Asked by Justice
A water through is to be constructed so that its cross-section is trapezium PQRS in which PQ=RS=6cm, QR=14cm and <SPQ=<PSR=x.
(a).Show that the area of trapezium is given as 84sinx+18sin2x
(b).Determine its perimeter in terms of cosx
(a).Show that the area of trapezium is given as 84sinx+18sin2x
(b).Determine its perimeter in terms of cosx
Answers
Answered by
Reiny
In my diagram, I have PS as the longer of the parallel sides.
From Q and R, draw perpendiculars to meet PS at M and N respectively. So I have two equal triangles and a rectangle.
In triangle PQM,
PM/6 = cosx ----> PM = 6cosx
QM/6 = sinx -----> QM = 6sinx
area = 2 triangles + rectangle
= 2(1/2)(PM) (QM) + 14QM
= 6cosx(6sinx) + 14(6sinx)
= 36 sinxcosx + 84 sinx
= 18sin (2x) + 84sinx , (recall sin (2x) = 2sinxcosx )
the perimeter is very easy, just add up the sides
You should see that PS = 2PM + 14
From Q and R, draw perpendiculars to meet PS at M and N respectively. So I have two equal triangles and a rectangle.
In triangle PQM,
PM/6 = cosx ----> PM = 6cosx
QM/6 = sinx -----> QM = 6sinx
area = 2 triangles + rectangle
= 2(1/2)(PM) (QM) + 14QM
= 6cosx(6sinx) + 14(6sinx)
= 36 sinxcosx + 84 sinx
= 18sin (2x) + 84sinx , (recall sin (2x) = 2sinxcosx )
the perimeter is very easy, just add up the sides
You should see that PS = 2PM + 14
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