Math, asked by answerqueen, 1 year ago

HEY ! ❤........................ quadrilateral ABCD IS A RECTANGLE . TAKING AD AS A DIAMETER,SEMICIRCLE AXD IS DRAWN WHICH INTERSECTS THE DIAGONAL BD AT X . IF AB = 12 CM AD = 9 CM FIND BD AND BX......THANKS ❤

Answers

Answered by amitnrw

AD = BC = 9 cm

AB = CD = 12 cm

${bd}^{2} = {ab}^{2} + {ad}^{2}$

${bd}^{2} = {12}^{2} + {9}^{2}$

${bd}^{2} = 144 + 81$

${bd}^{2} = 225$

bd = 15 cm

amitnrw: to find bx - Draw a parallel line to AB from point O (center of AD or semicricle) . This will bisect BD at K as O bisect AD . So BK=DK=15/2 . to find BX = BK + KX. now see triangle OXK where OX = 9/2cm (radius) and OK = 12/2 = 6 cm (Diagonal bisected so point in line to mid point of AB) Now Angle OBX = Angle ABD . now in Trinagle OBX we know two sides and one angle . using this you can calculate KX then BX can be find out.

amitnrw: @Answerqueen - if you have any doubt please let me know

Answered by sadik0406

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