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Please give me the answer for the 30 th one |
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Answer» er of shaded region = AB + BC + arc AC Note : @ = angle theta AB is the tangent to the circle and OA is radius...Here OA is perpendicular to AB=> AOB is a right angle TRIANGLE=> cos @ = OA / OB = r / OB=> cos @ = r / OB=> OB = r / cos @ = r × sec @ ( 1 / cos @ = sec @ )=> OB = r × sec @Now OB = OC + BC = r + BC ( OC = radius = r )=> OB = r + BC=> r × sec @ = r + BC => BC = r × sec @ - r Now in triangle AOB , we havetan @ = AB / OA = AB / r => AB = r × [email protected]Arc length AC =( @ / 360 ) × 2 × Pie × r Thus , perimeter of shaded region= AB + BC + arc AC= r tan @ + r sec @ - r + ( @ / 360 ) × 2 × Pie × r |
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