Question 1: In the given figure, E is any point on median AD of a ΔABC. Show that ar(ABE) = ar(ACE).
Answer:Median divides a triangle into two triangles of equal area because both the triangles have equal base and equal altitude.
Hence, ar(ABD) = ar(ACD)
Similarly, ar(BED) = ar(DEC)
If we remove ΔBEC, i.e. ΔBED + ΔDEC from this, then
ar(ABE) = ar(ACE) proved.
Question 2: In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = ¼ ar(ABC).
Answer: Let us use the figure in previous question to solve this problem.
ED will be the median for ΔBEC
Hence, ar(BED) = ar(CED)
Moreover, ar(BEC) = ½ ar(ABC)
From these equations, it is clear;
ar(BED) = ¼ ar(ABC)
Question 3: Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Answer: ABCD is a parallelogram in which diagonals AC and BD intersect at point O.
To Prove: ar(AOB) = ar(AOC) = ar(BOC) = ar(AOD)
As diagonals of a parallelogram bisect each other so, M and N are the mid-points of sides AD and BC respectively. This means; ar(ABNM) = ar(MNCD) = ½ ar(ABCD)
ar(ABO) = ½ ar(ABNM) because triangle on same base and with same height is half in area compared to parallelogram on same base and height.
Similarly, ar(DOC) = ½ ar(MNCD)
This means; ar(ABO) = ar(DOC) = ¼ ar(ABCD)
Similarly, following can be proved:
ar(AOD) = ar(BON) = ¼ ar(ABCD)
Hence, ar(AOB) = ar(BOC) = ar(DOC) = ar(AOD) = ¼ ar(ABCD)
Question 4: In the given figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar(ABD).
Answer: Since AO and BO bisect CD hence they are respective medians of triangles ACD and BCD.
Hence, ar(AOC) = ar(AOD)
Similarly, ar(COB) = ar(DOB)
Hence, ar(AOC) + ar(COB) = ar(AOD) + ar(DOB)
Or, ar(ABC) = ar(ABD) proved
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