NCERT Solutions for Exercise 6.3 Class 10 Maths Chapter 6 - Triangles

# NCERT Solutions for Exercise 6.3 Class 10 Maths Chapter 6 - Triangles

Edited By Ramraj Saini | Updated on Nov 21, 2023 01:31 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.3

NCERT Solutions for Exercise 6.3 Class 10 Maths Chapter 6 Triangles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 6.3 contains critical criteria or theorems for congruence such as Angle-Angle-Angle (AAA), Angle-Side-Angle (ASA), and Side-Side-Side (SSS). It contains 16 questions that cover all parts of the previously given theorems.

NCERT solutions for exercise 6.3 Class 10 Maths chapter 6 Triangles discusses the standards required to demonstrate the resemblance of triangles. 10th class Maths exercise 6.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths chapter 6 exercise 6.3

Triangles Class 10 Chapter 6 Exercise: 6.3

(i) $\angle A=\angle P=60 \degree$

$\angle B=\angle Q=80 \degree$

$\angle C=\angle R=40 \degree$

$\therefore \triangle ABC \sim \triangle PQR$ (By AAA)

So , $\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}$

(ii) As corresponding sides of both triangles are proportional.

$\therefore \triangle ABC \sim \triangle PQR$ (By SSS)

(iii) Given triangles are not similar because corresponding sides are not proportional.

(iv) $\triangle MNL \sim \triangle PQR$ by SAS similarity criteria.

(v) Given triangles are not similar because the corresponding angle is not contained by two corresponding sides

(vi) In $\triangle DEF$ , we know that

$\angle D+\angle E+\angle F=180 \degree$

$\Rightarrow 70 \degree+80 \degree+\angle F=180 \degree$

$\Rightarrow 150 \degree+\angle F=180 \degree$

$\Rightarrow \angle F=180 \degree-150 \degree=30 \degree$

In $\triangle PQR$ , we know that

$\angle P+\angle Q+\angle R=180 \degree$

$\Rightarrow 30 \degree+80 \degree+\angle R=180 \degree$

$\Rightarrow 110 \degree+\angle R=180 \degree$

$\Rightarrow \angle R=180 \degree-110 \degree=70 \degree$

$\angle Q=\angle P=70 \degree$

$\angle E=\angle Q=80 \degree$

$\angle F=\angle R=30 \degree$

$\therefore \triangle DEF\sim \triangle PQR$ ( By AAA)

Given : $\Delta ODC \sim \Delta OBA$ , $\angle BOC = 125 \degree$ and $\angle CDO = 70 \degree$

$\angle DOC+\angle BOC=180 \degree$ (DOB is a straight line)

$\Rightarrow \angle DOC+125 \degree=180 \degree$

$\Rightarrow \angle DOC=180 \degree-125 \degree$

$\Rightarrow \angle DOC=55 \degree$

In $\Delta ODC ,$

$\angle DOC+\angle ODC+\angle DCO=180 \degree$

$\Rightarrow 55 \degree+ 70 \degree+\angle DCO=180 \degree$

$\Rightarrow \angle DCO+125 \degree=180 \degree$

$\Rightarrow \angle DCO=180 \degree-125 \degree$

$\Rightarrow \angle DCO=55 \degree$

Since , $\Delta ODC \sim \Delta OBA$ , so

$\Rightarrow\angle OAB= \angle DCO=55 \degree$ ( Corresponding angles are equal in similar triangles).

In $\triangle DOC\, and\, \triangle BOA$ , we have

$\angle CDO=\angle ABO$ ( Alternate interior angles as $AB||CD$ )

$\angle DCO=\angle BAO$ ( Alternate interior angles as $AB||CD$ )

$\angle DOC=\angle BOA$ ( Vertically opposite angles are equal)

$\therefore \triangle DOC\, \sim \, \triangle BOA$ ( By AAA)

$\therefore \frac{DO}{BO}=\frac{OC}{OA}$ ( corresponding sides are equal)

$\Rightarrow \frac{OA }{OC} = \frac{OB }{OD }$

Hence proved.

Given : $\frac{QR }{QS } = \frac{QT}{PR}$ and $\angle 1 = \angle 2$

To prove : $\Delta PQS \sim \Delta TQR$

In $\triangle PQR$ , $\angle PQR=\angle PRQ$

$\therefore PQ=PR$

$\frac{QR }{QS } = \frac{QT}{PR}$ (Given)

$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$

In $\Delta PQS\, and\, \Delta TQR$ ,

$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$

$\angle Q=\angle Q$ (Common)

$\Delta PQS \sim \Delta TQR$ ( By SAS)

Given : $\angle$ P = $\angle$ RTS

To prove RPQ ~ $\Delta$ RTS.

In $\Delta$ RPQ and $\Delta$ RTS,

$\angle$ P = $\angle$ RTS (Given )

$\angle$ R = $\angle$ R (common)

$\Delta$ RPQ ~ $\Delta$ RTS. ( By AA)

Given : $\triangle ABE \cong \triangle ACD$

To prove ADE ~ $\Delta$ ABC.

Since $\triangle ABE \cong \triangle ACD$

$AB=AC$ ( By CPCT)

$AD=AE$ (By CPCT)

In $\Delta$ ADE and $\Delta$ ABC,

$\angle A=\angle A$ ( Common)

and

$\frac{AD}{AB}=\frac{AE}{AC}$ ( $AB=AC$ and $AD=AE$ )

Therefore, $\Delta$ ADE ~ $\Delta$ ABC. ( By SAS criteria)

To prove : $\Delta AEP \sim \Delta CDP$

In $\Delta AEP \, \, and\, \, \Delta CDP$ ,

$\angle AEP=\angle CDP$ ( Both angles are right angle)

$\angle APE=\angle CPD$ (Vertically opposite angles )

$\Delta AEP \sim \Delta CDP$ ( By AA criterion)

To prove : $\Delta ABD \sim \Delta CBE$

In $\Delta ABD \, \, and\, \, \Delta CBE$ ,

$\angle ADB=\angle CEB$ ( Both angles are right angle)

$\angle ABD=\angle CBE$ (Common )

$\Delta ABD \sim \Delta CBE$ ( By AA criterion)

To prove : $\Delta AEP \sim \Delta ADB$

In $\Delta AEP \, \, \, and\, \, \Delta ADB$ ,

$\angle AEP=\angle ADB$ ( Both angles are right angle)

$\angle A=\angle A$ (Common )

$\Delta AEP \sim \Delta ADB$ ( By AA criterion)

To prove : $\Delta PDC \sim \Delta BEC$

In $\Delta PDC \, \, and\, \, \, \Delta BEC$ ,

$\angle CDP=\angle CEB$ ( Both angles are right angle)

$\angle C=\angle C$ (Common )

$\Delta PDC \sim \Delta BEC$ ( By AA criterion)

To prove : $\Delta ABE \sim \Delta CFB$

In $\Delta ABE \, \, \, and\, \, \Delta CFB$ ,

$\angle A=\angle C$ ( Opposite angles of a parallelogram are equal)

$\angle AEB=\angle CBF$ ( Alternate angles of AE||BC)

$\Delta ABE \sim \Delta CFB$ ( By AA criterion )

To prove : $\Delta ABC \sim \Delta AMP$

In $\Delta ABC \, \, and\, \, \Delta AMP$ ,

$\angle ABC=\angle AMP$ ( Each $90 \degree$ )

$\angle A=\angle A$ ( common)

$\Delta ABC \sim \Delta AMP$ ( By AA criterion )

To prove :

$\frac{CA }{PA } = \frac{BC }{MP}$

In $\Delta ABC \, \, and\, \, \Delta AMP$ ,

$\angle ABC=\angle AMP$ ( Each $90 \degree$ )

$\angle A=\angle A$ ( common)

$\Delta ABC \sim \Delta AMP$ ( By AA criterion )

$\frac{CA }{PA } = \frac{BC }{MP}$ ( corresponding parts of similar triangles )

Hence proved.

To prove :

$\frac{CD}{GH} = \frac{AC}{FG}$

Given : $\Delta ABC \sim \Delta EGF$

$\angle A=\angle F,\angle B=\angle E\, \, and \, \, \angle ACB=\angle FGE,\angle ACB=\angle FGE$

$\therefore \angle ACD=\angle FGH$ ( CD and GH are bisectors of equal angles)

$\therefore \angle DCB=\angle HGE$ ( CD and GH are bisectors of equal angles)

In $\Delta ACD \, \, and\, \, \Delta FGH$

$\therefore \angle ACD=\angle FGH$ ( proved above)

$\angle A=\angle F$ ( proved above)

$\Delta ACD \sim \Delta FGH$ ( By AA criterion)

$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$

Hence proved.

To prove : $\Delta DCB \sim \Delta HGE$

Given : $\Delta ABC \sim \Delta EGF$

In $\Delta DCB \,\, \, and\, \, \Delta HGE$ ,

$\therefore \angle DCB=\angle HGE$ ( CD and GH are bisectors of equal angles)

$\angle B=\angle E$ ( $\Delta ABC \sim \Delta EGF$ )

$\Delta DCB \sim \Delta HGE$ ( By AA criterion )

To prove : $\Delta DCA \sim \Delta HGF$

Given : $\Delta ABC \sim \Delta EGF$

In $\Delta DCA \, \, \, and\, \, \Delta HGF$ ,

$\therefore \angle ACD=\angle FGH$ ( CD and GH are bisectors of equal angles)

$\angle A=\angle F$ ( $\Delta ABC \sim \Delta EGF$ )

$\Delta DCA \sim \Delta HGF$ ( By AA criterion )

To prove : $\Delta ABD \sim \Delta ECF$

Given: ABC is an isosceles triangle.

$AB=AC \, \, and\, \, \angle B=\angle C$

In $\Delta ABD \, \, and\, \, \Delta ECF$ ,

$\angle ABD=\angle ECF$ ( $\angle ABD=\angle B=\angle C=\angle ECF$ )

$\angle ADB=\angle EFC$ ( Each $90 \degree$ )

$\Delta ABD \sim \Delta ECF$ ( By AA criterion)

AD and PM are medians of triangles. So,

$BD=\frac{BC}{2}\, and\, QM=\frac{QR}{2}$

Given :

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{\frac{1}{2}BC}{\frac{1}{2}QR}=\frac{AD}{PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

In $\triangle ABD\, and\, \triangle PQM,$

$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

$\therefore \triangle ABD\sim \triangle PQM,$ (SSS similarity)

$\Rightarrow \angle ABD=\angle PQM$ ( Corresponding angles of similar triangles )

In $\triangle ABC\, and\, \triangle PQR,$

$\Rightarrow \angle ABD=\angle PQM$ (proved above)

$\frac{AB}{PQ}=\frac{BC}{QR}$

Therefore, $\Delta ABC \sim \Delta PQR$ . ( SAS similarity)

In, $\triangle ADC \, \, and\, \, \triangle BAC,$

$\angle ADC = \angle BAC$ ( given )

$\angle ACD = \angle BCA$ (common )

$\triangle ADC \, \, \sim \, \, \triangle BAC,$ ( By AA rule)

$\frac{CA}{CB}=\frac{CD}{CA}$ ( corresponding sides of similar triangles )

$\Rightarrow CA^2=CB\times CD$

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$ (given)

Produce AD and PM to E and L such that AD=DE and PM=DE. Now,

join B to E,C to E,Q to L and R to L.

AD and PM are medians of a triangle, therefore

QM=MR and BD=DC

PM=ML (By construction)

So, diagonals of ABEC bisecting each other at D,so ABEC is a parallelogram.

Similarly, PQLR is also a parallelogram.

Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$ (Given )

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{2.AD}{2.PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}$

$\Delta ABE \sim \Delta PQL$ (SSS similarity)

$\angle BAE=\angle QPL$ ...................1 (Corresponding angles of similar triangles)

Similarity, $\triangle AEC=\triangle PLR$

$\angle CAE=\angle RPL$ ........................2

$\angle BAE+\angle CAE=\angle QPL+\angle RPL$

$\angle CAB=\angle RPQ$ ............................3

In $\triangle ABC\, and\, \, \triangle PQR,$

$\frac{AB}{PQ}=\frac{AC}{PR}$ ( Given )

$\angle CAB=\angle RPQ$ ( From above equation 3)

$\triangle ABC\sim \triangle PQR$ ( SAS similarity)

CD = pole

AB = tower

In $\triangle ABE\, \, and\, \triangle CDF,$

$\angle CDF=\angle ABE$ ( Each $90 \degree$ )

$\angle DCF=\angle BAE$ (Angle of sun at same place )

$\triangle ABE\, \, \sim \, \triangle CDF,$ (AA similarity)

$\frac{AB}{CD}=\frac{BE}{QL}$

$\Rightarrow \frac{AB}{6}=\frac{28}{4}$

$\Rightarrow AB=42$ cm

Hence, the height of the tower is 42 cm.

$\Delta AB C \sim \Delta PQR$ ( Given )

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$ ............... ....1( corresponding sides of similar triangles )

$\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R$ ....................................2

AD and PM are medians of triangle.So,

$BD=\frac{BC}{2}\, and\, QM=\frac{QR}{2}$ ..........................................3

From equation 1 and 3, we have

$\frac{AB}{PQ}=\frac{BD}{QM}$ ...................................................................4

In $\triangle ABD\, and\, \triangle PQM,$

$\angle B=\angle Q$ (From equation 2)

$\frac{AB}{PQ}=\frac{BD}{QM}$ (From equation 4)

$\triangle ABD\, \sim \, \triangle PQM,$ (SAS similarity)

$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

## More About NCERT Solutions for Class 10 Maths Exercise 6.3

NCERT solutions Class 10 Maths chapter 6 exercise 6.3 Triangles contains well-written examples that demonstrate the key similarity theorems. Certain implications can be taken from the presented theories, such as the AA property, which states that if two angles of one triangle are equal to two angles of another triangle, then the two triangles are comparable (similar to each other) as the corresponding third angle of the triangle also becomes equal as the sum of the total angles of a triangle remains fixed to 180 degrees. Before moving on to the exercise questions, students must ensure that they have thoroughly practiced these examples. Furthermore, the chapter's highlights can be refreshed on a regular basis to keep the concepts fresh. Students can also access Triangles Class 10 Notes here and use them for quickly revision of the concepts related to Triangles.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 6.3

• If you go through the NCERT solution for Class 10 Maths chapter 6 exercise 6.3, we can easily revise the concepts of exercise 6.1 and 6.2 which deals with the measurement aspects of the sides of the triangle.
• Exercise 6.3 Class 10 Maths, is based on conditions for the concurrency of triangles.
• NCERT solutions for Class 10 Maths chapter 6 exercise 6.3 revolves around the different forms of similarity aspect of triangles.

Also, See:

## Subject Wise NCERT Exemplar Solutions

1. What is the core concept of Class 10 Maths chapter 6 exercise 6.3 ?

This exercise is about the criteria or theorems for congruence of triangles. Practice this class 10 ex 6.3 to command the concepts.

2. Name all the criteria for similarity of triangles according to NCERT solutions Class 10 Maths chapter 6 exercise 6.?

In this class 10 maths ex 6.3 following theorems are discussed in detail.

1. Angle-angle-angle

2. Angle-side-angle

3. Side-side-side

4. Angle-angle

5. Angle-angle-side

6. Side-angle-side

Go through this ex 6.3 class 10 to get deeper understanding of related concepts.

3. What is the total of a triangle's angels?

The sum of a triangle's angels is 180 degrees.

4. Is similarity a sign of congruence between the figures?

No, similarity simply refers to the shape of the figures; it does not imply equality.

5. If two corresponding angels of two triangles are same does this mean the triangles are similar?

Yes, because a triangle has three angles and their sum remains 180 degrees. So, if two angels are equal then the third also becomes equal.

6. If all the three sides of two triangles are in proportion, then does this means the triangles are similar?

Yes, if all the three sides of two triangles are in proportion, then does this means the triangles are similar. By side-side-side criteria.

7. What will happen if the ratio of the sides of two similar triangles becomes equal to 1?

If the ratio becomes equal to 1 then the triangles are congruent to each other.

8. For similar triangles, what is the purpose of side ratios?

We can find the length of the corresponding side of the other triangle using the side ratio and the length of any side. Practice 10th class maths exercise 6.3 answers to command the concepts.

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According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

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