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

a. ABCD is a rhombus

b. diagonals of ABCD are equal

c. diagonals of ABCD are equal and perpendicular

d. diagonals of ABCD are perpendicular.

**Solution:**

In the figure,

ABCD is a parallelogram

P, Q, R and S are the mid points of AB, BC, CD and DA

So PQRS is a square

PQ = QR = RS = PS … (1)

PR = SQ

As PR = BC and SQ = AB

AB = BC

So all sides of quadrilateral ABCD are equal

Quadrilateral ABCD is either a rhombus or a square

In Δ ADC, using the mid point theorem

SP || DB, SP = 1/2 DB … (2)

In Δ ABC,

PQ || AC, PQ = 1/2 AC … (3)

From the equation (1)

PS = PQ

From the equation (2) and (3)

1/2 DB = 1/2 AC

DB = AC

ABCD is a square as the diagonals are equal.

Here the diagonals are perpendicular also.

Therefore, the figure is a square only if diagonals of ABCD are equal and perpendicular.

**✦ Try This: **The figure formed by joining the mid-points of the sides of a quadrilateral EFGH, taken in order, is a square only if,

a. EFGH is a rhombus

b. diagonals of EFGH are equal

c. diagonals of EFGH are equal and perpendicular

d. diagonals of EFGH are perpendicular.

**☛ Also Check:** NCERT Solutions for Class 9 Maths Chapter 8

**NCERT Exemplar Class 9 Maths Exercise 8.1 Problem 11**

## The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, a. ABCD is a rhombus, b. diagonals of ABCD are equal, c. diagonals of ABCD are equal and perpendicular, d. diagonals of ABCD are perpendicular

**Summary:**

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if diagonals of ABCD are equal and perpendicular

**☛ Related Questions:**

- The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32º a . . . .
- Which of the following is not true for a parallelogram? ,a. opposite sides are equal ,b. opposite an . . . .
- D and E are the mid-points of the sides AB and AC respectively of ∆ABC. DE is produced to F. To prov . . . .

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