Formulae [2]
Perimeter = Sum of all side lengths
Area = Amount of space inside a flat shape
Theorems and Laws [2]
Figure, shows a sector of a circle, centre O, containing an angle θ°. Prove that perimeter of the shaded region is `r(tan θ + sec θ + (πtheta)/180^circ - 1)`.
Given angle subtended at centre of circle = ๐
∠OAB = 90° [At joint of contact, tangent is perpendicular to radius]
OAB is right angle triangle
Cos ๐ =`(adj.side)/(hypotenuse) =r/OB`⇒ ๐๐ต = ๐ sec ๐ … … (๐)
tan ๐ =`(opp.side)/(adju.side)=AB/r`⇒ ๐ด๐ต = ๐ tan ๐ … … . (๐๐)
Perimeter of shaded region = AB + BC + (CA arc)
= ๐ tan ๐ + (๐๐ต − ๐๐ถ) +`theta/360^@`× 2๐๐
= ๐ tan ๐ + ๐ sec ๐ − ๐ +`(pithetar)/180^@`
= ๐ (tan ๐ + sec ๐ +`(pitheta)/180^@`− 1)
Figure, shows a sector of a circle, centre O, containing an angle ๐°. Prove that area of the shaded region is `r^2/2(tanθ - (πθ)/180^circ)`.
Given angle subtended at centre of circle = ๐
∠OAB = 90° [At joint of contact, tangent is perpendicular to radius]
OAB is right angle triangle
Cos ๐ =`(adj.side)/(hypotenuse) =r/OB`⇒ ๐๐ต = ๐ sec ๐ … … (๐)
tan ๐ =`(opp.side)/(adju.side)=AB/r`⇒ ๐ด๐ต = ๐ tan ๐ … … . (๐๐)
Area of shaded region = (area of triangle) – (area of sector)
`= (1/2× OA × AB) −theta/360^@× pir^2`
`=1/2× r × r tan theta −r^2/2[theta/180^@× pi]`
=`r^2/2[tantheta −(pitheta)/180^@]`
Concepts [5]
- Basic Concepts in Mensuration
- Concept of Perimeter
- Circumference of a Circle
- Concept of Area
- Concept of Measurement Using a Basic Unit Area of a Square, Rectangle, Triangle, Parallelogram and Circle, Rings and Combined Figures.
