Quadrilaterals

Subject: Compulsory Maths

Find Your Query

Overview

A quadrilateral is a polygon with four edges (or sides) and four vertices or corners. Sometimes, the term quadrangle is used, by analogy with triangle, and sometimes tetragon for consistency with pentagon (5-sided), hexagon (6-sided) and so on.

Quadrilaterals

A closed plane figure formed by four line segments is a quadrilateral. A quadrilateral is also known as a polygon with four sides and four vertices or corners.

.

A quadrilateral ABCD has

  • four sides: AB, BC, CD and DA
  • four angles:∠A, ∠B,∠C and ∠D

There are many kinds of quadrilaterals. Such as: 

1. Parallelogram

Quadrilaterals having opposite sides parallel is known as a parallelogram.

In the figure AB ⁄⁄ CD and AD ⁄⁄ BC. So, ABCD is a parallelogram.

.

Theorems with parallelogram:

Theorem 1

The opposite sides of a parallelogram are congruent.

Verification:

Draw three parallelograms of different sizes as shown below:

.

Scholarships after +2 Abroad Studies Opportunities

Measure the sides and complete the table below:

Figure WZ XY Result WX ZY Result
(i)     WZ=XY     WX = ZY
(ii)            
(iii)            

Conclusion: Opposite sides of a parallelogram are equal.

Theorem 2

The opposite angles of a parallelogram are congruent.

Verification:

Draw three parallelograms of different sizes.

.

Measure the opposite angles and complete the table below:

Figure ∠W ∠Y Result ∠X ∠Z Result
(i)     ∠W =∠Y     ∠X =∠Z
(ii)            
(iii)            

Conclusion: The opposite angles of a parallelogram are congruent.

Theorem 3

The diagonals of a parallelogram bisect each other.

Verification:

Draw three parallelograms of different sizes. Join the diagonals WY and XZ.

.

Measure the segments of the diagonals and complete the table below:

Figure WO YO Result XO ZO Result
(i)     WO = YO     XO =ZO
(ii)            
(iii)            

Conclusion: Diagonals of the parallelogram bisect each other.

 2. Rectangle

The rectangle is a parallelogram with all angles 90o. Opposite sides are parallel and of equal length. It is also known as an equiangular parallelogram.

Diagonal created in a rectangle are also congruent.

.

Theorem

The diagonals of a rectangle are congruent.

Verification:

Draw three rectangles of different sizes. Join the diagonals WY and XZ.

.

Measure the diagonals WY and XZ with the ruler and complete the following table.

Figure WX XZ Result
(i)     WX = XZ
(ii)      
(iii)      

Conclusion: The diagonals of the rectangle are congruent.

 3. Square

Square is also a parallelogram with all sides and angles equal. It is also known as an equilateral and equiangular parallelogram. In another word, a square is a rectangle having adjacent sides equal. The diagonal of square bisects each other at right angles.

Theorem

The diagonals of a square bisect each other at right angles.

Verification:

Draw three squares of different sizes. Join the diagonals WY and XZ which intersect at O. Since a square is a parallelogram, the diagonals bisect each other i.e WO =YO and XO = ZO.

.

Measure the angles between the diagonals and complete the following table.

Figure ∠WOX ∠YOZ ∠WOZ ∠XOY Result
(i)         ∠WOX =∠YOZ =∠WOZ =∠XOY = 90°
(ii)          
(iii)          

Conclusion: The diagonals of a square bisect each other at right angles.

Things to remember
  • Quadrilateral just means 'four sides' (quad means four, lateral means side).
  • Quadrilateral are simple (not self-intersecting) or complex (self- intersecting) and also called crossed.
  • It includes every relationship which established among the people.
  • There can be more than one community in a society. Community smaller than society.
  • It is a network of social relationships which cannot see or touched.
  • common interests and common objectives are not necessary for society.
Questions and Answers

Solution:

Here, BP=CP ( the diagonal of a rectangle are equal and also bisect each other.)

In \(\triangle\)BPC, \(\angle\)PBC=x(say)

Then, x+x+124o=180o

2x=180o-124o=56o

x=\(\frac{56}{2}\)=28o

In \(\triangle\)ABC,\(\angle\)ABC+\(\angle\)ACB+\(\angle\)BAC=180o

90o+28o+\(\angle\)BAP=180o ( \(\therefore\)\(\angle\)BAC=\(\angle\)BAP)

\(\angle\)BAP=180o-118o=62o

Also,\(\angle\)CBD=\(\angle\)ADB (alternate angles)

So,\(\angle\)CBP=\(\angle\)ADP=28o

Solution:

Given,

QS = 15cm

TQ = SR

QR = TS

Now,

TR=QS =15cm, [diagonals of a rectangle are equal]

Again,

QP=\(\frac{1}{2}\) QS [half of diagonal]

=\(\frac{1}{2}\) \(\times\) 15cm [half of diagonals]

=7.5 cm

Also,

QP =PS = 7.5cm [Half of daigonal are equal]

And,

TP=\(\frac{1}{2}\) TR [half of diagonal]

=\(\frac{1}{2}\) \(\times\) 15cm [half of diagonals]

=7.5 cm

Also,

TP = PR = 7.5cm [Half of daigonal are equal]

\(\therefore\) TP = PR = QP = PS = 7.5cm

solution:

ABCD is a parallelogram.

Here,\(\angle\)ADC=70o

\(\angle\)DAB=x=?

Now,

\(\angle\)DAB+\(\angle\)ADC=180o [sum of co-interior angles of parallelogram is 180]

or, x+70o=180o

or, x=180o-70o

\(\therefore\) x =110o

Solution:

PQRS is a square which has diagonal QS=5 cm ,

We know that, diagonal of a square are equal. So, QS=PR

\(\therefore\) PR = 5 cm

Solution:

PQRS is a square in which diagonal are PR and QS. Angle of diagonal xo and yo.

We know that, the diagonals of a square bisects each other perpendicularly.

So, x= y=90o

Solution:

Given,

AB = 8cm, AD = 6cm, CD = xcm and BC = ycm

Now,

AB = CD and AD = BC [Oposite sides of rectangle are equal]

\(\therefore\) x = AB = 8cm and y = AD = 6cm

Solution:

Here,

Given,

\(\angle\)EHG = 90o

\(\angle\)HEF = xo

\(\angle\)EFG = yo

\(\angle\)FGH = zo

Now,

\(\angle\) HGF = \(\angle\)EFG = \(\angle\)FGH = 90o [Angles of rectangle are equal]

\(\therefore\) x = 90o, y = 90o and z = 90o

Solution:

Here, \(\angle\) A=\(\angle\)B=\(\angle\)C=\(angle\)=90o

So, All angle are equal and are 90oof rectangle.

Diagonal AC=BD=20.2 cm

So, Diagonal are equal in rectangle

AD=DC=10.1 cm and BO=CO=10 cm

so, diagonal of a rectangle bisects each other.

AB=CD=18 cm and BC=AD=9 cm

so, Opposite side of a rectangle is equal

 

Solution:

Steps of Construction:

Steps of Construction of Quadrilaterals
3

Step 1: Draw AB = 4.8 cm.

Step 2: With A as center and radius equal to 6 cm, draw an arc.

Step 3: With B as center and radius equal to 4.3 cm, draw another arc, cutting the previous arc at C.

Step4: Join BC.

Step 5: With A as center and radius equal to 4.2 cm, draw an arc.

Step 6: With C as center and radius equal to 3.6 cm, draw another arc, cutting the previous arc at D.

Step 7: Join AD and CD.

Then, ABCD is the required quadrilateral.

Solution:

Steps of Construction:

Steps of Construction of Quadrilaterals
3

Step 1: Draw AB = 4 cm.

Step 2: With A as center and radius equal to 5 cm, draw an arc.

Step 3: With B as center and radius equal to 3.8 cm, draw another arc, cutting the previous arc at C.

Step 4: Join BC.

Step 5: With A as center and radius equal to 3 cm, draw an arc.

Step 6: With B as center and radius equal to 4.6 cm draw another arc, cutting the previous arc at D.

Step 7: Join AD and CD.

Then, ABCD is the required quadrilateral.

Solution:

Steps of Construction:

Steps of Construction of Quadrilaterals

Step 1: Draw AB = 3.6 cm.

Step 2: Make ∠ABX = 80°.

Step 3: With B as center and radius equal to 4 cm, draw an arc, cutting BX at C.

Step 4: Make ∠BAY = 120°.

Step 5: With A as center and 5 cm as radius, draw an arc, cutting AY at D.

Step 6: Join CD.

Then, ABCD is the required quadrilateral.

Solution:

Steps of Construction:

Steps of Construction of Quadrilaterals

Step 1: Draw PQ = 4.5 cm.

Step 2: Make ∠PQX = 120°.

Step 3: With Q as center and radius 3.8 cm, draw an arc, cutting QX at R. Join QR.

Step 4: Make ∠QRY = 100°.

Step 5: ∠QPZ = 60° so that PZ and RY intersect each other at the point S.

Then, PQRS is the required quadrilateral.

solution:

Steps of Construction:

Steps of Construction of Quadrilaterals

Step 1: Draw AB = 3.8 cm.

Step 2: Make ∠ABX = 80°.

Step 3: From B, set off BC = 3.4 cm.

Step 4: With A as center and radius equal to 5 cm draw an arc.

Step 5: With C as center and radius equal to 4.5 cm, draw another arc, cutting the previous arc at D.

Step 5: Join AD and CD.

Then ABCD is the required quadrilateral.

solution:

Steps of Construction:

Steps of Construction of a Parallelogram

(i) Draw AB = 6 cm.

(ii) With A as center and radius 6.8 cm, draw an arc.

(iii) With B as center and radius 4.5 cm draw another arc, cutting the previous arc at C.

(iv) Join BC and AC.

(v) With A as center and radius 4.5 cm, draw an arc.

(vi) With C as center and radius 6 cm draw another arc, cutting the previously drawn arc at D.

(vii) Join DA and DC.

Then, ABCD is the required parallelogram.

Steps of Construction:

Steps of Construction of a Parallelogram

(i) Draw AB = 5.2 cm.

(ii) With A as center and radius 3.2 cm, draw an arc.

(iii) With B as center and radius 3 cm draw another arc, cutting the previous arc at O.

(iv) Join OA and OB.

(v) Produce AO to C such that OC = AO and produce BO to D such that OD = OB.

(vi) Join AD, BC and CD.

Then, ABCD is the required parallelogram.

Solution:

 Construct a Parallelogram



We know that the diagonals of a parallelogram bisect each other.
So, we may proceed according to the steps given below.

Steps of Construction:

(i) Draw AC = 5.4 cm.

(ii) Bisect AC at O.

(iii) Make ∠COX = 70° and produce XO to Y.

(iv) Set off OB = 1/2 (6.2) = 3.1 cm and OD = 1/2 (6.2) =3.1 cm as shown.

(v) Join AB, BC, CD and DA.


Then, ABCD is the required parallelogram.

Solution:

Steps of Construction:

Steps of Construction of Rectangle

(i) Draw BC = 5 cm.

(ii) Draw CX ⊥ BC.

(iii) With B as center and radius 6.2 cm draw an arc, cutting CX at D.

(iv) Join BD.

(v) With D as center and radius 5 cm, draw an arc.

(vi) With B as center and radius equal to CD draw another arc, cutting the previous arc at A.

(vii) Join AB and AD.

Then, ABCD is the required rectangle.

Solution:

Construction of Square

We know that the diagonals of a square bisect each other at right angles. So, we proceed according to the following steps.

Steps of Construction:

(i) Draw AC = 5.2 cm. (ii) Draw the right bisector XY of AC, meeting AC at O.

(iii) From O set off OB = 1/2 (5.2) = 2.6 cm along OY and OD = 2.6 cm along OX.

(iv) Join AB, BC, CD and DA.

Then, ABCD is the required square.

Solution:

Construction of Rhombus
Clearly, the adjacent angle = (180° - 65°) = 115°. So, we may proceed according to the steps given below.

Steps of Construction:

(i) Draw BC = 4.2 cm.

(ii) Make ∠CBX = 115° and ∠BCY = 65°.

(iii) Set off BA = 4.2 cm along BX and CD = 4.2 cm along CY.

(iv) Join AD.

Then, ABCD is the required rhombus.

Quiz

© 2019-20 Kullabs. All Rights Reserved.