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Jun 16, 2019
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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Jun 16, 2019
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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Jun 14, 2019
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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Jun 10, 2019
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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Jun 07, 2019
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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Jun 07, 2019
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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Jun 06, 2019
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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Jun 05, 2019
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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Jun 03, 2019
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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Jun 03, 2019
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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May 29, 2019
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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May 28, 2019
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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May 28, 2019
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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May 27, 2019
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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May 27, 2019
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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May 27, 2019
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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May 23, 2019
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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May 21, 2019
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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May 20, 2019
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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May 18, 2019
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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May 10, 2019
Volume and Surface Area of Cube | Diagonal a Cube | Total surface Area
What 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)
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May 10, 2019
Volume and Surface Area of Cuboid | Lateral Surface Area of a Cuboid
What 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.
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May 07, 2019
Solid Figures | Volume of a Solid Figure |Surface Area of Solid Figure
A 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.
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May 04, 2019
Riders Based on Pythagoras’ Theorem | Establishing Rides | Diagram
Here 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 of
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Apr 30, 2019
Criteria of Similarity between Triangles | SAS Criterion of Similarity
We 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.
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Apr 29, 2019
AA Criterion of Similarly on Quadrilateral | Alternate Angles
Here 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)
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Apr 25, 2019
Pythagoras’ Theorem | AA Criterion of Similarity | Proof with Diagram
The 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 is
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Apr 23, 2019
Applying Pythagoras’ Theorem | Proof by Pythagoras’ Theorem | Diagram
Applying 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 = 2QN
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Apr 23, 2019
Converse of Pythagoras’ Theorem | Sum of the Squares of Two Sides
If 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°
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Apr 20, 2019
Problems on Size Transformation | Area of the Terrace in the Model
Here 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 its
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Apr 20, 2019
Reduction Transformation | Centre of Reduction | Reduction Factor
We 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 \(\frac{X’Y’}{XY}\) = \(\frac{Y’Z’}{YZ}\) = \(\frac{Z’X’}{ZX}\) = k
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Apr 20, 2019
Enlargement Transformation | Centre of Enlargement |Enlargement Factor
We 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.
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Apr 13, 2019
Greater segment of the Hypotenuse = the Smaller Side of the Triangle
Here 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 the
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Apr 06, 2019
Application of Basic Proportionality Theorem | Internal Bisector
Here 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.
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Apr 04, 2019
Converse of Basic Proportionality Theorem | Proof with Diagram
Here 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: Statement
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Mar 28, 2019
Basic Proportionality Theorem | AA Criterion of Similarity | Diagram
Here 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.
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Mar 28, 2019
Similar Triangles | Congruency and Similarity of Triangles | Diagram
We 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 is
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Mar 25, 2019
AA Criterion of Similarity | Similarity on Right-angled Triangle
Here 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 one
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Mar 19, 2019
Properties of size Transformation |Enlargement|Reduction|Scale Factor
We 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.
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Mar 08, 2019
Proof By the Equal Intercepts Theorem | Line Joining the Midpoints
Here 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.
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Mar 08, 2019
Midpoint Theorem by using the Equal Intercepts Theorem |Proof |Diagram
Here 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.
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Mar 05, 2019
Problems on Equal Intercepts Theorem | Midpoint Theorem | Diagram
Here 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 the
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Feb 26, 2019
Equal Intercepts Theorem | Transversal makes Equal Intercepts
Intercept 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 lines
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Feb 25, 2019
Collinear Points Proved by Midpoint Theorem | Collinearity | Diagram
In ∆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.
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Feb 22, 2019
Midpoint Theorem on Right-angled Triangle | Proof | Statement | Reason
Here 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 PR
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Feb 21, 2019
Midsegment Theorem on Trapezium | Nonparallel Sides of a Trapezium
Here 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 the
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Feb 20, 2019
Midpoint Theorem on Trapezium | Converse of the Midpoint Theorem
PQRS 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. Construction
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Feb 13, 2019
Straight Line Drawn from the Vertex of a Triangle to the Base |Diagram
Here 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 of
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Feb 12, 2019
Four Triangles which are Congruent to One Another | Prove with Diagram
Here 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 ≅ LNM
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