Math 9th Exercise6.1 NBF
Math 9th Exercise6.1 NBF will be solved. Understanding the fundamentals of geometry is essential for mastering mathematics. One of the foundational topics in Class 9 Mathematics is Exercise 6.1, which focuses on lines, angles, and their properties. This exercise provides the basis for solving more advanced problems in geometry. In this blog post, we will explore the key concepts, definitions, and problem-solving techniques necessary for mastering Exercise 6.1.
Understanding Lines and Angles Exercise 6.1 primarily deals with different types of lines and angles, their relationships, and their properties. Before solving problems, let’s go over some essential definitions:
Key Concepts Covered:
- Types of Lines:
- Line Segment: A part of a line with two endpoints.
- Ray: A line that starts from a point and extends infinitely in one direction.
- Intersecting Lines: Lines that cross each other at a single point.
- Parallel Lines: Lines that never intersect and are always equidistant.
- Types of Angles:
- Acute Angle: Measures less than 90°.
- Right Angle: Measures exactly 90°.
- Obtuse Angle: Measures between 90° and 180°.
- Straight Angle: Measures exactly 180°.
- Reflex Angle: Measures between 180° and 360°.
- Angle Pairs and Their Properties:
- Complementary Angles: Two angles whose sum is 90°.
- Supplementary Angles: Two angles whose sum is 180°.
- Adjacent Angles: Two angles that share a common side and vertex but do not overlap.
- Linear Pair: Two adjacent angles that form a straight line, summing up to 180°.
- Vertically Opposite Angles: Angles formed by two intersecting lines, which are always equal.
Step-by-Step Problem Solving To help you understand how to solve problems in Exercise 6.1, let’s go through a sample question and its step-by-step solution.
Example Problem:
If two intersecting lines form an angle of 70°, find the measure of the vertically opposite angle.
Solution:
- Identify the given angle: 70°.
- Apply the property of vertically opposite angles:
- Vertically opposite angles are always equal.
- Conclusion:
- The required angle is 70°.
Another important type of problem involves complementary and supplementary angles.
Example Problem:
If two angles are supplementary and one angle is 110°, find the other angle.
Solution:
- Use the supplementary angle property:
- Sum of two supplementary angles = 180°.
- Subtract the given angle from 180°:
- 180° – 110° = 70°.
- Final Answer:
- The other angle is 70°.
Why is Exercise 6.1 Important? ✅ Builds a strong foundation in geometry, which is essential for advanced math topics.
✅ Enhances problem-solving skills by applying properties of angles and lines.
✅ Prepares students for exams by reinforcing important geometric concepts.
✅ Helps in real-life applications like architecture, engineering, and design.
Tips for Solving Exercise 6.1 Effectively 📌 Memorize the angle properties to quickly identify relationships.
📌 Use diagrams to visualize problems and make solving easier.
📌 Practice a variety of problems to strengthen your understanding.
📌 Review NCERT examples and past exam questions to get a clear grasp of the topic.
Conclusion Mastering Exercise 6.1 is crucial for understanding lines, angles, and their relationships. By learning key concepts, practicing problems, and applying the right techniques, students can gain confidence in geometry. Keep practicing, stay curious, and build a strong mathematical foundation for future success.
Q1 Convert the following measure of angles in seconds.
(i) :- $$5^\circ=\;{(\;5\times60)}’=\;300’\;\\={(\;300\times\;60)}”=18000”$$
(ii) :- $$30’\;=\;(\;30\times\;60)”\;=\;1800”$$
(iii) :- $$10^\circ30’=(10\times60+30)’=630’=(630\times60)”=37800”$$
(iv) :- $$20^\circ\;20’\;20″=(\;20\times3600\;)”+\;(\;20\times60\;)”+20”$$
= 72000″+ 1200″ + 20″
= 73220″
Q2 Convert the following measure of angle in minutes.
(i) :- $$70^\circ\;=\;(75\times60)’=4500’\;$$
(ii) :- $$120^\circ=(\frac{120}{60})’=\;2’\;$$
(iii) :- $$50^\circ40’=\;(\;50\times60\;)’+40’=\;3000+40’=\;3040’\;$$
(iv) :- $$30^\circ\;30’\;30″\\(30\times60)’+30′(\frac{30}{60})’$$
= 1800′ + 30′ + 0.5
= 1830 . 5
Q3 Convert the following measure of angle in degree and write the answer correct to 4 decimal places.
(i) :- $$135’=(\frac{135}{60})’=2.25^\circ$$
(ii) :- $$150’=(\frac{150}{3600})^\circ=0.0417^\circ$$
(iii) :- $$60^\circ\;60’=\;60^\circ+\;(\frac{60}{60})^\circ\\=\;60^\circ+1^\circ\;=\;61^\circ\\$$
(iv) :- $$45^\circ45’45”=(45+\frac{45}{60}+\frac{45}{3600})^\circ\\=\;(45+0.75+0.0125)^\circ\\=\;45.7625^\circ$$
