problem on a trapezoid having intersection of diagonals
This problem is hard for me. Can anyone help me? Thank you in advance.
Let ABCD be a trapezoid with the measure of base AB twice that of base DC, and let E be the point of intersection of diagonals. If the measure of diagonal AC is 11, then that of segment EC is equal to?
geometry
add a comment |
This problem is hard for me. Can anyone help me? Thank you in advance.
Let ABCD be a trapezoid with the measure of base AB twice that of base DC, and let E be the point of intersection of diagonals. If the measure of diagonal AC is 11, then that of segment EC is equal to?
geometry
1
triangle AEB is similar to the triangle CDE.
– runaround
Mar 21 '16 at 2:45
add a comment |
This problem is hard for me. Can anyone help me? Thank you in advance.
Let ABCD be a trapezoid with the measure of base AB twice that of base DC, and let E be the point of intersection of diagonals. If the measure of diagonal AC is 11, then that of segment EC is equal to?
geometry
This problem is hard for me. Can anyone help me? Thank you in advance.
Let ABCD be a trapezoid with the measure of base AB twice that of base DC, and let E be the point of intersection of diagonals. If the measure of diagonal AC is 11, then that of segment EC is equal to?
geometry
geometry
asked Mar 21 '16 at 2:36
user321645
1
triangle AEB is similar to the triangle CDE.
– runaround
Mar 21 '16 at 2:45
add a comment |
1
triangle AEB is similar to the triangle CDE.
– runaround
Mar 21 '16 at 2:45
1
1
triangle AEB is similar to the triangle CDE.
– runaround
Mar 21 '16 at 2:45
triangle AEB is similar to the triangle CDE.
– runaround
Mar 21 '16 at 2:45
add a comment |
1 Answer
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triangles ABE and CDE are similar, hence: $frac{AB}{DC}=frac{AE}{CE}=2$ and AE+EC=11. Hence, EC=$11/3$.
answered Mar 21 '16 at 2:47
Chip
97929
Thanks. I think that my brain did not work.
– user321645
Mar 21 '16 at 2:53
add a comment |
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triangles ABE and CDE are similar, hence: $frac{AB}{DC}=frac{AE}{CE}=2$ and AE+EC=11. Hence, EC=$11/3$.
answered Mar 21 '16 at 2:47
Chip
97929
Thanks. I think that my brain did not work.
– user321645
Mar 21 '16 at 2:53
add a comment |
triangles ABE and CDE are similar, hence: $frac{AB}{DC}=frac{AE}{CE}=2$ and AE+EC=11. Hence, EC=$11/3$.
answered Mar 21 '16 at 2:47
Chip
97929
Thanks. I think that my brain did not work.
– user321645
Mar 21 '16 at 2:53
add a comment |
triangles ABE and CDE are similar, hence: $frac{AB}{DC}=frac{AE}{CE}=2$ and AE+EC=11. Hence, EC=$11/3$.
answered Mar 21 '16 at 2:47
Chip
97929
triangles ABE and CDE are similar, hence: $frac{AB}{DC}=frac{AE}{CE}=2$ and AE+EC=11. Hence, EC=$11/3$.
answered Mar 21 '16 at 2:47
Chip
97929
answered Mar 21 '16 at 2:47
Chip
97929
answered Mar 21 '16 at 2:47
Chip
97929
answered Mar 21 '16 at 2:47
Chip
97929
97929
Thanks. I think that my brain did not work.
– user321645
Mar 21 '16 at 2:53
add a comment |
Thanks. I think that my brain did not work.
– user321645
Mar 21 '16 at 2:53
Thanks. I think that my brain did not work.
– user321645
Mar 21 '16 at 2:53
Thanks. I think that my brain did not work.
– user321645
Mar 21 '16 at 2:53
add a comment |
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triangle AEB is similar to the triangle CDE.
– runaround
Mar 21 '16 at 2:45