English

The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______. - Mathematics

Advertisements
Advertisements

Question

The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______.

Options

  • a rhombus

  • a rectangle

  • a square

  • any parallelogram

MCQ
Fill in the Blanks
Advertisements

Solution

The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is a rectangle.

Explanation:

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.


Join AC, RP and SQ

In ∆ABC,

P is midpoint of AB and Q is midpoint of BC

∴ By midpoint theorem,

PQ || AC and PQ = `1/2` AC  ...(1)

Similarly,

In ∆DAC,

S is midpoint of AD and R is midpoint of CD

∴ By midpoint theorem,

SR || AC and SR = `1/2` AC  ...(2)

From (1) and (2),

PQ || SR and PQ = SR

⇒ PQRS is a parallelogram

ABQS is a parallelogram

⇒ AB = SQ

PBCR is a parallelogram

⇒ BC = PR

⇒ AB = PR  ...[∵ BC = AB, sides of rhombus]

⇒ SQ = PR

∴ Diagonals of the parallelogram are equal.

Hence, it is a rectangle.

shaalaa.com
  Is there an error in this question or solution?
Chapter 8: Quadrilaterals - Exercise 8.1 [Page 74]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 9
Chapter 8 Quadrilaterals
Exercise 8.1 | Q 9. | Page 74

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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 the given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.

[Hint: DN || QM]


In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.


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)


A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that

CQ = `[1]/[4]`AC. PQ produced meets BC at R.

Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB


In trapezium ABCD, AB is parallel to DC; P and Q are the mid-points of AD and BC respectively. BP produced meets CD produced at point E.

Prove that:

  1. Point P bisects BE,
  2. PQ is parallel to AB.

D, E, and F are the mid-points of the sides AB, BC, and CA respectively of ΔABC. AE meets DF at O. P and Q are the mid-points of OB and OC respectively. Prove that DPQF is a parallelogram.


Use the following figure to find:
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm
(iv) DF, if CG = 11 cm.


In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find ∠FDB if ∠ACB = 115°.


In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: A is the mid-point of PQ.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×