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Question Number 112060 by Aina Samuel Temidayo last updated on 06/Sep/20
In a trapezium, ABCD, with AB parallel to CD. If M is the midpoint of line segment AD and P is a point on line BC such that MP is perpendicular to BC. Show that, we need only the lengths of line segments MP and BC to calculate the area ABCD.
$$\mathrm{In}\:\mathrm{a}\:\mathrm{trapezium},\:\mathrm{ABCD},\:\mathrm{with}\:\mathrm{AB} \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{CD}.\:\mathrm{If}\:\mathrm{M}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of} \\ $$$$\mathrm{line}\:\mathrm{segment}\:\mathrm{AD}\:\mathrm{and}\:\mathrm{P}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on} \\ $$$$\mathrm{line}\:\mathrm{BC}\:\mathrm{such}\:\mathrm{that}\:\mathrm{MP}\:\mathrm{is}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{BC}.\:\mathrm{Show}\:\mathrm{that},\:\mathrm{we}\:\mathrm{need}\:\mathrm{only}\:\mathrm{the} \\ $$$$\mathrm{lengths}\:\mathrm{of}\:\mathrm{line}\:\mathrm{segments}\:\mathrm{MP}\:\mathrm{and}\:\mathrm{BC} \\ $$$$\mathrm{to}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{ABCD}. \\ $$
Answered by 1549442205PVT last updated on 06/Sep/20
Commented by Aina Samuel Temidayo last updated on 06/Sep/20
Thanks. Please what about the solution?
$$\mathrm{Thanks}.\:\mathrm{Please}\:\mathrm{what}\:\mathrm{about}\:\mathrm{the} \\ $$$$\mathrm{solution}? \\ $$
Commented by 1549442205PVT last updated on 06/Sep/20
Denote by G midpoint of BC Suppose known MP=m,BC=a,then S_(BMC) =((BC.MP)/2)=((am)/2).We prove that S_(ABCD) =2S_(BMC) .Indeed,Denote byI,H orthogonal projections of points B and C on MG respectively.Then S_(CMG) = ((MG.CH)/2),S_(BMG) =((MG.BI)/2) S_(BMC) =S_(CMG) +S_(BMG) =((MG(BI+CH))/2)(1) Since M,G are midpoints of BC and AD ,MG is midline of the trapezium ABCD.Since AB//CD,BI+CH equal to altitude h of the trapezium . Therefore,MG=((AB+CD)/2),BI+CH=h and S_(ABCD) =(((AB+CD).h)/2)=MG.h(2) From (1) and (2) we infer S_(ABCD) =2S_(BMC) =a.m .That thing show that only need know the length of the line segments BC and MP we can calculate the area of the trapezium
$$\mathrm{Denote}\:\mathrm{by}\:\mathrm{G}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{BC}\: \\ $$$$\mathrm{Suppose}\:\mathrm{known}\:\mathrm{MP}=\mathrm{m},\mathrm{BC}=\mathrm{a},\mathrm{then} \\ $$$$\mathrm{S}_{\mathrm{BMC}} =\frac{\mathrm{BC}.\mathrm{MP}}{\mathrm{2}}=\frac{\mathrm{am}}{\mathrm{2}}.\mathrm{We}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{S}_{\mathrm{ABCD}} =\mathrm{2S}_{\mathrm{BMC}} .\mathrm{Indeed},\mathrm{Denote}\:\mathrm{byI},\mathrm{H} \\ $$$$\mathrm{orthogonal}\:\mathrm{projections}\:\mathrm{of}\:\mathrm{points}\:\mathrm{B}\:\mathrm{and} \\ $$$$\mathrm{C}\:\mathrm{on}\:\mathrm{MG}\:\mathrm{respectively}.\mathrm{Then}\:\mathrm{S}_{\mathrm{CMG}} = \\ $$$$\frac{\mathrm{MG}.\mathrm{CH}}{\mathrm{2}},\mathrm{S}_{\mathrm{BMG}} =\frac{\mathrm{MG}.\mathrm{BI}}{\mathrm{2}} \\ $$$$\mathrm{S}_{\mathrm{BMC}} =\mathrm{S}_{\mathrm{CMG}} +\mathrm{S}_{\mathrm{BMG}} =\frac{\mathrm{MG}\left(\mathrm{BI}+\mathrm{CH}\right)}{\mathrm{2}}\left(\mathrm{1}\right) \\ $$$$\mathrm{Since}\:\mathrm{M},\mathrm{G}\:\mathrm{are}\:\mathrm{midpoints}\:\mathrm{of}\:\mathrm{BC}\:\mathrm{and} \\ $$$$\mathrm{AD}\:,\mathrm{MG}\:\mathrm{is}\:\mathrm{midline}\:\mathrm{of}\:\mathrm{the}\:\mathrm{trapezium} \\ $$$$\mathrm{ABCD}.\mathrm{Since}\:\mathrm{AB}//\mathrm{CD},\mathrm{BI}+\mathrm{CH}\:\mathrm{equal} \\ $$$$\mathrm{to}\:\mathrm{altitude}\:\mathrm{h}\:\mathrm{of}\:\mathrm{the}\:\mathrm{trapezium}\:. \\ $$$$\mathrm{Therefore},\mathrm{MG}=\frac{\mathrm{AB}+\mathrm{CD}}{\mathrm{2}},\mathrm{BI}+\mathrm{CH}=\mathrm{h} \\ $$$$\mathrm{and}\:\mathrm{S}_{\mathrm{ABCD}} =\frac{\left(\mathrm{AB}+\mathrm{CD}\right).\mathrm{h}}{\mathrm{2}}=\mathrm{MG}.\mathrm{h}\left(\mathrm{2}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{infer} \\ $$$$\mathrm{S}_{\mathrm{ABCD}} =\mathrm{2S}_{\mathrm{BMC}} =\mathrm{a}.\mathrm{m}\:.\mathrm{That}\:\mathrm{thing}\:\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{only}\:\mathrm{need}\:\mathrm{know}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{line}\:\mathrm{segments}\:\mathrm{BC}\:\mathrm{and}\:\mathrm{MP}\:\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{trapezium} \\ $$
Commented by Aina Samuel Temidayo last updated on 06/Sep/20
Please check my other questions out. I need help please.
$$\mathrm{Please}\:\mathrm{check}\:\mathrm{my}\:\mathrm{other}\:\mathrm{questions}\:\mathrm{out}.\:\mathrm{I} \\ $$$$\mathrm{need}\:\mathrm{help}\:\mathrm{please}. \\ $$
Commented by Aina Samuel Temidayo last updated on 06/Sep/20
Please how did you know S_(ABCD ) =2S_(BMC) ?
$$\mathrm{Please}\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{know} \\ $$$$\mathrm{S}_{\mathrm{ABCD}\:} =\mathrm{2S}_{\mathrm{BMC}} ? \\ $$
Commented by 1549442205PVT last updated on 06/Sep/20
I proved above!It follows from(1)&(2)
$$\mathrm{I}\:\mathrm{proved}\:\mathrm{above}!\mathrm{It}\:\mathrm{follows}\:\mathrm{from}\left(\mathrm{1}\right)\&\left(\mathrm{2}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 06/Sep/20
Ok. Thanks. Please check the question I just posted.
$$\mathrm{Ok}.\:\mathrm{Thanks}.\:\mathrm{Please}\:\mathrm{check}\:\mathrm{the} \\ $$$$\mathrm{question}\:\mathrm{I}\:\mathrm{just}\:\mathrm{posted}. \\ $$

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