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
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1 Answers
39well the area of the triangles PMQ and MSR can be written as
1/2PMMQSinθ=16
and
1/2MSMRSinθ=25
now area of quadrilateral PQRS=25+16+ 1/2MQMRSinθ+1/2PMMSSinθ
substituting for MQSinθ and MSSinθ 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
