Advertisements
Advertisements
Question
The medians QM and RN of ΔPQR intersect at O. Prove that: area of ΔROQ = area of quadrilateral PMON.
Advertisements
Solution

Join MN. Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side; so, MN || QR
Clearly, ΔQMN and ΔRNM are on the same base MN and between the same parallel lines.
Therefore, area(ΔQMN) = area(ΔRNM)
⇒ Area(ΔQMN) - area(ΔONM) = area(ΔRNM) - area(ΔONM)
⇒ Area)ΔQON) = area (ΔROM) ......(i)
We know that a median of a triangle divides it into two triangles of equal areas.
Therefore, area(ΔQMR) = area(ΔPQM)
⇒ area(ΔROQ) + area(ΔROM) = area(quad, PMON) + area(ΔQON)
⇒ area(ΔROQ) + area(ΔROM) = area(quad. PMON) + area(ΔROM) ...(from (i))
⇒ area(ΔROQ) = area(quad. PMON).
APPEARS IN
RELATED QUESTIONS
SN and QM are perpendiculars to the diagonal PR of parallelogram PQRS.
Prove that:
(i) ΔSNR ≅ ΔQMP
(ii) SN = QM
PQRS is a parallelogram. T is the mid-point of RS and M is a point on the diagonal PR such that MR = `(1)/(4)"PR"`. TM is joined and extended to cut QR at N. Prove that QN = RN.
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 the given figure, PQRS is a parallelogram in which PA = AB = Prove that: SAQB is a parallelogram.
Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
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 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 the given figure, PQ ∥ SR ∥ MN, PS ∥ QM and SM ∥ PN. Prove that: ar. (SMNT) = ar. (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.
