Guest clover Posted June 16, 2004 Posted June 16, 2004 Please use the attachment. Find the shaded area of the triangle. i've made some changes and i think the diagram cann't be any more clear than this. attachment.doc
aommaster Posted June 16, 2004 Posted June 16, 2004 The diagram is not that clear. The numbers are obviously lengths, but, looking at the side 8 and 18, is the 8 length part of the 18, so the total length is 18, or is the total length 26?
Dave Posted June 16, 2004 Posted June 16, 2004 Looks like it's the latter case to me. I think the easiest way would be using similar triangles (although it's quite quick to do it by integration). Look at my diagram that I've included. We know that the triangles A and B are similar to their respective larger triangles. (sorry they don't intersect the point on both big triangles but MS Word sucks). From this you can work out the areas of the smaller triangles (hint: consider the length of the hypoteneuse), the point at which the two lines intersect and the area of the square. Add the areas together and you've got your answer.
winfred01 Posted June 16, 2004 Posted June 16, 2004 Looks like it's the latter case to me. I think the easiest way would be using similar triangles (although it's quite quick to do it by integration). Look at my diagram that I've included. We know that the triangles A and B are similar to their respective larger triangles. (sorry they don't intersect the point on both big triangles but MS Word sucks). From this you can work out the areas of the smaller triangles (hint: consider the length of the hypoteneuse)' date=' the point at which the two lines intersect and the area of the square. Add the areas together and you've got your answer.[/quote'] This is what I did... Assuming THE angle is a right angle and 6.8.16.18 are coordinates... and the diagram starts from 0... then...we will have enough infomation... first, I noticed that 16/8 and 18/3 (the gradients) are both whole numbers... If we consider the two hypotenuses as two linear equations y=mx+c and work out the intersection point (I).... then finding the shaded area is a piece of cake...
aommaster Posted June 16, 2004 Posted June 16, 2004 Look at my diagram that I've included Where is the diagram? Is it in the same file? Also, how can you be sure about the lengths problem that I had come across?: The diagram is not that clear. The numbers are obviously lengths, but, looking at the side 8 and 18, is the 8 length part of the 18, so the total length is 18, or is the total length 26?
Dave Posted June 16, 2004 Posted June 16, 2004 Where is the diagram? Is it in the same file? Whoops, forgot to add it. But it's the same as the one above anyway. Also, how can you be sure about the lengths problem that I had come across? It looks fairly obvious to me, don't know about anyone else. They look like they're included in the lengths.
aommaster Posted June 16, 2004 Posted June 16, 2004 looked to me like that as well, but, it would give totally wrong aswers if it wasn't!
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