Saved Bookmarks
| 1. |
A chord of circle of radius 10cm subtent angle 60 find the area of major and minor segment |
| Answer» \tWe know that area of minor segment\t= Area of minor sector OAB - Area of\xa0ΔOAB\t{tex}\\because \\text { area of } \\triangle \\mathrm{OAB}=\\frac{1}{2}(O A)(O B) \\sin \\angle A O B{/tex}\t{tex}=\\frac{1}{2}(O A)(O B)\\left(\\because \\angle A O B=90^{\\circ}\\right){/tex}\tArea of sector = {tex}\\frac{\\theta}{360} \\pi r^{2}{/tex}\t= {tex}\\frac 14{/tex}(3.14) (100) - 50 = 25(3.14) - 50 = 78.50 - 50 = 28.5 cm2\tArea of major segment = Area of the circle - Area of minor segment\t= {tex}\\pi{/tex}(10)2 - 28.5\t= 100(3.14) - 28.5\t= 314 - 28. 5 = 285.5 cm2 | |