Subject: Compulsory Maths

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.

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.

• 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:

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.

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

so, diagonal of a rectangle bisects each other.

so, Opposite side of a rectangle is equal

Solution:

Steps of Construction:

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:

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:

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:

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:

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:

(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:

(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:

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:

(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.

Then, ABCD is the required rectangle.

Solution:

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:

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.