# Math Blog

### Triangles with Equal Areas on the Same Base have Equal Corresponding..

Here we will prove that triangles with equal areas on the same base have equal corresponding altitudes (or are between the same parallels). Given: PQR and SQR are two triangles on the same base QR, and ar(∆PQR) = ar(∆SQC). Also, PN and SM are their corresponding altitudes.

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### Triangles on the Same Base & between Same Parallels are Equal in Area

Here we will prove that triangles on the same base and between the same parallels are equal in area. Given: PQR and SQR are two triangles on the same base QR and are between the same parallel lines QR and MN, i.e., P and S are on MN. To prove: ar(∆PQR) = ar(∆SQR)

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### Area of a Triangle is Half that of a Parallelogram on the Same Base

Here we will prove that the area of a triangle is half that of a parallelogram on the same base and between the same parallels. Given: PQRS is a parallelogram and PQM is a triangle with the same base PQ, and are between the same parallel lines PQ and SR.

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### Area of a Parallelogram is Equal to that of a Rectangle Between ......

Here we will prove that the area of a parallelogram is equal to that of a rectangle on the same base and of the same altitude, that is between the same parallel lines. Given: PQRS is a parallelogram and PQ MN is a rectangle on the same base PQ and between the same parallel

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### Parallelogram on the Same Base and Between the Same Parallel Lines

Here we will prove that parallelogram on the same base and between the same parallel lines are equal in area. Given: PQRS and PQMN are two parallelograms on the same base PQ and between same parallel lines PQ and SM. To prove: ar(parallelogram PQRS) = ar(parallelogram PQMN).

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### Diagonal of a Parallelogram Divides it into Two Triangles of EqualArea

Here we will prove that every diagonal of a parallelogram divides it into two triangles of equal area. Given: PQRS is a parallelogram in which PQ ∥ SR and SP ∥ RQ. PR is a diagonal of the parallelogram. To prove: ar(∆PSR) = ar(∆RQP). Proof: Statement 1. ∠SPR = ∠PRQ. 2. ∠SRP

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### Base and Height (Altitude) in a Triangle and a Parallelogram | Diagram

We will discuss here about the Base and height (altitude) in a triangle and a parallelogram. In ∆PQR, any side may be taken as the base. If QR is taken as the base then the perpendicular PM on QR is the corresponding altitude (height) of the triangle. In the parallelogram

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### Area of a Closed Figure |Measurement of Area |Area Axiom for Rectangle

We will discuss here about the area of a closed figure, measurement of area, area axiom for rectangle, area axiom for congruent figures and addition axiom for area. The measure of the reason bounded by a closed figure in a plane is called its area. In the following the areas

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### Bisectors of the Angles of a Parallelogram form a Rectangle | Diagram

Here we will prove that the bisectors of the angles of a parallelogram form a rectangle. Given: PQRS is a parallelogram in which PQ ∥ SR and SP ∥ RQ. The bisectors of ∠P, ∠Q, ∠R and ∠S are PJ, QK, RL and SM respectively which enclose the quadrilateral JKLM. To prove: JKLM is

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### Conditions for Classification of Quadrilaterals and Parallelograms

We will discuss here about Conditions for classification of quadrilaterals and parallelograms. On the basis of the above definitions, theorems and converse propositions we conclude the following. 1. A quadrilateral is a parallelogram if any one of the following holds.

### Diagonals of a Parallelogram are Equal & Intersect at Right Angles

Here we will prove that if in a parallelogram the diagonals are equal in length and intersect at right angles, the parallelogram will be a square. Given: PQRS is a parallelogram in which PQ ∥ SR, PS ∥ QR and diagonal PR ⊥diagonal QS. To prove: PQRS is a square, i.e., PQ

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### Diagonals of a Square are Equal in Length & they Meet at Right Angles

Here we will prove that in a square, the diagonals are equal in length and they meet at right angles. Given: PQRS is a square in which PQ = QR = RS = SP, and ∠QPS = ∠PQR = ∠QRS = ∠RSP = 90°. To prove: PR = QS and PR ⊥ QS Proof: Statement 1. In ∆SPQ and ∆RQP, (i) SP = QR

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### A Parallelogram, whose Diagonals are of Equal Length, is a Rectangle

Here we will prove that a parallelogram, whose diagonals are of equal length, is a rectangle. Given: PQRS is a parallelogram in which PQ ∥ SR, PS ∥ QR and PR = QS. To prove: PQRS is a parallelogram, i.e., in the parallelogram PQRS, one angle, say ∠QPS = 90°. Proof: In ∆PQR

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### In a Rectangle the Diagonals are of Equal Lengths | Proof | Diagram

Here we will prove that in a rectangle the diagonals are of equal lengths. Given: PQRS is rectangle in which PQ ∥ SR, PS ∥ QR and ∠PQR = ∠QRP = ∠RSP = ∠SPQ = 90°. To prove: The diagonals PR and QS are equal. Proof: Statement In ∆PQR and ∆RSP 1.∠QPR = ∠SRP 2. ∠QRP = ∠SPR

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### A Parallelogram whose Diagonals Intersect at Right Angles is a Rhombus

Here we will prove that a parallelogram, whose diagonals intersect at right angles, is a rhombus. Given: PQRS is a parallelogram in which PQ ∥ SR, PS ∥ QR and ∠QOR = ∠POQ = ∠ROS = ∠POS = 90°. To prove: PQRS is a rhombus, i.e., PQ = QR = RS = SP. Proof: In ∆PQR and ∆RSP,

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### A Rhombus is a Parallelogram whose Diagonals Meet at Right Angles

Here we will prove that a rhombus is a parallelogram whose diagonals meet at right angles. Given: PQRS is a rhombus. So, by definition, PQ = QR = RD = SP. Its diagonals PR and QS intersect at O. To prove: (i) PQRS is a parallelogram. (ii) ∠POQ = ∠QOR = ∠ROS = ∠SOP = 90°.

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### Pair of Opposite Sides of a Parallelogram are Equal and Parallel

Here we will discuss about one of the important geometrical property of parallelogram. A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel Given: PQRS is a quadrilateral in which PQ = SR and PQ ∥ SR. To prove: PQRS is a parallelogram.

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### A Quadrilateral is a Parallelogram if its Diagonals Bisect each Other

Here we will discuss about a quadrilateral is a parallelogram if its diagonals bisect each other. Given: PQRS is a quadrilateral whose diagonals PR and QS bisect each other at O, i.e., OP = OR and OQ = OS. To prove: PQRS is a parallelogram. Proof: In ∆OPQ and ∆ORS, OP = OR

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### Diagonals of a Parallelogram Bisect each Other | Diagonals Bisect each

Here we will discuss about the diagonals of a parallelogram bisect each other. In a parallelogram, diagonals bisect each other and each diagonal bisects the parallelogram into two congruent triangles. Given: PQRS is a parallelogram in which PQ ∥ SR and PS ∥ QR. Its diagonals

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### Opposite Angles of a Parallelogram are Equal | Related Solved Examples

Here we will discuss about the opposite angles of a parallelogram are equal. In a parallelogram, each pair of opposite angles are equal. Given: PQRS is a parallelogram in which PQ ∥ SR and QR ∥ PS To prove: ∠P = ∠R and ∠Q = ∠S Construction: Join PR and QS. Proof: Statement:

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### Opposite Sides of a Parallelogram are Equal | Solved Examples

Here we will discuss about the opposite sides of a parallelogram are equal in length. In a parallelogram, each pair of opposite sides are of equal length. Given: PQRS is a parallelogram in which PQ ∥ SR and QR ∥ PS. To prove: PQ = SR and PS = QR. Construction: Join PR

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### Concept of Parallelogram |Quadrilateral| Rectangle| Rhombus| Trapezium

Here we will discuss about the concept of parallelogram. Quadrilateral: A rectilinear figure enclosed by four line segments is called a quadrilateral. In the adjoining figures, we have two quadrilaterals PQRS, each enclosed by four line segments PQ, QR, RS and SP which

### What is Rectilinear Figure? | What is Diagonal of a Polygon? | Polygon

What is rectilinear figure? A plane figure whose boundaries are line segments is called a rectilinear figure. A rectilinear figure may be closed or open. Polygon: A closed plane figures whose boundaries are line segments is called a polygon. The line segments are called its

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### Sum of the Exterior Angles of an n-sided Polygon | Solved Examples

Here we will discuss the theorem of the sum of all exterior angles of an n-sided polygon and sum related example problems. 2. If the sides of a convex polygon are produced in the same order, the sum of all the exterior angles so formed is equal to four right angles.

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### Sum of the Interior Angles of an n-sided Polygon | Related Problems

Here we will discuss the theorem of sum of the interior angles of an n-sided polygon and some related example problems. The sum of the interior angles of a polygon of n sides is equal to (2n - 4) right angles. Given: Let PQRS .... Z be a polygon of n sides.

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### Area and Perimeter of Combined Figures | Circle | Triangle |Square

Here we will solve different types of problems on finding the area and perimeter of combined figures. 1. Find the area of the shaded region in which PQR is an equilateral triangle of side 7√3 cm. O is the centre of the circle. (Use π = $$\frac{22}{7}$$ and √3 = 1.732.)

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### Area of a Circular Ring | Radius of the Outer Circle and Inner Circle

Here we will discuss about the area of a circular ring along with some example problems. The area of a circular ring bounded by two concentric circle of radii R and r (R > r) = area of the bigger circle – area of the smaller circle = πR^2 - πr^2 = π(R^2 - r^2)

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### Area and Perimeter of a Semicircle | Solved Example Problems | Diagram

Here we will discuss about the area and perimeter of a semicircle with some example problems. Area of a semicircle = $$\frac{1}{2}$$ πr$$^{2}$$ Perimeter of a semicircle = (π + 2)r. Solved example problems on finding the area and perimeter of a semicircle

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### Area and Circumference of a Circle |Area of a Circular Region |Diagram

Here we will discuss about the area and circumference (Perimeter) of a circle and some solved example problems. The area (A) of a circle or circular region is given by A = πr^2, where r is the radius and, by definition, π = circumference/diameter = 22/7 (approximately).

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### Perimeter and Area of Trapezium | Geometrical Properties of Trapezium

Here we will discuss about the perimeter and area of a trapezium and some of its geometrical properties. Area of a trapezium (A) = 1/2 (sum of parallel sides) × height = 1/2 (a + b) × h Perimeter of a trapezium (P) = sum of parallel sides + sum of oblique sides

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### Perimeter and Area of Regular Hexagon | Solved Example Problems

Here we will discuss about the perimeter and area of a Regular hexagon and some example problems. Perimeter (P) = 6 × side = 6a Area (A) = 6 × (area of the equilateral ∆OPQ)

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### Perimeter and Area of Irregular Figures | Solved Example Problems

Here we will get the ideas how to solve the problems on finding the perimeter and area of irregular figures. The figure PQRSTU is a hexagon. PS is a diagonal and QY, RO, TX and UZ are the respective distances of the points Q, R, T and U from PS. If PS = 600 cm, QY = 140 cm

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### Perimeter and Area of Quadrilateral | Solved Example Problems |Diagram

Here we will discuss about the perimeter and area of a quadrilateral and some example problems. In the quadrilateral PQRS, PR is a diagonal, QM ⊥ PR and SN ⊥ PR. Then, area (A) of the quadrilateral PQRS = area of ∆PQR + area of ∆SPR = (1/2 × QM × PR) + (1/2 × SN × PR)

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### Perimeter and Area of Parallelogram | Geometrical Properties | Diagram

Here we will discuss about the perimeter and area of a parallelogram and some of its geometrical properties. Perimeter of a parallelogram (P) = 2 (sum of the adjacent sides) = 2 × a + b. Area of a parallelogram (A) = base × height = b × h. Some geometrical properties

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### Perimeter and Area of Rhombus | Geometrical Properties of Rhombus

Here we will discuss about the perimeter and area of a rhombus and some of its geometrical properties. Perimeter of a rhombus (P) = 4 × side = 4a Area of a rhombus (A) = 1/2 (Product of the diagonals) = 1/2 × d$$_{1}$$ × d$$_{2}$$ Some geometrical properties of a rhombus

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### Perimeter and Area of Mixed Figures |Rectangular Field |Triangles Area

Here we will discuss about the Perimeter and area of mixed figures. The length and breadth of a rectangular field is 8 cm and 6 cm respectively. On the shorter sides of the rectangular field two equilateral triangles are constructed outside. Two right-angled isosceles

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### Perimeter and Area of a Square | Geometrical Properties of a Square

Here we will discuss about the perimeter and area of a square and some of its geometrical properties. Perimeter of a square (P) = 4 × side = 4a Area of a square (A) = (side)^2 = a^2 Diagonal of a square (d) = √2a Side of a square (a) = √A = P/4

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### Perimeter and Area of a Triangle | Some Geometrical Properties

Here we will discuss about the perimeter and area of a triangle and some of its geometrical properties. Perimeter of a triangle (P) = sum of the sides = a + b + c Semiperimeter of a triangle (s) = 1/2(a + b + c). Area of a triangle (A) = 1/2 × base × altitude = 1/2ah

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### Perimeter and Area of a Rectangle | Perimeter of a rectangle | Diagram

Here we will discuss about the perimeter and area of a rectangle and some of its geometrical properties. Perimeter of a rectangle (P) = 2(length + breadth) = 2(l + b) Area of a rectangle (A) = length × breadth = l × b Diagonal of a rectangle (d) = sqrt(l^2 + b^2)

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### Perimeter and Area of Plane Figures | Definition of Perimeter and Area

A plane figure is made of line segments or arcs of curves in a plane. It is a closed figure if the figure begins and ends at the same point. We are familiar with plane figures like squares, rectangles, triangles and circles. Definition of Perimeter: The perimeter (P) of a

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### Volume of Cube | Formula for Finding the Volume of a Cube | Diagram

Here we will learn how to solve the application problems on Volume of cube using the formula. Formula for finding the volume of a cube Volume of a Cube (V) = (edge)3 = a3; where a = edge A cubical wooden box of internal dimensions 1 m × 1 m × 1 m is made of 5 cm thick

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### Problems on Right Circular Cylinder | Application Problem | Diagram

Problems on right circular cylinder. Here we will learn how to solve different types of problems on right circular cylinder. 1. A solid, metallic, right circular cylindrical block of radius 7 cm and height 8 cm is melted and small cubes of edge 2 cm are made from it.

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### Hollow Cylinder | Volume |Inner and Outer Curved Surface Area |Diagram

We will discuss here about the volume and surface area of Hollow Cylinder. The figure below shows a hollow cylinder. A cross section of it perpendicular to the length (or height) is the portion bounded by two concentric circles. Here, AB is the outer diameter and CD is the

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### Volume of Cuboid | Formula for Finding the Volume of a Cuboid |Diagram

Here we will learn how to solve the application problems on Volume of cuboid using the formula. Formula for finding the volume of a cuboid Volume of a Cuboid (V) = l × b × h; Where l = Length, b = breadth and h = height. 1. A field is 15 m long and 12 m broad. At one corner

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### Lateral Surface Area of a Cuboid | Are of the Four Walls of a Room

Here we will learn how to solve the application problems on lateral surface area of a cuboid using the formula. Formula for finding the lateral surface area of a cuboid Area of a Rooms is example of cuboids. Are of the four walls of a room = sum of the four vertical

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### Right Circular Cylinder | Lateral Surface Area | Curved Surface Area

A cylinder, whose uniform cross section perpendicular to its height (or length) is a circle, is called a right circular cylinder. A right circular cylinder has two plane faces which are circular and curved surface. A right circular cylinder is a solid generated by the

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### Cross Section | Area and Perimeter of the Uniform Cross Section

The cross section of a solid is a plane section resulting from a cut (real or imaginary) perpendicular to the length (or breadth of height) of the solid. If the shape and size of the cross section is the same at every point along the length (or breadth or height) of the

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### Cylinder | Formule for the Volume and the Surface Area of a Cylinder

A solid with uniform cross section perpendicular to its length (or height) is a cylinder. The cross section may be a circle, a triangle, a square, a rectangle or a polygon. A can, a pencil, a book, a glass prism, etc., are examples of cylinders. Each one of the figures shown

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### Volume and Surface Area of Cube and Cuboid |Volume of Cuboid |Problems

Here we will learn how to solve the problems on Volume and Surface Area of Cube and Cuboid: 1. Two cubes of edge 14 cm each are joined end to end to form a cuboid. Find the volume and the total surface area of the cuboid. Solution: The volume of the cuboid = 2 × volume

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