Question 608760
Hi, there,
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Do you know which sides of the rectangular pane contain the points X and Y? The fraction will differ depending upon where X and Y are located.
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Here is one solution to your problem. If the points X and Y are on different sides of the rectangle with respect to point C the fraction may be different. However the solution method is the same.
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Make a drawing to follow along.
Let X be the midpoint of side AB
Let Y be the midpoint of side AD
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Draw triangle XYC.
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Let W be the width of the rectangle. So W=AB=CD.
Let L be the length of the rectangle. So L=AD=BC.
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The area of the rectangle is W*L.
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Now find the areas of the three right triangles and subtract those area from the rectangle area. This is the area of triangle XYC.
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Triangle AXY:
base = 1/2 * W
height = 1/2 * L
area:
{{{A[1]=(1/2)*(1/2) (W)*(1/2)(L)}}}
{{{A[1]=(1/8)WL}}}
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Triangle YCD:
base= W
height = 1/2 * L
area:
{{{A[2]=(1/2)*(W)*(1/2)( L)}}}
{{{A[2]=(1/4)WL}}}
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Triangle XBC:
base = 1/2 * W
height = L
area:
{{{A[3]=(1/2)*(1/2)(W)*(L)}}}
{{{A[3]=(1/4)WL}}}
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Area of triangle XYC 
{{{A=WL - ((1/8)WL+(1/4)WL+(1/4)WL)}}}
{{{A=WL-(5/8)WL}}}
{{{A=(3/8)WL}}}
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Fraction of triangle area over rectangle area:
{{{((3/8)WL)/(WL)=3/8}}}
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Hope this helps! Feel free to email via gmail if you have questions,
Ms.Figgy
math.in.the.vortex@gmail.com