 NCERT Solutions Class 10 Maths Chapter 6 Triangles – Here are all the NCERT solutions for Class 10 Maths Chapter 6. This solution contains questions, answers, images, explanations of the complete Chapter 6 titled Triangles of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 6 Triangles. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 6 Triangles in one place.

## NCERT Solutions Class 10 Maths Chapter 6 Triangles

Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 10 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 6 Triangles , Maths, Class 10.

 Class 10 Subject Maths Book Mathematics Chapter Number 6 Chapter Name Triangles

### NCERT Solutions Class 10 Maths chapter 6 Triangles

Class 10, Maths chapter 6, Triangles solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Triangles

Q.1: Fill in the blanks using the correct word given in brackets :
(i) All circles are ________________. (congruent, similar)
(ii) All squares are________________ . (similar, congruent)
(iii) All______________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are____________ .(equal, proportional)
Ans : (i) Similar
(ii) Similar
(iii) Equilateral
(iv) (a) Equal
(b) Proportional 
Q.2: Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
Ans : (i) Two equilateral triangles with sides 1 cm and 2 cm. Two squares with sides 1 cm and 2 cm. (ii) Trapezium and square Q.3: State whether the following quadrilaterals are similar or not: Ans : Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e 1:2, but their corresponding angles are not equal.
Q.1: In Figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). Ans : (i) Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
$$\begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{1.5}{3}=\frac{1}{x}} \\ {x=\frac{3 \times 1}{1.5}} \\ {x=2} \\ {\therefore \mathrm{EC}=2 \mathrm{cm}}\end{array}$$
(ii) It is given that DE || BC
By using basic proportionality theorem, we obtain
$$\begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{x}{7.2}=\frac{1.8}{5.4}} \\ {x=\frac{1.8 \times 7.2}{5.4}} \\ {x=2.4} \\ {\therefore \mathrm{AD}=2.4 \mathrm{cm}}\end{array}$$ 
Q.2: E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF || QR :
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Ans : (i) Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
$$\frac{P E}{E Q}=\frac{3.9}{3}=1.3$$
$$\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3.6}{2.4}=1.5$$
Hence, $$\frac{\mathrm{PE}}{\mathrm{EQ}} \neq \frac{\mathrm{PF}}{\mathrm{FR}}$$
Therefore, EF is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
$$\frac{P E}{E Q}=\frac{4}{4.5}=\frac{8}{9}$$
$$\frac{P F}{F R}=\frac{8}{9}$$
Hence $$\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}$$
$$Therefore, EF is parallel to QR.$$
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
$$\frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}$$
$$\frac{\mathrm{PF}}{\mathrm{PR}}=\frac{0.36}{2.56}=\frac{9}{64}$$
$$\begin{array}{l}{\text { Hence, } \frac{\mathrm{PE}}{\mathrm{PQ}}=\frac{\mathrm{PF}}{\mathrm{PR}}} \\ {\text { Therefore, EF is parallel to QR. }}\end{array}$$ 
Q.3: In Figure, if LM || CB and LN || CD, prove that
$$\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$$ Ans : In the given figure, LM || CB
By using basic proportionality theorem, we obtain
$$\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AL}}{\mathrm{AC}}$$ ..(i)
Similarly, LN || CD
$$\therefore \frac{\mathrm{AN}}{\mathrm{AD}}=\frac{\mathrm{AL}}{\mathrm{AC}}$$
(ii)
From (i) and (ii), we obtain
$$\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}$$ 
Q.4: In Figure, DE || AC and DF || AE. Prove that
$$\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$$ Ans : In $$\triangle A B C, D E\|A C$$
$$\therefore \frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BE}}{\mathrm{EC}}$$ (Basic Proportionality Theorem) (i) In $$\triangle B A E, D F\|A E$$
$$\therefore \frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BF}}{\mathrm{FE}} \quad(\text { Basic Proportionality Theorem) }$$ (ii)
From (i) and (ii), we obtain
$$\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}$$ 
Q.5: In Figure, DE || OQ and DF || OR. Show that EF || QR. Ans : In $$\triangle \mathrm{POQ}, \mathrm{DE}\|\mathrm{OQ}$$
$$\therefore \frac{P E}{E Q}=\frac{P D}{D O} \quad(\text { Basic proportionality theorem })$$ (i) In $$\triangle \mathrm{POR}, \mathrm{DF}\|\mathrm{OR}$$
$$\therefore \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{\mathrm{PD}}{\mathrm{DO}} \quad$$ (Basic proportionality theorem)
(ii)
From (i) and (ii), we obtain
$$\frac{P E}{E Q}=\frac{P F}{F R}$$
$$\therefore \mathrm{EF}\|\mathrm{OR} \quad \text { (Converse of basic proportionality theorem) }$$ Q.6: In Figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. Ans : In $$\triangle \mathrm{POQ}, \mathrm{AB}\|\mathrm{PQ}$$
$$\therefore \frac{\mathrm{O} \mathrm{A}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}} \quad$$ (Basic proportionality theorem) (i) In $$\triangle \mathrm{POR}, \mathrm{AC}\|\mathrm{PR}$$
$$\therefore \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}} \quad$$ (By basic proportionality theorem) (ii)
From (i) and (ii), we obtain.
$$\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}$$
$$\therefore \mathrm{BC}\|\mathrm{QR}$$ (By the converse of basic proportionality theorem) Q.7: Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Ans : Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ || BC
By using basic proportionality theorem, we obtain
$$\frac{\mathrm{AQ}}{\mathrm{QC}}=\frac{\mathrm{AP}}{\mathrm{PB}}$$
$$\frac{\mathrm{AQ}}{\mathrm{QC}}=\frac{1}{1} \quad(\mathrm{P} \text { is the mid-point of } \mathrm{AB} . \therefore \mathrm{AP}=\mathrm{PB})$$
$$\Rightarrow A Q=Q C$$
Or, Q is the mid-point of AC. 
Q.8: Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Ans : Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.
I.e., AP = PB and AQ = QC
It can be observed that
$$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{1}{1}$$
and $$\frac{\mathrm{AQ}}{\mathrm{QC}}=\frac{1}{1}$$
$$\therefore \frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AQ}}{\mathrm{QC}}$$
Hence, by using basic proportionality theorem, we obtain
$$\mathrm{PQ}\|\mathrm{BC}$$ 
Q.9: ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that
$$\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$$
Ans : Draw a line EF through point O, such that EF || CD
In $$\triangle A D C, E O\|C D$$
By using basic proportionality theorem, we obtain
$$\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}}$$ ...(1)
In $$\triangle \mathrm{ABD}, \text { OE }\|\mathrm{AB}$$
So, by using basic proportionality theorem, we obtain
$$\frac{\mathrm{ED}}{\mathrm{AE}}=\frac{\mathrm{OD}}{\mathrm{BO}}$$
$$\Rightarrow \frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{OD}}$$... (2)
From equations (1) and (2), we obtain
$$\begin{array}{l}{\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{BO}}{\mathrm{OD}}} \\ {\Rightarrow \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OD}}}\end{array}$$ 
Q.10: The diagonals of a quadrilateral ABCD intersect each other at the point O such that $$\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$$. Show that ABCD is a trapezium.
Ans : Let us consider the following figure for the given question. Draw a line OE || AB In $$\triangle A B D, O E\|A B$$
By using basic proportionality theorem, we obtain
$$\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BO}}{\mathrm{OD}}$$ (1)
However, it is given that
$$\frac{\mathrm{AO}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$$ (2)
From equations (1) and (2), we obtain
$$\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{AO}}{\mathrm{OC}}$$
$$\Rightarrow \text { EO }\|\text { DC }[\mathrm{By} \text { the converse of basic proportionality theorem] }$$
$$\Rightarrow A B\|O E\| D C$$
$$\begin{array}{l}{\Rightarrow A B\|C D} \\ {\therefore A B C D \text { is a trapezium. }}\end{array}$$ 
Q.1: State which pairs of triangles in Figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form : Ans : (i) $$\angle A=\angle P=60^{\circ}$$
$$\angle B=\angle Q=80^{\circ}$$
$$\angle C=\angle R=40^{\circ}$$
Therefore, $$\triangle \mathrm{ABC}-\triangle \mathrm{PQR}[\mathrm{By} \text { AAA similarity criterion }]$$
$$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{CA}}{\mathrm{PQ}}$$

(ii) $$\therefore \triangle \mathrm{ABC}-\Delta \mathrm{QRP} \quad[\text { By SSS similarity criterion }]$$

(iii) The given triangles are not similar as the corresponding sides are not proportional.

(iv) The given triangles are not similar as the corresponding sides are not proportional.

(v) THe given triangles are not similar as the corresponding sides are not proportional.

(vi) In $$\Delta \mathrm{DEF}$$
$$\angle D+\angle E+\angle F=180^{\circ}$$
( Sum of the measures of the angles of a triangle is $$180^{\circ}$$. )
$$70^{\circ}+80^{\circ}+\angle F=180^{\circ}$$
$$\angle F=30^{\circ}$$
Similarly, In $$\triangle \mathrm{PQR}$$
$$\angle P+\angle Q+\angle R=180^{\circ}$$
( Sum of the measures of the angles of a triangle is $$180^{\circ}$$. )
$$\begin{array}{l}{\angle P+80^{\circ}+30^{\circ}=180^{\circ}} \\ {\angle P=70^{\circ}}\end{array}$$
In $$\triangle D E F \text { and } \Delta P Q R,$$
$$\angle D=\angle P\left(E a c h 70^{\circ}\right)$$
$$\angle E=\angle Q\left(E a c h 80^{\circ}\right)$$
$$\angle F=\angle R\left(\text { Each } 30^{\circ}\right)$$
$$\therefore \triangle \mathrm{DEF} \sim \triangle \mathrm{PQR}[\mathrm{By} \text { AAA similarity criterion }]$$ 
Q.2: In Figure, ∆ ODC ~ ∆ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB. Ans : DOB is a straight line.
$$\therefore \angle D O C+\angle C O B=180^{\circ}$$
$$\begin{array}{l}{\Rightarrow \angle D O C=180^{\circ}-125^{\circ}} \\ {=55^{\circ}}\end{array}$$
In $$\Delta D O C$$
$$\angle D C O+\angle C D O+\angle D O C=180^{\circ}$$
( Sum of the measures of the angles of a triangle is $$180^{\circ}$$. )
$$\begin{array}{l}{\Rightarrow \angle D C O+70^{\circ}+55^{\circ}=180^{\circ}} \\ {\Rightarrow \angle D C O=55^{\circ}}\end{array}$$
It is given that $$\triangle O D C-\triangle O B A$$
$$\therefore \angle O A B=\angle O C D[\text { Corresponding angles are equal in similar triangles.] }$$
$$\Rightarrow \angle O A B=55^{\circ}$$ 
Q.3: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $$\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$$
Ans : In $$\triangle D O C \text { and } \triangle B O A$$,
$$\angle \mathrm{CDO}=\angle \mathrm{ABO}[\text { Alternate interior angles as } \mathrm{AB} \| \mathrm{CD}]$$
$$\angle \mathrm{DCO}=\angle \mathrm{BAO}[\text { Alternate interior angles as } \mathrm{AB} \| \mathrm{CD}]$$
$$\angle D O C=\angle B O A[\text { Vertically opposite angles }]$$
$$\therefore \triangle D O C-\triangle B O A$$ [AAA similarity criterion]
$$\therefore \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}}$$ [Corresponding sides are proportional]
$$\Rightarrow \frac{\mathrm{O} \mathrm{A}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$$ 
Q.4: In Figure, $$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$$ and $$\angle 1=\angle 2$$. Show that $$\Delta \mathrm{PQS}-\Delta \mathrm{TQR}$$. Ans : In $$\Delta P Q R, \angle P Q R=\angle P R Q$$
$$\therefore \mathrm{PQ}=\mathrm{PR}(\mathrm{i})$$
Given,
$$\frac{\mathrm{Q} \mathrm{R}}{\mathrm{OS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$$
Using (i), we obtain
$$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}}$$ (ii)
In $$\triangle \mathrm{PQS} \text { and } \Delta \mathrm{TQR}$$
$$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}} \quad[\mathrm{Using}(i i)]$$
$$\angle Q=\angle Q$$
$$\therefore \triangle \mathrm{PQS} \sim \triangle \mathrm{TQR}$$ [SAS similarity criterion] 
Q.5: S and T are points on sides PR and QR of ∆ PQR such that ∠ P = ∠ RTS. Show that ∆ RPQ ~ ∆ RTS.
Ans : In $$\triangle \mathrm{RPQ} \text { and } \Delta \mathrm{RST}$$,
$$\angle \mathrm{RTS}=\angle \mathrm{QPS}(\text { Given })$$
$$\begin{array}{l}{\angle R=\angle R(\text { Common angle })} \\ {\therefore \triangle R P Q-\Delta R T S(B y \text { AA similarity criterion) }}\end{array}$$ 
Q.6: In Figure, if ∆ ABE ≅ ∆ ACD, show that ∆ ADE ~ ∆ ABC. Ans : It is given that $$\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}$$
$$\therefore A B=A C[B y C P C T](1)$$
$$\begin{array}{l}{\text { And, } A D=A E[B Y C P C T](2)} \\ {\text { In } \triangle A D E \text { and } \triangle A B C}\end{array}$$
$$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}[\text { Dividing equation }(2) \mathrm{by}(1)]$$
$$\angle A=\angle A[\text { Common angle }]$$
$$\therefore \triangle \mathrm{ADE}-\triangle \mathrm{ABC}[\mathrm{By} \text { SAS similarity criterion] }$$ 
Q.7: In Figure, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that:
(i) ∆AEP ~ ∆ CDP
(ii) ∆ABD ~ ∆ CBE
(iv) ∆ PDC ~ ∆ BEC Ans : (i) In $$\triangle A E P \text { and } \triangle C D P.$$
$$\angle A E P=\angle C D P\left(E a c h 90^{\circ}\right)$$
$$\angle \mathrm{APE}=\angle \mathrm{CPD}(\text { Vertically opposite angles })$$
Hence, by using AA similarity criterion,
$$\triangle \mathrm{AEP} \sim \Delta \mathrm{CDP}$$

(ii) In $$\triangle \mathrm{ABD} \text { and } \Delta \mathrm{CBE}$$
$$\angle A D B=\angle C E B\left(E a c h 90^{\circ}\right)$$
$$\angle A B D=\angle C B E(\text { Common })$$
$$\begin{array}{l}{\text { Hence, by using AA similarity criterion, }} \\ {\triangle A B D-\Delta C B E}\end{array}$$

(iii) In $$\triangle \mathrm{AEP} \text { and } \triangle \mathrm{ADB}$$ ,
\begin{aligned} \angle A E P &=\angle A D B\left(E a c h 90^{\circ}\right) \\ \angle P A E &=\angle D A B(\text { Common }) \end{aligned}
Hence, y using AA similarity criterion,
$$\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$$
(iv) In $$\Delta \mathrm{PDC} \text { and } \Delta \mathrm{BEC}$$
\begin{aligned} \angle \mathrm{PDC} &=\angle \mathrm{BEC}\left(\text { Each } 90^{\circ}\right) \\ \angle \mathrm{PCD} &=\angle \mathrm{BCE}(\text { Common angle }) \end{aligned}
Hence, by using AA similarity criterion,
$$\triangle P D C-\triangle B E C$$ 
Q.8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ ABE ~ ∆ CFB.
Ans : In $$\triangle \mathrm{ABE} \text { and } \Delta \mathrm{CFB}$$,
$$\angle \mathrm{A}=\angle \mathrm{C}(\text { Opposite angles of a parallelogram) }$$
$$\angle A E B=\angle C B F(\text { Alternate interior angles as } A E \| B C)$$
$$\therefore \triangle \mathrm{ABE}-\Delta \mathrm{CFB}(\mathrm{By} \text { AA similarity criterion) }$$ 
Q.9: In Figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) $$\Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$$
(ii) $$\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$$ Ans : In $$\triangle A B C \text { and } \triangle A M P,$$
$$\angle A B C=\angle A M P\left(E a c h 90^{\circ}\right)$$
$$\angle A=\angle A(\text { Common })$$
$$\therefore \triangle \mathrm{ABC}-\triangle \mathrm{AMP}(\mathrm{By} \text { AA similarity criterion) }$$
$$\Rightarrow \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}} \quad$$ ( Corresponding sides of similar triangles are proportional) 
Q.10: CD and GH are respectively the bisectors of ∠ACB and ∠ EGF such that D and H lie on sides AB and FE of ∆ ABC and ∆ EFG respectively. If ∆ABC ~ ∆ FEG, show that:
(i) $$\frac{C D}{G H}=\frac{A C}{F G}$$
(ii) $$\Delta \mathrm{DCB}-\Delta \mathrm{HGE}$$
(iii) $$\Delta \mathrm{DCA}-\Delta \mathrm{HGF}$$
Ans : It is given that $$\triangle \mathrm{ABC}-\Delta \mathrm{FEG}$$
$$\therefore \angle A=\angle F, \angle B=\angle E, \text { and } \angle A C B=\angle F G E$$
$$\begin{array}{l}{\angle A C B=\angle F G E} \\ {\therefore \angle A C D=\angle F G H(\text { Angle bisector })}\end{array}$$
$$\begin{array}{l}{\text { And, } \angle \mathrm{DCB}=\angle \mathrm{HGE} \text { (Angle bisector) }} \\ {\text { In } \triangle \mathrm{ACD} \text { and } \Delta \mathrm{FGH} \text { , }}\end{array}$$
$$\begin{array}{l}{\angle \mathrm{A}=\angle \mathrm{F}(\text { Proved above) }} \\ {\angle \mathrm{ACD}=\angle \mathrm{FGH}(\text { Proved above })}\end{array}$$
$$\therefore \triangle A C D \sim \Delta F G H(B y \text { AA similarity criterion) }$$
$$\Rightarrow \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}$$
In $$\triangle D C B \text { and } \Delta H G E,$$
$$\begin{array}{l}{\angle \mathrm{DCB}=\angle \mathrm{HGE}(\text { Proved above })} \\ {\angle \mathrm{B}=\angle \mathrm{E}(\text { Proved above })}\end{array}$$
$$\triangle \triangle D C B \sim \Delta H G E(B y \text { AA similarity criterion) }$$
$$\begin{array}{l}{\text { In } \triangle D C A \text { and } \Delta H G F} \\ {\angle A C D=\angle F G H(\text { Proved above })}\end{array}$$
$$\angle A=\angle F(\text { Proved above })$$
$$\therefore \triangle \mathrm{DCA} \sim \Delta \mathrm{HGF}(\text { By AA similarity criterion) }$$ 
Q.11: In Figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥AC, prove that ∆ ABD ~ ∆ ECF. Ans : It is given that ABC is an isosceles triangle.
$$\begin{array}{l}{\therefore \mathrm{AB}=\mathrm{AC}} \\ {\Rightarrow \angle \mathrm{ABD}=\angle \mathrm{ECF}} \\ {\text { In } \triangle \mathrm{ABD} \text { and } \Delta \mathrm{ECF}}\end{array}$$
$$\angle A D B=\angle E F C\left(E a c h 90^{\circ}\right)$$
$$\angle B A D=\angle C E F(\text { Proved above })$$
$$\therefore \triangle \mathrm{ABD} \sim \Delta \mathrm{ECF}(\mathrm{By} \text { using } \mathrm{AA} \text { similarity criterion })$$ 
Q.12: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆ PQR (see Figure). Show that ∆ ABC ~ ∆ PQR. Ans : Median divides the opposite side.
$$\therefore$$ $$\mathrm{BD}=\frac{\mathrm{BC}}{2} \text { and } \mathrm{QM}=\frac{\mathrm{QR}}{2}$$
$$\begin{array}{l}{\text { Given that, }} \\ {\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}}\end{array}$$
$$\begin{array}{l}{\Rightarrow \frac{A B}{P Q}=\frac{\frac{1}{2} B C}{\frac{1}{2} Q R}=\frac{A D}{P M}} \\ {\Rightarrow \frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}}\end{array}$$
In $$\triangle A B D \text { and } \Delta P Q M$$
$$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}$$ (Proved above)
$$\triangle \triangle \mathrm{ABD} \sim \triangle \mathrm{PQM}(\mathrm{By} \text { SSS similarity criterion })$$
$$\Rightarrow \angle A B D=\angle P Q M(\text { Corresponding angles of similar triangles) }$$
$$\begin{array}{l}{\text { In } \triangle A B C \text { and } \Delta P Q R \text { , }} \\ {\angle A B D=\angle P Q M \text { (Proved above) }}\end{array}$$
$$\begin{array}{l}{\frac{A B}{P Q}=\frac{B C}{Q R}} \\ {\therefore \triangle A B C \sim \Delta P Q R(B y \text { SAS similarity criterion) }}\end{array}$$ 
Q.13: D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that $$\mathrm{CA}^{2}$$ = CB.CD.
Ans : In $$\triangle A D C \text { and } \Delta B A C$$,
\begin{aligned} \angle A D C &=\angle B A C(\text { Given }) \\ \angle A C D &=\angle B C A(\text { Common angle }) \end{aligned}
$$\therefore \triangle A D C \sim \Delta B A C(B y \text { AA similarity criterion) }$$
We know that corresponding sides of similar triangles are in proportion.
$$\therefore \frac{\mathrm{CA}}{\mathrm{CB}}=\frac{\mathrm{CD}}{\mathrm{CA}}$$
$$\Rightarrow \mathrm{CA}^{2}=\mathrm{CB} \times \mathrm{CD}$$ 
Q.14: Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆ PQR.
Ans : Given that,
$$\frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P M}$$
Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L. We know that medians divide opposite sides.
Therefore, BD = DC and QM = MR
Also, AD = DE (By construction)
And, PM = ML(By construction)
In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
$$\therefore A C=B E \text { and } A B=E C$$ (Opposite sides of a parallelogram are equal)
Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR
It was given that
$$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}$$
$$\Rightarrow \frac{A B}{P Q}=\frac{B E}{Q L}=\frac{2 A D}{2 P M}$$
$$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BE}}{\mathrm{QL}}=\frac{\mathrm{AE}}{\mathrm{PL}}$$
$$\therefore \triangle \mathrm{ABE} \sim \triangle \mathrm{PQL}$$ (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
$$\therefore \angle B A E=\angle Q P L \ldots(1)$$
Similarly, it can be proved that $$\triangle \mathrm{AEC} \sim \triangle \mathrm{PLR}$$ and $$\angle C A E=\angle R P L \ldots(2)$$
Adding equation (1) and (2), we obtain
$$\angle B A E+\angle C A E=\angle Q P L+\angle R P L$$
$$\Rightarrow \angle C A B=\angle R P Q \ldots(3)$$
In $$\triangle A B C \text { and } \Delta P Q R_{\prime}$$
$$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}$$ (Given)
$$\angle C A B=\angle R P Q[U \text { sing equation }(3)]$$
$$\therefore \triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}(\mathrm{By} \text { SAS similarity criterion) }$$ 
Q.15: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Ans : Let AB and CD be a tower and a pole respectively.
Let the shadow of BE and DF be the shadow of AB and CD respectively.
At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.
Therefore, $$\angle D C F=\angle B A E$$
And, $$\angle \mathrm{DFC}=\angle \mathrm{BEA}$$
$$\angle \mathrm{CDF}=\angle \mathrm{ABE}(\text { Tower and pole are vertical to the ground) }$$
$$\therefore \triangle \mathrm{ABE} \sim \triangle \mathrm{CDF}$$ (AAA similarity criterion)
$$\Rightarrow \frac{\mathrm{AB}}{\mathrm{CD}}=\frac{\mathrm{BE}}{\mathrm{DF}}$$
$$\Rightarrow \frac{\mathrm{AB}}{6 \mathrm{m}}=\frac{28}{4}$$
$$\Rightarrow A B=42 \mathrm{m}$$
Therefore, the height of the tower will be 42 metres. 
Q.16: If AD and PM are medians of triangles ABC and PQR, respectively where ∆ ABC ~ ∆ PQR, prove that $$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}$$
Ans : It is given that $$\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$$
We know that the corresponding sides of similar triangles are in proportion.
$$\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}}$$ …(1)
Since AD and PM are medians, they will divide their opposite sides.
$$\mathrm{BD}=\frac{\mathrm{BC}}{2} \text { and } \mathrm{QM}=\frac{\mathrm{QR}}{2}$$ …(3)
From equations (1) and (3), we obtain
$$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}$$ …(4)
In $$\triangle \mathrm{ABD} \text { and } \Delta \mathrm{PQM}$$
$$\angle B=\angle Q[U \sin g \text { equation }(2)]$$
$$\frac{A B}{P Q}=\frac{B D}{Q M}$$ [Using equation (4)]
$$\therefore \triangle \mathrm{ABD} \sim \Delta \mathrm{PQM}$$ (By SAS similarity criterion)
$$\Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}$$ 
Q.1: Let ∆ ABC ~ ∆ DEF and their areas be, respectively, $$64 \mathrm{cm}^{2} \text { and } 121 \mathrm{cm}^{2}$$  . If EF = 15.4 cm, find BC.
Ans : It is given that $$\triangle \mathrm{ABC}-\Delta \mathrm{DEF}$$ .
$$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DEF})}=\left(\frac{\mathrm{AB}}{\mathrm{DE}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{DF}}\right)^{2}$$
Given that,
$$\mathrm{EF}=15.4 \mathrm{cm}$$,
\begin{aligned} \operatorname{ar}(\Delta \mathrm{ABC}) &=64 \mathrm{cm}^{2} \\ \operatorname{ar}(\Delta \mathrm{DEF}) &=121 \mathrm{cm}^{2} \end{aligned}
$$\therefore \frac{\operatorname{ar}(\mathrm{ABC})}{\operatorname{ar}(\mathrm{DEF})}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)^{2}$$
$$\Rightarrow\left(\frac{64 \mathrm{cm}^{2}}{121 \mathrm{cm}^{2}}\right)=\frac{\mathrm{BC}^{2}}{(15.4 \mathrm{cm})^{2}}$$
$$\Rightarrow \frac{\mathrm{BC}}{15.4}=\left(\frac{8}{11}\right) \mathrm{cm}$$
$$\Rightarrow \mathrm{BC}=\left(\frac{8 \times 15.4}{11}\right) \mathrm{cm}=(8 \times 1.4) \mathrm{cm}=11.2 \mathrm{cm}$$ 
Q.2: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Ans : Since AB || CD,
$$\therefore \angle O A B=\angle O C D \text { and } \angle O B A=\angle O D C(\text { Alternate interior angles) }$$
In $$\triangle \mathrm{AOB} \text { and } \Delta \mathrm{COD} ,$$
$$\angle \mathrm{AOB}=\angle \mathrm{COD}$$ (Vertically opposite angles)
$$\angle O A B=\angle O C D(\text { Alternate interior angles })$$
$$\angle O B A=\angle O D C(\text { Alternate interior angles })$$
$$\therefore \triangle \mathrm{AOB} \sim \Delta \mathrm{COD}(\mathrm{By} \text { AAA similarity criterion })$$
$$\therefore \frac{\operatorname{ar}(\Delta \mathrm{AOB})}{\operatorname{ar}(\Delta \mathrm{COD})}=\left(\frac{\mathrm{AB}}{\mathrm{CD}}\right)^{2}$$
Since $$A B=2 C D$$
$$\therefore \frac{\operatorname{ar}(\Delta \mathrm{AOB})}{\operatorname{ar}(\Delta \mathrm{COD})}=\left(\frac{2 \mathrm{CD}}{\mathrm{CD}}\right)^{2}=\frac{4}{1}=4 : 1$$ 
Q.3: In Figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
$$\frac{\operatorname{ar}(\mathrm{ABC})}{\mathrm{ar}(\mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}$$ Ans : Let us draw two perpendicular AP and DM on line BC. We know that area of a triangle = 1/2 x Base x Height
$$\therefore \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DBC})}=\frac{\frac{1}{2} \mathrm{BC} \times \mathrm{AP}}{\frac{1}{2} \mathrm{BC} \times \mathrm{DM}}=\frac{\mathrm{AP}}{\mathrm{DM}}$$
In $$\triangle \mathrm{APO} \text { and } \triangle \mathrm{DMO}$$
$$\angle A P O=\angle D M O\left(E a c h=90^{\circ}\right)$$
$$\angle \mathrm{AOP}=\angle \mathrm{DOM}(\text { Vertically opposite angles) }$$
$$\therefore \triangle \mathrm{APO} \sim \triangle \mathrm{DMO}(\mathrm{By} \text { AA similarity criterion) }$$
$$\therefore \frac{A P}{D M}=\frac{A O}{D O}$$
$$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}}$$ 
Q.4: If the areas of two similar triangles are equal, prove that they are congruent.
Ans : Let us assume two similar triangles are $$\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$$
$$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)^{2}$$ (1)
Given that, ar $$(\Delta \mathrm{ABC})=\text { ar }(\Delta \mathrm{PQR})$$
$$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=1$$
Putting this value in equation (1), we obtain
$$\mathrm{I}=\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)^{2}$$
$$\Rightarrow \mathrm{AB}=\mathrm{PQ}, \mathrm{BC}=\mathrm{QR}, \text { and } \mathrm{AC}=\mathrm{PR}$$
$$\therefore \Delta \mathrm{ABC} \cong \Delta \mathrm{PQR} \quad(\text { By SSS congruence criterion })$$ 
Q.5: D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC.
Ans : $$\therefore \mathrm{DE}\left\|\mathrm{AC} \text { and } \mathrm{DE}=\frac{1}{2} \mathrm{AC}\right.$$
In $$\triangle B E D \text { and } \Delta B C A,$$
$$\angle B E D=\angle B C A$$ (Corresponding angles)
$$\angle \mathrm{BDE}=\angle \mathrm{BAC}$$ (Corresponding angles)
$$\angle \mathrm{EBD}=\angle \mathrm{CBA}$$ ( Common angles)
$$\therefore \triangle B E D \sim \Delta B C A$$ (AAA similarity criterion)
$$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{BED})}{\operatorname{ar}(\Delta \mathrm{BCA})}=\frac{1}{4}$$
$$\Rightarrow \operatorname{ar}(\Delta \mathrm{BED})=\frac{1}{4} \operatorname{ar}(\Delta \mathrm{BCA})$$
Similarly, $$\operatorname{ar}(\Delta \mathrm{CFE})=\frac{1}{4} \operatorname{ar}(\mathrm{CBA}) \text { and ar }(\Delta \mathrm{ADF})=\frac{1}{4} \operatorname{ar}(\Delta \mathrm{ABC})$$
Also $$(\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{ABC})-[\operatorname{ar}(\Delta \mathrm{BED})+\operatorname{ar}(\Delta \mathrm{CFE})+\operatorname{ar}(\Delta \mathrm{ADF})]$$
$$\Rightarrow \operatorname{ar}(\Delta \mathrm{DEF})=\operatorname{ar}(\Delta \mathrm{ABC})-\frac{3}{4} \operatorname{ar}(\Delta \mathrm{ABC})=\frac{1}{4} \operatorname{ar}(\Delta \mathrm{ABC})$$
$$\Rightarrow \frac{\operatorname{ar}(\Delta \mathrm{DEF})}{\operatorname{ar}(\Delta \mathrm{ABC})}=\frac{1}{4}$$ 
Q.6: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Ans : Let us assume two similar triangles as $$\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$$. Let AD and PS be the medians of these triangles.
$$\because \triangle A B C \sim \Delta P Q R$$
$$\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}$$ ..(1)
$$\angle A=\angle P, \angle B=\angle Q, \angle C=\angle R \ldots(2)$$
Since AD and PS are medians,
$$\therefore B D=D C=\frac{B C}{2}$$
And, $$Q S=S R=\frac{Q R}{2}$$
Equation (1) becomes
$$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QS}}=\frac{\mathrm{AC}}{\mathrm{PR}}$$ …(3)
In $$\triangle A B D \text { and } \Delta P Q S,$$
$$\angle B=\angle Q[\text { Using equation }(2)]$$
And, $$\frac{A B}{P Q}=\frac{B D}{Q S}[U s i n g \text { equation }(3)]$$
$$\therefore \triangle \mathrm{ABD} \sim \Delta \mathrm{PQS}$$ (SAS similarity criterion)
Therefore, it can be said that
$$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QS}}=\frac{\mathrm{AD}}{\mathrm{PS}}$$ …(4)
$$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)^{2}$$
From equations (1) and (4), we may find that
$$\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}=\frac{A D}{P S}$$
And hence,
$$\frac{\operatorname{ar}(\Delta \mathrm{ABC})}{\operatorname{ar}(\Delta \mathrm{PQR})}=\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)^{2}$$ 
Q.7: Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Ans : Let ABCD be a square of side a.
Therefore, its diagonal = $$\sqrt{2} a$$
Two desired equilateral triangles are formed as $$\triangle \mathrm{ABE} \text { and } \triangle \mathrm{DBF}$$
Two desired equilateral triangles are formed as $$\triangle \mathrm{ABE} \text { and } \Delta \mathrm{DBF}$$.
Side of an equilateral triangle, $$\triangle \mathrm{ABE}$$, described on one of its sides = a
Side of an equilateral triangle, $$\triangle \mathrm{DBF}$$ , described on one of its diagonals  $$=\sqrt{2} a$$
We know that equilateral triangles have all its angles as $$60^{\circ}$$ and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
$$\frac{\text { Area of } \Delta \mathrm{ABE}}{\text { Area of } \Delta \mathrm{DBF}}=\left(\frac{a}{\sqrt{2} a}\right)^{2}=\frac{1}{2}$$ 
Q.8: Tick the correct answer and justify :
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
Ans : We know that equilateral triangles have all its angles as $$60^{\circ}$$ and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
Let side of $$\Delta \mathrm{ABC}=x$$
Therefore, side of $$\Delta B D E=\frac{x}{2}$$
$$\therefore \frac{\operatorname{area}(\Delta \mathrm{ABC})}{\operatorname{area}(\Delta \mathrm{BDE})}=\left(\frac{x}{\frac{x}{2}}\right)^{2}=\frac{4}{1}$$
Hence, the correct answer is (C). 
Q.9: Tick the correct answer and justify :
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
Ans : If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.
It is given that the sides are in the ratio 4:9.
Therefore, ratio between areas of these triangles  $$=\left(\frac{4}{9}\right)^{2}=\frac{16}{81}$$
Hence, the correct answer is (D). 
Q.1: Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
Ans : (i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides, we will obtain 49, 576, and 625.
49 + 576 = 625
Or, $$7^{2}+24^{2}=25^{2}$$
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
We know that the longest sides of a right triangle is the hypotenuse.
Therefore, the length of the hypotenuse of this triangle is 25 cm.

(ii) It is given that sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will obtain 9, 64, and 36.
However, 9 + 36 $$\neq 64$$
Or, $$3^{2}+6^{2} \neq 8^{2}$$
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Hence, it is not a right triangle
(iii) Given that sides are 50 cm, 80 cm. And 100 cm.
Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.
However, 2500 + 6400 $$\neq 10000$$
Or, $$50^{2}+80^{2} \neq 100^{2}$$
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.
(iv) Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will obtain 169, 144, and 24.
Clearly, 144 + 25 = 169.
Or, $$12^{2}+5^{2}=13^{2}$$
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
We know that the longest side of a right triangle is the hypotenuse.
Therefore, the length of the hypotenuse of this triangle is 13 cm. 
Q.2: PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that $$\mathrm{PM}^{2}$$ = QM . MR
Ans : Let $$\angle \mathrm{MPR}=x$$
In $$\Delta \mathrm{MPR}$$
$$\begin{array}{l}{\angle M R P=180^{\circ}-90^{\circ}-x} \\ {\angle M R P=90^{\circ}-x}\end{array}$$
\begin{aligned} \text { Similarly, in } \Delta \mathrm{MPQ} \\ \angle \mathrm{MPQ} &=90^{\circ}-\angle \mathrm{MPR} \\ &=90^{\circ}-x \end{aligned}
$$\begin{array}{l}{\angle \mathrm{MQP}=180^{\circ}-90^{\circ}-\left(90^{\circ}-x\right)} \\ {\angle \mathrm{MOP}=x}\end{array}$$
In $$\triangle \mathrm{QMP} \text { and } \Delta \mathrm{PMR} ,$$
$$\begin{array}{l}{\angle \mathrm{MPQ}=\angle \mathrm{MRP}} \\ {\angle \mathrm{PMQ}=\angle \mathrm{RMP}} \\ {\angle \mathrm{MQP}=\angle \mathrm{MPR}}\end{array}$$
$$\therefore \triangle \mathrm{QMP}-\mathrm{APMR} \quad(\text { By AAA similarity criterion })$$
$$\begin{array}{l}{\Rightarrow \frac{\mathrm{QM}}{\mathrm{PM}}=\frac{\mathrm{MP}}{\mathrm{MR}}} \\ {\Rightarrow \mathrm{PM}^{2}=\mathrm{QM} \times \mathrm{MR}}\end{array}$$ 
Q.3: In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that
(i) $$\mathrm{AB}^{2}=\mathrm{BC} . \mathrm{BD}$$
(ii) $$\mathrm{AC}^{2}=\mathrm{BC} \cdot \mathrm{DC}$$
(iii) $$\mathrm{AD}^{2}=\mathrm{BD} . \mathrm{CD}$$ Ans : (i) In $$\triangle \mathrm{ADB} \text { and } \triangle \mathrm{CAB}$$,
$$\angle D A B=\angle A C B \quad\left(\text { Each } 90^{\circ}\right)$$
$$\angle \mathrm{ABD}=\angle \mathrm{CBA} \quad(\text { Common angle })$$
$$\therefore \triangle \mathrm{ADB}-\triangle \mathrm{CAB}$$ (AA similarity criterion)
$$\Rightarrow \frac{A B}{C B}=\frac{B D}{A B}$$
$$\Rightarrow A B^{2}=C B \times B D$$

(ii) Let $$\angle \mathrm{CAB}=x$$
In $$\Delta \mathrm{CBA} ,$$
$$\begin{array}{l}{\angle C B A=180^{\circ}-90^{\circ}-x} \\ {\angle C B A=90^{\circ}-x}\end{array}$$
Similarly, In $$\triangle \mathrm{CAD} ,$$
\begin{aligned} \angle \mathrm{CAD} &=90^{\circ}-\angle \mathrm{CAB} \\ &=90^{\circ}-x \end{aligned}
$$\begin{array}{l}{\angle C D A=180^{\circ}-90^{\circ}-\left(90^{\circ}-x\right)} \\ {\angle C D A=x}\end{array}$$
In $$\triangle \mathrm{CBA} \text { and } \Delta \mathrm{CAD}$$
\begin{aligned} \angle \mathrm{CBA} &=\angle \mathrm{CAD} \\ \angle \mathrm{CAB} &=\angle \mathrm{CDA} \end{aligned}
$$\angle A C B=\angle D C A \quad\left(\operatorname{Each} 90^{\circ}\right)$$
$$\therefore \Delta \mathrm{CBA} \sim \Delta \mathrm{CAD} \quad(\text { By AAA rule })$$
$$\Rightarrow \frac{\mathrm{AC}}{\mathrm{DC}}=\frac{\mathrm{BC}}{\mathrm{AC}}$$
$$\Rightarrow \mathrm{AC}^{2}=\mathrm{DC} \times \mathrm{BC}$$

(iii) In $$\triangle D C A \text { and } \Delta D A B,$$
\begin{aligned} \angle \mathrm{DCA} &=\angle \mathrm{DAB}\left(\mathrm{Each} 90^{\circ}\right) \\ \angle \mathrm{CDA} &=\angle \mathrm{ADB}(\text { Common angle }) \end{aligned}
$$\therefore \Delta \mathrm{DCA} \sim \Delta \mathrm{DAB} \quad$$ (AA similarity criterion)
$$\Rightarrow \frac{\mathrm{DC}}{\mathrm{DA}}=\frac{\mathrm{DA}}{\mathrm{DB}}$$
$$\Rightarrow A D^{2}=B D \times C D$$ 
Q.4: ABC is an isosceles triangle right angled at C. Prove that $$\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$$.
Ans : Given that $$\triangle \mathrm{ABC}$$ is an isosceles triangle.
$$\therefore A C=C B$$
Applying Pythagoras theorem in $$\triangle \mathrm{ABC}$$ (i.e., right-angled at point C), we obtain
$$\mathrm{AC}^{2}+\mathrm{CB}^{2}=\mathrm{AB}^{2}$$
$$\Rightarrow \mathrm{AC}^{2}+\mathrm{AC}^{2}=\mathrm{AB}^{2}$$ $$(A C=C B)$$
$$\Rightarrow 2 \mathrm{AC}^{2}=\mathrm{AB}^{2}$$ 
Q.5: ABC is an isosceles triangle with $$\mathrm{AC}=\mathrm{BC} . \text { If } \mathrm{AB}^{2}=2 \mathrm{AC}^{2}$$.  prove that ABC is a right triangle.
Ans : Given that,
$$\mathrm{AB}^{2}=2 \mathrm{AC}^{2}$$
$$\Rightarrow \mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{AC}^{2}$$
$$\Rightarrow \mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2} \quad(\mathrm{As \ } \mathrm{AC}=\mathrm{BC})$$
The triangle is satisfying the pythagoras theorem.
Therefore, the given triangle is a right-angled triangle. 
Q.6: ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Ans : Let AD be the altitude in the given equilateral triangle, $$\triangle \mathrm{ABC}$$.
We know that altitude bisects the opposite side.
$$\therefore B D=D C=a$$
In $$\triangle \mathrm{ADB}$$
Applying pythagoras theorem, we obtain
$$\mathrm{AD}^{2}+\mathrm{DB}^{2}=\mathrm{AB}^{2}$$
$$\begin{array}{l}{\Rightarrow \mathrm{AD}^{2}+a^{2}=(2 a)^{2}} \\ {\Rightarrow \mathrm{AD}^{2}+a^{2}=4 a^{2}} \\ {\Rightarrow \mathrm{AD}^{2}=3 a^{2}} \\ {\Rightarrow \mathrm{AD}=a \sqrt{3}}\end{array}$$
In an equilateral triangle, all the altitudes are equal in length.
Therefore, the length of each altitude will be $$\sqrt{3} a$$. 
Q.7: Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Ans : In $$\triangle \mathrm{AOB}, \Delta \mathrm{BOC}, \Delta \mathrm{COD}, \triangle \mathrm{AOD}$$
Applying Pythagoras theorem, we obtain
$$\mathrm{AB}^{2}=\mathrm{AO}^{2}+\mathrm{OB}^{2}$$ …(1)
$$\mathrm{BC}^{2}=\mathrm{BO}^{2}+\mathrm{OC}^{2}$$ ….(2)
$$\mathrm{CD}^{2}=\mathrm{CO}^{2}+\mathrm{OD}^{2}$$ …(3)
$$\mathrm{AD}^{2}=\mathrm{AO}^{2}+\mathrm{OD}^{2}$$ …(4)
Adding all these equations, we obtain
$$\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}+\mathrm{AD}^{2}=2\left(\mathrm{AO}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}+\mathrm{OD}^{2}\right)$$
$$=2\left(\left(\frac{\mathrm{AC}}{2}\right)^{2}+\left(\frac{\mathrm{BD}}{2}\right)^{2}+\left(\frac{\mathrm{AC}}{2}\right)^{2}+\left(\frac{\mathrm{BD}}{2}\right)^{2}\right)$$
( (Diagonals bisect each other) \)
$$=2\left(\frac{(A C)^{2}}{2}+\frac{(B D)^{2}}{2}\right)$$
$$=(A C)^{2}+(B D)^{2}$$ 
Q.8: In Figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥AC and OF ⊥AB. Show that (i) $$\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{OD}^{2}-\mathrm{OE}^{2}-\mathrm{OF}^{2}=\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{CE}^{2}$$
(ii) $$A F^{2}+B D^{2}+C E^{2}=A E^{2}+C D^{2}+B F^{2}$$
Ans : (i) Applying Pythagoras theorem in $$\triangle \mathrm{AOF}$$, we obtain
$$\mathrm{OA}^{2}=\mathrm{OF}^{2}+\mathrm{AF}^{2}$$
Similarly, in $$\triangle B O D$$,
$$\mathrm{OB}^{2}=\mathrm{OD}^{2}+\mathrm{BD}^{2}$$
Similarly, in $$\Delta \mathrm{COE}$$,
$$\mathrm{OC}^{2}=\mathrm{OE}^{2}+\mathrm{EC}^{2}$$
$$\begin{array}{l}{\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}=\mathrm{OF}^{2}+\mathrm{AF}^{2}+\mathrm{OD}^{2}+\mathrm{BD}^{2}+\mathrm{OE}^{2}+\mathrm{EC}^{2}} \\ {\mathrm{OA}^{2}+\mathrm{OB}^{2}+\mathrm{OC}^{2}-\mathrm{OD}^{2}-\mathrm{OE}^{2}-\mathrm{OF}^{2}=\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{EC}^{2}}\end{array}$$
(ii) From the above result,
$$\begin{array}{l}{\mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{EC}^{2}=\left(\mathrm{OA}^{2}-\mathrm{OE}^{2}\right)+\left(\mathrm{OC}^{2}-\mathrm{OD}^{2}\right)+\left(\mathrm{OB}^{2}-\mathrm{OF}^{2}\right)} \\ {\therefore \mathrm{AF}^{2}+\mathrm{BD}^{2}+\mathrm{EC}^{2}=\mathrm{AE}^{2}+\mathrm{CD}^{2}+\mathrm{BF}^{2}}\end{array}$$ 
Q.9: A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Ans : Let OA be the wall and AB be the ladder.
Therefore, by Pythagoras theorem,
$$\mathrm{AB}^{2}=\mathrm{OA}^{2}+\mathrm{BO}^{2}$$
$$(10 \mathrm{m})^{2}=(8 \mathrm{m})^{2}+\mathrm{OB}^{2}$$
$$100 \mathrm{m}^{2}=64 \mathrm{m}^{2}+\mathrm{OB}^{2}$$
$$\begin{array}{l}{\mathrm{OB}^{2}=36 \mathrm{m}^{2}} \\ {\mathrm{OB}=6 \mathrm{m}}\end{array}$$
Therefore, the distance of the foot of the ladder from the base of the wall is 6 m. 
Q.10: A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Ans : Let OB be the pole and AB be the wire.
By Pythagoras theorem,
$$\mathrm{AB}^{2}=\mathrm{OB}^{2}+\mathrm{OA}^{2}$$
$$(24 \mathrm{m})^{2}=(18 \mathrm{m})^{2}+\mathrm{OA}^{2}$$
$$\mathrm{OA}^{2}=(576-324) \mathrm{m}^{2}=252 \mathrm{m}^{2}$$
$$\mathrm{OA}=\sqrt{252} \mathrm{m}=\sqrt{6 \times 6 \times 7} \mathrm{m}=6 \sqrt{7} \mathrm{m}$$
Therefore, the distance from the base is $$6 \sqrt{7} \mathrm{m}$$. 
Q.11: An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after $$1 \frac{1}{2}$$ hours?
Ans : Distance travelled by the plane flying towards north in $$1 \frac{1}{2} \mathrm{hrs}$$
$$=1,000 \times 1 \frac{1}{2}=1,500 \mathrm{km}$$
Similarly, distance travelled vu the plane flying towards west in $$1 \frac{1}{2} \mathrm{hrs}$$
$$=1,200 \times 1 \frac{1}{2}=1,800 \mathrm{km}$$
Let these distances be represented by OA and OB respectively.
Applying Pythagoras as theorem,
Distance between these planes after $$1 \frac{1}{2} h r s, A B=\sqrt{O A^{2}+O B^{2}}$$
$$\begin{array}{l}{=\left(\sqrt{(1,500)^{2}+(1,800)^{2}}\right) \mathrm{km}=(\sqrt{2250000+3240000}) \mathrm{km}} \\ {=(\sqrt{5490000}) \mathrm{km}=(\sqrt{9 \times 610000}) \mathrm{km}=300 \sqrt{61} \mathrm{km}}\end{array}$$
Therefore, the distance between these planes will be $$300 \sqrt{61} \mathrm{km}$$ after $$1 \frac{1}{2} \mathrm{hrs}$$ 
Q.12: Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Ans : Let CD and AB be the poles of height 11 m and 6 m.
Therefore, Cp = 11 - 6 = 5 m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for $$\triangle \mathrm{APC}$$, we obtain
$$\begin{array}{l}{\mathrm{AP}^{2}+\mathrm{PC}^{2}=\mathrm{AC}^{2}} \\ {(12 \mathrm{m})^{2}+(5 \mathrm{m})^{2}=\mathrm{AC}^{2}} \\ {\mathrm{AC}^{2}=(144+25) \mathrm{m}^{2}=169 \mathrm{m}^{2}} \\ {\mathrm{AC}=13 \mathrm{m}}\end{array}$$
Therefore, the distance between their tops is 13 m. 
Q.13: D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $$\mathrm{AE}^{2}+\mathrm{BD}^{2}=\mathrm{AB}^{2}+\mathrm{DE}^{2}$$.
Ans : Applying Pythagoras theorem in $$\triangle \mathrm{ACE}$$, we obtain
$$\mathrm{AC}^{2}+\mathrm{CE}^{2}=\mathrm{AE}^{2}$$ …(1)
Applying Pythagoras theorem in $$\triangle B C D$$ , we obtain
$$\mathrm{BC}^{2}+\mathrm{CD}^{2}=\mathrm{BD}^{2}$$ ..(2)
Using equation (1) and equation (2), we obtain
$$\mathrm{AC}^{2}+\mathrm{CE}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}=\mathrm{AE}^{2}+\mathrm{BD}^{2}$$ …(3)
Applying Pythagoras theorem in $$\Delta \mathrm{CDE} ,$$ we obtain
$$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{CB}^{2}$$
Putting the values in equation (3), we obtain
$$\mathrm{DE}^{2}+\mathrm{AB}^{2}=\mathrm{AE}^{2}+\mathrm{BD}^{2}$$ 
Q.14: The perpendicular from A on side BC of a ∆ ABC intersects BC at D such that DB = 3 CD (see Figure). Prove that $$2 \mathrm{AB}^{2}=2 \mathrm{AC}^{2}+\mathrm{BC}^{2}$$. Ans : Applying Pythagoras theorem for $$\triangle \mathrm{ACD}$$, we obtain
$$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}$$
$$\mathrm{AD}^{2}=\mathrm{AC}^{2}-\mathrm{DC}^{2}$$ …(1)
Applying Pythagoras theorem in $$\triangle \mathrm{ABD}$$, we obtain
$$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{DB}^{2}$$
$$\mathrm{AD}^{2}=\mathrm{AB}^{2}-\mathrm{DB}^{2}$$ …(2)
From equation (1) and equation (2), we obtain
$$\mathrm{AC}^{2}-\mathrm{DC}^{2}=\mathrm{AB}^{2}-\mathrm{DB}^{2} \quad \ldots$$ (3)
It is given that 3DC = DB
$$\therefore \mathrm{DC}=\frac{\mathrm{BC}}{4} \text { and } \mathrm{DB}=\frac{3 \mathrm{BC}}{4}$$
Putting these values in equation (3), we obtain,
$$\mathrm{AC}^{2}-\left(\frac{\mathrm{BC}}{4}\right)^{2}=\mathrm{AB}^{2}-\left(\frac{3 \mathrm{BC}}{4}\right)^{2}$$
$$\mathrm{AC}^{2}-\frac{\mathrm{BC}^{2}}{16}=\mathrm{AB}^{2}-\frac{9 \mathrm{BC}^{2}}{16}$$
$$\begin{array}{l}{16 \mathrm{AC}^{2}-\mathrm{BC}^{2}=16 \mathrm{AB}^{2}-9 \mathrm{BC}^{2}} \\ {16 \mathrm{AB}^{2}-16 \mathrm{AC}^{2}=8 \mathrm{BC}^{2}} \\ {2 \mathrm{AB}^{2}=2 \mathrm{AC}^{2}+\mathrm{BC}^{2}}\end{array}$$ 
Q.15: In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that $$9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2}$$.
Ans : Let the side of the equilateral triangle be a, and AE be the altitude of $$\triangle \mathrm{ABC}$$
$$\therefore B E=E C=\frac{B C}{2}=\frac{a}{2}$$
And, $$A E=\frac{a \sqrt{3}}{2}$$
Given that, BD = 1/3 BC
$$\therefore B D=\frac{a}{3}$$
$$\mathrm{DE}=\mathrm{BE}-\mathrm{BD}=$$ $$\frac{a}{2}-\frac{a}{3}=\frac{a}{6}$$
Applying Pythagoras theorem in $$\triangle \mathrm{ADE}$$, we obtain
$$\mathrm{AD}^{2}=\mathrm{AE}^{2}+\mathrm{DE}^{2}$$
$$\mathrm{AD}^{2}=\left(\frac{a \sqrt{3}}{2}\right)^{2}+\left(\frac{a}{6}\right)^{2}$$
$$=\left(\frac{3 a^{2}}{4}\right)+\left(\frac{a^{2}}{36}\right)$$
$$=\frac{28 a^{2}}{36}$$
$$=\frac{7}{9} \mathrm{AB}^{2}$$
$$\Rightarrow 9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2}$$ 
Q.16: In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Ans : Let the side of the equilateral triangle be a, and AE be the altitude of $$\triangle \mathrm{ABC}$$
$$\therefore B E=E C=\frac{B C}{2}=\frac{a}{2}$$
Applying Pythagoras theorem in $$\triangle \mathrm{ABE}$$, we obtain
$$A B^{2}=A E^{2}+B E^{2}$$
$$a^{2}=\mathrm{AE}^{2}+\left(\frac{a}{2}\right)^{2}$$
$$\mathrm{AE}^{2}=a^{2}-\frac{a^{2}}{4}$$
$$\mathrm{AE}^{2}=\frac{3 a^{2}}{4}$$
$$4 \mathrm{AE}^{2}=3 \mathrm{a}^{2}$$
$$\Rightarrow 4 \times(\text { Square of altitude })=3 \times$$ (Square of one side) 
Q.17: Tick the correct answer and justify : In ∆ABC, AB = $$6 \sqrt{3}$$ cm, AC = 12 cm and BC = 6 cm. The angle B is :
(A) 120°
(B) 60°
(C) 90°
(D) 45°
Ans : Given that, AB = $$=6 \sqrt{3} \mathrm{cm}$$, AC = 12 cm, and BC = 6 cm
It can be observed that
$$\begin{array}{l}{A B^{2}=108} \\ {A C^{2}=144}\end{array}$$
$$\begin{array}{l}{\text { And, } B C^{2}=36} \\ {A B^{2}+B C^{2}=A C^{2}}\end{array}$$
The given triangle, $$\triangle \mathrm{ABC}$$, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
$$\therefore \angle B=90^{\circ}$$
Hence, the correct answer is (C) 
Q.1: In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that $$\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}$$ Ans : Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T,
Given that, PS is the angle bisector of $$\angle \mathrm{QPR}$$.
$$\angle \mathrm{QPS}=\angle \mathrm{SPR} \ldots(1)$$
By construction,
$$\angle S P R=\angle P R T(A S P S \| T R) \ldots(2)$$
$$\angle Q P S=\angle Q T R(A S P S| | T R) \ldots(3)$$
Using these equations, we obtain
$$\begin{array}{l}{\angle \mathrm{PRT}=\angle \mathrm{QTR}} \\ {\therefore \mathrm{PT}=\mathrm{PR}}\end{array}$$
By construction,
PS || TR
By using basic proportionality theorem for $$\Delta \mathrm{QTR}$$,
$$\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{QP}}{\mathrm{PT}}$$
$$\Rightarrow \frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}$$ (PT =TR) 
Q.2: In Figure, D is a point on hypotenuse AC of ∆ABC, such that BD ⊥AC, DM ⊥ BC and DN ⊥ AB. Prove that :
(i) $$\mathrm{DM}^{2}=\mathrm{DN} \cdot \mathrm{MC}$$
(ii) $$\mathrm{DN}^{2}=\mathrm{D} \mathrm{M} . \mathrm{AN}$$ Ans : (i) Let us join DB. We have, DN || CB, DM || AB, and $$\angle B=90^{\circ}$$
$$\therefore \mathrm{DMBN}$$ is a rectangle.
$$\therefore \mathrm{DN}=\mathrm{MB} \text { and } \mathrm{DM}=\mathrm{NB}$$
The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.
$$\therefore \angle C D B=90^{\circ}$$
$$\Rightarrow \angle 2+\angle 3=90^{\circ} \ldots(1)$$
In $$\triangle \mathrm{CDM}$$,
$$\begin{array}{l}{\angle 1+\angle 2+\angle \mathrm{DMC}=180^{\circ}} \\ {\Rightarrow \angle 1+\angle 2=90^{\circ} \ldots(2)}\end{array}$$
In $$\triangle \mathrm{DMB}$$,
$$\begin{array}{l}{\angle 3+\angle D M B+\angle 4=180^{\circ}} \\ {\Rightarrow \angle 3+\angle 4=90^{\circ} \ldots(3)}\end{array}$$
From equation (1) and (2), we obtain
$$\angle 2=\angle 4$$
In $$\triangle \mathrm{D} \mathrm{CM} \text { and } \triangle \mathrm{BDM}$$,
$$\begin{array}{l}{\angle 1=\angle 3 \text { (Proved above) }} \\ {\angle 2=\angle 4(\text { Proved above })}\end{array}$$
$$\therefore \triangle \mathrm{DCM} \sim \triangle \mathrm{BDM}(\text { AA similarity criterion) }$$
$$\Rightarrow \frac{\mathrm{BM}}{\mathrm{DM}}=\frac{\mathrm{DM}}{\mathrm{MC}}$$
$$\Rightarrow \frac{\mathrm{DN}}{\mathrm{DM}}=\frac{\mathrm{DM}}{\mathrm{MC}} \quad(\mathrm{BM}=\mathrm{DN})$$
$$\Rightarrow \mathrm{DM}^{2}=\mathrm{D} \mathrm{N} \times \mathrm{MC}$$
(ii) In right triangle DBN,
$$\angle 5+\angle 7=90^{\circ} \ldots(4)$$
$$In right triangle DAN,$$
$$\angle 6+\angle 8=90^{\circ} \ldots(5)$$
D is the foot of the perpendicular drawn from B to AC.
$$\therefore \angle A D B=90^{\circ}$$
$$\Rightarrow \angle 5+\angle 6=90^{\circ} \ldots(6)$$
From equation (5) and (6), we obtain
$$\angle 8=\angle 5$$
In $$\triangle \mathrm{DNA} \text { and } \triangle \mathrm{B} \mathrm{ND}$$,
$$\begin{array}{l}{\angle 6=\angle 7 \text { (Proved above) }} \\ {\angle 8=\angle 5 \text { (Proved above) }}\end{array}$$
$$\therefore \triangle \mathrm{DNA} \sim \Delta \mathrm{BND}$$ (AA similarity criterion)
$$\Rightarrow \frac{\mathrm{AN}}{\mathrm{DN}}=\frac{\mathrm{DN}}{\mathrm{NB}}$$
$$\begin{array}{l}{\Rightarrow \mathrm{DN}^{2}=\mathrm{AN} \times \mathrm{NB}} \\ {\Rightarrow \mathrm{DN}^{2}=\mathrm{AN} \times \mathrm{DM}(\mathrm{AS} \mathrm{NB}=\mathrm{DM})}\end{array}$$ 
Q.3: In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that $$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}+2 \mathrm{BC} . \mathrm{BD}$$ Ans : Applying Pythagoras theorem in $$\triangle \mathrm{ADB}$$, we obtain
$$A B^{2}=A D^{2}+D B^{2} \ldots(1)$$
Applying Pythagoras theorem in $$\triangle \mathrm{ACD}$$, we obtain
$$\begin{array}{l}{A C^{2}=A D^{2}+D C^{2}} \\ {A C^{2}=A D^{2}+(D B+B C)^{2}} \\ {A C^{2}=A D^{2}+D B^{2}+B C^{2}+2 D B \times B C}\end{array}$$
$$A C^{2}=A B^{2}+B C^{2}+2 D B \times B C[\text { Using equation }(1)]$$ 
Q.4: In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that $$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}-2 \mathrm{BC} \cdot \mathrm{BD}$$. Ans : Applying Pythagoras theorem in $$\triangle \mathrm{ADB}$$, we obtain
$$\begin{array}{l}{A D^{2}+D B^{2}=A B^{2}} \\ {\Rightarrow A D^{2}=A B^{2}-D B^{2} \ldots(1)}\end{array}$$
Applying Pythagoras theorem in $$\triangle \mathrm{ADC}$$, we obtain
$$\mathrm{AD}^{2}+\mathrm{DC}^{2}=\mathrm{AC}^{2}$$
$$\begin{array}{l}{\mathrm{AB}^{2}-\mathrm{BD}^{2}+\mathrm{DC}^{2}=\mathrm{AC}^{2}[\text { Using equation }(1)]} \\ {\mathrm{AB}^{2}-\mathrm{BD}^{2}+(\mathrm{BC}-\mathrm{BD})^{2}=\mathrm{AC}^{2}} \\ {\mathrm{AC}^{2}=\mathrm{AB}^{2}+(\mathrm{BC}-\mathrm{BD})^{2}=\mathrm{AC}^{2}} \\ {\mathrm{AC}^{2}=\mathrm{AB}^{2}-\mathrm{BD}^{2}+\mathrm{BC}^{2}+\mathrm{BD}^{2}-2 \mathrm{BC} \times \mathrm{BD}} \\ {=\mathrm{AB}^{2}+\mathrm{BC}^{2}-2 \mathrm{BC} \times \mathrm{BD}}\end{array}$$ 
Q.5: In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :
(i) $$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \cdot \mathrm{DM}+\left(\frac{\mathrm{BC}}{2}\right)^{2}$$
(ii) $$\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DM}+\left(\frac{\mathrm{BC}}{2}\right)^{2}$$
(iii) $$\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$$ Ans : (i) Applying Pythagoras theorem in $$\triangle \mathrm{AMD}$$, we obtain
$$A M^{2}+M D^{2}=A D^{2} \ldots(1)$$
Applying Pythagoras theorem in $$\triangle \mathrm{AMC}$$, we obtain
$$\begin{array}{l}{\mathrm{AM}^{2}+\mathrm{MC}^{2}=\mathrm{AC}^{2}} \\ {\mathrm{AM}^{2}+(\mathrm{MD}+\mathrm{DC})^{2}=\mathrm{AC}^{2}} \\ {\left(\mathrm{AM}^{2}+\mathrm{MD}^{2}\right)+\mathrm{DC}^{2}+2 \mathrm{MD} \cdot \mathrm{DC}=\mathrm{AC}^{2}}\end{array}$$
$$A D^{2}+D C^{2}+2 M D . D C=A C^{2}[\text { Using equation }(1)]$$
Using the result, $$\quad \mathrm{DC}=\frac{\mathrm{BC}}{2},$$, we obtain
$$\mathrm{AD}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}+2 \mathrm{MD} \cdot\left(\frac{\mathrm{BC}}{2}\right)=\mathrm{AC}^{2}$$
$$\mathrm{AD}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}+\mathrm{MD} \times \mathrm{BC}=\mathrm{AC}^{2}$$

(ii) Applying Pythagoras theorem in $$\triangle A B M$$, we obtain
$$A B^{2}=A M^{2}+M B^{2}$$
$$\begin{array}{l}{=\left(A D^{2}-D M^{2}\right)+M B^{2}} \\ {=\left(A D^{2}-D M^{2}\right)+(B D-M D)^{2}} \\ {=A D^{2}-D M^{2}+B D^{2}+M D^{2}-2 B D \times M D} \\ {=A D^{2}+B D^{2}-2 B D \times M D}\end{array}$$
$$=\mathrm{AD}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}-2\left(\frac{\mathrm{BC}}{2}\right) \times \mathrm{MD}$$
$$=\mathrm{AD}^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}-\mathrm{BC} \times \mathrm{MD}$$

(iii) Applying Pythagoras theorem in $$\triangle \mathrm{ABM}$$, we obtain
$$A M^{2}+M B^{2}=A B^{2} \dots(1)$$
Applying Pythagoras theorem in $$\triangle \mathrm{AMC}$$, we obtain
$$\mathrm{AM}^{2}+\mathrm{MC}^{2}=\mathrm{AC}^{2} \ldots(2)$$
Adding equations (1) and (2), we obtain
$$\begin{array}{l}{2 \mathrm{AM}^{2}+\mathrm{MB}^{2}+\mathrm{MC}^{2}=\mathrm{AB}^{2}+\mathrm{AC}^{2}} \\ {2 \mathrm{AM}^{2}+(\mathrm{BD}-\mathrm{DM})^{2}+(\mathrm{MD}+\mathrm{DC})^{2}=\mathrm{AB}^{2}+\mathrm{AC}^{2}} \\ {2 \mathrm{AM}^{2}+\mathrm{BD}^{2}+\mathrm{DM}^{2}-2 \mathrm{BD} \cdot \mathrm{DM}+\mathrm{MD}^{2}+\mathrm{DC}^{2}+\mathrm{AC}^{2}} \\ {2 \mathrm{AM}^{2}+2 \mathrm{MD}^{2}+\mathrm{BD}^{2}+\mathrm{DC}^{2}+2 \mathrm{MD}(-\mathrm{BD}+\mathrm{DC})=\mathrm{AB}^{2}+\mathrm{AC}^{2}}\end{array}$$
$$2\left(\mathrm{AM}^{2}+\mathrm{MD}^{2}\right)+\left(\frac{\mathrm{BC}}{2}\right)^{2}+\left(\frac{\mathrm{BC}}{2}\right)^{2}+2 \mathrm{MD}\left(-\frac{\mathrm{BC}}{2}+\frac{\mathrm{BC}}{2}\right)=\mathrm{AB}^{2}+\mathrm{AC}^{2}$$
$$2 \mathrm{AD}^{2}+\frac{\mathrm{BC}^{2}}{2}=\mathrm{AB}^{2}+\mathrm{AC}^{2}$$ 
Q.6: Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Ans : Let ABCD be a parallelogram.
Let us draw perpendicular DE on extended side AB, and AF on side DC.
Applying Pythagoras theorem in $$\triangle D E A$$, we obtain
$$\mathrm{DE}^{2}+\mathrm{EA}^{2}=\mathrm{D} \mathrm{A}^{2} \ldots(i)$$
Applying Pythagoras theorem in $$\triangle D E A$$, we obtain
$$\mathrm{DE}^{2}+\mathrm{EA}^{2}=\mathrm{DA}^{2} \ldots$$ (i)
Applying Pythagoras theorem in $$\triangle \mathrm{DEB}$$, we obtain
$$\begin{array}{l}{\mathrm{DE}^{2}+\mathrm{EB}^{2}=\mathrm{DB}^{2}} \\ {\mathrm{DE}^{2}+(\mathrm{EA}+\mathrm{AB})^{2}=\mathrm{DB}^{2}} \\ {\left(\mathrm{DE}^{2}+\mathrm{EA}^{2}\right)+\mathrm{AB}^{2}+2 \mathrm{EA} \times \mathrm{AB}=\mathrm{DB}^{2}} \\ {\mathrm{DA}^{2}+\mathrm{AB}^{2}+2 \mathrm{E} \mathrm{A} \times \mathrm{AB}=\mathrm{DB}^{2} \ldots \text { (ii) }}\end{array}$$
Applying Pythagoras theorem in $$\triangle \mathrm{ADF}$$, we obtain
$$\mathrm{AD}^{2}=\mathrm{AF}^{2}+\mathrm{FD}^{2}$$
Applying Pythagoras theorem in $$\triangle \mathrm{AFC}$$, we obtain
$$A C^{2}=A F^{2}+F C^{2}$$
$$\begin{array}{l}{=\mathrm{AF}^{2}+(\mathrm{DC}-\mathrm{FD})^{2}} \\ {=\mathrm{AF}^{2}+\mathrm{DC}^{2}+\mathrm{FD}^{2}-2 \mathrm{DC} \times \mathrm{FD}} \\ {=\left(\mathrm{AF}^{2}+\mathrm{FD}^{2}\right)+\mathrm{DC}^{2}-2 \mathrm{DC} \times \mathrm{FD}} \\ {\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}-2 \mathrm{DC} \times \mathrm{FD} \ldots(i i)}\end{array}$$
Since ABCD is a parallelogram,
$$\begin{array}{l}{A B=C D \ldots(i v)} \\ {\text { And, } B C=A D \ldots(v)}\end{array}$$
In $$\triangle \mathrm{DE} \mathrm{A} \text { and } \triangle \mathrm{ADF}$$
\begin{aligned} \angle D E A &=\angle A F D\left(B o t h 90^{\circ}\right) \\ \angle E A D &=\angle A D F(E A \| D F) \end{aligned}
$$A D=A D(\text { Common })$$
$$\therefore \Delta E A D \cong \triangle F D A$$ (AAS congruence criterion)
$$\Rightarrow E A=D F \ldots(v i)$$
Adding equations (i) and (iii), we obtain
$$\begin{array}{l}{\mathrm{DA}^{2}+\mathrm{AB}^{2}+2 \mathrm{EA} \times \mathrm{AB}+\mathrm{AD}^{2}+\mathrm{DC}^{2}-2 \mathrm{DC} \times \mathrm{FD}=\mathrm{DB}^{2}+\mathrm{AC}^{2}} \\ {\mathrm{DA}^{2}+\mathrm{AB}^{2}+\mathrm{AD}^{2}+\mathrm{DC}^{2}+2 \mathrm{EA} \times \mathrm{AB}-2 \mathrm{DC} \times \mathrm{FD}=\mathrm{DB}^{2}+\mathrm{AC}^{2}} \\ {\mathrm{BC}^{2}+\mathrm{AB}^{2}+\mathrm{AD}^{2}+\mathrm{DC}^{2}+2 \mathrm{E} \mathrm{A} \times \mathrm{AB}-2 \mathrm{AB} \times \mathrm{EA}=\mathrm{DB}^{2}+\mathrm{AC}^{2}}\end{array}$$
[Using equations (iv) and (vi)]
$$A B^{2}+B C^{2}+C D^{2}+D A^{2}=A C^{2}+B D^{2}$$ 
Q.7: In Figure, two chords AB and CD intersect each other at the point P. Prove that :
(i) ∆APC ~ ∆ DPB
(ii) AP . PB = CP . DP Ans : (i) In $$\triangle A P C \text { and } \Delta D P B$$,
$$\angle \mathrm{APC}=\angle \mathrm{DPB}$$ (Vertically opposite angles)
$$\angle \mathrm{CAP}=\angle \mathrm{BDP}$$ ( Angles in the same segment for chord CB)
$$\triangle \mathrm{APC} \sim \Delta \mathrm{DPB}$$ (By AA similarity criterion)
(ii) We have already proved that
$$\triangle \mathrm{APC} \sim \Delta \mathrm{DPB}$$
We know that the corresponding sides of similar triangles are proportional.
$$\therefore \frac{\mathrm{AP}}{\mathrm{DP}}=\frac{\mathrm{PC}}{\mathrm{PB}}=\frac{\mathrm{CA}}{\mathrm{BD}}$$
$$\Rightarrow \frac{A P}{D P}=\frac{P C}{P B}$$
$$\therefore \mathrm{AP} . \mathrm{PB}=\mathrm{PC} . \mathrm{DP}$$ 
Q.8: In Figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) ∆ PAC ~ ∆ PDB
(ii) PA . PB = PC . PD Ans : (i) In $$\triangle \mathrm{PAC} \text { and } \Delta \mathrm{PDB}$$,
$$\angle P=\angle P(\text { Common })$$
$$\begin{array}{l}{\angle \mathrm{PAC}=\angle \mathrm{PDB} \text { (Exterior angle of a cyclic quadrilateral is } \angle \mathrm{PCA}=\angle \mathrm{PBD} \text { equal to the }} \\ {\text { opposite interior angle) }}\end{array}$$
$$\therefore \triangle \mathrm{PAC} \sim \Delta \mathrm{PDB}$$
(ii) We know that the corresponding sides of similar triangles are proportional.
$$\therefore \frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{AC}}{\mathrm{DB}}=\frac{\mathrm{PC}}{\mathrm{PB}}$$
$$\Rightarrow \frac{\mathrm{PA}}{\mathrm{PD}}=\frac{\mathrm{PC}}{\mathrm{PB}}$$
$$\therefore \text { PA.PB }=\mathrm{PC.PD}$$ 
Q.9: In Figure, D is a point on side BC of ∆ ABC such that $$\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}}$$ Prove that AD is the bisector of ∠ BAC. Ans : Let us extend BA to P such that AP =AC, Join PC. It is given that,
$$\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AC}}$$
$$\Rightarrow \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{AP}}{\mathrm{AC}}$$
By using the converse of basic proportionality theorem, we obtain AD || PC
$$\mathrm{AD}\|\mathrm{PC}$$
$$\Rightarrow \angle B A D=\angle A P C(\text { Corresponding angles }) \ldots$$ (1)
And $$\angle D A C=\angle A C P$$ (Alternate interior angles)...(2) By construction, we have
$$A P=A C$$
$$\Rightarrow \angle A P C=\angle A C P \ldots(3)$$
On comparing equations (1), (2), and (3), we obtain
$$\angle B A D=\angle A P C$$
$$\Rightarrow \mathrm{AD}$$ is the bisector of the angle BAC. 
Q.10: Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds? Ans : Let AB be the height of the tip of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod.
Then, AC is the length of the string.
AC can be found by applying Pythagoras theorem in $$\triangle \mathrm{ABC}$$
$$A C^{2}=A B^{2}+B C^{2}$$
$$A B^{2}=(1.8 m)^{2}+(2.4 m)^{2}$$
$$\begin{array}{l}{A B^{2}=(3.24+5.76) m^{2}} \\ {A B^{2}=9.00 m^{2}}\end{array}$$
$$\Rightarrow A B=\sqrt{9} \mathrm{m}=3 \mathrm{m}$$
Thus, the length of the string out is 3 m.
She pulls the string at the rate of 5 cm per second.
Therefore, string pulled in 12 seconds = 12 x 5 = 60 cm = 0.6 m. 

Did you find NCERT Solutions Class 10 Maths chapter 6 Triangles helpful? If yes, please comment below. Also please like, and share it with your friends!

### NCERT Solutions Class 10 Maths chapter 6 Triangles- Video

You can also watch the video solutions of NCERT Class10 Maths chapter 6 Triangles here.

Video – will be available soon.

If you liked the video, please subscribe to our YouTube channel so that you can get more such interesting and useful study resources.

### Download NCERT Solutions Class 10 Maths chapter 6 Triangles In PDF Format

You can also download here the NCERT Solutions Class 10 Maths chapter 6 Triangles in PDF format.