Advertisements
Advertisements
प्रश्न
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if ______.
विकल्प
PQRS is a rectangle
PQRS is a parallelogram
diagonals of PQRS are perpendicular
diagonals of PQRS are equal
Advertisements
उत्तर
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if diagonals of PQRS are perpendicular.
Explanation:

Since, diagonals of rectangle are equal
∴ AC = BD
⇒ PQ = QR
∴ PQRS is a rhombus
Diagonals of a rhombus are perpendicular.
APPEARS IN
संबंधित प्रश्न
ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.
In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.
In Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.

ABC is a triang D is a point on AB such that AD = `1/4` AB and E is a point on AC such that AE = `1/4` AC. Prove that DE = `1/4` BC.
In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:
PR = `[1]/[2]` ( AB + CD)

In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.
In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: AB, if DC = 8 cm and PQ = 9.5 cm
The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the midpoints of quadrilateral ABCD is a rectangle.
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF || AB and EF = `1/2` (AB + CD).
[Hint: Join BE and produce it to meet CD produced at G.]
