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  • The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to

(A)  \frac{1}{2}ar(ABC)
(B)  \frac{1}{3}ar(ABC)
(C)  \frac{1}{4}ar(ABC)

(D)  ar(ABC)

Answers (1)

best_answer

Answer: [A]

Solution. 
 Let the given triangle ABC be shown as below with medians AE, BF, CD.

D is the mid-point of AB, E is the mid-point of BC, F is the mid-point of AC
Mid-point theorem states that the line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side
So,  DF=\frac{1}{2}BC

DF = BE = EC …(i)
DE=\frac{1}{2}AC
DE = AF = FC …(ii)
FE=\frac{1}{2}AB
FE = AD = DB …(iii)
In \triangle ADF and \triangle FEC
AD = FE (from iii)
DF = EC (from i)
AF = FC (given)
\triangle ADF\cong \triangle FEC (SSS congruence)
Similarly we can prove that,
\triangle ADF \cong \triangle FEC \cong \triangle DBE \cong \triangle EFD
So the bigger triangle is divided into 4 smaller congruent triangles.
So area of \triangle ABC = 4 (area of one smaller triangle) …(iv)
If we consider the parallelograms formed then we can see that one \parallel gmconsists of two such triangles.
\parallel BDFE = \triangle DBE + \triangle EFD
\parallel DFCE = \triangle FEC + \triangle EFD
\parallel ADEF = \triangle ADF + \triangle EFD
So area of every parallelogram = 2 (area of one smaller triangle) …(v)
Using (iv) and (v),
area of every \parallel gm =\frac{1}{2}ar(ABC)

Therefore option (A) is correct.

 

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