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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1204.
Is PL⊥AC and PK⊥BD.Area (PBD)=(BD.PK)/2=(AC.OL)/2.Is PA^2-PB^2=2.AC.OL=8^2-5^2=64-25=39.Therefore (PBD)=(2.AC.OL)/4=39/4.
Let S be the area of ∆BPD and let p be the length of the side of thesquare. Draw EPF parallel to AB,DC with E being on AD and F on BC. DrawGPH parallel to AD, BC with G being on AB and H being on DC.Let PF = n so that EP = p-n. Let PH = m so that GP = p-mThen S = p2 /2 – pm/2 – pn/2……(1)Further using Pythagorasm2 + n2 = 52…….(2) and(p-n)2 + (p-m)2 = 82….(3)From (2) we have 2p2 - 2p(m+n) + m2 + n2 = 64 or dividing both sides by4,p2 /2 – pm/2 – pn/2 = 16 – 25/4 using (2)From (1), LHS = S hence S = 16 – 25/4 = 9 ¾ or 9.75Sumith PeirisMoratuwaSri Lanka
In (2) and (3) 52 is 5 squared and 82 is 8 squared etc.
If we want to compute the area of Tr. APC or any other triangle in the diagram we will need to know the value of p, the side of the squareFurther sqrt2 X p < 13Or we need to know either of m or n in my proof above
Name MN // AB, KL//BC (drown from P) Name x=AM, y=MP, AB=ax²+y²=64, (a-x)²+(a-y)²=25 => 2ax+2ay-2a²=39 (1)S(DBP)=S(DBC)-S(DPC)-S(CPB)=a²-a(a-x)-a(a-y)=ax+ay-a²From (1) S=39/4
The lazy method: since only AP and PC are given, answer must be independent from P position. So let's say AP and PC are aligned. Then area(BPD) = (8+5).(8-5)/2/2=39/4 :-)