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Practice the questions given in the worksheet on application problems on expansion of powers of binomials and trinomials. 1. Use (a ± b)^2 = a^2 ± 2ab + b2 to evaluate the following: (i) (3.001)^2 (ii) (5.99)^2 (iii) 1001 × 999 (iv) 5.63 × 5.63 + 11.26 × 2.37 + 2.37 × 2.37

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Here we will solve different types of application problems on expansion of powers of binomials and trinomials. 1. Use (x ± y)^2 = x^2 ± 2xy + y^2 to evaluate (2.05)^2. Solution: (2.05)^2 = (2 + 0.05)^2 = 2^2 + 2 × 2 × 0.05 + (0.05)^2 = 4 + 0.20 + 0.0025 = 4.2025.

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We will discuss here about the expansion of (x + a)(x + b)(x + c). (x + a)(x + b)(x + c) = (x + a){(x + b)(x + c)} = (x + a){x\(^{2}\) + (b + c)x + bc} = x{x\(^{2}\) + (b + c)x + bc} + a{x\(^{2}\) + (b + c)x + bc} = x\(^{3}\) + (b + c)x\(^{2}\) + bcx + ax\(^{2}\) + a(b + c)x

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We will discuss here about the expansion of (a + b + c)(a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca). (a + b + c)(a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca) = a(a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca) + b(a\(^{2}\) + b\(^{2}\) + c\(^{2}\) –ab – bc – ca)

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We will discuss here about the expansion of (a ± b)(a\(^{2}\) ∓ ab + b\(^{2}\)). (a + b)(a\(^{2}\) - ab + b\(^{2}\)) = a(a\(^{2}\) - ab + b\(^{2}\)) + b(a\(^{2}\) - ab + b\(^{2}\)) = a\(^{3}\) - a\(^{2}\)b + ab\(^{2}\) + ba\(^{2}\) - ab\(^{2}\) + b\(^{3}\) =

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We will discuss here about the expansion of (a ± b)\(^{3}\). (a + b)\(^{3}\) = (a + b) ∙ (a + b)\(^{2}\) = (a + b)(a\(^{2}\) + 2ab + b\(^{2}\)) = a(a\(^{2}\) + 2ab + b\(^{2}\)) + b(a\(^{2}\) + 2ab + b\(^{2}\))=a\(^{3}\)+2a\(^{2}\)b+ab\(^{2}\)+ba\(^{2}\)+2ab\(^{2}\)+b\(^{3}\)

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Here we will express a^2 + b^2 + c^2 – ab – bc – ca as sum of squares. If a, b, c are real numbers then (a – b)^2, (b – c)^2 and (c – a)^2 are positive as square of every real number is positive. So, a^2 + b^2 + c^2 – ab – bc – ca is always positive.

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Practice the questions given in the worksheet on simplification of (a + b)(a – b). 1. Simplify by applying standard formula. (i) (5x – 9)(5x + 9) (ii) (2x + 3y)(2x – 3y) (iii) (a + b – c)(a – b + c) (iv) (x + y – 3)(x + y + 3) (v) (1 + a)(1 – a)(1 + a^2)

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Practice the questions given in the worksheet on completing square. Write the following as a perfect square. (i) 4X^2 + 4X + 1 (ii) 9a^2 – 12ab + 4b^2 (iii) 1 + 6/a + 9/a^2 2. Indicate the perfect squares among the following. Express each of the perfect squares as the square

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Here we will learn how to completing a square.Problems on completing a square 1. What should be added to the polynomial 4m^2 + 8m so that it becomes perfect square? Solution: 4m^2 + 8m = (2m)^2 + 2 ∙ (2m) ∙ 2

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Practice the questions given in the worksheet on expansion of (x ± a)(x ± b). 1. (i) Find the product using standard formula. (i) (x + 2)(x + 5) (ii) (a – 4)(a – 7) (iii) (x + 1)(x – 8) (iv) (a – 3)(a + 2) (v) (3x + 1)(3x + 2) (vi) (4x – y)(4x + 2y) 2. Find the product.

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Practice the questions given in the worksheet on expanding of (a ± b ± c)^2 and its corollaries. 1. Expand the squares of the following trinomials. (i) a + 2b + 3c (ii) 2x + 3y + 4z (iii) x + 2y – 3z (iv) 3a – 4b – c (v) 1 – x - \(\frac{1}{x}\) (vi) 1 – a – a^2. 2. Simplify:

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Practice the questions given in the worksheet on expansion of (a ± b)^2 and its corollaries. 1. Expand the squares of the following: (i) 4x + y (ii) 5a + 3b (iii) 2x + \(\frac{1}{x}\) 2. Expand the following: (i) (x – 2y)^2 (ii) (3y – 2z)^2 (iii) (3x - \(\frac{1}{3x}\))^2

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We will discuss here about the Simplification of (a + b)(a – b). (a + b)(a – b) = a(a – b) + b(a – b) = a\(^{2}\) - ab + ba - b\(^{2}\) = a\(^{2}\) - b\(^{2}\) Thus, we have (a + b)(a - b) = a\(^{2}\) - b\(^{2}\) Solved Examples on Simplification of (a + b)(a – b) 1.

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We will discuss here about the expansion of (x ± a)(x ± b) (x + a)(x + b) = x(x + b) + a (x + b) = x^2 + xb + ax + ab = x^2 + (b + a)x + ab (x - a)(x - b) = x(x - b) - a (x - b) = x^2 - xb - ax + ab = x^2 - (b + a)x + ab (x + a)(x - b) = x(x - b) + a (x - b) = x^2 - xb

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We will discuss here about the expansion of (a ± b ± c)^2. (a + b + c)^2 = {a + (b + c)}^2 = a^2 + 2a(b + c) + (b + c)^2 = a^2 + 2ab + 2ac + b^2 + 2bc + c^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = sum of squares of a, b, c + 2(sum of the products of a, b, c taking two at a ti

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

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

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

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

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A binomial is an algebraic expression which has exactly two terms, for example, a ± b. Its power is indicated by a superscript. For example, (a ± b)2 is a power of the binomial a ± b, the index being 2. A trinomial is an algebraic expression which has exactly three terms

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

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

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

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Here we will prove that if each diagonal of a quadrilateral divides it in two triangles of equal area then prove that the quadrilateral is a parallelogram. Solution: Given: PQRS is a quadrilateral whose diagonals PR and QS cut at O such that ar(∆PQR) = ar(∆PSR), and

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Here we will prove that the area of a rhombus is equal to half the product of its diagonals. Solution: Given: PQRS is a rhombus whose diagonals are PR and QS. The diagonals intersect at O. To prove: ar(rhombus PQRS) = 1/2 ×PR × QS. Statement ar(∆RSQ) = 1/2 ×Base × Altitude

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Here we will prove that the area of the triangle formed by joining the middle points of the sides of a triangle is equal to one-fourth area of the given triangle. Solution: Given: X, Y and Z are the middle points of sides QR, RP and PQ respectively of the triangle PQR.

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Here we will learn how to solve different types of problems on finding area of triangle and parallelogram. 1. In the figure, XQ ∥ SY, PS ∥ QR, XS ⊥ SY, QY ⊥ SY and QY = 3 cm. Find the areas of ∆MSR and parallelogram PQRS. Solution: ar(∆MSR) = 1/2 × ar(rectangle of SR

Continue reading "Problems on Finding Area of Triangle and Parallelogram | With Diagram"

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

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

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

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

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

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

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

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

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

Continue reading "Bisectors of the Angles of a Parallelogram form a Rectangle | Diagram"

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

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

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

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

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

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

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

Continue reading "What is Rectilinear Figure? | What is Diagonal of a Polygon? | Polygon"

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

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

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

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

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

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