Table of Contents

## Areas and Perimeters of Plane Figures:

### Mensuration:

Mensuration is that branch of practical Mathematics that deals with the length of lines, areas of surfaces, and volumes of solids. Sound knowledge of it is essential for engineering students in designing, constructing roads and bridges, preparing estimates and plans, etc.

*Mensuration may be divided further into two branches:*

(1) ** Plane Mensuration- **It is that branch of mensuration that deals with finding out the lengths of sides, perimeters, and areas of plane figures of different shapes.

(2) ** Solid Mensuration-** It is that branch of mensuration that deals with finding out the surface, areas, and volumes of solid objects of different shapes.

### Units of Length, Area, and Volume:

If we take any unit of length and make a square on it, then the area of this square will be that unit of an area corresponding to the given unit of length. If a cube is described as taking this square as a base, the volume of this cube will be that unit of volume corresponding to the given unit of length. For example, the square meter and the cubic meter correspond to the linear meter.

### Rectangle:

It is a plane figure bounded by four sides, having the opposite sides equal and all four angles as right angles.

Thus in the above figure, AB and BC are taken as the length and breadth respectively of the rectangle ABCD. The length and breadth taken together are called the dimensions of the rectangle.

AC and BD are called diagonals and the sum of the four sides of it is called its perimeter.

(1) ** Area of a Rectangle-** Let ABCD be the rectangle whose length is 5m and breadth is 3m.

Divide AB into 5 equal parts and BC into 2 equal parts, so that each of these will represent a foot. Through the points of division of AB, draw straight lines parallel to BC and through those of BC parallel to AB. Now this rectangle is divided into 5 columns, each containing 3 squares and each square represents a square foot.

∴ Number of squares in the rectangle ABCD = 3 x 5 = 15 wherein one square represents 1 sq. meter.

∴ 15 squares represent 15 sq. ms.

∴ Area of the rectangle ABCD = 15 sq. ms. or 15m^{2}

Hence Area of any rectangle = Length x Breadth |

(2) *Perimeter-*

The perimeter of the rectangle ABCD = AB + BC + CD + DA = (5 + 3 + 5 + 3) m = 2 (5 + 3) m

Hence Perimeter of any rectangle = 2 (Length + Breadth) |

### Square:

A square is a rectangle with all sides equal.

In the case of a square, length is equal to breadth. Let the length of the side be l.

∴ Area of the square = l x l = l^{2} = (side)^{2} |

Perimeter of the square = 2 (l + l) = 4l

Hence Perimeter of any square = 4 (side) |

**Corollary:** Area of a square in terms of its diagonal.

From the right-angled △ ABC,AC ^{2} = AB^{2} + BC^{2} = l^{2} + l^{2} = 2l^{2}(Diagonal) ^{2} = 2 (side)^{2}Let A be the area of square ABCD ∴ A = l ^{2}But we know that (Diagonal) ^{2} = 2l^{2}∴ 2A = (Diagonal) ^{2}∴ A = (Diagonal) ^{2}/2Hence the area of any square = (Diagonal)^{2}/2 |

### Triangle:

A plane figure bounded by three straight lines is called a triangle.

Any angular point of the triangle may be taken as its ** vertex** and the side opposite to that vertex is called the

**of the triangle.**

*base*The perpendicular drawn from any vertex upon the opposite side of the triangle is called its ** height** or

**.**

*altitude*It has three sides and three angles. Triangles are classified based on the length of their sides and the size of their angles.

*There are three types of triangles based on their sides:*

(1) ** Scalene Triangle:** A scalene triangle is a triangle in which all three sides have different lengths.

(2) ** Isosceles Triangle:** An isosceles triangle is a triangle in which two sides have the same length, and the third side has a different length.

(3) ** Equilateral Triangle:** An equilateral triangle is a triangle in which all three sides have the same length.

*There are three types of triangles based on their angles:*

(1) ** Acute Triangle:** An acute triangle is a triangle in which all three angles are acute angles (less than 90 degrees).

(2) ** Obtuse Triangle:** An obtuse triangle is a triangle in which one angle is an obtuse angle (greater than 90 degrees).

(3) ** Right Triangle:** A right triangle is a triangle in which one angle is a right angle (exactly 90 degrees).

*Area of a Triangle having given its base and altitude:*

Let ABC be the triangle with BC as base and AD as altitude. Let BC and AD measure ‘ a‘ units and ‘h‘ units respectively.Construct a rectangle BCEF on the side BC and the same altitude. Now we know that a diagonal of a rectangle divides it into two equal triangles. ∴ Area of △ ABD = (1/2) Area of rectangle AFBD ……….(i)Also Area of △ ADC = (1/2) Area of rectangle ADCE ……….(ii)Adding (i) and (ii) we get, Area of △ ABD + Area of △ ADC = 1/2 (Area of rectangle AFBD + Area of rectangle ADCE)∴ Area of △ ABC = 1/2 (Area of rectangle BCEF) ⇒ Area of △ ABC = 1/2 (BC x BF)⇒ Area of △ ABC = 1/2 (BC x AD) [∵ BF = AD]⇒ Area of △ ABC = (1/2) ah sq. unitsHence Area of any triangle = Half the product of its base and altitude. |

**Corollary 1:** ** Right-angled Triangle-** A triangle having one angle equal to one right angle is called a right-angled triangle.

Let ABC be the right-angled triangle having right-angled at B. Here base BC = a and height BA = c. ∴ Area of right-angled △ ABC = (1/2) BC x BA = (1/2) a x c = (1/2) ac sq. unitsHence the Area of any right-angled triangle = Half the product of the sides containing the right angle. |

**Corollary 2:** When the sides containing the right angle are equal, it is called an Isosceles right-angled triangle.

In this isosceles right-angled triangle ACB, AC = BC or b = a ∴ Area of isosceles right-angled △ ACB = (1/2) BC x CA⇒ Area of isosceles right-angled △ ACB = (1/2) a x b = (1/2) a x a = (1/2) b x b [∵ a = b]⇒ Area of isosceles right-angled △ ACB = (1/2) a^{2} = (1/2) b^{2} sq. units Hence the area of an isosceles right-angled △ = (1/2) (Square of one of the sides containing the right angle). |

**Corollary 3:** Area of an equilateral triangle having given its side.

Let ABC be an equilateral triangle. Let each side be equal to ‘I‘ units of length.From A draw AD ⊥ BC. ∴ BD = DC = l/2 Now by applying Pythagoras’ theorem to the right-angled △ ADB, we get,(AB) ^{2} = (AD)^{2} + (DB)^{2}or l ^{2} = (AD)^{2} + (l/2)^{2}or (AD) ^{2} = l^{2} – (l/2)^{2} = l^{2} – l^{2}/4 = 3l^{2}/4Taking the square root of both sides we get, AD = (√3/2) l ∴ Area of △ ABC = (1/2) BC x AD⇒ Area of △ ABC = (1/2) l x (√3/2) l = (√3/4) l^{2}∴ Area of an equilateral triangle = (√3/4) (side)^{2} |

**Corollary 4:** Area of a triangle when its three sides are given.

**Area of the △ ABC = √[s (s – a) (s – b) (s -c )]**

Where ‘s’ stands for half the sum of three sides.

### Quadrilateral:

A plane figure bounded by four straight lines is called a quadrilateral.

The line joining the opposite corners is called it’s diagonal.

Thus ABCD is a quadrilateral with AC and BD as diagonals.

### Parallelogram:

It is a quadrilateral in which both the pairs of opposite sides are parallel and equal.

The diagonals are the straight lines joining its opposite corners. For example, AC and BD are the diagonals of the parallelogram ABCD.

The height or altitude of the parallelogram is the length of the perpendicular drawn from any angular point on the opposite side. This opposite side is called it’s base.

*Area of a parallelogram having given its base and height:*

Let ABCD be the parallelogram whose base is ‘b‘ and height is ‘h‘.From D and C draw DE and CF perpendiculars upon AB. Produce AB to meet CF in F. Now AD = BC ∠DEA = ∠CFB and ∠DAE = ∠CBF ∴ △s AED and CFB are ≡∴ Area of △ AED = Area of △ CFB ……….(i)Now Area of parallelogram ABCD = Area of △ AED + Area of DEBC⇒ Area of parallelogram ABCD = Area of △ DEBC + Area of △ CFB ……….[∵ of (i)]⇒ Area of parallelogram ABCD = Area of rectangle DEFC = DC X DE = b x h sq. units Hence, the area of any parallelogram = Product of base and height. |

### Rhombus:

If all the sides of the parallelogram are equal, but its angles are not right angles, the parallelogram is called a rhombus.

The diagonals of a rhombus are the right bisectors of each other.

*Area of a rhombus having been given its diagonals:*

Let ABCD be the rhombus with diagonals AC and BD bisecting each other at right angles at E. Let AC and BD measure d _{1} and d_{2} units of length.Area of the rhombus ABCD = Area of △ ABC + Area of △ ACD⇒ Area of the rhombus ABCD = (1/2) (AC x BE) + (1/2) (AC x DE) ⇒ Area of the rhombus ABCD = (1/2) (BE + ED) x AC = (1/2) BD x AC ⇒ Area of the rhombus ABCD = (1/2) d _{1} x d_{2} = (1/2) d_{1}d_{2} sq. unitsHence Area of any Rhombus = (1/2) x Product of its diagonals |

### Trapezium (or Trapezoid):

A quadrilateral is called a trapezium or trapezoid when its two opposite sides are parallel but unequal.

The perpendicular distance between the parallel sides is called the height of trapezium.

*Area of the trapezium having given the parallel sides and perpendicular distance between them:*

Let ABCD be the trapezium with parallel sides AB and DC. Draw CF perpendicular from C upon AB. ∴ CF is the perpendicular distance between the parallel sides. Join A and C. Let AB, DC, and CF measure a, b, and h units of length respectively. △s ABC and ADC have the same height.Area of Trapezium ABCD = Area of △ ABC + Area of △ ADC⇒ Area of Trapezium ABCD = (1/2) AB x CF + (1/2) DC x DE ⇒ Area of Trapezium ABCD = (1/2) (AB + DC) CF [∵ DE = CF] ⇒ Area of Trapezium ABCD = (1/2) (a + b) h sq. units Hence the area of any trapezium = (1/2) (Sum of its parallel sides) x (perpendicular distance between them). |

### Polygon:

A plane figure bounded by four or more than four straight lines is called a polygon. When all the sides and the angles of a polygon are equal. it is called a ** regular polygon**.

*Circum-radius and in-radius of a regular polygon:*

Let A, B, C, D, E, and F………..be the corners of a regular polygon of n sides. Let the bisectors of ∠BCD and ∠CDE meet at O. Join O with all the corners. The O is called the center of the circle which circumscribes the given regular polygon. The radius i.e. OA or OB or OC etc., is called the circum-radius and is denoted by R. From O draw OL ⊥ BC, then OL is called the in-radius of the circle which is inscribed in the given polygon. This in-radius is denoted by r. |

*Area of the n-sided regular polygon in terms of its side and in-radius:*

Let ABCD……… be the regular polygon of n sides. Let O be the center and ON be the in-radius of the inscribed circle. Let the side of the polygon and in-radius measure ‘ a‘ and ‘r‘ units of length respectively.By joining O with the ‘ n‘ corners of the regular n-sided polygon, the polygon is divided into n equal △s like OAB.∴ Area of polygon ABCD…….. = Sum of area of n equal △sor Area of the polygon ABCD…….. = n Area of △OAB ………(i)Now Area of △OAB = (1/2) (AB x ON) = (1/2) ar sq. unitsPutting the value of the area of △OAB in (i) we get,Area of Polygon ABCD……… = n x (1/2) ar sq. units = (1/2) nar sq. units Corollary: The perimeter of n-sided regular polygon = n x sideHere side = a ∴ Perimeter = na Area of polygon = (1/2) (na) r Hence the area of polygon = (1/2) (perimeter) x (inradius) sq, units |

*Area of the n-sided regular polygon in terms of its side and**the radius of circumscribed circle:*

Here we are given AB = a units Let OA = R units From the right-angled △ONAON ^{2} = OA^{2} – AN^{2}⇒ ON ^{2} = R^{2} – (a/2)^{2}⇒ ON ^{2} = R^{2} – (a^{2}/4)or ON = √[R ^{2} – (a^{2}/4)]Area of polygon ABCD…….. = n. Area of △OAB⇒ Area of polygon ABCD…….. = (1/2) n x ON x AB ⇒ Area of polygon ABCD…….. = (1/2) na √[R ^{2} – (a^{2}/4)] sq. unitsHence the area of a regular polygon = (1/2) perimeter x √[(circum-radius)^{2} – (side/2)^{2}] |

**Corollary:** Also from Trigonometry we know that:

(1) Area of an n-sided regular polygon in terms of its side = [n (side)^{2}/4] cot π/n sq. units

(2) Area of an n-sided regular polygon in terms of its inradius = (n x r^{2}) tan π/n sq. units

(3) Area of an n-sided regular polygon in terms of its circum-radius = [(1/2) nR^{2} sin 2π/n

*Note:*

- A polygon with 5 sides is called a pentagon.
- A polygon with 6 sides is called a hexagon.
- A polygon with 7 sides is called a heptagon.
- A polygon with 8 sides is called an octagon.
- A polygon with 9 sides is called a nonagon.
- A polygon with 10 sides is called a decagon.
- A polygon with 11 sides is called a hendecagon (also undecagon or endecagon)
- A polygon with 12 sides is called a dodecagon.
- A polygon with 13 sides is called a tridecagon.
- A polygon with 14 sides is called a tetradecagon.
- A polygon with 15 sides is called a pentadecagon or quindecagon.

### Circle:

A circle is a plane figure bounded by a curved line, all the points of which are equidistant from a fixed point within the figure.

The bounding curved line is called the ** circumference of the circle** and the fixed point is called the

**.**

*center of this circle*The ** radius** is the straight line joining the center with any point on the circumference.

The diameter of the circle is the straight line passing through the center and terminated both ways by the circumference. Thus in the circle ABC, OB is the radius, AC is the diameter and O is the center.

*Circumference of the circle or Perimeter of the circle**having been given its radius:*

Let the radius of the circle with center ‘O‘ be ‘r‘ units of length and AC be the diameter.Since the circumference of a circle has a constant ratio with the diameter of the same circle. ∴ Circumference/Diameter = constant But the value of the constant ratio is denoted by π. ∴ Circumference/2r = π or Circumference = 2πr Hence the circumference of a circle = 2 x π x radius units. |

*Area of a circle having been given its radius:*

Let the radius of the circle measure ‘r‘ units of length and center be ‘C‘.Let AB be one of the sides of the n-sided regular polygon inscribed in the circle. From C draw CL ⊥ AB. Now, if the number of sides of the polygon be increased indefinitely, the perimeter of the polygon will coincide with the circumference of the circle and the area of the polygons will be the area of the given circle. Also, the radius of the inscribed circle will be the Radius of the given circle. Now we know that Area of a regular polygon = (1/2) (perimeter) x (radius of the inscribed circle) ∴ Area of the circle = (1/2) (circumference of the circle) x (radius of the circle) ⇒ Area of the circle = (1/2) x 2πr x r ⇒ Area of the circle = πr ^{2} sq. unitsHence the area of any circle = π (radius)^{2}. |

### Sector:

It is the portion of a circle enclosed by two radii and the arc intercepted between them.

The angle contained by two radii is called the angle of the sector.

Thus in the circle ABD, CAB is the sector and ∠ACB is the angle of the sector.

*Area of a sector in terms of the radius of the circle and the angle of the sector:*

Let CAB be the sector of the circle whose radius is ‘r‘ and the center is ‘C‘.Let the angle of the sector be x°.Produce BC to meet the circumference in D. ∴ ∠BCD = 180° We know that the areas of the sectors are proportional to their angles. or Area of the sector CAB = (∠ACB/∠DCB) x Area of semi-circle BAD ⇒ Area of the sector CAB = (x°/180°) x (1/2) πr ^{2} = (x°/360°) πr^{2} sq. unitsHence the area of any sector = (Angle of the Sector/360°) x [Area of the circle whose sector it is] |

** Corollary 1:** Also the area of the sector = (θ/2π) x πr

^{2}= (1/2) θr

^{2}where θ is the angle of the sector measured in radius and r is the radius of the circle whose sector it is.

** Corollary 2:** Also from trigonometry, we know that l/r = θ where l is the arc of the sector.

∴ l = rθ

Area of the sector = (1/2) r^{2}θ = (1/2) (rθ) (r) = (1/2) lr sq. units

### Segment of the Circle:

It is the portion of a circle bounded by a chord and the portion of the circumference which is cut by the chord.

Thus in the circle AEBF, AEB is the segment of the circle and DE is the height of the segment.

Chord AB divides the circle into two segments. Segment AEB is the minor segment while segment AFB is the major segment.

*Area of a segment:*

Area of segment AEB = Area of sector CAEB – Area of △ CABArea of segment AFB = Area of sector CAFB + Area of △ CABAlso the area of a segment of a circle = (4/3) (height) √[(1/4) (chord) ^{2} + (2/5) (height)^{2}] an area of segment AEB of the circle AEBF = (4/3) (DE) √[(1/4) (AB)For example-^{2} + (2/5) (DE)^{2}] |

*Length of an arc:*

Length of an arc AEB = (8 x AE – AB)/3, where AE is the chord of semi-arc AE.

### Similar Figures:

Figures are said to be similar when they are of the same shape but of different sizes.

*Properties of similar figures:*

Similar figures have several properties that distinguish them from other types of figures. Some of the key properties of similar figures include:

(1) ** Corresponding angles:** Similar figures have corresponding angles that are equal. This means that if two figures are similar, then the angles of one figure will have the same measures as the corresponding angles of the other figure.

(2) ** Corresponding sides:** Similar figures have corresponding sides that are proportional. This means that if two figures are similar, then the ratio of the lengths of any two corresponding sides will be the same for both figures.

(3) ** Scale factor:** The ratio of the lengths of corresponding sides of similar figures is called the scale factor. This is the same for all pairs of corresponding sides.

(4) ** Area:** The ratio of the areas of similar figures is equal to the square of the scale factor. This means that if the scale factor is k, then the ratio of the areas of the two similar figures is k².

(5) ** Perimeter:** The ratio of the perimeters of similar figures is also equal to the scale factor. This means that if the scale factor is k, then the ratio of the perimeters of the two similar figures is also k.

These properties are important for understanding the relationships between similar figures and for solving problems involving scaling, proportions, and geometric transformations.

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