Advertisements
Advertisements
प्रश्न
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.
उत्तर
In ΔABD,
BO = OD
⇒ O is the mid-point of BD
⇒ AO is a median.
⇒ ar(ΔAOB) = ar(ΔAOD) ..........(i)
In ΔCBD, O is the mid-point of BD
⇒ CO is a median.
⇒ ar(ΔCOB) = ar(ΔCOD) ..........(ii)
Adding (i) and (ii)
ar(ΔAOB) 6 ar(ΔCOB) = ar(ΔAOD) + ar(ΔCOD)
Therefore, ar(ΔABC) = ar(ΔADC).
APPEARS IN
संबंधित प्रश्न
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.
ABCD is a rectangle with ∠ADB = 55°, calculate ∠ABD.
P is a point on side KN of a parallelogram KLMN such that KP : PN is 1 : 2. Q is a point on side LM such that LQ : MQ is 2 : 1. Prove that KQMP is a parallelogram.
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
PMRN is a parallelogram.
In the given figure, PQRS is a trapezium in which PQ ‖ SR and PS = QR. Prove that: ∠PSR = ∠QRS and ∠SPQ = ∠RQP
Prove that the diagonals of a kite intersect each other at right angles.
In the given figure, AB ∥ SQ ∥ DC and AD ∥ PR ∥ BC. If the area of quadrilateral ABCD is 24 square units, find the area of quadrilateral PQRS.
In ΔPQR, PS is a median. T is the mid-point of SR and M is the mid-point of PT. Prove that: ΔPMR = `(1)/(8)Δ"PQR"`.