What kind of quadrilateral is formed when the mid points of the sides of the following are joined a rectangle?

Geometry is derived from a Greek word that means ‘Earth Measurement’. It is a branch of mathematics and is concerned with the properties of space i.e., a visual study of shapes, position of figures, patterns, sizes, etc. Geometry is a subject of countless developments, so there exist many types. They are Euclidean Geometry, Non-Euclidean Geometry, Algebraic Geometry, Riemannian Geometry, and Symplectic Geometry.

Quadrilateral

A quadrilateral can be separated into two words Quad means four and lateral means side. So a quadrilateral is a closed figure with four sides. It had four vertices. The sides of a quadrilateral are equal/ unequal/ parallel/ irregular which leads to various geometric figures. Example: Square, Rectangle, Rhombus, Parallelogram, Trapezium, etc.

The midpoint of a side divides a side of any figure into two equal parts (length-wise). In a quadrilateral, there will be a midpoint for each side i.e., Four mid-points.

There are a few factors that determine the shape formed by joining the midpoints of a quadrilateral. Those factors are the kind of quadrilateral, diagonal properties, etc. These factors affect the shape formed by joining the midpoints in a given quadrilateral. Let’s look into the various scenarios to get a better understanding.

Solution:

Let’s consider a Quadrilateral ABCD, Find the shape of figure formed by joining the mid points.

Here A, B, C, D are vertices of quadrilateral 

P, Q, R, S are midpoints of sides AB, AC, BD and CD respectively.

As midpoint divides a side into equal parts, AP = PB and same will applied to all sides.

BC is the diagonal of the quadrilateral which form two triangles ABC and BCD.

Consider the Triangle BCD

CB is parallel to SR (CB || SR)

According to Mid-point theorem 

SR = CB/2

So CB||SR and SR = CB/2 ⇢ (1)

Same as in Triangle ABC

QP||CB and QP = CB/2 ⇢ (2)

From (1) & (2)

SR||QP and SR = QP

As one pair of opposite sides are equal in length and parallel to each other, The resultant figure by joining the midpoints of a quadrilateral become a parallelogram.

Sample Questions

Question 1: Consider the rhombus ABCD which is also a kind of Quadrilateral. Find the shape of the figure formed by joining the midpoints.

Solution:

Let ABCD is a rhombus and P,Q,R,S are mid points of sides AB, BC, CD, DA respectively.

In Triangle ABD we have:

PS || BD & PS = BD/2            ….(1) 

(According to midpoint theorem.)

In Triangle BDC we have

QR||BD & QR=BD/2               ….(2) 

(According to mid point theorem.)

From equations (1) & (2) we get,

PS || QR

So PQRS is a parallelogram

As diagonals of rhombus bisect each other at 90° (Right angles)

Diagonals are perpendicular to each other

AC ⊥ BD

As PS || QR & PQ || SR & AC⊥ BD 

PQ is also perpendicular to QR.

PQ ⊥ QR (∠PQR = 90°)

Hence PQRS is a Rectangle.

So, the figure formed by combining the midpoints of rhombus forms a rectangle.

Question 2: If the figure formed by joining the midpoint of a quadrilateral is square only if, do explain the condition.

Solution:

Let ABCD is a rhombus and P, Q, R, S are mid points of sides AB, BC, CD, DA respectively. 

Given PQRS is a Square.

Such that PQ = QR = RS = SP               ….(1)

Also diagonals in a square are of equal length i.e., PR = SQ

But PR = BC & SQ = AB

Hence AB = BC (as PR = SQ)

So all sides of quadrilateral are equal.

The given quadrilateral must be a square or rhombus.

In Triangle ADC we have

RS||AC & RS = AC/2                            ….(2) 

(According to midpoint theorem.)

In Triangle BDC we have

QR || BD & QR = BD/2                      …..(3) 

(According to mid point theorem. 0

From equation (1),

RS = QR

So, AC/2 = BD/2

AC = BD

Thus, the length of diagonals of quadrilateral are same, So ABCD is a square with diagonals perpendicular to each other.

Question 3: What is the figure formed by joining the midpoints of a parallelogram.

Solution:

The given figure is Parallelogram ABCD with midpoints P, Q, R, S for the sides AB, BD, CD, AC respectively.

 In Triangle ABC we have,

PS || BC & PS = BC/2                 ….(1) 

(According to midpoint theorem.)

In Triangle BDC we have:

QR || BC & QR = BC/2            ….(2) 

(According to mid point theorem.)

From equation (1) & (2) we get,

PS || QR

So, PQRS is a parallelogram.

Question 4: What is the figure formed by joining the midpoints of a quadrilateral whose diagonals are of length equal?

Answer: 

Rhombus which is a two-dimensional plane figure that is closed. It is regarded as a peculiar parallelogram, and it has its own identity as a quadrilateral due to its unique features. Because all of its sides are the same length, a rhombus is also known as an equilateral quadrilateral. The name ‘rhombus’ is derived from the ancient Greek word ‘rhombos,’ which literally means “to spin.”

Question 5: What is the figure formed by joining the midpoints of a quadrilateral whose diagonals are perpendicular but not of equal length?

Answer: 

Rectangle, which is a quadrilateral with equal angles on all sides and equal and parallel opposite sides. There are numerous rectangle items in our environment. Each rectangle shape has two distinct dimensions: length and width. The length and width of a rectangle are defined as the longer side and the shorter side, respectively.

If joining the midpoint of a rhombus forms a rectangle, what is the shape of the quadrilateral formed by joining midpoints of a rectangle? If you guessed it is a rhombus, you are correct, as we will now show.

Problem

Strategy

As we have done before, in order to save us some work, we will rely on what we have already proven. A rectangle is a quadrilateral, so connecting its midpoints creates a parallelogram. To prove this parallelogram Is a rectangle, we need to show that all of its sides are equal.

Since this quadrilateral is a parallelogram, we already know that the opposite sides are equal, as this is a property of parallelograms. All that remains to be shown, then, is that two adjacent sides of the parallelogram are equal.

Sides EH and GH are two such adjacent sides, and they are also the sides of tow triangles, ΔEHA and ΔGHD. So let's try to show those two triangles are congruent, proving that |EH|=|GH| as corresponding sides in congruent triangles.

To do so, we will rely on the fact that ABCD is a rectangle, and use the properties of rectangles: the opposite sides are equal (since a rectangle is a parallelogram), and the interior angles of a rectangle are all right angles (by definition).

And if the opposite sides are equal, then so are their halves. Since E and G are both midpoints, |AE|=|DG|, as halves of equal sides. H is also a midpoint, so |AH|=|HD|.

The angles of the rectangle are all equal and all measure 90°, so ∠EAH ≅ ∠GDH. We have now shown that  ΔEHA and ΔGHD are congruent by the Side-Angle-Side postulate, and as a result |EH|=|GH| as corresponding sides in congruent triangles. And since |EH|=|FG| and |EF|=|HG|, it follows that all the edges are equal and EFGH is a rhombus.

Proof

(1) ABCD is a rectangle //Given(2) |AE|=|EB| //Given(3) |BF|=|FC| //Given(4) |CG|=|GD| //Given(5) |DH|=|HA| //Given(6) |AB|=|DC| //(1), Opposite sides of a rectangle are equal(7) ½|AB|=½|DC| //Division Property of Equality(8) |AE|=|DG| //(6), (7)(9) ∠EAH ≅ ∠GDH //All interior angles of a rectangle are equal(10)  ΔEHA ≅ ΔGHD //(5), (8), (9) , Side-Angle-Side postulate(11) |EH|=|GH| //Corresponding sides in congruent triangles (CPCTC)

(12) EFGH is a parallelogram //connecting midpoints of a quadrilateral creates a parallelogram.

(15) EFGH is a rhombus //(14), definition of a rhombus