
$\Rightarrow \frac{6}{10}=\frac{A B}{6}$
$\Rightarrow A B=\frac{36}{10}$
⇒ AB = 3.6cm

Question 14

In the adjoining figure, AD is the bisector of ∠BAC. If AB = 10 cm, AC = 6 cm, BC = 12 cm, find BD.

Sol :
Given: AD is the bisector of ∠BAC
and by Angle-Bisector theorem which states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides.
$\therefore \frac{\mathrm{CD}}{\mathrm{DB}}=\frac{\mathrm{AC}}{\mathrm{AB}}$
$\Rightarrow \frac{\mathrm{CD}}{\mathrm{BC}-\mathrm{CD}}=\frac{6}{10}$
$\Rightarrow \frac{\mathrm{CD}}{\mathrm{CD}\left(\frac{\mathrm{BC}}{\mathrm{CD}}-1\right)}=\frac{6}{10}$
$\Rightarrow \frac{\mathrm{BC}}{\mathrm{CD}}-1=\frac{10}{6}$
$\Rightarrow \frac{\mathrm{BC}}{\mathrm{CD}}=\frac{10}{6}+1$
$\Rightarrow \frac{12}{\mathrm{CD}}=\frac{10+6}{6}$
$\Rightarrow \frac{12}{\mathrm{CD}}=\frac{16}{6}$
$\Rightarrow \mathrm{CD}=\frac{12 \times 6}{16}$
$\Rightarrow \mathrm{CD}=\frac{9}{2}=4.5 \mathrm{cm}$
And BC – CD = DB
⇒ 12 – 4.5 = DB
⇒ DB = 7.5cm

Question 15

In EABC, AD is the bisector of ∠A. If AB = 3.5 cm, AC = 4.2 cm, DC = 2.4 cm. Find BD.

Sol :

Given: AD is the bisector of ∠A
and by Angle-Bisector theorem which states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides.
$\therefore \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$
$\Rightarrow \frac{\mathrm{BD}}{2.4}=\frac{3.5}{4.2}$
$\Rightarrow \mathrm{BD}=\frac{3.5 \times 2.4}{4.2}$
⇒ BD = 2cm

S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1