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 quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:

  1. SR || AC and SR = `1/2AC`
  2. PQ = SR
  3. PQRS is a parallelogram.


ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.


In trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that: AB + DC = 2EF.


In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P.
Prove that:
(i) BP = 2AD
(ii) O is the mid-point of AP.


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 triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use the intercept Theorem to show that MN bisects AD.


In a parallelogram ABCD, M is the mid-point AC. X and Y are the points on AB and DC respectively such that AX = CY. Prove that:
(i) Triangle AXM is congruent to triangle CYM, and

(ii) XMY is a straight line.


In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: PQ, if AB = 12 cm and DC = 10 cm.


In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:

ST = `(1)/(3)"LS"`


In ∆ABC, AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are respectively the mid-points of AB and BC, determine the length of DE.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×