Advertisements
Advertisements
प्रश्न
ABCD is a trapezium in which side AB is parallel to side DC. P is the mid-point of side AD. IF Q is a point on the Side BC such that the segment PQ is parallel to DC, prove that PQ = `(1)/(2)("AB" + "DC")`.
Advertisements
उत्तर
PQ || DC ⇒ OQ || DC ||AB
Therefore, Q and O are mid-points of BC and BD respectively.
In ΔABD,
P and O are mid-points of AD and BD respectively
⇒ OP = `(1)/(2)"AB"` ........(i)
In ΔBCD,
Q and O are mid-points of BC and BD respectively
⇒ OQ = `(1)/(2)"CD"` ........(ii)
Adding (i) and (ii)
OP + OQ = `(1)/(2)"AB" + (1)/(2)"CD"`
⇒ PQ = `(1)/(2)("AB" + "CD")`.
APPEARS IN
संबंधित प्रश्न
ABCD is a quadrilateral P, Q, R and S are the mid-points of AB, BC, CD and AD. Prove that PQRS is a parallelogram.
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.
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.
Prove that the diagonals of a square are equal and perpendicular to each other.
Prove that the diagonals of a parallelogram divide it into four triangles of equal area.
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 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"`.
