मराठी

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

Advertisements
Advertisements

प्रश्न

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

पर्याय

  • PQRS is a rhombus

  • PQRS is a parallelogram

  • diagonals of PQRS are perpendicular

  • diagonals of PQRS are equal

MCQ
रिकाम्या जागा भरा

उत्तर

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if diagonals of PQRS are equal.

Explanation:

Given, the quadrilateral ABCD is a rhombus.

So, sides AB, BC, CD and AD are equal.

Now, in ΔPQS, we have

D and C are the mid-points of PQ and PS.

So, `DC = 1/2 QS`   [By mid-point theorem]  ...(i)

Similarly, in ΔPSR, `BC = 1/2 PR`   [By mid-point theorem]  ...(ii)

As BC = DC  ...[Since, ABCD is a rhombus]

∴ `1/2 QS = 1/2 PR`  ...[From equations (i) and (ii)]

⇒ QS = PR

Hence, diagonals of PQRS are equal.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Quadrilaterals - Exercise 8.1 [पृष्ठ ७३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 9
पाठ 8 Quadrilaterals
Exercise 8.1 | Q 5. | पृष्ठ ७३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC.


ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.


In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.

 


Fill in the blank to make the following statement correct:

The figure formed by joining the mid-points of consecutive sides of a quadrilateral is           


In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E.
Show that BE is also a median.


Prove that the straight lines joining the mid-points of the opposite sides of a quadrilateral bisect each other.


Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 3DF = EF


AD is a median of side BC of ABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF: AC = 1 : 3.


The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, ______.


P, Q, R and S are respectively the mid-points of the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. Prove that PQRS is a rhombus.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×