# Math Blog

### 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 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 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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### 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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### 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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### 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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### 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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### 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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### 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)

### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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### 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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