anonymous
  • anonymous
Need Help with Geometry and Ratio!!!!
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
In the diagram, AX = 10, XB=5, AY=4 and YC=8. What is the ratio of the area of AXY to the area of ABC? Express your answer as a common fraction.
anonymous
  • anonymous
|dw:1327715430695:dw|
Directrix
  • Directrix
Are the triangles given to be similar? If so, what is the correspondence of vertices? Would you check to determine if the diagram is correctly labeled? Thanks.

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anonymous
  • anonymous
The diagram (seems) not correct.
Directrix
  • Directrix
I agree. I think that the SAS Similarity Theorem is needed for solution but the given data doesn't satisfy the theorem's hypothesis. Triangles are similar if two sides in one triangle are in the same proportion to the corresponding sides in the other, and the included angles are equal.
anonymous
  • anonymous
this is all the given information
anonymous
  • anonymous
I could draw the diagram again to clarify.
anonymous
  • anonymous
|dw:1327717724546:dw|
Directrix
  • Directrix
Are both the black segments of length h? It seems that they would differ in length. And, are the segments labeled h perpendicular to segment AB?
anonymous
  • anonymous
How come the black segments be equal in length? or am I misunderstanding ?!
anonymous
  • anonymous
yeah sorry, they don't have the same length and i believe they are perpendicular. That was the hint from my teacher.
anonymous
  • anonymous
|dw:1327720461436:dw|
anonymous
  • anonymous
are you sure that this are the only given?
anonymous
  • anonymous
yes
anonymous
  • anonymous
I think it has something to do with ratio
anonymous
  • anonymous
i have here only assumptions. .
anonymous
  • anonymous
ok
anonymous
  • anonymous
the area of ACB-Ayx= area of CyxB the area of ACB-CyxB= area of Ayx
anonymous
  • anonymous
Yes, but how do I find the area?
anonymous
  • anonymous
this is what I've got 4/12 = p/h 4h/12=p |dw:1327722363321:dw|
anonymous
  • anonymous
but i am not sure how to find ABC.
anonymous
  • anonymous
your problem is ABC? find area of ACB: sinA=h/(8+4) therefor h=12sinA area = 1/2bh = 1/2(10+5)(12sinA) ; substitute the value of h in the formula = 1/2(15)(12sinA) therefor: the area of ACB is 90sinA
anonymous
  • anonymous
Can you clarify sinA?
anonymous
  • anonymous
no it,s not (8+4) for h, it's 10+5 , and (10+5) is 8+4, but anyway the answer is still 90 sinA
anonymous
  • anonymous
sin A will be find if there's a given measure for the angle. .
anonymous
  • anonymous
ok
anonymous
  • anonymous
thanks
anonymous
  • anonymous
you're welcome. .

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