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shruti sharma

· started a discussion

· 1 Months ago

. (Question: 207368.0 )

explanation?

Question:
The vertex A of \(\triangle\) ABC is joined to a point D on BC. If E is the midpoint of AD, then area of \(\triangle\) BEC is eqaul to  –
Options:
A) \(\cfrac{1}{3}\) are of \(\triangle\) ABC
B)

\(\cfrac{1}{4}\) are of \(\triangle\) ABC

C) \(\cfrac{1}{2}\) are of \(\triangle\) ABC
D) \(\cfrac{1}{6}\) are of \(\triangle\) ABC
Solution:
Ans: (c)


In the figure, vertex A of  ΔABC is joined to a point D on the side BC. The mid-point of AD is E. then ar(ΔBEC)=1/2 ar(ΔABC)

Given: E is any point on median AD of a Δ ABC.

To Prove: area(Δ ABC) = area(Δ ACE).

Proof: In Δ ABC,

AD is a median.

∴ area(Δ ABD) = area(Δ ACD) ….(1)

In Δ EBC,

ED is a median.

∴ area(Δ EBD) = area(Δ ECD) …..(2)

Subtracting (2) from (1), we get

area(Δ ABD) – area(Δ EBD) = area(Δ ACD) – area(Δ ECD)

⇒ area(Δ ABE) = area(Δ ACE).

Using this we can prove ar(ΔBEC)=1/2 ar(ΔABC).

Knowledge Expert

· commented

· 1 Months ago

Hi Shruti, its theorem.....

In the figure, vertex A of ΔABC is joined to a point D on the side BC. The mid-point of AD is E. then ar(ΔBEC)=1/2 ar(ΔABC)
Given: E is any point on median AD of a Δ ABC.

To Prove: area(Δ ABC) = area(Δ ACE).

Proof: In Δ ABC,
AD is a median.

∴ area(Δ ABD) = area(Δ ACD) ….(1)
In Δ EBC,
ED is a median.

∴ area(Δ EBD) = area(Δ ECD) …..(2)

Subtracting (2) from (1), we get

area(Δ ABD) – area(Δ EBD) = area(Δ ACD) – area(Δ ECD)

⇒ area(Δ ABE) = area(Δ ACE).
Using this we can prove ar(ΔBEC)=1/2 ar(ΔABC).
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