Subject: Compulsory Maths

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.

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:

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:

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.

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.

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.

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

Diagonal created in a rectangle are also congruent.

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.

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.

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.

- 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.

In the figure, ABCD is a rectangle.If \(\angle\)BPC=124^{o}, Calculate \(\angle\)BAP and \(\angle\)ADP.

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+124^{o}=180^{o}

2x=180^{o}-124^{o}=56^{o}

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

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

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

\(\angle\)BAP=180^{o}-118^{o}=62^{o}

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

So,\(\angle\)CBP=\(\angle\)ADP=28^{o}

In the figure, QRST is a rectangle.If QS=15 cm, find TR, QP, PS, TP and PR.

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

Find the value of x from the given figure.

solution:

ABCD is a parallelogram.

Here,\(\angle\)ADC=70^{o}

\(\angle\)DAB=x=?

Now,

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

or, x+70^{o}=180^{o}

or, x=180^{o}-70^{o}

\(\therefore\) x =110^{o}

Find the value of diagonal PR from the given figure.

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

Find the value of x and y from the given figure.

Solution:

PQRS is a square in which diagonal are PR and QS. Angle of diagonal x^{o} and y^{o}.

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

So, x= y=90^{o}

Find the value of x and y from the given figure.

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

Find the value of x, y and z from the given figure.

Solution:

Here,

Given,

\(\angle\)EHG = 90^{o}

\(\angle\)HEF = x^{o}

\(\angle\)EFG = y^{o}

\(\angle\)FGH = z^{o}

Now,

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

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

Construct Rectangle having length(l)=18 cm and breadth(b)=9 cm and examine it.

Solution:

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

So, All angle are equal and are 90^{o}of 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

Construct a quadrilateral ABCD in which AB = 4.8 cm, BC = 4.3 cm, CD = 3.6 cm, AD = 4.2 cm and diagonal AC = 6 cm.

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.

Construct a quadrilateral ABCD in which AB = 4 cm BC = 3.8 cm, AD = 3 cm, diagonal AC = 5 cm and diagonal BD = 4.6 cm.

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.

Construct a quadrilateral ABCD in which AB = 3.6 cm, ∠ABC = 80°, BC = 4 cm, ∠BAD = 120° and AD = 5 cm.

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.

Construct a quadrilateral PQRS in which PQ = 4.5 cm ∠PQR = 120°, QR = 3.8 cm, ∠QRS = 100° and ∠QPS = 60°.

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.

Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4cm, CD = 4.5 cm, AD = 5cm and ∠B = 80°.

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.

Construct a parallelogram ABCD in which AB = 6 cm, BC = 4.5 cm and diagonal AC = 6.8 cm.

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.

Construct a parallelogram, one of whose sides is 5.2 cm and whose diagonals are 6 cm and 6.4 cm.

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.

Construct a parallelogram whose diagonals are 5.4 cm and 6.2 cm and an angle between them is 70°.

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.

Construct a rectangle ABCD in which side BC = 5 cm and diagonal BD = 6.2 cm.

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.

(vii) Join AB and AD.

Then, ABCD is the required rectangle.

Construct a square ABCD, each of whose diagonals is 5.2 cm.

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.

Construct a rhombus with side 4.2 cm and one of its angles equal to 65°.

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.

(iv) Join AD.

Then, ABCD is the required rhombus.

© 2019-20 Kullabs. All Rights Reserved.