हिंदी

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if ______.

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

MCQ
रिक्त स्थान भरें
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.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Quadrilaterals - Exercise 8.1 [पृष्ठ ७३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
अध्याय 8 Quadrilaterals
Exercise 8.1 | Q 4. | पृष्ठ ७३

वीडियो ट्यूटोरियलVIEW ALL [1]

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.]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×