Class 10 Math Chapter 6 Triangles Solutions

Here, Class 10 Math Chapter 6 Triangles Solutions are provided. Subject matter experts have created these solutions to help students in their board exam. They have created solutions in such a way that students can revise the entire solutions in very little time and can score good marks in the board exam. Every solution is provided in detail and in very simple language.

Class 10 Math Chapter 6 Triangles Solutions are very important for board exam. Solving these Triangles solutions will boost the confidence and students can score better in exam. Solutions are provided according to latest NCERT syllabus. Experts created these step-by-step solutions, which will definitely be helpful for students.

Triangles Solutions

Exercise 6.1

Exercise 6.2

Exercise 6.3

Chapter 7: Coordinate Geometry Solutions

We have given solutions of all exercise of class 10 math chapter 6. These solutions will be helpful for students understanding core concept of Triangles.

To understand the chapter thoroughly, it is essential to practice all the questions provided within it, as well as any additional questions. Solving previous years’ question papers also proves to be very helpful in exam preparation. You can also find other study material resources on our website.

Important Questions for NCERT Chapter 6: Triangles

Question:

EandFarepointsonthesidesPQandPRrespectivelyofaPQR.ForeachoftheE and F are points on the sides PQ and PR respectively of a \triangle PQR. For each of the
followingcases,statewhetherEFQR.following cases, state whether EF \parallel QR.
PE=3.9 cm,EQ=3 cm,PF=3.6 cm and FR=2.4 cmPE = 3.9 \text{ cm}, EQ = 3 \text{ cm}, PF = 3.6 \text{ cm} \text{ and } FR = 2.4 \text{ cm}

Answer

Given that, PE=3.9 cm,EQ=3 cm,PF=3.6 cm,FR=2.4 cm\text{Given that, } PE = 3.9 \text{ cm}, EQ = 3 \text{ cm}, PF = 3.6 \text{ cm}, FR = 2.4 \text{ cm}
PEEQ=3.93=1.3\frac{PE}{EQ} = \frac{3.9}{3} = 1.3
PFFR=3.62.4=1.5\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5
Hence, PEEQPFFR\text{Hence, } \frac{PE}{EQ} \neq \frac{PF}{FR}
Therefore, EF is not parallel to QR.\text{Therefore, EF is not parallel to QR.}

Question:

EandFarepointsonthesidesPQandPRrespectivelyofaPQR.ForeachoftheE and F are points on the sides PQ and PR respectively of a \triangle PQR. For each of the
followingcases,statewhetherEFQR.following cases, state whether EF \parallel QR.
𝐏𝐄=𝟒 cm,𝐐𝐄=𝟒.𝟓 cm,𝐏𝐅=𝟖 cm and 𝐑𝐅=𝟗 cm\mathbf{PE = 4 \text{ cm}, QE = 4.5 \text{ cm}, PF = 8 \text{ cm} \text{ and } RF = 9 \text{ cm}}

Answer

PE=4 cm,QE=4.5 cm,PF=8 cm,RF=9 cmPE = 4 \text{ cm}, QE = 4.5 \text{ cm}, PF = 8 \text{ cm}, RF = 9 \text{ cm}
PEEQ=44.5=89\frac{PE}{EQ} = \frac{4}{4.5} = \frac{8}{9}
PFFR=89\frac{PF}{FR} = \frac{8}{9}
Hence, PEEQ=PFFR\text{Hence, } \frac{PE}{EQ} = \frac{PF}{FR}
Therefore, EF is parallel to QR.\text{Therefore, EF is parallel to QR.}

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