मराठी

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

Advertisements
Advertisements

प्रश्न

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.

बेरीज
Advertisements

उत्तर

Given: In a square ABCD, P, Q, R and S are the mid-points of AB, BC, CD and DA, respectively.

To show: PQRS is a square.

Construction: Join AC and BD.


Proof: Since, ABCD is a square.

∴ AB = BC = CD = AD

Also, P, Q, R and S are the mid-points of AB, BC, CD and DA, respectively.

Then, in ΔADC, SR || AC

And SR = `1/2`AC   [By mid-point theorem] ...(i)

In ΔABC, PQ || AC

And PQ = `1/2`AC  ...(ii)

From equations (i) and (ii),

SR || PQ and SR = PQ = `1/2`AC  ...(iii)

Similarly, SP || BD and BD || RQ

∴ SP || RQ and SP = `1/2`BD

And RQ = `1/2`BD

∴ SP = RQ = `1/2`BD

Since, diagonals of a square bisect each other at right angle.

∴ AC = BD

⇒ SP = RQ = `1/2`AC   ...(iv)

From equations (iii) and (iv),

SR = PQ = SP = RQ  ...[All side are equal]

Now, in quadrilateral OERF,

OE || FR and OF || ER

∴ ∠EOF = ∠ERF = 90°

Hence, PQRS is a square.

Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Quadrilaterals - Exercise 8.4 [पृष्ठ ८२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 9
पाठ 8 Quadrilaterals
Exercise 8.4 | Q 11. | पृष्ठ ८२

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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 a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of

the triangle formed by joining the mid-points of the sides of this triangle. 


The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is rectangle.


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.


ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and Ac respectively. Prove that EFGH is a rhombus.


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


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.


In parallelogram ABCD, P is the mid-point of DC. Q is a point on AC such that CQ = `(1)/(4)"AC"`. PQ produced meets BC at R. Prove that

(i) R is the mid-point of BC, and

(ii) PR = `(1)/(2)"DB"`.


ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: The line drawn through G and parallel to FE and bisects DA.


Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×