Monday, April 4, 2016

Geometry Problem 1204: Square, Triangle, Area

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.

Geometry Problem 1204: Square, Triangle, Area

6 comments:

  1. 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.

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  2. Let S be the area of ∆BPD and let p be the length of the side of the
    square. 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

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    Replies
    1. In (2) and (3) 52 is 5 squared and 82 is 8 squared etc.

      Delete
    2. 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 square

      Further sqrt2 X p < 13

      Or we need to know either of m or n in my proof above

      Delete
  3. Name MN // AB, KL//BC (drown from P) Name x=AM, y=MP, AB=a
    x²+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

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  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 :-)

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