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CBSE:10th Math :Chapter Triangles : Solved Concept... : If a line divides any two sides of a triangles proportionally then the line is parallel to the third side Statement:   If a line divides... Also try: X Similar Triangle test paper-1 Download File X Similar Triangle test paper-2 Download File more Download

JSUNIL'S X maths Chapter: 02 Polynomials Test paper :

ASSIGNMENT AREAS RELATED TO CIRCLES CLASS X : 1. The radius of the circle is 3 m. What is the circumference of another circle, whose area is 49 times that of the first? 2. Two circles touch externally. The sum of their areas is 130 p sq. cm and the distance between their centres is 14 cm. Find the radii of the circles. 3. A wire when bent in the form of an equilateral triangle encloses an area of 121 √3 cm2 . If the same wire is bent in the form of a circle, find the area of the circle. 4. The area enclosed between the two concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle. 5. A wheel of diameter 42 cm, makes 240 revolutions per minute. Find : (i) the total distance covered by the wheel in one minute. (ii) the speed of the wheel in km/hr. Read more » 10th Maths SA-2 Chapter  Quick links Quadratic Equations Circles Co-ordinate Geometry Arithmetic Progression

10th maths Extra score questions Pair of Linear Equation in Two Variable HOTS By JSUNIL TUTORIAL 1 mark questions :- Q. 1 Is the pair of linear equations consistent: - x + y = 3 ; 3x – 2y = 4 . Q. 2 Find out whether the lines representing the following pair of linear equations are parallel or coincide:- 6x – 3y +10 = 0 ; 2x – y + 9 = 0 Q. 3 Write the value of k for which the following pair of linear equations has unique solution:- x + k y + 6 = 0 ; 2x + 3y + 8 = 0 Q. 4 Write the value of k for which the following pair of linear equations has no solution: - 4x +y =11 ; k x + 3y =5 Q. 5 Write the value of k for which the system of equations has infinite solutions:- 2x – 3y +10 = 0 ; 3x – ky +15 = 0 Q. 6 Does (2, –3) lie on the linear equation 3x – 2y + 5 = 0 Read more »

MATH ADDA "10TH MATHS BY JSUNIL": 10th maths polynomial extrascore questions : Math Adda Chapter: POLYNOMIAL YEAR-2011-2012 LEVEL-I 1. The zeroes of the polynomial 2x 2 -3x-2 are a. ...

MATH ADDA "10TH MATHS BY JSUNIL": 10th Number system solved important question : 1. Prove that one of every three consecutive integers is divisible by 3. Ans: n,n+1,n+2 be three consecutive positive integers We k...

CBSE MATH STUDY: Problem related to factorization,ratiolization : Problem-1 If the values of a – b and ab are 6 and 40 respectively, find the values of a 2 + b 2 and (a + b) 2 . Solution: a 2 + b 2 = (a – b) 2 + 2ab = 6 2 + 2(40) = 36 + 80 = 116 (a + b) 2 = (a – b) 2 + 4ab = 6 2 + 4(40)= 36 + 160 = 196 Problem-2. If (x + p)(x + q) = x 2 – 5x – 300, find the value of p 2 + q 2 . Solution: By product formula, we have (x + p) (x + q) = x 2 + (p + q)x + pq. So, by comparison, we get p + q = –5, pq = –300. Now, we have p 2 + q 2 = (p + q) 2 – 2 pq = (–5) 2 –2(–300) = 25 + 600 = 625. Problem-3. If (x + a)(x + b)(x + c) ≡ x 3 – 6x 2 + 11x – 6, find the value of a 2 + b 2 + c 2 . Solution: From the product formula, we have (x+a)(x+b)(x+c) = x 3 + (a + b + c)x 2 + (ab + bc + ca)x + abc. Comparing, we get a + b + c = –6, ab + bc + ca = 11, abc = –6. ∴ a 2 + b 2 + c 2 = (a + b + c) 2 –2 (ab + bc + ca) = (– 6) 2 – 2(11) = 36 – 22 = 14. Problem-4 If a+b=2 and a2+b2=8,find a3+b3 and a4+b4. CB

1. The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Justify your answer No. According to Euclid’s division lemma, a = 3q + r, where 0 ≤ r < 3 and r is an integer. Therefore, the values of r can be 0, 1 or 2. 2. Can the number 6n, n being a natural number, end with the digit 5? Give reasons. : No, because 6 n = (2 × 3) n = 2 n × 3 n , so the only primes in the factorization of 6 n are 2 and 3, and not 5. Hence, it cannot end with the digit 5. 3. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer. No, because an integer can be written in the form 4q, 4q+1, 4q+2, 4q+3. 4. “The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons. True, because n (n+1) will always be even, as one out of n or (n+1) must be even 5. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or f

Trigonometric Ratios of 30 0 and 60 0 Let D ABC be an equilateral triangle whose sides have length a Draw AD ^ BC, then D bisects the side BC. So, BD =  DC=  a/2 and < BAD = < DAC = 30 0 . Now,  in right triangle ADB, < BAD = 30 0 and BD=  a/2 So, AB 2 =   AD 2 +  BD 2 a 2   = AD 2 + (a/2) 2 AD  = √ 3/2 a   Hence, we can find the trigonometric ratios of angle 30 0 from the right triangle BAD

The word trigonometry is a derivation from the Greek language and means measurement of triangles. Hipparchus , a Greek astronomer and mathematician developed the subject trigonometry.  The first trigonometric table was compiled by him. He is now known as “the Father of Trigonometry”. The side that is opposite to the right angle is called the Hypotenuse . This is the longest side in a right triangle. The side that is opposite to the angle i is called the Opposite sid e.

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Preparation of Question Bank of classes IX & X for Term - I (September 2011) Class X New Marking scheme CBSE sample paper 10th Polynomials polynomials grade 10 test paper Polynomials-linear equation test paper Polynomial test paper for class 10 10th Math Test Paper (15)

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1. In  Δ  PQR, given that S is a point on PQ such that ST II  QR and PS/SQ=3/5 If PR = 5.6 cm, then find PT. 2. In  Δ  ABC, AE is the external bisector of <A, meeting BC produced at E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then find CE. 3. P and Q are points on sides AB and AC respectively, of  Δ ABC. If AP = 3 cm,PB = 6 cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ. 4. The image of a tree on the film of a camera is of length 35 mm, the distance from the lens to the film is 42 mm and the distance from the lens to the tree is 6 m. How tall is the portion of the tree being photographed? 5. D is the midpoint of the side BC of  Δ  ABC. If P and Q are points on AB and on AC such that DP bisects <BDA and DQ bisects <ADC, then prove that PQ II BC. 6. If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. 7. If a straight line divides any two sides of a triangle in the same ratio,

Section -A 1.The pair of equations x + 2 y + 5 = 0 and –3 x – 6 y + 1 = 0 have (A) a unique solution (B) exactly two solutions        (C) infinitely many solutions (D) no solution 2. If a pair of linear equations is consistent, then the lines will be (A) parallel      (B) always coincident             (C) intersecting or coincident             (D) always intersecting 3. The pair of equations y = 0 and y = –7 has (A) one solution          (B) two solutions        (C) infinitely many solutions              (D) no solution 4.If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is (A) 1                                        (B)1/2                          (C) 2                                                   (D) 3 5. Given that sinα =1/2 and cosβ =1/2 , then the value of (α + β) is (A) 0°              (B) 30°                                    (C) 60°                                                            (D) 90° Section - B 1.For which values of p and q , will

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Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides. Given that ABCD be a parallelogram. Construction : We have to Draw AF and DE perpendicular on side DC and on extended side AB to use Pythagoras theorem In ΔDEA, DE2 + EA2 = DA2 … (i) Similarly, In ΔDEB, DE2 + EB2 = DB2 DE2 + (EA + AB)2 = DB2 ( EB =EA + AB) DE2 + EA2 + AB2 + 2EA × AB == DB2 (DE2 + EA2) + AB2 + 2EA × AB = DB2 DA2 + AB2 + 2EA × AB = DB2 … (ii)( Using -----1) In ΔAFC, AC2 = AF2 + FC2 = AF2 + (DC − FD)2 = AF2 + DC2 + FD2 − 2DC × FD = (AF2 + FD2) + DC2 − 2DC × FD =AC2 = AD2 + DC2 − 2DC × FD ( using AD2 = AF2 + FD2)… (iii) Now,given that  ABCD is a parallelogram, AB = CD … (iv) And, BC = AD … (v) Look In ΔDEA and ΔADF, ∠DEA = ∠AFD (Both 90°) ∠EAD = ∠ADF (EA || DF -Alternate angles ) AD = AD (Common) ∴ ΔEAD cong. ΔFDA (AAS congruence criterion) ⇒ EA = DF … (vi) Adding equations (i) and (iii), we obtain DA2 +

Real_numbers_test_paper.1 Download File Real_numbers_test_paper-2 Download File Real_number_test_paper (solved)-3 Download File 10th_real_number_objectives_questions.pdf Download File Polynomial_cbse_test_paper-1 Download File Linear equations 2-variables_MCQ-test paper-1 Download File Linear_equations_in_two_variables_test paper-2 Download File Linear_equations_in_two_variables_test paper-3 Download File Linear_equations_in_two_variables_test paper-4 Download File Linear_equations_in_two_variables_test paper-5 Download File Linear_equations_in_two_variables_test paper-6 Download File Triangle test paper-1 Download File Triangle test paper-2 Download File Triangle test paper-3 Download File    For more test paper  Click Me