Introduction

 

What is the Perimeter of a Triangle?

 

The perimeter of a triangle is the total length of its boundary, calculated by adding up the lengths of its three sides. It’s a simple yet vital concept in mensuration and geometry that lays the foundation for more advanced topics.

 

Introduction

 

The concept of perimeter is fundamental in geometry and mensuration. Whether you are planning a DIY project, fencing your garden, or helping your child with math homework, understanding of how to calculate the perimeter is an essential skill. Here, in this article, we will break down the process of finding the perimeter, cover formulas for various shapes, and provide practical examples to help you get a solid grasp on this concept.

 

Pythagoras (who died about 500 BC) was a Greek philosopher. He was a geometer and a great mystic, who with his followers, seems to have been the first to take mathematics seriously as a study in its own right as opposed to being a collection of formulae for practical calculation. The Pythagoreans are credited with the discovery of the well-known “Pythagoras Theorem” on right-angled triangles.

 

Theorem is a geometrical statement which is true and can be proved by using known facts and axioms. Generally, we prove geometrical theorems by two methods, one method is experimental verification and another method is theoretical proof.

 

Some of the triangle theorems and their experimental verification and/or theoretical proofs are given below:

 

A closed bounded geometrical figure with three or more sides is called a polygon. There are specific names of the polygon according to the number of sides in the polygon. They are as follows: 

Vector Geometry: We can study different properties and relations relating to geometry with the help of vectors. Such a study is known as the Vector Geometry.

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Here are 12 geometrical theorems and their proofs by vector method.

Tangents of a Circle

A line which is drawn from an external point of a circle and meets the circle at one and only point on the circumference is called a tangent of a circle. 

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The point A where tangent touches the circle is called the point of contact.

In the figure, TAN is a line which meets the circle at a point A. So TAN is a tangent and the point A is point of contact.

In the same figure, a line TB cuts the circle at two points D and E. So the line TB is called a secant.

 

Cyclic Quadrilateral

 

A quadrilateral having its all four vertices on the circumference of a circle is called a cyclic quadrilateral.

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ABCD, in the given figure, is a cyclic quadrilateral.

 

Concyclic Points

 

The points lying on the circumference of the circle are called concyclic points.

In the given circle, A, B, C, D are concyclic points.

 

 

Theorems on Cyclic Quadrilateral and Their Proofs:

 

THEOREM 1:

The sum of opposite angles of a cyclic quadrilateral is always 180°.

 

PROOFS:

Given: PQRS is a cyclic quadrilateral.

To Prove:    (i) ∠QPS + ∠QRS = 180°

(ii) ∠PQR + ∠PSR = 180°

Proof:

Statements                         Reasons

1.     ∠QPS = ½ Arc QRS -----> Inscribed angle is equal to the half of its opposite arc.

2.     ∠QRS = ½ Arc QPS -----> Inscribed angle is equal to half of its opposite arc.

3.     ∠QPS + ∠QRS = ½ Arc QRS + ½ Arc QPS -----> Adding statements 1 and 2.

or, ∠QPS + ∠QRS = ½ Arc (QRS + QPS)

or, ∠QPS + ∠QRS = ½ Circle PQRS

or, ∠QPS + ∠QRS = ½ 360°

or, ∠QPS + ∠QRS = 180°

4.     ∠PQR + ∠PSR = 180° -----> Same as above.

 

Hence proved.

 

 

THEOREM 2:

In a cyclic quadrilateral, the exterior angle is equal to its opposite interior angle.

 

PROOFS:

Given: ABCD is a cyclic quadrilateral with exterior angle ∠DCE.

To Prove: ∠DCE = ∠BAD

 

Proof:

Statements                         Reasons

1.     ∠DCE + ∠BCD = 180° -----> Opposite angles of a cyclic quadrilateral are supplementary.

2.     ∠BAD + ∠BCD = 180° -----> Linear pair of angles.

3.     ∠DCE + ∠BCD = ∠BAD + ∠BCD -----> From statements 1 and 2.

4.     ∠DCE = ∠BAD -----> Cancelling common angle ∠BCD.

 

Hence proved.

 

Problems on Cyclic Quadrilaterals

 

Worked Out Examples

 

Example 1: Find ∠COD in the given figure.

Solution: From the figure,

110° + ∠ODC = 180° [Opposite angles of a cyclic quadrilateral]

or,     ∠ODC = 180° – 110°

or,     ∠ODC = 70°

 

∠ODC = ∠OCD = 70° [Being OC = OD, radii of same circle]

 

∠COD + ∠ODC + ∠OCD = 180° [Sum of angles of DCOD]

or,     ∠COD + 70° + 70° = 180°

or,     ∠COD + 140° = 180°

or,     ∠COD = 180° – 140°

or,     ∠COD = 40°  Ans.

 

 

Example 2: Calculate the value of x and y in the given figure.

Solution: From the figure,

x = ½ ∠BOD [Inscribed angle is half of central angle]

or,     x = ½ 160°

or,     x = 80°

 

y + x = 180° [Opposite angles of a cyclic quadrilateral]

or,     y + 80° = 180°

or,     y = 180° – 80°

or,     y = 100°

 

Hence, x = 80° and y = 100°  Ans.

 

Example 3: Find the angle x in the given figure.

Solution: From the figure,

∠CAD = 56° [Inscribed angle on same arc CD]

∠BAD = 96° [Exterior angle of a cyclic quadrilateral]

 

Now,

         x = ∠BAD – ∠CAD

            = 96° – 56°

            = 40°  Ans.

 

 

Example 4: In the adjoining figure, ABCD is a cyclic quadrilateral, side CD is produced to the point E, where BC = DE. If CA bisects ∠BCD prove that DACE is an isosceles triangle.

 

Solution: Here,

Given: ABCD is a cyclic quadrilateral. BC = DE and ∠BCA = ∠ACD.

To Prove: DACE is an isosceles triangle.

 

Proof:

     Statements                Reasons

1.     ∠BCA = ∠ACD -----> Given.

2.     Arc AB = arc AD -----> Arcs subtended by equal inscribed angles.

3.     AB =AD ------> Equal chords cut off equal arcs in a circle.

4.     In DABC and DADE

i.       AB = AD (S) -----> From statement 3

ii.    ∠ABC = ∠ADE (A) -----> Exterior angle of a cyclic quadrilateral

iii. BC = DE (S) -----> Given.

5.     DABC ≅ DADE -----> By SAS axiom

6.     AC = AE -----> Corresponding sides of ≅ triangles.

7.     ACE is an isosceles triangle -----> Being AC = AE.

 

Hence proved.

 

Example 5: In the given figure, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = SR.

 

Solution: Here,

Given: PQRS is a cyclic quadrilateral. ∠RPS = ∠SPT.

To Prove: SQ = SR.

Proof:

     Statements               Reasons

1.     ∠SPT = ∠RPS -----> From given.

2.     ∠SPT = ∠QRS -----> Exterior angle of cyclic quadrilateral PQRS.

3.     ∠SQR = ∠RPS -----> Inscribed angle on same arc  RS.

4.     ∠SQR = ∠QRS -----> From statements 1, 2 and 3.

5.     SQ = SR -----> Being base angles of DSQR equal, statement 4.

 

Hence proved.

 

 

If you have any question or problems regarding the Cyclic Quadrilaterals, you can ask here, in the comment section below.

 

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