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-usMathMon, 20 May 2019 16:54:17 -0400Mon, 20 May 2019 16:54:17 -0400math-only-math.comMay 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 20, 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 theMon, 20 May 2019 15:05:01 -0400May 18, 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 theSat, 18 May 2019 19:08:23 -0400May 18, 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 theSat, 18 May 2019 18:51:03 -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 16, 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 verticalThu, 16 May 2019 18:41:14 -0400May 16, 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 thickThu, 16 May 2019 18:19:28 -0400May 16, 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 cornerThu, 16 May 2019 17:50:14 -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 -0500Feb 22, Midpoint Theorem on Right-angled Triangle | Proof | Statement | Reason
https://www.math-only-math.com/midpoint-theorem-on-right-angled-triangle.html454cf9f12bd6d66471b7dc8e7d075837Here we will prove that in a right-angled triangle the median drawn to the hypotenuse is half the hypotenuse in length. Solution: In ∆PQR, ∠Q = 90°. QD is the median drawn to hypotenuse PRFri, 22 Feb 2019 14:28:35 -0500Feb 21, Midsegment Theorem on Trapezium | Nonparallel Sides of a Trapezium
https://www.math-only-math.com/midsegment-theorem-on-trapezium.htmlea1ba5508578c7d93b1cb0a56fd2d579Here we will prove that the line segment joining the midpoints of the nonparallel sides of a trapezium is half the sum of the lengths of the parallel sides and is also parallel to them. Solution: Given: PQRS is a trapezium in which PQ ∥ RS. U and V are theThu, 21 Feb 2019 11:55:27 -0500Feb 20, Midpoint Theorem on Trapezium | Converse of the Midpoint Theorem
https://www.math-only-math.com/midpoint-theorem-on-trapezium.html2ed1b4ed31c52c1aaaa9259688806f1fPQRS is a trapezium in which PQ ∥ RS. T is the midpoint of QR. TU is drawn parallel to PQ which meets PS at U. Prove that 2TU = PQ + RS. Given: PQRS is a trapezium in which PQ ∥ RS. T is the midpoint of QR. TU ∥ PQ and TU meets PS at U. To prove: 2TU = PQ + RS. ConstructionWed, 20 Feb 2019 14:10:14 -0500Feb 13, Straight Line Drawn from the Vertex of a Triangle to the Base |Diagram
https://www.math-only-math.com/straight-line-drawn-from-the-vertex-of-a-triangle-to-the-base.html5a7bd7ddca3ce151ea2158526d799c57Here we will prove that any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other two sides of the triangle. Solution: Given: Q and R are the midpoints of the sides XY and XZ respectively ofWed, 13 Feb 2019 16:56:46 -0500Feb 12, Four Triangles which are Congruent to One Another | Prove with Diagram
https://www.math-only-math.com/four-triangles-which-are-congruent-to-one-another.htmlb78f37ceb8c9ddf73c2a89707d1a7ce3Here we will show that the three line segments which join the middle points of the sides of a triangle, divide it into four triangles which are congruent to one another. Solution: Given: In ∆PQR, L, M and N are the midpoints of QR, RP and PQ respectively. To prove ∆PMN ≅ LNMTue, 12 Feb 2019 14:48:34 -0500Feb 11, Midpoint Theorem Problem | Midpoint Theorem | Converse of Midpoint
https://www.math-only-math.com/midpoint-theorem-problem.html48e0fed8d299d2b83419230864cda284Here we will learn how to solve different types of midpoint theorem problem. In the adjoining figure, find (i) ∠QPR, (ii) PQ if ST = 2.1 cm. Solution: In ∆PQR, S and T are the midpoints of PR and QR respectively. Therefore, ST = \(\frac{1}{2}\)PQ and ST ∥ PQ.Mon, 11 Feb 2019 16:57:42 -0500Feb 6, Converse of Midpoint Theorem | Proof of Converse of Midpoint Theorem
https://www.math-only-math.com/converse-of-midpoint-theorem.htmlc4783fcbf0bd33cb2983cc47b1b0d1e8The straight line drawn through the midpoint of one side of a triangle parallel to another bisects the third side. Given: In ∆PQR, S is the midpoint of PQ, and ST is drawn parallel to QR. To prove: ST bisects PR, i.e., PT = TR. Construction: Join SU where U is the midpointWed, 6 Feb 2019 15:35:00 -0500Feb 4, Midpoint Theorem |AAS & SAS Criterion of Congruency Prove with Diagram
https://www.math-only-math.com/midpoint-theorem.htmledc5bae19b5ff24210dbdd2c7d00d6e8The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it. Given: A triangle PQR in which S and T are the midpoint of PQ and PR respectively. To prove: ST ∥ QR and ST = 1/2QR Construction: Draw RU ∥ QP such that Mon, 4 Feb 2019 18:03:57 -0500Jan 31, Sum of Four Sides of a Quadrilateral Exceeds the Sum of the Diagonals
https://www.math-only-math.com/sum-of-the-four-sides-of-a-quadrilateral-exceeds-the-sum-of-the-diagonals.htmlab7054bd88cc2b38c503998ba8d039acHere we will prove that in any quadrilateral the sum of the four sides exceeds the sum of the diagonals. Solution: Given: ABCD is a quadrilateral; AC and BD are its diagonals. To prove: (AB + BC + CD + DA) > (AC + BD). Proof: Statement 1. In ∆ADB, (DA + AB) > BD.Thu, 31 Jan 2019 17:33:31 -0500Jan 30, Sum Of Any Two Sides Is Greater Than Twice The Median | Proof|Diagram
https://www.math-only-math.com/sum-of-any-two-sides-is-greater-than-twice-the-median.html8c474089523850f86387527ada70bad6Here we will prove that in a triangle the sum of any two sides is greater than twice the median which bisects the remaining side. Solution: Given: In ∆XYZ, XP is the median that bisects YZ at P. To prove: (XY + XZ) > 2XP. Construction: Produce XP to Q such that XP = PQ.Wed, 30 Jan 2019 15:25:30 -0500Jan 28, Problem on Inequalities in Triangle | Solution with Diagram
https://www.math-only-math.com/problem-on-inequalities-in-triangle.html2ba41d9f39f1759a5ba55682521e4305Here we will solve the problem on inequalities in triangle. Let XYZ be a triangle in which XM bisects ∠YXZ. Prove that XY is greater than YM. As XM bisects ∠YXZ, we have ∠YXZ = ∠MXZ ............ (i) Also, in ∆XMZ, ∠XMY > ∠MXZ, as an exterior angle of a triangle is alwaysMon, 28 Jan 2019 17:37:40 -0500Jan 25, Comparison of Sides and Angles in a Triangle | Geometrical Property
https://www.math-only-math.com/comparison-of-sides-and-angles-in-a-triangle.htmla5df1799899b1fcf68cf77f97336c682Here we will solve different types of problems on comparison of sides and angles in a triangle. 1. In ∆XYZ, ∠XYZ = 35° and ∠YXZ = 63°. Arrange the sides of the triangle in the descending order of their lengths. ∠XZY = 180° - (∠XYZ + ∠YXZ) = 180° - (35° + 63°) = 180° - 98°Fri, 25 Jan 2019 17:50:07 -0500Jan 24, Perpendicular is the Shortest Theorem | Inequalities in Triangle
https://www.math-only-math.com/perpendicular-is-the-shortest.html04126b0e87197ea134955f432f1feca6Here we will prove that of all the straight lines that can be drawn to a straight line from a given point outside it, the perpendicular is the shortest. Given: XY is a straight line and O is a point outside it. OP is perpendicular to XY and OZ is an oblique. To Prove: OP <OZThu, 24 Jan 2019 15:11:09 -0500Jan 23, The Sum of any Two Sides of a Triangle is Greater than the Third Side
https://www.math-only-math.com/the-sum-of-any-two-sides-of-a-triangle-is-greater-than-the-third-side.html7c11a4bcf72d7939569a7c78e8511b3eHere we will prove that the sum of any two sides of a triangle is greater than the third side. Given: XYZ is a triangle. To Prove: (XY + XZ) > YZ, (YZ + XZ) > XY and (XY + YZ) > XZ Construction: Produce YX to P such that XP = XZ. Join P and Z. Statement 1. ∠XZP = ∠XPZ.Wed, 23 Jan 2019 16:56:18 -0500Jan 21, Greater Side has the Greater Angle Opposite to It | Triangle Inequalit
https://www.math-only-math.com/greater-side-has-the-greater-angle-opposite-to-it.html81bcaa3a346e3e0ca6e8e4e045b58794Here we will prove that if two sides of a triangle are unequal, the greater side has the greater angle opposite to it. Given: In ∆XYZ, XZ > XY To prove: ∠XYZ > ∠XZY. Construction: From XZ, cut off XP such that XP equals XY. Join Y and P. Proof: Statement 1. In ∆XYP, ∠XYP =Mon, 21 Jan 2019 17:34:10 -0500Jan 21, Greater Angle has the Greater Side Opposite to It | Prove with Diagram
https://www.math-only-math.com/greater-angle-has-the-greater-side-opposite-to-it.htmlc9bb7e32e300815e6fbd026b2668d358Here we will prove that if two angles of a triangle are unequal, the greater angle has the greater side opposite to it. Given: In ∆XYZ, ∠XYZ > ∠XZY To Prove: XZ > XY Proof: Statement 1. Let us assume that XZ is not greater than XY. Then XZ must be either equal to or lessMon, 21 Jan 2019 17:31:49 -0500Jan 17, Theorem on Isosceles Triangle | Proof Involving Isosceles Triangles
https://www.math-only-math.com/theorem-on-isosceles-triangle.html961080ef6cd77c9951d89cf6d26eaf08Here we will prove that the equal sides YX and ZX of an isosceles triangle XYZ are produced beyond the vertex X to the points P and Q such that XP is equal to XQ. QY and PZ are joined. Show that QY is equal to PZ. Solution: In ∆XYZ, XY = XZ. YX and XZ are produced to P andThu, 17 Jan 2019 14:31:59 -0500Jan 16, Points on the Base of an Isosceles Triangle | Prove with Diagram
https://www.math-only-math.com/base-of-an-isosceles-triangle.htmlda0f6a7391b900b99823c2596e958c8eHere we will prove that if two given points on the base of an isosceles triangle are equidistant from the extremities of the base, show that they are also equidistance from the vertex. Solution: Given: In the isosceles ∆XYZ, XY = XZ, M and N points on the base YZ such thatWed, 16 Jan 2019 14:29:49 -0500Jan 15, Lines Joining the Extremities of the Base of an Isosceles Triangle
https://www.math-only-math.com/straight-lines-joining-the-extremities-of-the-base-of-an-isosceles-triangle.html90841070873238102c830d64747bf4d8Here we will show that the straight lines joining the extremities of the base of an isosceles triangle to the midpoints of the opposite sides are equal. Solution: Given: In ∆XYZ, XY = XZ, M and N are the midpoints of XY and XZ respectively.Tue, 15 Jan 2019 16:56:51 -0500Jan 10, Problem on Two Isosceles Triangles on the Same Base | Proof | Diagram
https://www.math-only-math.com/problem-on-two-isosceles-triangles-on-the-same-base.html25fce0483d3d3ae7bb12c1b608c46db8Here we will prove that ∆PQR and ∆SQR are two isosceles triangles drawn on the same base QR and on the same side of it. If P and S be joined, prove that each of the angles ∠QPR and ∠QSR will be divided by the line PS into two equal parts.Thu, 10 Jan 2019 16:36:07 -0500Jan 9, Problems on Properties of Isosceles Triangles | Find x° and y°
https://www.math-only-math.com/problems-on-properties-of-isosceles-triangles.html75bbd1c3208f5102ffcfaa27ddec6a6dHere we will solve some numerical problems on the properties of isosceles triangles Find x° from the given figures. In ∆XYZ, XY = XZ. Therefore, ∠XYZ = ∠XZY = x°. Now, ∠YXZ + ∠XYZ + XZY = 180° ⟹ 84° + x° + x° = 180° ⟹ 2x° = 180° - 84° ⟹ 2x° = 96°Wed, 9 Jan 2019 17:34:31 -0500Jan 7, Three Angles of an Equilateral Triangle are Equal | Axis of Symmetry
https://www.math-only-math.com/three-angles-of-an-equilateral-triangle.html2e421cde09b3209f6da124228ce7a00fHere we will prove that if the three angles of a triangle are equal, it is an equilateral triangle. Given: In ∆XYZ, ∠YXZ = ∠XYZ = ∠XZY. To prove: XY = YZ = ZX. Proof: Statement 1. XY = ZX. 2. XY = YZ. 3. XY = YZ = ZX. (Proved) Reason 1. Sides opposite to equal angles ∠XZYMon, 7 Jan 2019 13:13:36 -0500Jan 5, Sides Opposite to the Equal Angles of a Triangle are Equal | Diagram
https://www.math-only-math.com/sides-opposite-to-the-equal-angles-of-a-triangle-are-equal.html1eb193fbbb41ede3607c33cc6cc025d4Here we will prove that the sides opposite to the equal angles of a triangle are equal. Given: In ∆ABC, ∠XYZ = ∠XZY. To prove: XY = XZ. Construction: Draw the bisector XM of ∠YXZ so that it meets YZ at M. Proof: Statement 1. In ∆XYM and ∆XZM, (i) ∠XYM = XZM (ii) ∠YXM = ∠ZXMSat, 5 Jan 2019 15:16:59 -0500Jan 3, The Three Angles of an Equilateral Triangle are Equal | With Diagram
https://www.math-only-math.com/the-three-angles-of-an-equilateral-triangle-are-equal.html83aa43b02c0d02be891e3a333d04daa7Here we will prove that the three angles of an equilateral triangle are equal. Given: PQR is an equilateral triangle. To prove: ∠QPR = ∠PQR = ∠ PRQ. Proof: Statement 1. ∠QPR = ∠PQR 2. ∠PQR = ∠ PRQ. 3. ∠QPR = ∠PQR = ∠ PRQ. (Proved). Reason 1. Angles opposite to equal sides QRThu, 3 Jan 2019 17:19:18 -0500Jan 2, Equal Sides of an Isosceles Triangle are Produced, the Exterior Angles
https://www.math-only-math.com/equal-sides-of-an-isosceles-triangle.htmla3c095c9319d6d02b9d55a1d504b1b1aHere we will prove if the equal sides of an isosceles triangle are produced, the exterior angles are equal. Given: In the isosceles triangle PQR, the equal sides PQ and PR are produced to S and T respectively. To prove: ∠RQS = ∠QRT. Proof: Statement 1. ∠PQR = ∠PRQ 2. ∠RQSWed, 2 Jan 2019 14:23:00 -0500Jan 1, Angles Opposite to Equal Sides of an Isosceles Triangle are Equal
https://www.math-only-math.com/angles-opposite-to-equal-sides-of-an-isosceles-triangle-are-equal.htmld8f4149932935093a97f036d8997d0e8Here we will prove that in an isosceles triangle, the angles opposite to the equal sides are equal. Solution: Given: In the isosceles ∆XYZ, XY = XZ. To prove ∠XYZ = ∠XZY. Construction: Draw a line XM such that it bisects ∠YXZ and meets the side YZ at M. Proof: StatementTue, 1 Jan 2019 16:41:15 -0500Dec 31, Application of Congruency of Triangles | Isosceles triangle Proved
https://www.math-only-math.com/application-of-congruency-of-triangles.htmle3caaa91c392d548571f78342909ef46Here we will prove some Application of congruency of triangles. PQRS is a rectangle and POQ an equilateral triangle. Prove that SRO is an isosceles triangle. Solution: Given: PQRS is a rectangle. POQ is an equilateral triangle to prove ∆SOR is an isosceles triangle. Proof:Mon, 31 Dec 2018 15:47:37 -0500Dec 28, Prove that the Bisectors of the Angles of a Triangle Meet at a Point
https://www.math-only-math.com/bisectors-of-the-angles-of-a-triangle-meet-at-a-point.html2850dc977aeb1d1e1cbf8fd4225a33dbHere we will prove that the bisectors of the angles of a triangle meet at a point. Solution: Given In ∆XYZ, XO and YO bisect ∠YXZ and ∠XYZ respectively. To prove: OZ bisects ∠XZY. Construction: Draw OA ⊥ YZ, OB ⊥ XZ and OC ⊥ XY. Proof: Statement 1. In ∆XOC and ∆XOBFri, 28 Dec 2018 17:45:43 -0500Dec 27, Point on the Bisector of an Angle | Corresponding Parts of a Triangles
https://www.math-only-math.com/point-on-the-bisector-of-an-angle.html3c7d75335af6158ef3c26b7858d4c5ceHere we will prove that any point on the bisector of an angle is equidistant from the arms of that angle. Solution: Given OZ bisects ∠XOY and PM ⊥ XO and PN ⊥ OY. To prove PM = PN. Proof: Statement 1. In ∆OPM and ∆OPN, (i) ∠MOP = ∠NOP. (ii) ∠OMP = ∠ONP = 90°Thu, 27 Dec 2018 17:08:14 -0500