| 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
(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
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
| 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 |
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