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 :-)

ReplyDelete