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A trapezium is divided into 4 triangles by its diagonals. Prove the ares of the coloured triangles are equal. Thank you

Try to imagine that the coloured triangles are the left and right triangled in the trapezium, meeting vertically opposite, instead of the ones above and below

2007-11-29 20:49:49 · 3 answers · asked by little zebulon 4 in Science & Mathematics Mathematics

3 answers

you will see that there are two big triangles on the same base (one of the parallel sides) and between the parallel sides. they are formed by both diagonals independently. sinces their bases and heights are equal, their areas are equal.
they will have a common triangle whose area is common to both of them. since their areas are equal and they have a common trialngle, the areas of the triangles not common to both of them (the coloured triangles) are equal.

2007-11-29 21:15:19 · answer #1 · answered by Bhaskar 4 · 0 0

Given trapezium ABCD with BC ║ AD, AOB = COD iff AB = CD. The same is true of ABC and BCD

2007-11-30 05:39:08 · answer #2 · answered by Helmut 7 · 0 0

http://i4.tinypic.com/81srnvb.jpg
in ∆ ABC & ∆ CBD
∵AD // CB & CB is a common base
∴area of ∆ ABC = area of ∆ CBD
∵area of ∆ ABC = area o ∆f EBA + area of ∆ EBC
& Area of ∆ CBD =are of ∆ CED +area of ∆ EBC

∴area of ∆ AEB = area of ∆ CEB

2007-11-30 05:03:46 · answer #3 · answered by mbdwy 5 · 1 0

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