# NCERT Solutions For Class 10 Mathematics Chapter 6 Exercise 6.3

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## NCERT Class 10 Chapter 6 Exercise 6.3 Triangle Solutions

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NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 deal with problems in which various theorems about the similarity of triangles must be applied to solve them. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 covers all aspects of this subject, giving students plenty of opportunity to practise preparing for the exam. Extramarks experts have prepared NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 online using the latest CBSE curriculum with clear step-by-step solutions. This will make students proficient in solving similar problems independently. Subjects like science, Mathematics, and English will become easy to learn if students have access to the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  problem-solving, and other subject solutions.

### Access NCERT Solutions For Mathematics Chapter 6 – Triangle Solutions

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### CBSE Class 10 Mathematics Chapter 6 Exercise 6.3 Solutions

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### Topic Covered In Class 10 Mathematics Chapter 6 Exercise 6.3

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are based on the topic of Similarity between different Triangles. Criteria for similarity of triangles: Two triangles are said to be similar if their corresponding angles are congruent and their corresponding sides are also proportional.

### Different Theorems Covered in Exercise 6.3

Theorem 1: If in two triangles the corresponding angles are congruent, then their corresponding sides have the same ratio. So it can be said that the two triangles are similar. Students should refer to the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 for answers.

Theorem 2: If in two triangles, the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are congruent. Hence, it can be said that the two triangles are similar. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 provides all the answers related to this theorem.

Theorem 3: If an angle of a triangle is equal to an angle of another triangle and the sides including these angles are proportional to each other, then the two triangles are similar. For solutions, students can look at the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

### Class 10 Mathematics Exercise 6.3 Solutions

Extramarks provides the answers to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 to assist students in their time of need.They can visit the website of Extramarks to get access to all the answers in the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

#### Exercise 6.3 Lesson 10 Question 1

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are the answers to questions in the Mathematics Class 10 Chapter 6 Exercise 6.3 is based on the similarity of pairs of triangles. A set of pairs of triangles is given, and students need to determine which pairs are similar and also mention the criteria by which one defines them as congruent triangles. Students must also give the correct notation to describe similar triangles.

#### Exercise 6.3 Class 10 Question 2

In the second question of the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, students have to use the relationship between adjacent angles and the similarity of triangles to find the angle of two similar triangles formed by the intersection of  two intersecting lines and two parallel lines.

#### Exercise 6.3 Class 10 Question 3

In NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, students are given a trapezium ABCD, and they have to prove that for the intersection O the ratio of the lengths of the diagonals  is AO/OC = OB/OD. They need to apply the theorem to the alternate inscribed angle and the vertical opposite angle to solve this problem.

#### Exercise 6.3 Class 10 Question 4

In this question, a triangle is given, which has two interior triangles. By applying the theorem that if the sides of a triangle are proportional, then their corresponding angles are congruent, one can prove that the two triangles are similar. The answer is available in the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3

#### Exercise 6.3 Class 10 Question 5

This is a simple question where two angles of two triangles are congruent (criteria AA), so students can prove that the given triangles are similar. Students are advised to refer to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 for answers.

#### Exercise 6.3 Class 10 Question 6

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 use knowledge of similar triangles and CPCT (corresponding parts of similar triangles) to prove that the given triangles are similar.

#### Exercise 6.3 Class 10 Question 7

In this question, the altitudes of two triangles are given, and students have to prove that the triangles formed by the intersection of the altitudes are congruent. Students will use the concept of vertically opposite angles and common angles to solve this question. The NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 answer this question as well.

#### Exercise 6.3 Class 10 Question 8

This question has a parallelogram ABCD, and a line from point B that intersects DC at F and extends to point E. Point D is extended until it meets E, forming the triangle AEB. Using the opposite angle property of parallelograms and the concept of alternating angles, students can take the help of the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 to prove that △ABE and △CFB are similar.

#### Exercise 6.3 Class 10 Question 9

This is again a simple question where two right triangles have a common side and students have to prove that both triangles are similar triangles. They can easily prove the common angle and the corresponding sides of the triangle. Similarity and the concept of similarity of angles with the help of NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

#### Exercise 6.3 Class 10 Question 10

Question 10 of this exercise is related to the bisector theorem and the similar angle substitution criterion to prove that two triangles formed by bisectors of an angle are congruent. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 can help students understand how to correctly answer the question.

#### Exercise 6.3 Class 10 Question 11

In this problem, an isosceles triangle is given, one of its sides is stretched, and a perpendicular is dropped from the point of extension to the opposite side of the triangle. Students must demonstrate that the triangle formed by stretching and the triangle formed by dropping perpendicular to the base of the original triangle are congruent. The properties of isosceles triangles and the alternating angle criterion given in NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are used to solve this problem.

#### Exercise 6.3 Class 10 Question 12

There are two triangles ABC and PQR in this problem, and it is known that the medians of the two triangles and their adjacent sides are proportional. Students have to prove that triangles ABC and PQR are similar triangles. The median dividing the opposite side, using this concept with SSS (for congruent triangles) and corresponding angles of similar triangles given in NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, this problem can be solved.

#### Exercise 6.3 Class 10 Question 13

In triangle ABC, draw line AD on side BC  and ∠ADC is equal to ∠BAC. Students have to prove that CA² = CB.CD. This is done using the concepts of common angle, alternative angle and corresponding angle from similar triangle rules. Students can refer to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 for answers.

#### Exercise 6.3 Class 10 Question 14

This question is similar to question 12 in that the sides of two triangles and their medians are proportional, and students have to prove that the two triangles are similar. This question is solved by extending the medians of two triangles by equal lengths, then using the rules for parallelograms, SSS, SAS, and corresponding angles from the Congruent Triangle Standard given on NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  to prove that the triangles are similar.

#### Exercise 6.3 Class 10 Question 15

It is a question about a vertical pillar and its shadow with the shadow of a tower. Given the height of the column and the length of the shade,  students need to find the height of the tower. Using the AAA (angle-angle-angle) congruence theorem, one can deduce the height of the tower. They can also take the help of NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3

#### Exercise 6.3 Class 10 Question 16

Given two similar triangles, the student must prove that the ratios of the sides of the triangle and their medians are equal. By using the properties of the mean and the corresponding angle criterion given in NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, they can solve this problem.

### Key Features Of NCERT Solutions For Class 10 Mathematics Exercise 6.3

Extramarks provides a lot of courses to students and these solutions are easily accessible to them on the Extramarks website.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are prepared in simple language for students to understand it.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are prepared by the subject experts of Extramarks.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 give accurate answers to all the questions in NCERT textbooks.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 will help the students with a quick revision.

### NCERT Solutions For Class 10 Mathematics Chapter 6 All Other Exercises

Exercise 6.1: It comprises Three Questions and Answers, and these are Short Answer Type Questions. Students can access the solutions of this exercise like NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 on Extramarks

Exercise 6.2: It consists of 10 Questions and Answers including 9 Short Answers and 1 long answer which are available in resources like NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 and others provided by Extramarks

Exercise 6.4: It consists of 9 Questions and Answers with 7 Short and 2 long Answers. Extramarks provide the answers to these questions in a similar format as the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

Exercise 6.5: It comprises 17 Questions and Answers i.e, 15 short answers and 2 long answers. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 gives a blueprint of the kinds of answers that will be helpful for solving the questions.

Exercise 6.6: It consists of 10 Questions and Answers. 5 Short answers and 5 long answers. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 is an example of the kind of answers that Extramarks provides for this exercise.

Q.1 State which pairs of triangles in the following figures 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.

$\begin{array}{l}\text{(i)}\\ \text{Corresponding angles are equal in the given triangles. i.e.,}\\ \angle \mathrm{A}=\angle \mathrm{P}=60\mathrm{°},\\ \angle \mathrm{B}=\angle \mathrm{Q}=80\mathrm{°}\\ \text{and}\\ \angle \mathrm{C}=\angle \mathrm{R}=40\mathrm{°}.\\ \text{Therefore, by AAA similarity criterion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta PQR}.\\ \text{(ii)}\\ \text{Ratios of the corresponding sides of the given pair of}\\ \text{triangles are equal.}\\ \text{i.e.,}\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{CA}}{\mathrm{PQ}}=\frac{1}{2}\\ \text{Therefore, by SSS similarity criterion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta QRP}.\\ \text{(iii)}\\ \text{The given pair of triangles are not similar as their}\\ \text{corresponding sides are not proportional.}\\ \\ \text{(iv)}\\ \text{The given pair of triangles are not similar as their}\\ \text{corresponding sides are not proportional.}\\ \text{(v)}\\ \text{The given pair of triangles are not similar as their}\\ \text{corresponding sides are not proportional.}\\ \text{(vi)}\\ \text{In}\mathrm{\Delta DEF},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{D}+\angle \mathrm{E}+\angle \mathrm{F}=180\mathrm{°}\\ ⇒70\mathrm{°}+80\mathrm{°}+\angle \mathrm{F}=180\mathrm{°}\\ ⇒\angle \mathrm{F}=180\mathrm{°}-150\mathrm{°}=30\mathrm{°}\\ \text{In}\mathrm{\Delta PQR},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180\mathrm{°}\\ ⇒\angle \mathrm{P}+80\mathrm{°}+30\mathrm{°}=180\mathrm{°}\\ ⇒\angle \mathrm{P}=180\mathrm{°}-110\mathrm{°}=70\mathrm{°}\\ \text{We find out that corresponding angles are equal in}\mathrm{\Delta DEF}\text{}\\ \text{and}\mathrm{\Delta PQR}.\text{i.e.,}\\ \angle \mathrm{D}=\angle \mathrm{P}=70\mathrm{°},\\ \angle \mathrm{E}=\angle \mathrm{Q}=80\mathrm{°}\\ \text{and}\\ \angle \mathrm{F}=\angle \mathrm{R}=30\mathrm{°}.\\ \text{Therefore, by AAA similarity criterion,}\mathrm{\Delta DEF}~\mathrm{\Delta PQR}.\end{array}$

Q.2 In the following figure,

$\Delta \text{ODC}\sim \Delta \text{OBA,}\angle \text{BOC}=125°\text{and}\angle \text{CDO}=70°.\text{Find}\angle \text{DOC,}\angle \text{DCO and}\angle \text{OAB}\text{.}$

Ans.

$\begin{array}{l}\mathrm{DOB}\text{is a straight line.}\\ \therefore \angle \text{BOC}+\angle \mathrm{D}\text{OC}=180\mathrm{°}\\ ⇒\angle \text{DOC}=180\mathrm{°}-\angle \mathrm{B}\text{OC}=180\mathrm{°}-125\mathrm{°}=55\mathrm{°}\\ \text{In}\mathrm{\Delta }\text{DOC,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{D}\text{OC}+\angle \text{DC}\mathrm{O}+\angle \text{O}\mathrm{D}\text{C}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}}\angle \mathrm{D}\text{CO}=180\mathrm{°}-\angle \mathrm{D}\text{OC}-\angle \text{O}\mathrm{D}\text{C}\\ ⇒\text{\hspace{0.17em}}\angle \mathrm{D}\text{CO}=180\mathrm{°}-55\mathrm{°}-70\mathrm{°}=55\mathrm{°}\\ \text{It is given that}\mathrm{\Delta }\text{ODC}~\mathrm{\Delta }\text{OBA and so corresponding angles}\\ \text{in these triangles are equal.}\\ \therefore \angle \text{OAB}=\angle \text{DCO}=55\mathrm{°}\end{array}$

Q.3

$\begin{array}{l}\text{Diagonals AC and BD of a trapezium ABCD with AB}\parallel \text{DC intersect each other at the point O. Using}\\ \text{a similarity criterion for two triangles, show that}\frac{\text{OA}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}.\end{array}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{DOC and}\mathrm{\Delta }\text{BOA,}\\ \angle \text{CDO}=\angle \text{ABO [Alternate interior angles as AB}\parallel \text{CD]}\\ \angle \text{DCO}=\angle \text{BAO [Alternate interior angles as AB}\parallel \text{CD]}\\ \angle \text{DOC}=\angle \text{BOA [Vertically opposite angles]}\\ \therefore \mathrm{\Delta }\text{DOC}~\mathrm{\Delta }\text{BOA [AAA similarity critarion]}\\ ⇒\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}}\text{[Corresponding sides are proportional]}\\ ⇒\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\end{array}$

Q.4

$\text{In the following figure,}\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\text{and}\angle \text{1}=\angle 2.\text{Show that}\Delta \text{PQS}\sim \Delta \text{TQR}\text{.}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{PQR,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{1}=\angle 2\text{(Given)}\\ ⇒\text{PQ}=\text{PR}...\text{(1)}\\ \text{It is given that}\\ \text{}\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\\ ⇒\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{QP}}\text{[From (1), PR}=\text{QP] \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \angle \text{Q is common in}\mathrm{\Delta }\text{PQS and}\mathrm{\Delta }\text{TQR and the sides including}\\ \text{this angle in both triangles are proportional as shown in}\\ \text{equation (2).}\\ \text{Therefore, by SAS critarion,}\mathrm{\Delta }\text{PQS}~\mathrm{\Delta }\text{TQR}.\end{array}$

Q.5

$\text{S and T are points on sides PR and QR of}\mathrm{\Delta }\text{PQR such that}\angle \text{P}=\angle \text{RTS. Show that}\mathrm{\Delta }\text{RPQ}~\mathrm{\Delta }\text{RTS.}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{RPQ and}\mathrm{\Delta }\text{RTS,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{P}=\angle \mathrm{T}\text{(Given)}\\ \text{and}\angle \mathrm{R}=\angle \mathrm{R}\text{}\\ \text{Therefore, by AA criterion,}\mathrm{\Delta }\text{RPQ}~\mathrm{\Delta }\text{RTS}.\end{array}$

Q.6

$\text{In the following figure, if}\mathrm{\Delta }\text{ABE}\cong \mathrm{\Delta }\text{ACD, show that}\mathrm{\Delta }\text{ADE}~\mathrm{\Delta }\text{ABC.}$

Ans.

$\begin{array}{l}\text{It is given that}\mathrm{\Delta }\text{ABE}\cong \mathrm{\Delta }\text{ACD.}\\ \text{Therefore, by CPCT,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}=\text{AC}\\ \text{and AD}=\text{AE}\\ \text{So,}\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\\ \text{In}\mathrm{\Delta }\text{ADE and}\mathrm{\Delta }\text{ABC,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}=\angle \mathrm{A}\text{}\\ \text{and}\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\\ \text{Therefore, by SAS criterion,}\mathrm{\Delta }\text{ADE}~\mathrm{\Delta }\text{ABC}.\end{array}$

Q.7

$\begin{array}{l}\text{In the following figure, altitudes AD and CE of}\mathrm{\Delta }\text{ABC intersect each other at the point P. Show that:}\\ \text{(i)}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{CDP}\\ \text{(ii)}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{CBE}\\ \text{(iii)}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{ADB}\\ \text{(iv)}\mathrm{\Delta }\text{PDC}~\mathrm{\Delta }\text{BEC}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{In}\mathrm{\Delta }\text{AEP and}\mathrm{\Delta }\text{CDP,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{APE}=\angle \mathrm{CPD}\text{(Vertically opposite angles)}\\ \text{and}\angle \text{AEP}=\angle \text{CDP}=90\mathrm{°}\text{}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{CDP}.\\ \text{(ii)}\\ \text{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{CBE,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ADB}=\angle \text{CEB}=90\mathrm{°}\\ \text{and}\angle \text{ABD}=\angle \text{CBE}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{CBE}.\\ \text{(iii)}\\ \text{In}\mathrm{\Delta }\text{AEP and}\mathrm{\Delta }\text{ADB,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AEP}=\angle \text{ADB}=90\mathrm{°}\\ \text{and}\angle \text{PAE}=\angle \text{BAD}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{ADB}.\\ \text{(iv)}\\ \text{In}\mathrm{\Delta }\text{PDC and}\mathrm{\Delta }\text{BEC,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PDC}=\angle \text{BEC}=90\mathrm{°}\\ \text{and}\angle \text{DCP}=\angle \text{ECB}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{PDC}~\mathrm{\Delta }\text{BEC}.\end{array}$

Q.8

$\begin{array}{l}\text{E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD}\\ \text{at F. Show that}\mathrm{\Delta }\text{ABE}~\mathrm{\Delta }\text{CFB.}\end{array}$

Ans.

$\begin{array}{l}\text{In the above figure, ABCD is a parallelogram in which AB}\parallel \text{DC.}\\ \text{Opposite angles in parallelogram are equal. Therefore,}\\ \angle \text{EAB}=\angle \text{BCF}\\ \text{Also,}\\ \text{}\angle \text{ABE}=\angle \text{CFB [Alternate interior angles]}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ABE}~\mathrm{\Delta }\text{CFB}.\end{array}$

Q.9

$\begin{array}{l}\text{In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively.}\\ \text{Prove that:}\\ \text{(i)}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{AMP}\\ \text{(ii)}\frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{In}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{AMP,}\\ \angle \text{ABC}=\angle \text{AMP}=90\mathrm{°}\\ \text{and,}\\ \text{}\angle \text{CAB}=\angle \text{PAM}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{AMP}.\\ \text{(ii)}\\ \text{Corresponding sides are proportional in similar triangles.}\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{AMP}\\ ⇒\frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}\end{array}$

Q.10

$\begin{array}{l}\text{CD and GH are respectively the bisectors of}\mathrm{\Delta }\text{ACB and}\mathrm{\Delta }\text{EGF such that D and H lie on sides AB and FE}\\ \text{of}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{EFG respectively. If}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{FEG, show that:}\\ \text{(i) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCB}~\mathrm{\Delta }\text{HGE}\\ \text{(iii)\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCA}~\mathrm{\Delta }\text{HGF}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{FEG.}\\ \angle \text{A}=\angle \text{F,}\angle \text{B}=\angle \text{E,}\angle \text{BCA}=\angle \text{EGF}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\angle \text{BCA}=\frac{1}{2}\angle \text{EGF}\\ ⇒\angle \text{ACD}=\angle \text{FGH and}\angle \text{DCB}=\angle \text{HGE}\\ \therefore \text{By AA similarity critarion,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCA}~\mathrm{\Delta }\text{HGF and}\mathrm{\Delta }\text{DCB}~\mathrm{\Delta }\text{HGE}\\ ⇒\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}\end{array}$

Q.11

$\begin{array}{l}\text{In the following figure, E is a point on side CB produced of an isosceles triangle ABC}\\ \text{with AB}=\text{AC}\text{. If AD}\perp \text{BC and EF}\perp \text{AC, prove that}\Delta \text{ABD}\sim \Delta \text{ECF}\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{ECF,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BDA}=\angle \text{CFE}=90\mathrm{°}\\ \text{and}\angle \text{ABD}=\angle \text{ECF [AB}=\text{AC}⇒\angle \text{B}=\angle \text{C]}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{ECF}.\end{array}$

Q.12

$\begin{array}{l}\text{Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and}\\ \text{median PM of}\mathrm{\Delta }\text{PQR (see the following figure). Show that}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR.}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that AD and PM are medians}.\text{Therefore,}\\ \frac{1}{2}\text{BC}=\text{BD}=\text{DC}\\ \text{and}\\ \frac{1}{2}\text{QR}=\text{QM}=\text{MR}\\ \text{It is given that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AD}}{\text{PM}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AB}}{\text{PQ}}=\frac{\frac{1}{2}\text{BC}}{\frac{1}{2}\text{QR}}=\frac{\text{AD}}{\text{PM}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}=\frac{\text{AD}}{\text{PM}}\text{}\\ \text{Therefore, by SSS similarity critarion,}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{PQM and so}\\ \text{in}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{PQR,}\angle \text{B}=\angle \text{Q and}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}.\\ \text{Therefore, by SAS similarity critarion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR.}\end{array}$

Q.13

$\text{D is a point on the side BC of a triangle ABC such that}\angle \text{ADC}=\angle {\text{BAC. Show that CA}}^{\text{2}}=\text{CB}·\text{CD.}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{BAC and}\mathrm{\Delta }\text{ADC,}\\ \angle \text{ADC}=\angle \text{BAC (Given)}\\ \angle \text{ACD}=\angle \text{BCA (Common angle)}\\ \therefore \mathrm{\Delta }\text{ADC}~\mathrm{\Delta }\text{BAC (By AA similarity critarion)}\\ \text{We know that corresponding sides of similar triangles}\\ \text{are proportional.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{CA}}{\text{CD}}=\frac{\text{CB}}{\text{CA}}\\ ⇒{\text{\hspace{0.17em}\hspace{0.17em}CA}}^{2}=\text{CB}\cdot \text{CD}\end{array}$

Q.14

$\begin{array}{l}\text{Sides AB and AC and median AD of a triangle ABC}\\ \text{are respectively proportional to sides PQ and PR and}\\ \text{median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that,}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}=\frac{\text{AD}}{\text{PM}}\\ \text{We extend AD and PM up to point E and L respectively,}\\ \text{such that AD}=\text{DE and PM}=\text{ML. We join B to E, C to E,}\\ \text{Q to L and R to L.}\end{array}$

$\begin{array}{l}\text{It is given that AD and PM are medians}.\text{Therefore,}\\ \frac{1}{2}\text{BC}=\text{BD}=\text{DC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\text{QR}=\text{QM}=\text{MR}\\ \text{In quadrilateral ABEC, diagonals AE and BC bisect each}\\ \text{other at point D. Therefore, quadrilateral ABEC is a}\\ \text{parallelogram.}\\ \therefore \text{AC}=\text{BE and AB}=\text{EC}\\ \text{Similarly, we can prove that quadrilateral PQLR is}\\ \text{a parallelogram and PR}=\text{QL, PQ}=\text{LR.}\\ \text{It is given that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}=\frac{\text{AD}}{\text{PM}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QL}}=\frac{\text{2AD}}{\text{2PM}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QL}}=\frac{\text{AE}}{\text{PL}}\text{}\\ \text{Therefore, by SSS similarity critarion,}\mathrm{\Delta }\text{ABE}~\mathrm{\Delta }\text{PQL and so}\\ \text{in}\mathrm{\Delta }\text{ABE and}\mathrm{\Delta }\text{PQL,}\\ \angle \text{BAE}=\angle \text{QPL}...\text{(1)}\\ \text{Similarly, we can show that}\mathrm{\Delta }\text{AEC}~\mathrm{\Delta }\text{PLR and}\\ \angle \text{CAE}=\angle \text{RPL}...\text{(2)}\\ \text{Adding (1) and (2), we get}\\ \angle \text{BAE}+\angle \text{CAE}=\angle \text{QPL}+\angle \text{RPL}\\ \angle \text{CAB}=\angle \text{RPQ}\\ \text{Also, we have}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}\\ \text{Therefore, by SAS similarity critarion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR.}\end{array}$

Q.15

$\begin{array}{l}\text{A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower}\\ \text{casts a shadow 28 m long. Find the height of the tower.}\end{array}$

Ans.

$\begin{array}{l}\text{Let CD is a pole and DF is its shadow}\text{.}\\ \text{Let AB is a tower and BE is its shadow}\text{.}\\ \text{At the same time in a day, the sun rays will fall on pole}\\ \text{and tower at the same angle}\text{.}\\ \text{Therefore,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{​}\text{​}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle \text{DCF}=\angle \text{BAE}\\ \text{Also,}\text{​}\angle \text{CDF}=\angle \text{ABE}=90°\\ \therefore \Delta \text{ABE}~\Delta \text{CDF [By AA similarity critarion]}\\ \text{Corresponding sides are proportional in similar triangles}\text{.}\\ \text{Therefore,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{AB}}{\text{CD}}=\frac{\text{BE}}{\text{DF}}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{AB}}{\text{6}}=\frac{\text{28}}{\text{4}}=7\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AB}=42\\ \text{Therefore, the height of the tower is 42 m.}\end{array}$

Q.16

$\begin{array}{l}\text{If AD and PM are medians of triangles ABC and PQR, respectively where}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR, prove that}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}.\end{array}$

Ans.

$\begin{array}{l}\text{It is given that AD and PM are medians}.\text{Therefore,}\\ \frac{1}{2}\text{BC}=\text{BD}=\text{DC and}\frac{1}{2}\text{QR}=\text{QM}=\text{MR}\\ \text{Also,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR}\\ \text{Therefore,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AC}}{\text{PR}}\text{and}\angle \text{A}=\angle \text{P,}\angle \text{B}=\angle \text{Q,\hspace{0.17em}\hspace{0.17em}}\angle \text{C}=\angle \text{R}\\ \text{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{PQM,}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\frac{1}{2}\text{BC}}{\frac{1}{2}\text{QR}}=\frac{\text{BD}}{\text{QM}}\text{and}\angle \text{B}=\angle \text{Q}\\ \therefore \text{}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{PQM [By SAS similarity critarion]}\\ ⇒\frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}\text{}\end{array}$

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### 1. Where can I find NCERT solutions for Class 10 Mathematics Exercise 6.3?

Students can find NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 on the Extramarks website or application. Students can download NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 from the website as they are written in an easy-to-understand, detailed format with topic-specific explanations. All NCERT questions are provided by Extramarks experts. They can browse through these solutions to answer any questions or concerns they may have.

### 2. Do students have to practice all the questions from the NCERT Solutions?

Yes, it is important to practice all questions in the Class 10 Mathematics Exercise 6.3 NCERT Answers. Each question provided is an application of a different concept, and practising these questions will help students to fully understand each concept. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  can be downloaded for free from his website or app at Extramarks. Extramarks provide NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7 and NCERT Solutions Class 6

### 3. How many examples are there of Class 10 Mathematics Chapter 6 Exercise 6.3?

Exercise 6.3 in Mathematics Class 10 has a total of 16 sample questions. These examples focus on problems that require the application of various triangle theorems and triangle congruence. The answers to these questions can be found in the NCERT solution for Class 10 Mathematics, available at Extramarks. Extramarks team of experts has structured their answers to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 concisely and in detail. Also, the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are fully based on the CBSE curriculum.

### 4. How long does it take students to complete Class 10 Maths Chapter 6 Exercise 6.3?

The time to complete NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  will vary for each student. It may take several hours or a day or two to complete the entire exercise. Students may be able to complete the exercises in less time if the concepts are very clear to them. Therefore, to assist students, Extramarks provides access to handpicked answers of NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  to these questions. Students can examine them and clear their doubts. All resources are available in the Extramarks app.

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