Math Blog
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Newly added pages can be seen from this page. Keep visiting to this page so that you will remain updated.en-usMathSun, 18 Aug 2019 16:19:26 -0400Sun, 18 Aug 2019 16:19:26 -0400math-only-math.comAug 18, Triangles on the Same Base & between Same Parallels are Equal in Area
https://www.math-only-math.com/triangles-on-the-same-base-and-between-the-same-parallels-are-equal-in-area.html3d65fece0d51ab8dad34dcc5dfc300d8Here 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)Sun, 18 Aug 2019 16:19:21 -0400Aug 17, Area of a Triangle is Half that of a Parallelogram on the Same Base
https://www.math-only-math.com/area-of-a-triangle-is-half-that-of-a-parallelogram-on-the-same-base-and-between-the-same-parallels.html52671967f84c9a51ae2645ad5fc4d4caHere 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.Sat, 17 Aug 2019 17:00:18 -0400Aug 14, Area of a Parallelogram is Equal to that of a Rectangle Between ......
https://www.math-only-math.com/area-of-a-parallelogram-is-equal-to-that-of-a-rectangle-between-the-same-parallel-lines.html5e86da61c46e3782249c83dd30dd8a3bHere 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 parallelWed, 14 Aug 2019 11:55:57 -0400Aug 13, Parallelogram on the Same Base and Between the Same Parallel Lines
https://www.math-only-math.com/parallelogram-on-the-same-base-and-between-the-same-parallel-lines-are-equal-in-area.htmla0e7fe8fd3f951c8d348f7e7ee27b7faHere 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).Tue, 13 Aug 2019 16:17:25 -0400Aug 11, Diagonal of a Parallelogram Divides it into Two Triangles of EqualArea
https://www.math-only-math.com/every-diagonal-of-a-parallelogram-divides-it-into-two-triangles-of-equal-area.htmlccccf7e508789cf471850bacd7c4d37eHere 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. ∠SRPSun, 11 Aug 2019 13:10:55 -0400Aug 11, Base and Height (Altitude) in a Triangle and a Parallelogram | Diagram
https://www.math-only-math.com/base-and-height-in-a-triangle-and-a-parallelogram.html7a3a8459165d946b4d253db00f289233We 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 parallelogramSun, 11 Aug 2019 12:07:03 -0400Aug 8, Area of a Closed Figure |Measurement of Area |Area Axiom for Rectangle
https://www.math-only-math.com/area-of-a-closed-figure.html1d41e2a516c37581287f6e2181890794We 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 areasThu, 8 Aug 2019 17:43:18 -0400Aug 6, Bisectors of the Angles of a Parallelogram form a Rectangle | Diagram
https://www.math-only-math.com/bisectors-of-the-angles-of-a-parallelogram-form-a-rectangle.htmle157bbf5124d9fbd72f7a7de5b17bab8Here 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 isTue, 6 Aug 2019 16:18:18 -0400Aug 2, Conditions for Classification of Quadrilaterals and Parallelograms
https://www.math-only-math.com/classification-of-quadrilaterals-and-parallelograms.htmlc41587777493d51d3fff2e256895f446We 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.Fri, 2 Aug 2019 15:29:34 -0400Aug 1, Diagonals of a Parallelogram are Equal & Intersect at Right Angles
https://www.math-only-math.com/parallelogram-will-be-a-square.htmlc239f7bc1a75321cc40db5e6f92725d8Here 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., PQThu, 1 Aug 2019 16:50:40 -0400Aug 1, Diagonals of a Square are Equal in Length & they Meet at Right Angles
https://www.math-only-math.com/diagonals-of-a-square-are-equal-in-length-and-they-meet-at-right-angles.html999cc80626505b351b6b2c70c15e94a2Here 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 = QRThu, 1 Aug 2019 11:10:59 -0400Jul 29, A Parallelogram, whose Diagonals are of Equal Length, is a Rectangle
https://www.math-only-math.com/a-parallelogram-whose-diagonals-are-of-equal-length-is-a-rectangle.htmlb1f04147ae9b244d05694d39f27a1941Here 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 ∆PQRMon, 29 Jul 2019 16:42:35 -0400Jul 25, In a Rectangle the Diagonals are of Equal Lengths | Proof | Diagram
https://www.math-only-math.com/in-a-rectangle-the-diagonals-are-of-equal-lengths.html5383babc1985e7d26a6aa4b5eeb2b7f9Here 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 = ∠SPRThu, 25 Jul 2019 15:45:55 -0400Jul 25, A Parallelogram whose Diagonals Intersect at Right Angles is a Rhombus
https://www.math-only-math.com/a-parallelogram-whose-diagonals-intersect-at-right-angles-is-a-rhombus.html16f5ddd6d104751b1f5b5da51c17c1ccHere 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,Thu, 25 Jul 2019 15:07:54 -0400Jul 23, A Rhombus is a Parallelogram whose Diagonals Meet at Right Angles
https://www.math-only-math.com/a-rhombus-is-a-parallelogram-whose-diagonals-meet-at-right-angles.htmle62cfb3b5b2d921b3380bbd941894ca6Here 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°.Tue, 23 Jul 2019 13:22:42 -0400Jul 16, Pair of Opposite Sides of a Parallelogram are Equal and Parallel
https://www.math-only-math.com/pair-of-opposite-sides-of-a-parallelogram-are-equal-and-parallel.html8f16ce62bfe73e65784fa3a855a68d5bHere 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.Tue, 16 Jul 2019 17:25:47 -0400Jul 15, A Quadrilateral is a Parallelogram if its Diagonals Bisect each Other
https://www.math-only-math.com/a-quadrilateral-is-a-parallelogram-if-its-diagonals-bisect-each-other.html135e7b7c04ff496035756b1d3770af54Here 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 = ORMon, 15 Jul 2019 16:06:12 -0400Jul 15, Diagonals of a Parallelogram Bisect each Other | Diagonals Bisect each
https://www.math-only-math.com/diagonals-of-a-parallelogram-bisect-each-other.html9cbbc2ced76b29141b62d992d0005cd9Here 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 diagonalsMon, 15 Jul 2019 15:03:06 -0400Jul 10, Opposite Angles of a Parallelogram are Equal | Related Solved Examples
https://www.math-only-math.com/opposite-angles-of-a-parallelogram-are-equal.htmld514b8b7a64661180d3d25c39285acb0Here 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:Wed, 10 Jul 2019 14:32:43 -0400Jul 9, Opposite Sides of a Parallelogram are Equal | Solved Examples
https://www.math-only-math.com/opposite-sides-of-a-parallelogram-are-equal.html677aa4b6b32c8665a0f3fefd9b42a7cdHere 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 PRTue, 9 Jul 2019 15:12:17 -0400Jul 8, Concept of Parallelogram |Quadrilateral| Rectangle| Rhombus| Trapezium
https://www.math-only-math.com/concept-of-parallelogram.html4cc41004c4cec39e072024140d4c5735Here 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 whichMon, 8 Jul 2019 15:50:35 -0400Jul 2, What is Rectilinear Figure? | What is Diagonal of a Polygon? | Polygon
https://www.math-only-math.com/rectilinear-figures.htmld7d800525ad90423facb299fe7751c2aWhat 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 itsTue, 2 Jul 2019 11:22:50 -0400Jul 1, Sum of the Exterior Angles of an n-sided Polygon | Solved Examples
https://www.math-only-math.com/sum-of-the-exterior-angles-of-an-n-sided-polygon.html832e84da81d222512dd06ca636c59b78Here 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.Mon, 1 Jul 2019 15:42:25 -0400Jun 27, Sum of the Interior Angles of an n-sided Polygon | Related Problems
https://www.math-only-math.com/sum-of-the-interior-angles-of-an-n-sided-polygon.html7f5ba301dd97f6232f4dc5e72bf60bbaHere 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.Thu, 27 Jun 2019 13:54:03 -0400Jun 24, Area and Perimeter of Combined Figures | Circle | Triangle |Square
https://www.math-only-math.com/area-and-perimeter-of-combined-figures.html002dafd39d36c9fb563ac310df1366f1Here 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.)Mon, 24 Jun 2019 15:25:25 -0400Jun 20, Area of a Circular Ring | Radius of the Outer Circle and Inner Circle
https://www.math-only-math.com/area-of-a-circular-ring.html65322023d3a1bcba2f339b8b5913c13dHere 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) Thu, 20 Jun 2019 14:08:20 -0400Jun 20, Area and Perimeter of a Semicircle | Solved Example Problems | Diagram
https://www.math-only-math.com/area-and-perimeter-of-a-semicircle.htmldf26838355a6ac181bdc4d7ee42028e5Here 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 semicircleThu, 20 Jun 2019 14:05:46 -0400Jun 20, Area and Circumference of a Circle |Area of a Circular Region |Diagram
https://www.math-only-math.com/area-and-circumference-of-a-circle.html6e7db2773c79d1d782d7d8c89ebb5ba4Here 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).Thu, 20 Jun 2019 10:52:06 -0400Jun 17, Perimeter and Area of Trapezium | Geometrical Properties of Trapezium
https://www.math-only-math.com/perimeter-and-area-of-trapezium.html0345f24ce43b41ff87c52013de824cd5Here 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 sidesMon, 17 Jun 2019 11:05:56 -0400Jun 16, Perimeter and Area of Regular Hexagon | Solved Example Problems
https://www.math-only-math.com/perimeter-and-area-of-regular-hexagon.htmlbccda82be2b77663497bcadd1e879e18Here 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)Sun, 16 Jun 2019 11:12:28 -0400Jun 16, Perimeter and Area of Irregular Figures | Solved Example Problems
https://www.math-only-math.com/perimeter-and-area-of-irregular-figures.htmlfe8d04f4b5658e5a6550841ef23b166bHere 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 cmSun, 16 Jun 2019 04:14:21 -0400Jun 14, Perimeter and Area of Quadrilateral | Solved Example Problems |Diagram
https://www.math-only-math.com/perimeter-and-area-of-quadrilateral.htmlebe3bb45ba5ac3a8dcfbfcf42a51da23Here 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) Fri, 14 Jun 2019 00:04:13 -0400Jun 7, Perimeter and Area of Parallelogram | Geometrical Properties | Diagram
https://www.math-only-math.com/perimeter-and-area-of-parallelogram.htmla9144e91b5e8eabf5a35e823827f5fd9Here 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 propertiesFri, 7 Jun 2019 15:52:03 -0400Jun 7, Perimeter and Area of Rhombus | Geometrical Properties of Rhombus
https://www.math-only-math.com/perimeter-and-area-of-rhombus.html34a94069c3e70ad163db68e595436f5eHere 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 rhombusFri, 7 Jun 2019 15:16:16 -0400Jun 6, Perimeter and Area of Mixed Figures |Rectangular Field |Triangles Area
https://www.math-only-math.com/perimeter-and-area-of-mixed-figures.html00498ada2a5a817f30cea7df98425712Here 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 isoscelesThu, 6 Jun 2019 18:01:57 -0400Jun 5, Perimeter and Area of a Square | Geometrical Properties of a Square
https://www.math-only-math.com/perimeter-and-area-of-a-square.html13e2b75522e94536ed3fad33800efb8fHere 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/4Wed, 5 Jun 2019 17:36:59 -0400Jun 3, Perimeter and Area of a Triangle | Some Geometrical Properties
https://www.math-only-math.com/perimeter-and-area-of-a-triangle.htmleb31d02addb2d631139cbc0550100c3cHere 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 = 1/2ahMon, 3 Jun 2019 19:00:38 -0400Jun 3, Perimeter and Area of a Rectangle | Perimeter of a rectangle | Diagram
https://www.math-only-math.com/perimeter-and-area-of-a-rectangle.htmlf8cf731445f531db29cf10e280fc4023Here 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)Mon, 3 Jun 2019 18:38:56 -0400May 29, Perimeter and Area of Plane Figures | Definition of Perimeter and Area
https://www.math-only-math.com/perimeter-and-area-of-plane-figures.htmlad62fa6d3c85afa6c8ece67594eb7652A 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 aWed, 29 May 2019 05:50:38 -0400May 28, Volume of Cube | Formula for Finding the Volume of a Cube | Diagram
https://www.math-only-math.com/volume-of-cube.html20777d7eb8027e0b1372864fcc8cea70Here 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 thickTue, 28 May 2019 19:31:48 -0400May 28, Problems on Right Circular Cylinder | Application Problem | Diagram
https://www.math-only-math.com/problems-on-right-circular-cylinder.html876624c9573c91ae8db6f5ba0384918eProblems 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.Tue, 28 May 2019 18:19:57 -0400May 27, Hollow Cylinder | Volume |Inner and Outer Curved Surface Area |Diagram
https://www.math-only-math.com/hollow-cylinder.html6b38fabf9f6225e79f75ae314577b98cWe 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 theMon, 27 May 2019 19:00:25 -0400May 27, Volume of Cuboid | Formula for Finding the Volume of a Cuboid |Diagram
https://www.math-only-math.com/volume-of-cuboid.html8a790ad4b4e18e2bd8059a4231599640Here 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 cornerMon, 27 May 2019 18:50:24 -0400May 27, Lateral Surface Area of a Cuboid | Are of the Four Walls of a Room
https://www.math-only-math.com/lateral-surface-area-of-a-cuboid.htmlfe230b946467320126881205e1eaeb2aHere 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 verticalMon, 27 May 2019 18:36:45 -0400May 23, Right Circular Cylinder | Lateral Surface Area | Curved Surface Area
https://www.math-only-math.com/right-circular-cylinder.html38b47b86eca768ffe7b1c63cd0f4bb45A 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 theThu, 23 May 2019 19:23:10 -0400May 21, Cross Section | Area and Perimeter of the Uniform Cross Section
https://www.math-only-math.com/cross-section.html8976081d04b6c9e396a4782de8cf0e5fThe 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 theTue, 21 May 2019 14:41:03 -0400May 20, Cylinder | Formule for the Volume and the Surface Area of a Cylinder
https://www.math-only-math.com/cylinder.html99a40f4b0cfadcecff99e82b8d198c89A 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 shownMon, 20 May 2019 16:54:12 -0400May 18, Volume and Surface Area of Cube and Cuboid |Volume of Cuboid |Problems
https://www.math-only-math.com/volume-and-surface-area-of-cube-and-cuboid.html992a75245ddf327369f1021e4a79f07cHere 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 Sat, 18 May 2019 16:36:15 -0400May 10, Volume and Surface Area of Cube | Diagonal a Cube | Total surface Area
https://www.math-only-math.com/volume-and-surface-area-of-cube.html206c668a944539ee7b7a1a9f9729de34What is Cube? A cuboid is a cube if its length, breadth and height are equal. In a cube, all the faces are squares which are equal in area and all the edges are equal. A dice is an example of a cube. Volume of a Cube (V) = (edge)^3 = a^3 Total surface Area of a Cube (S)Fri, 10 May 2019 19:02:32 -0400May 10, Volume and Surface Area of Cuboid | Lateral Surface Area of a Cuboid
https://www.math-only-math.com/volume-and-surface-area-of-cuboid.html6927c987ad82c41ddceb0c56d2c6f4a6What is Cuboid? A cuboid is a solid with six rectangular plane faces, for example, a brick or a matchbox. Each of these is made up of six plane faces which are rectangular. Remember that since a square is a special case of a rectangle, a cuboid may have square faces too.Fri, 10 May 2019 18:35:59 -0400May 7, Solid Figures | Volume of a Solid Figure |Surface Area of Solid Figure
https://www.math-only-math.com/solid-figures.html826dad0c79ca1d324c09920413d65c55A figure made up of a number of plane or curved faces is a solid figure. Bricks, matchboxes, talcum powder containers and rooms are all examples of solid figures. A solid figure has three dimensions while a plane figure has only two dimensions.Tue, 7 May 2019 18:57:57 -0400May 4, Riders Based on Pythagoras’ Theorem | Establishing Rides | Diagram
https://www.math-only-math.com/riders-based-on-pythagoras-theorem.html82e76f84ba0b5e73fe08682642970086Here we will solve different types of examples on establishing riders based on Pythagoras’ theorem. 1. In the quadrilateral PQRS the diagonals PR and QS intersects at a right angle. Prove that PQ^2+ RS^2 = PS^2 + QR^2. Solution: Let the diagonals intersect at O, the angle ofSat, 4 May 2019 19:14:26 -0400Apr 30, Criteria of Similarity between Triangles | SAS Criterion of Similarity
https://www.math-only-math.com/criteria-of-similarity-between-triangles.html8c551ef3e60f5f98e617fca38d3324a2We will discuss here about the different criteria of similarity between triangles with the figures. 1. SAS criterion of similarity: If two triangles have an angle of one equal to an angle of the other and the sides including them are proportional, the triangles are similar.Tue, 30 Apr 2019 18:09:40 -0400Apr 29, AA Criterion of Similarly on Quadrilateral | Alternate Angles
https://www.math-only-math.com/aa-criterion-of-similarly-on-quadrilateral.html4d7a1a7d338a3d8076587e926c59927dHere we will prove that in the quadrilateral ABCD, AB ∥ CD. Prove that OA × OD = OB × OC. Solution: Proof: 1. In ∆ OAB and ∆OCD, (i) ∠AOB = ∠COD (ii) ∠OBA = ∠ODC. 2. ∆ OAB ∼ ∆OCD. 3. Therefore, OA/OC = OB/OD ⟹ OA × OD = OB × OC. (Proved)Mon, 29 Apr 2019 10:51:21 -0400Apr 25, Pythagoras’ Theorem | AA Criterion of Similarity | Proof with Diagram
https://www.math-only-math.com/pythagoras-theorem.html68a65bcf28a6df94a766081fefd0c430The lengths of the sides of a right-angled triangle have a special relationship between them. This relation is widely used in many branches of mathematics, such as mensuration and trigonometry. Pythagoras’ Theorem: In a right-angled triangle, the square on the hypotenuse isThu, 25 Apr 2019 19:00:13 -0400Apr 23, Applying Pythagoras’ Theorem | Proof by Pythagoras’ Theorem | Diagram
https://www.math-only-math.com/applying-pythagoras-theorem.htmlbef57c7b8f8a0155713be0c65e658dcbApplying Pythagoras’ theorem we will prove the problem given below. ∆PQR is right-angle at Q. M and N are the midpoints of PQ and QR respectively. Prove that PN^2 + RM^2 = 5MN^2. Solution: Given In ∆PQR, ∠PQR = 90°. PM = MQ and QN = NR Therefore, PQ = 2MQ and QR = 2QNTue, 23 Apr 2019 19:02:15 -0400Apr 23, Converse of Pythagoras’ Theorem | Sum of the Squares of Two Sides
https://www.math-only-math.com/converse-of-pythagoras-theorem.html3f0d126389c7bb667bafec5cce950e85If in a triangle the sum of the squares of two sides is equal to the square of the third side then the triangle is a right-angled triangle, the angle between the first two sides being a right angle. Given In the ∆XYZ, XY\(^{2}\) + YZ\(^{2}\) = XZ\(^{2}\) To prove ∠XYZ = 90°Tue, 23 Apr 2019 17:17:48 -0400Apr 20, Problems on Size Transformation | Area of the Terrace in the Model
https://www.math-only-math.com/problems-on-size-transformation.htmlcac646dffd063ec2ad7babc914e1eaa6Here we will solve different types of problems on size transformation. 1. A map of a rectangular park is drawn to a scale of 1 : 5000. (i) Find the actual length of the park if the length of the same in the map is 25 cm. (ii) If the actual width of the park is 1 km, find itsSat, 20 Apr 2019 16:28:41 -0400Apr 20, Reduction Transformation | Centre of Reduction | Reduction Factor
https://www.math-only-math.com/reduction-transformation.html9c470d8444d45d52bd33eebbcfe8af21We will discuss here about the similarity on Reduction transformation. In the figure given below ∆X’Y’Z’ is a reduced image of ∆XYZ. The two triangles are similar. Here also the triangles are equiangular and \(\frac{X’Y’}{XY}\) = \(\frac{Y’Z’}{YZ}\) = \(\frac{Z’X’}{ZX}\) = kSat, 20 Apr 2019 15:58:11 -0400Apr 20, Enlargement Transformation | Centre of Enlargement |Enlargement Factor
https://www.math-only-math.com/enlargement-transformation.html22bbdeeda296be67ffeea261500f354eWe will discuss here about the similarity on enlargement transformation. Cut out some geometrical figures like triangles, quadrilaterals, etc., from a piece of cardboard. Hold these figures, one-by-one, between a point source of light and a wall. Sat, 20 Apr 2019 15:37:27 -0400Apr 13, Greater segment of the Hypotenuse = the Smaller Side of the Triangle
https://www.math-only-math.com/greater-segment-of-the-hypotenuse-is-equal-to-the-smaller-side-of-the-triangle.htmlee1c5fab745c12f0c1c13a90eec31d0dHere we will prove that if a perpendicular is drawn from the right-angled vertex of right-angled triangle to the hypotenuse and if the sides of the right-angled triangle are in continued proportion, the greater segment of the hypotenuse is equal to the smaller side of theSat, 13 Apr 2019 18:32:39 -0400Apr 6, Application of Basic Proportionality Theorem | Internal Bisector
https://www.math-only-math.com/application-of-basic-proportionality-theorem.html9c495758e8077e4367781aa682c24519Here we will prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. Given: XP is the internal bisector of ∠YXZ, intersecting YZ at P.Sat, 6 Apr 2019 15:36:56 -0400Apr 4, Converse of Basic Proportionality Theorem | Proof with Diagram
https://www.math-only-math.com/converse-of-basic-proportionality-theorem.html01bc3a694fe7073100bc2118d8001ae7Here we will prove converse of basic proportionality theorem. The line dividing two sides of a triangle proportionally is parallel to the third side. Given: In ∆XYZ, P and Q are points on XY and XZ respectively, such that XP/PY = XQ/QZ. To prove: PQ ∥ YZ Proof: StatementThu, 4 Apr 2019 17:25:30 -0400Mar 28, Basic Proportionality Theorem | AA Criterion of Similarity | Diagram
https://www.math-only-math.com/basic-proportionality-theorem.html38b0f0921c29ed904fb73e2fa0855d01Here we will learn how to prove the basic proportionality theorem with diagram. A line drawn parallel to one side of a triangle divides the other two sides proportionally. Given In ∆XYZ, P and Q are points on XY and XZ respectively, such that PQ ∥ YZ. To prove XP/PY = XQ/QZ.Thu, 28 Mar 2019 17:32:55 -0400Mar 28, Similar Triangles | Congruency and Similarity of Triangles | Diagram
https://www.math-only-math.com/similar-triangles.htmla29be216e23486f3417f2bbef150ce45We will discuss here about the similar triangles. If two triangles are similar then their corresponding angles are equal and corresponding sides are proportional. Here, the two triangles XYZ and PQR are similar. So, ∠X = ∠P, ∠Y = ∠Q, ∠Z = ∠R and XY/PQ=YZ/QR = XZ/PR. ∆XYZ isThu, 28 Mar 2019 14:28:51 -0400Mar 25, AA Criterion of Similarity | Similarity on Right-angled Triangle
https://www.math-only-math.com/aa-criterion-of-similarity.html85c46fbb513b40b9a7445c99fa8b943aHere we will prove the theorems related to AA Criterion of Similarity on Quadrilateral. 1. In a right-angled triangle, if a perpendicular is drawn from the right-angled vertex to the hypotenuse, the triangles on each side of it are similar to the whole triangle and to oneMon, 25 Mar 2019 13:53:23 -0400Mar 19, Properties of size Transformation |Enlargement|Reduction|Scale Factor
https://www.math-only-math.com/properties-of-size-transformation.html0bca71df3ee8efef5966a47299f28fc3We will discuss here about the different properties of size transformation. 1. The shape of the image is the same as that of the object. 2. If the scale factor of the transformation is k then each side of the image is k times the corresponding side of the object.Tue, 19 Mar 2019 17:34:08 -0400Mar 8, Proof By the Equal Intercepts Theorem | Line Joining the Midpoints
https://www.math-only-math.com/proof-by-the-equal-intercepts-theorem.htmld172495b241c6f0cccf15a9650a83b50Here we will prove that in the given ∆XYZ, M and N are the midpoints of XY and XZ respectively. T is any point on the base YZ. Prove that MN bisects XT. Solution: Given: In ∆XYZ, XM = MY and XN = NZ. MN cuts XT at U. To prove: XU = UT. Construction: Through X, draw PQ ∥ YZ.Fri, 8 Mar 2019 16:10:23 -0500Mar 8, Midpoint Theorem by using the Equal Intercepts Theorem |Proof |Diagram
https://www.math-only-math.com/midpoint-theorem-by-using-the-equal-intercepts-theorem.html1ca75e3e796b8fcd37827022f6c20921Here we will prove that converse of the Midpoint Theorem by using the Equal Intercepts Theorem. Solution: Given: P is the midpoint of XY in ∆XYZ. PQ ∥ YZ. To prove: XQ = QZ. Construction: Through X, draw MN ∥ YZ. Proof: Statement 1. PQ ∥ YZ. 2. MN ∥ PQ ∥ YZ.Fri, 8 Mar 2019 14:48:00 -0500Mar 5, Problems on Equal Intercepts Theorem | Midpoint Theorem | Diagram
https://www.math-only-math.com/problems-on-equal-intercepts-theorem.html3cf41d8dd2bee0bd1512e467332a738bHere we will solve different types of problems on Equal Intercepts Theorem. 1. In the given figure, MN ∥ KL ∥ GH and PQ = QR. If ST = 2.2 cm, find SU. Solution: The transversal PR makes equal intercepts, PQ and QR, on the three parallel lines MN, KL and GH. Therefore, by theTue, 5 Mar 2019 18:24:33 -0500Feb 26, Equal Intercepts Theorem | Transversal makes Equal Intercepts
https://www.math-only-math.com/equal-intercepts-theorem.html35643909306d36b35563a2e4240fd236Intercept In the figure given above, XY is a transversal cutting the line L1 and L2 at P and Q respectively. The line segment PQ is called the intercept made on the transversal XY by the lines L1 and L2. If a transversal makes equal intercepts on three or more parallel linesTue, 26 Feb 2019 14:36:59 -0500Feb 25, Collinear Points Proved by Midpoint Theorem | Collinearity | Diagram
https://www.math-only-math.com/collinear-points-proved-by-midpoint-theorem.htmla7112a3501587e6b991cf82a092a86c6In ∆XYZ, the medians ZM and YN are produced to P and Q respectively such that ZM = MP and YN = NQ. Prove that the points P, X and Q are collinear, and X is the midpoint of PQ. Solution: Given: In ∆XYZ, the points M and N are the midpoints of XY and XZ respectively.Mon, 25 Feb 2019 14:45:16 -0500