**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.

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### Question & Answer

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.4: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) Let AD = x cm It is given that DE || BC By using basic proportionality theorem, we obtain \( \begin{array}{l}{\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}} \\ {\frac{x}{7.2}=\frac{1.8}{5.4}} \\ {x=\frac{1.8 \times 7.2}{5.4}} \\ {x=2.4} \\ {\therefore \mathrm{AD}=2.4 \mathrm{cm}}\end{array} \)

Q.5: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 \( \frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64} \) \( \frac{\mathrm{PF}}{\mathrm{PR}}=\frac{0.36}{2.56}=\frac{9}{64} \) \( \begin{array}{l}{\text { Hence, } \frac{\mathrm{PE}}{\mathrm{PQ}}=\frac{\mathrm{PF}}{\mathrm{PR}}} \\ {\text { Therefore, EF is parallel to QR. }}\end{array} \)

## NCERT / CBSE Book for Class 10 Maths

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### All NCERT Solutions Class 10

- NCERT Solutions for Class 10 English
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- NCERT Solutions for Class 10 Sanskrit

### All NCERT Solutions

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