Advertisements
Advertisements
प्रश्न
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
∠P = 100°, ∠Q = 80°, ∠R = 95°
Advertisements
उत्तर
We have a quadrilateral named PQRS, with diagonals PR and QS intersecting at O.
∠P =100° , ∠Q = 80° ,∠R = 100°
By angle sum property of a quadrilateral, we get:
∠P + ∠Q + ∠R +∠S = 360°
100° + 80° + 100° +∠S = 360°
280° +∠S 360°
∠S = 80°
Clearly, ∠P = ∠R
And ∠Q = ∠S
Thus we have PQRS a quadrilateral with opposite angles are equal.
Therefore,
PQRS is a parallelogram.
APPEARS IN
संबंधित प्रश्न
In Fig., below, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.

In a parallelogram ABCD, determine the sum of angles ∠C and ∠D .
ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.
P and Q are the points of trisection of the diagonal BD of a parallelogram AB Prove that CQ is parallel to AP. Prove also that AC bisects PQ.
In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Find AB : AD.
We get a rhombus by joining the mid-points of the sides of a
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
In the given Figure, if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF. (Use congruent property of triangles)
Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.
