Advertisements
Advertisements
Question
Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
Advertisements
Solution
The diagonals of a parallelogram bisect each other.
Therefore, O is the mid-point of AC and BD.
Bo is the median in ΔABC.
Therefore, it will divide it into two triangles of equal areas
∴ ar(ΔAOB) = ar(ΔBOC)......(i)
In ΔBCD, CO is the median.
∴ ar(ΔBOC) = ar(ΔCOD).......(ii)
Similarly, ar(ΔCOD) = ar(ΔAOD)......(iii)
From (i), (ii) and (iii)
ar(ΔAOB) = ar(ΔBOC) = ar(ΔCOD) = ar(ΔAOD)
Hence, diagonals of a parallelogram divide it into foor triangles of equal areas.
APPEARS IN
RELATED QUESTIONS
ABCD is a parallelogram. P and Q are mid-points of AB and CD. Prove that APCQ is also a parallelogram.
ABCD is a parallelogram. P and T are points on AB and DC respectively and AP = CT. Prove that PT and BD bisect each other.
In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that
RN and RM trisect QS.
In a parallelogram PQRS, M and N are the midpoints of the opposite sides PQ and RS respectively. Prove that
MN bisects QS.
In the given figure, PQRS is a parallelogram in which PA = AB = Prove that: SA ‖ QB and SA = QB.
In the given figure, PQRS is a trapezium in which PQ ‖ SR and PS = QR. Prove that: ∠PSR = ∠QRS and ∠SPQ = ∠RQP
The diagonals AC and BC of a quadrilateral ABCD intersect at O. Prove that if BO = OD, then areas of ΔABC an ΔADC area equal.
PQRS is a parallelogram and O is any point in its interior. Prove that: area(ΔPOQ) + area(ΔROS) - area(ΔQOR) + area(ΔSOP) = `(1)/(2)`area(|| gm PQRS)
In ΔABC, the mid-points of AB, BC and AC are P, Q and R respectively. Prove that BQRP is a parallelogram and that its area is half of ΔABC.
In the figure, ABCD is a parallelogram and CP is parallel to DB. Prove that: Area of OBPC = `(3)/(4)"area of ABCD"`
