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Answer:
1. Sol.
Given : AD is a medium of Δ ABC and E is any point on AD.
To prove : ar (ABE) = ar (ACE)
Proof : Since AD is the median of Δ ABC
Therefore ar (ABD) = ar (ACD) ... (1)
Also , ED is the median of Δ EBC
Therefore ar (BED) = ar (CED) ... (2)
Subtracting (2) from (1) , we get
ar (ABD) – ar (BED) = ar (ACD) – ar (CED)
⇒ ar (ABE) = ar (ACE).
Q.2 In a triangle ABC, E is the mid- point of median AD. Show that ar (BED) = 14 ar (ABC).
Sol.
Given : A Δ ABC, E is the mid- point of the median AD.
To prove : ar (BED) = 1/4 ar (ABC)
Proof : Since AD is a median of Δ ABC and median divides a triangle into two triangles of equal area.
Therefore ar (ABD) = ar (ADC)
⇒ ar (ABD) = 12 ar (ABC) ... (1)
In Δ ABD, BE is the median
Therefore ar(BED) = ar (BAE) ... (2)
⇒ ar(BED)=12ar(ABD) [ar(BAE)=12ar(ABD) ]
⇒ ar (BED) = 12×12 ar (ABC) [Using (1)]
⇒ ar (BED) = 14 ar (ABC) .
Q.3 Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Sol.
Given : A parallelogram ABCD.
To prove : The diagonals AC and BD divide the || gm ABCD into four triangles of equal area.
Construction : Draw BL ⊥ AC.
Proof : Since ABCD is a || gm and so its diagonals AC and BD bisect each other at O.
Therefore AO = OC and BO = OD
Now, ar (AOB) = 12 × AO × BL
ar (OBC) = 12 × OC × BL
But AO = OC
Therefore ar (AOB) = ar (OBC)
Similarly, we can show that
ar (OBC) = ar (OCD) ; ar (OCD) = ar(ODA) ;
ar (ODA) = ar (OAB) ; ar (OAB) = ar (OBC)
ar (OCD) = ar (ODA)
Thus , ar (OAB) = ar (OBC) = ar(OCD) = ar (OAD)
Q.4 In 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).
Sol.
Given : ABC and ABD are two triangles on the same base AB. A line segment CD is bisected by AB at O i.e., OC = OD.
To prove : ar (ABC) = ar (ABD).
Proof : In Δ ACD, we have
OC = OD [Given]
Therefore AO is the median.
Therefore ar (AOC) = ar (AOD) [Since median divides a Δ in two Δs of equal area]
Similarly, in Δ BCD, BO is the median
Therefore ar (BOC) = ar (BOD)
Adding (1) and (2), we get
ar (AOC) + ar (BOC) = ar (AOD) + ar (BOD)
⇒ ar (ABC) = ar (ABD)
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