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}
\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
(i)
\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
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
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} =………
(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}
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}
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}
\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.
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
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}
or BD = 3.6 cm
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 cmor 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
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,
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
=\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}
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.
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,
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}
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
Sol :
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
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