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A, B and C are three points on a straight line, not
necessarily equidistant with B being between A and C. Three
semicircles are drawn on the same side of the line with AB,
BC and AC as the diameters. BD is perpendicular to the line
ABC, and D lies on the semicircle AC.
If the funny shaped diagram between the three semicircles
has an area of 1000 square cms, find the length of BD.
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The length of BD is 35.68 cms
There are 3 right-angled triangles - ABD, CBD and ADC.
From ABD, AB^2 + BD^2 = AD^2 ------ I
From CBD, CB^2 + BD^2 = CD^2 ------ II
From ADC, AD^2 + CD^2 = AC^2 ------ III
Adding I and II,
AB^2 + BC^2 + 2*BD^2 = AD^2 + CD^2 ------ IV
FROM III and IV
AB^2 + BC^2 + 2*BD^2 = AC^2
AB^2 + BC^2 + 2*BD^2 = (AB+CB)^2
2*BD^2 = 2*AB*CB
BD^2 = AB*CB
BD = SQRT(AB*CB)
Given that funny shaped diagram beween three semicircles has
an area of 1000 square cms.
[PI/2 * (AC/2)^2] - [PI/2 * (AB/2)^2] - [PI/2 * (BC/2)^2] =
1000
PI/8 * [AC^2 - AB^2 - BC^2] = 1000
PI * [(AB+BC)^2 - AB^2 - BC^2] = 8000
PI * [2*AB*BC] = 8000
AB * BC = 4000/PI
Hence BD = SQRT(4000/PI) = 35.68 cms
where PI = 3.141592654
Hence, the length of BD is 35.68 cms.
| Is This Answer Correct ? | 0 Yes | 2 No |
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