krit.club logo

4-gons (Quadrilaterals) - Analyse central symmetry and transformations in quadrilaterals

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Central Symmetry in Parallelograms: A parallelogram possesses point symmetry (central symmetry) about the point of intersection of its diagonals. Every point PP on the boundary has a corresponding point P′P' such that the intersection point OO is the midpoint of PP′PP'.

A parallelogram showing diagonals intersecting at point O, demonstrating central symmetry.
•

Rotational Symmetry: A square has rotational symmetry of order 4 about its center (point of intersection of diagonals), while a rectangle and a rhombus have rotational symmetry of order 2.

Square showing 90 degree rotational symmetry about the center.
•

Mid-point Theorem Transformation: If you join the mid-points of the sides of any quadrilateral, the resulting figure is always a parallelogram. This transformation simplifies complex quadrilateral properties into known parallelogram properties.

A general quadrilateral with its midpoints connected to form an inscribed parallelogram.
•

Reflection Symmetry: Line symmetry occurs in specific quadrilaterals. An isosceles trapezoid has one line of symmetry (joining midpoints of parallel sides), while a rhombus has two (its diagonals).

📐Formulae

∠A+∠B+∠C+∠D=360∘\angle A + \angle B + \angle C + \angle D = 360^\circ

Mid-point length: PQ=12 (Base side length)\text{Mid-point length: } PQ = \frac{1}{2} \text{ (Base side length)}

Area of Parallelogram=base×height\text{Area of Parallelogram} = \text{base} \times \text{height}

Area of Rhombus=12×d1×d2\text{Area of Rhombus} = \frac{1}{2} \times d_1 \times d_2

💡Examples

Problem 1:

In a quadrilateral ABCDABCD, the angles are in the ratio 3:5:9:133:5:9:13. Find all the angles of the quadrilateral.

Solution:

Let the angles be 3x3x, 5x5x, 9x9x, and 13x13x. By the angle sum property of a quadrilateral: 3x+5x+9x+13x=360∘3x + 5x + 9x + 13x = 360^\circ 30x=360∘30x = 360^\circ x=360∘30=12∘x = \frac{360^\circ}{30} = 12^\circ The angles are: 3×12∘=36∘3 \times 12^\circ = 36^\circ 5×12∘=60∘5 \times 12^\circ = 60^\circ 9×12∘=108∘9 \times 12^\circ = 108^\circ 13×12∘=156∘13 \times 12^\circ = 156^\circ

Explanation:

We use the property that the sum of all interior angles in any quadrilateral is 360∘360^\circ to solve for the unknown variable xx.

Problem 2:

Prove that the diagonals of a parallelogram divide it into four triangles of equal area and demonstrate central symmetry.

Solution:

Let ABCDABCD be a parallelogram with diagonals ACAC and BDBD intersecting at OO. Since diagonals bisect each other, AO=OCAO = OC and BO=ODBO = OD. In △AOB\triangle AOB and △COD\triangle COD:

  1. AO=COAO = CO (Diagonals bisect each other)
  2. BO=DOBO = DO (Diagonals bisect each other)
  3. ∠AOB=∠COD\angle AOB = \angle COD (Vertically opposite angles) △AOB≅△COD\triangle AOB \cong \triangle COD (by SAS congruence). Similarly, △BOC≅△DOA\triangle BOC \cong \triangle DOA. The point OO acts as the center of symmetry because every point PP on side ABAB has a corresponding point P′P' on side CDCD such that OO is the midpoint of PP′PP'.

Explanation:

The bisection of diagonals is the geometric basis for central symmetry in parallelograms. A 180∘180^\circ rotation about OO maps the parallelogram onto itself.

Problem 3:

A quadrilateral is formed by joining the mid-points of the sides of a rectangle ABCDABCD. Identify the type of quadrilateral formed.

Solution:

Let P,Q,R,SP, Q, R, S be the mid-points of sides AB,BC,CD,DAAB, BC, CD, DA of rectangle ABCDABCD. Join ACAC. In △ABC\triangle ABC, PQ∥ACPQ \parallel AC and PQ=12ACPQ = \frac{1}{2}AC (Mid-point theorem). In △ADC\triangle ADC, SR∥ACSR \parallel AC and SR=12ACSR = \frac{1}{2}AC. Thus, PQ∥SRPQ \parallel SR and PQ=SRPQ = SR, making PQRSPQRS a parallelogram. Since ABCDABCD is a rectangle, its diagonals are equal (AC=BDAC = BD). Using the mid-point theorem for BDBD, we find PS=QR=12BDPS = QR = \frac{1}{2}BD. Since AC=BDAC = BD, then PQ=QR=RS=SPPQ = QR = RS = SP. A parallelogram with all sides equal is a rhombus.

Explanation:

The mid-point theorem is applied to show that the resulting figure is a parallelogram, and the specific properties of the parent shape (rectangle diagonals being equal) determine that the result is a rhombus.

Problem 4:

Show that the quadrilateral formed by the internal angle bisectors of a parallelogram ABCDABCD is a rectangle.

Parallelogram with interior angle bisectors intersecting to form a small rectangle PQRS.

Solution:

In △ASD,∠DAS+∠ADS=12∠A+12∠D\text{In } \triangle ASD, \angle DAS + \angle ADS = \frac{1}{2} \angle A + \frac{1}{2} \angle D ∵∠A+∠D=180∘ (consecutive interior angles),\because \angle A + \angle D = 180^\circ \text{ (consecutive interior angles),} ∠DAS+∠ADS=12(180∘)=90∘\angle DAS + \angle ADS = \frac{1}{2}(180^\circ) = 90^\circ By triangle angle sum, ∠ASD=180∘−90∘=90∘\text{By triangle angle sum, } \angle ASD = 180^\circ - 90^\circ = 90^\circ

Similarly, all interior angles of the quadrilateral PQRSPQRS are 90∘90^\circ, making it a rectangle.

Explanation:

Since the sum of adjacent angles in a parallelogram is 180∘180^\circ, the sum of their halves is 90∘90^\circ. The angle bisectors form triangles where the third angle must be 90∘90^\circ. By symmetry, this applies to all four intersections.

Problem 5:

In △ABC\triangle ABC, DD, EE, and FF are the mid-points of sides ABAB, BCBC, and CACA respectively. Show that △ABC\triangle ABC is divided into four congruent triangles by joining DD, EE, and FF.

Triangle ABC with midpoints D, E, F joined to form four smaller triangles.

Solution:

By the Mid-point Theorem: DF∥BC and DF=12BC=BE=ECDF \parallel BC \text{ and } DF = \frac{1}{2} BC = BE = EC DE∥AC and DE=12AC=AF=FCDE \parallel AC \text{ and } DE = \frac{1}{2} AC = AF = FC EF∥AB and EF=12AB=AD=DBEF \parallel AB \text{ and } EF = \frac{1}{2} AB = AD = DB In quadrilateral ADEFADEF, DE∥AFDE \parallel AF and EF∥ADEF \parallel AD, so ADEFADEF is a parallelogram. The diagonal DFDF divides it into two congruent triangles: △ADF≅△FED\triangle ADF \cong \triangle FED. Similarly, △BDE≅△FED\triangle BDE \cong \triangle FED and △CEF≅△FED\triangle CEF \cong \triangle FED. Thus, all four triangles are congruent.

Explanation:

Joining midpoints creates four smaller triangles. Using the Mid-point Theorem, we identify three parallelograms within the large triangle. Since a diagonal divides a parallelogram into two congruent triangles, all four small triangles are proved congruent to the central one.