Question 19 - End-of-Chapter Exercises - Chapter 5 Class 9 - I’m Up and Down, and Round and Round (Ganita Manja

Transcript

Question 19 A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle. We have a regular hexagon inscribed in a circle with center O and radius r. Since it's a regular hexagon, all 6 of its sides are equal in length, and all its interior angles are equal. Let's draw lines from the center O to each of the 6 vertices. These lines are all radii of the circle, so their length is r. This splits the hexagon into 6 identical triangles. Since a full circle is 360°, the central angle for one triangle is (𝟑𝟔𝟎° )/𝟔 = 60° In Yellow Triangle Sides on left and right are radius So, the sides are equal And, we know that Angle opposite equal sides are equal Thus, Both base angles would be equal, and it would be ∴ Base angle = ((𝟏𝟖𝟎° − 𝟔𝟎°))/𝟐 = (120° )/2 = 60° Thus, all three angles are 60° ∴ Yellow triangle is an equilateral triangle So, all its sides would be same ∴ Length of the hexagon's side = r Thus, all three angles are 60° ∴ Yellow triangle is an equilateral triangle So, all its sides would be same ∴ Length of the hexagon's side = r Distance to the Center We need to find distance of each side from the centre of the circle. We need to find OM, i.e. d Since Perpendicular to the chord bisects the chord ∴ Base is split into two equal parts with length 𝒓/𝟐 In right angled green triangle By Pythagoras theorem 𝒓^𝟐=𝒅^𝟐+(𝒓/𝟐)^𝟐 𝑟^2=𝑑^2+𝑟^2/4 𝑟^2−𝑟^2/4=𝑑^2 (3𝑟^2)/4=𝑑^2 𝒅^𝟐=(𝟑𝒓^𝟐)/𝟒 𝑑=√((3𝑟^2)/4) 𝑑=√((3𝑟^2)/2^2 ) 𝒅=√𝟑/𝟐 𝒓 Thus, Distance from the center to any side is 𝑑=√3/2 𝑟