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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.
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.
The diagram (seems) not correct.
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.
this is all the given information
I could draw the diagram again to clarify.
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?
How come the black segments be equal in length? or am I misunderstanding ?!
yeah sorry, they don't have the same length and i believe they are perpendicular. That was the hint from my teacher.
are you sure that this are the only given?
I think it has something to do with ratio
i have here only assumptions. .
the area of ACB-Ayx= area of CyxB the area of ACB-CyxB= area of Ayx
Yes, but how do I find the area?
this is what I've got 4/12 = p/h 4h/12=p |dw:1327722363321:dw|
but i am not sure how to find ABC.
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
Can you clarify sinA?
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
sin A will be find if there's a given measure for the angle. .
you're welcome. .