Area of triangle is 1/4 of triangle abc in midpoint theorem
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It would be 1/4th of the area ABC, i.e, 5cmˆ2.
Explanation:
There is a property of lines joining mid-points of two sides of a triangle.
Property - Line joining the mid-points of two sides is always parallel to the third side.
There is a Theorem for the above.
Coming back to the original question. If you let length of the sides to be AB, BC and AC. Then,
AD = BD = 1/2AB
BE = CE = 1/2BC
AF = CF = 1/2AC
From the property, we can say that DE is parallel to AC. Similarly, EF is parallel to AB and DF is parallel to BC.
If you draw the figure and see, you will notice that 3 parallelograms, AFED, FDEC and FDBE.
You can thus equate the parallel sides and we get FD = EB, EF = BD and DE = CF which are halves of the parallel sides.
Now if you check the sides of the four triangles, all of will have the same dimensions. Thus total area of ABC is sum of the areas of the 4 triangles.
Due to the same dimensions, all of them will have the same area.
So area of DEF is 1/4 area of ABC
Explanation:
There is a property of lines joining mid-points of two sides of a triangle.
Property - Line joining the mid-points of two sides is always parallel to the third side.
There is a Theorem for the above.
Coming back to the original question. If you let length of the sides to be AB, BC and AC. Then,
AD = BD = 1/2AB
BE = CE = 1/2BC
AF = CF = 1/2AC
From the property, we can say that DE is parallel to AC. Similarly, EF is parallel to AB and DF is parallel to BC.
If you draw the figure and see, you will notice that 3 parallelograms, AFED, FDEC and FDBE.
You can thus equate the parallel sides and we get FD = EB, EF = BD and DE = CF which are halves of the parallel sides.
Now if you check the sides of the four triangles, all of will have the same dimensions. Thus total area of ABC is sum of the areas of the 4 triangles.
Due to the same dimensions, all of them will have the same area.
So area of DEF is 1/4 area of ABC
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