Question 23 - End-of-Chapter Exercises - Chapter 5 Class 9 - I’m Up and Down, and Round and Round (Ganita Manja

Transcript

Question 23 Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA. Let’s do it step by step Step 1 of 6 The Setup Let's start with a circle with center and a random point located somewhere inside it. First, we draw a line segment connecting the center to the point A. Previous Next StepStep 2 of 6 2. The Perpendicular Chord Now, let's draw a specific chord passing through point : the one that is exactly perpendicular to the line segment OA . Let's label the endpoints of this chord and . This is our candidate for the shortest chord. Previous Next StepStep 3 of 6 3. A Random Chord To prove PQ is the shortest, we must compare it to any other chord passing through A. Let's draw another random chord through A, and label its endpoints R and S . Previous Next StepStep 4 of 6 4. Distance to the New Chord In geometry, the distance from a center to a chord is measured by a perpendicular line. Let's drop a perpendicular from the center to our new chord RS. Let it meet the chord at point . So, OM is the shortest distance to chord RS. Previous Next StepStep 5 of 6 5. The Right Triangle Look closely at the triangle we've formed: . Because OM is perpendicular to RS, the angle at M is . This means is a right-angled triangle! Previous Next StepStep 6 of 6 6. The Proof In the right triangle , the side is the hypotenuse (the longest side). Therefore, the leg OM must be shorter than . Rule: The closer a chord is to the center, the longer it is. Since distance OM < distance OA , chord RS is closer to the center than chord PQ . Thus, PQ . Since RS was ANY other chord, PQ must be the shortest possible chord!