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07 Aug 2006, 07:58
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Triangle ABC was cut from construction paper that is white on one side and grey on the other. It is folded along line FH, which is parallel to line AB so that a two-coloured shape is made, as shown above. If the area of triangle IDE is one quarter of that of triangle ABC, what is the length of FH?
(1) The length of AB is 20
(2) Triangle ABC is an isosceles triangle with an area of 400.
(1) The length of AB is 20
(2) Triangle ABC is an isosceles triangle with an area of 400.
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07 Aug 2006, 08:51
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A for me.
I will try to put my explanation the best I can.
The area of a triangle ABC with base as BC is denoted as
1/2 * base * height.
Now if I draw a line parallel to BC joining the midpoints of the sides AB (E) and AC (F) then it would also bisect the height of the triangle. Also the length of the line EF would be half of BC.
So now the height of the triangle AEF is half the height of the triangle ABD and the base is also half of BC. So the Area of the triangle AEF is 1/4 of the triangle ABC.
In the given problem we can say that DE is half the length of AB and the height of the triangle is 1/2 the height of the triangle ABC.
Now we are left with the other half of the height of the triangle ABC (Half of which is behind the fold and half of which above the fold. ) So with the similar analogy we can say that the distance between the base BC and FH is 1/4 the height of the triangle.
Applying the concept of proportionality, the length of FH should be 3/4 ( 1/2 + 1/4 ) of that of BC. Hence SUFF. (This is applicable for all the triangles )
Statement 2 - We can have multiple isosceles triangles with the given area. Hence INSUFF.
Hope I am clear with my explanation.
I will try to put my explanation the best I can.
The area of a triangle ABC with base as BC is denoted as
1/2 * base * height.
Now if I draw a line parallel to BC joining the midpoints of the sides AB (E) and AC (F) then it would also bisect the height of the triangle. Also the length of the line EF would be half of BC.
So now the height of the triangle AEF is half the height of the triangle ABD and the base is also half of BC. So the Area of the triangle AEF is 1/4 of the triangle ABC.
In the given problem we can say that DE is half the length of AB and the height of the triangle is 1/2 the height of the triangle ABC.
Now we are left with the other half of the height of the triangle ABC (Half of which is behind the fold and half of which above the fold. ) So with the similar analogy we can say that the distance between the base BC and FH is 1/4 the height of the triangle.
Applying the concept of proportionality, the length of FH should be 3/4 ( 1/2 + 1/4 ) of that of BC. Hence SUFF. (This is applicable for all the triangles )
Statement 2 - We can have multiple isosceles triangles with the given area. Hence INSUFF.
Hope I am clear with my explanation.
07 Aug 2006, 14:39
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It is definitely not B as I can think of any number of isosceles triangles with different heights and same areas.
A makes sense as the length of the segment AB helps and with similar triangles ABC and IDE we can determine length of FH. I won't do actual calculations as it is not required.
A makes sense as the length of the segment AB helps and with similar triangles ABC and IDE we can determine length of FH. I won't do actual calculations as it is not required.
