Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Monday, April 4, 2016
Geometry Problem 1204: Square, Triangle, Area
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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.
ReplyDeleteTherefore (PBD)=(2.AC.OL)/4=39/4.
Let S be the area of ∆BPD and let p be the length of the side of the
ReplyDeletesquare. Draw EPF parallel to AB,DC with E being on AD and F on BC. Draw
GPH 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-m
Then S = p2 /2 – pm/2 – pn/2……(1)
Further using Pythagoras
m2 + 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 by
4,
p2 /2 – pm/2 – pn/2 = 16 – 25/4 using (2)
From (1), LHS = S hence S = 16 – 25/4 = 9 ¾ or 9.75
Sumith Peiris
Moratuwa
Sri Lanka
In (2) and (3) 52 is 5 squared and 82 is 8 squared etc.
DeleteIf 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 square
DeleteFurther sqrt2 X p < 13
Or 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=a
ReplyDeletex²+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-a²
From (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 :-)
ReplyDeleteI hope to who posted his solution explain it by sketch and more clear in steps for us to understand his solution .
ReplyDelete