quadrilateral PQRS......

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the diagonals PR & QS of the quadrilatrl PQRS intersect at M and the areas of the triangles MPQ and MRS are 16 cm² and 25 cm² respectivly. let the min. possible area of the quadrilateral PQRS be A cm² .then the sum of the squares of the digits in A is..........??

ans: 65

#Mathematics #Co-ordinate Geometry

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1 Answers

39

Dr.House ·2010-10-12 13:31:06

well the area of the triangles PMQ and MSR can be written as

1/2PMMQSinÎ¸=16

and

1/2MSMRSinÎ¸=25

now area of quadrilateral PQRS=25+16+ 1/2MQMRSinÎ¸+1/2PMMSSinÎ¸

substituting for MQSinÎ¸ and MSSinÎ¸ from above 2 formulae
we get

area=
16MR/PM + 25PM/MR +16+25

now 16MR/PM+25PM/MR is of the form 16x+25/x which ahs a minimum at 5/4 and the minimum is 40

so minimum area of PQRS is 40+16+25=81

so A=81

so sum of digits of A=8²+1²=65

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