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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.



A, B and C are three points on a straight line, not necessarily equidistant with B being between A ..

Answer / guest

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.

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