#CBSEClass10Maths

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Chapter 1 Real Numbers | Class 10 Maths | Support Material 2022-23 Solution | CBSE Board Exam 2024

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DIVISION OF A LINE SEGENT

DIVISION OF A LINE SEGENT :

This content is part of NCERT Solutions for Class 10 Maths Chapter 11 Constructions

           In order to divide a line segment internally is a given ratio m: n, where both m and n are positive integers,

 we follow the following steps:

Step of construction :

           (i)         Draw a line segment AB of given length by using a ruler.

(ii)         Draw and ray AX making an acute angle with AB.

(iii)        Along AX mark off (m + n) points A1, A2,..., Am+n such that AA1 = A1A2 = ....=Am+n+Am+n.

(iv)        Join B Am+n

(v)        Through the point Am draw a line parallel to Am+n B by making an angle equal to at Am.

Suppose this line meets AB at a point P.

           The point P so obtained is the required point which divides AB internally in the ratio m : n.

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NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals

This content is part of NCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals

GENERAL PROPERTIES OF METALS AND NON-METALS :

 7.3 (a) Electronic Configuration of Metals :

 The atoms of metals have 1 to 3 electrons in their outermost shells. For example, all the alkali metals have one electron in their outermost shells (lithium 2, 1; sodium 2,8,1: potassium 2,8,8,1 etc.)

Sodium, magnesium and aluminum are metals having 1,2 and 3 electrons respectively in their valence shells. Similarly, other metals have 1 to 3 electron in their outermost shells.

It may be noted that hydrogen and helium are exception because hydrogen is a non-metal having only  electron in the outermost shell (K shell) of its atom and helium is also a non-metal having 2 electron in the outermost shell (K shell).

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NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

This content is part of NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes

MENSURATION (SOLID FIGURES) :

           If any figure such as cuboids, which has three dimensions length, width and height are height are known as three dimensional figures. Where as rectangle has only two dimensional i.e., length and width. Three dimensional figures have volume in addition to areas of surface from which these soils figures are formed.

           Some of the main solid figures are:

 13.3     (a) Cuboid:

             Total Surface Area (T.S.A.) : The area of surface from which cuboid is formed. There are six faces (rectangular), eight vertices and twelve edges n a cuboid.

           (i)Total Surface Area (T.S.A.)  

             (ii) Lateral Surface Area (L.A.A.) =

             (or Area of 4 walls)                   =

             (iii)       Volume of Cuboid = (Area of base) × height

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NCERT Solutions for Class 10 Maths Chapter 10 Circles

This content is part of NCERT Solutions for Class 10 Maths Chapter 10 Circles

9.4        THEORM :

 Statement :       Lengths of two tangents drawn from an external point to a circle are equal.

Given:              AP and AQ are two tangents drawn from a point A to a circle C (O, r).

 To prove :        AP = AQ

 Construction :  Join OP, OQ and OA.

 Proof :              In AOQ and APO    

                       [Tangent at any point of a circle is perp. to radius through the point of contact]

AO = AO           [Common]

OQ = OP          [Radius]

So, by R.H.S. criterion of congruency

          AQ = AP           [By CPCT]                                 Hence Proved.

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NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions: PROPERTIES OF A.P.

This content is part of NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic

PROPERTIES OF A.P. :

(A)         For any real numbers a and b, the sequence whose nth term is an = an + b is always an A.P. with common difference ‘a’ (i.e. coefficient of term containing n)

               (B)          If any nth term of sequence is a linear expression in n then the given sequence is an A.P.

(C)         If a constant term is added to or subtracted from each term of an A.P. then the resulting sequence is also an A.P. with the same common difference.

(D)         If each term of a given A.P. is multiplied or divided by a non-zero constant K, then the resulting sequence is also an A.P. with common difference Kd or respectively. Where d is the common difference of the given  A.P.

(E)          In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of 1st and last term.

               (F)           If three numbers a,b,c are in A.P., then 2b = a + c.

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NCERT Solutions for Class 10 Maths Chapter 14 Statistics

This content is part of NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Ex.1     The mean of marks scored by 100 students was found to be 40. Later on its was discovered that a score of

 56 was misread as 83. Find the correct mean.

 Sol.      n = 100, = 40

            Incorrect value of

           Now, Correct value of 4000 - 83 + 83 = 3970

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions: DIVISION OF A LINE SEGMENT

This content is part of NCERT Solutions for Class 10 Maths Chapter 11 Constructions

ALTERNATIVE METHOD FOR DIVISION OF A LINE SEGMENT INTERNALLY IN A GIVEN RATIO :

           Use the following steps to divide a given line segment AB internally in a given ration m : n, where m and

 natural members.

Steps of Construction :

             (i)         Draw a line segment AB of given length.

             (ii)         Draw any ray AZ making an acute angle with AB.

             (iii)        Draw a ray BY, on opposite side of AX, parallel to AX making an angle equal to .

             (iv)        Mark off a points A1, A2,....Am’ on AX and n points B1, B2,...Bn on BY such that AA1 = A1A2 = ...... =

Am-1 Am = B1B2 = ....Bn-1Bn.

             (v)        Join AmBn. Suppose it intersect AB at P.

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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations : NATURE OF ROOTS

This content is part of NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

NATURE OF ROOTS :

Consider the quadratic equation, ax2 + bx + c = 0 having as its roots and b2 - 4ac is called discriminate of roots of quadratic equation. It is denoted by D or D.

               Roots of the given quadratic equation may be

               (i) Real and unequal         (ii) Real and equal             (iii) Imaginary and unequal.

               Let the roots of the quadratic equation ax2 + bx + c = 0 (where ) be and then

                                        .....(i)

               and                  ....(ii)

The nature of roots depends upon the value of expression ‘b2 - 4ac’ with in the square root sign. This is known as discriminate of the given quadratic equation.

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NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

This content is part of NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Ex.2     A vertical tower stands on a horizontal plane and is surmounted by vertical flag staff of height 5 meters. At a point on the plane, the angle of elevation of the bottom and the top of the flag staff are respectively 300 and 600 find the height of tower.                                                                                                         [CBSE-2006]

Sol.      Let AB be the tower of height h metre and BC be the height of flag staff surmounted on the tower, Let the point of the place be D at a distance x meter from the foot of the tower in ABD

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Chapter 8 Introduction to Trigonometry | Class 10 Maths | Support Material 2022-23 Solution | CBSE Board Exam 2023

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CIRCLE

CIRCLE

This content is part of NCERT Solutions for Class 10 Maths Chapter 10 Circles

           A circle is the locus of a points which moves in a plane in such a way that its distance from a fixed point

 remains constant.

 9.2        SECANT AND TANGENT :

                  Secant to a circle is a line which intersects the circle in two distinct points.

                  A tangent to a circle is a line that intersects the circle in exactly one point.

 9.3        THEOREM :

Statement :       A tangent to a circle i perpendicular to the radius through the point of contact.

Given :             A circle C (O, r) and a tangent AB at a point P.

 To prove :        OP AB

 Construction :  Take any points Q, other than P on the tangent AB. Join OQ. Suppose OQ meets the circle at R.

 Proof:              Among all line segments joining the point O to a point on AB, the shorted one is perpendicular to

 AB. So, to prove that OP AB, it is sufficient to prove that OP is shorter than any other segment

 joining O to any point of AB.

Clearly  OP = OR

Now,     OQ  OR + RQ

      OQ > OR

      OQ > OP (OP = OR)

Thus, OP is shorter than any other segment joining O to any point of AB.

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NCERT Solutions for Class 10 Maths Chapter 3 CONDITIONS FOR SOLVABILITY

This content is part of NCERT Solutions for Class 10 Maths Chapter 3 Pairs of Linear Equations in Two Variables

CONDITIONS FOR SOLVABILITY (OR CONSISTENCY) OF SYSTEM OF EQUATIONS:

2.2          (a) Unique Solution :

Two lines a1 + b1y + c1 = 0 and a2x + b2y + c2 = 0, if the denominator a1b2 - a2b1 0 then the given system of equations have unique solution (i.e. only one solution) and solutions are said to be consistent.

               \     a1b2 - a2b1 0         Þ            

2.2 (b)  No Solution :

Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, if the denominator a1b2 - a2b1 = 0 then the given system of equations have no solution and solutions are said to be consistent.

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry: CENTROID OF A TRIANGLE

This content is part of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

CENTROID OF A TRIANGLE :

Prove that the coordinates of the triangle whose vertices are (x1, y1), (x2, y2) and (y3, y3) are . Also, deduce that the medians of a triangle are concurrent.

               Proof :

Let A(x1, y1, B(x2, y2) and C(x3, y3) be the vertices of DABC whose medians are AD, BE and CF respectively. So. D,E and F are respectively the mid-points of BC, CA and AB.

               Coordinates of D are . Coordinates of a point dividing AD in the ratio 2 : 1 are

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NCERT Solutions for Class 10 Maths Chapter 15 Probability

This content is part of NCERT Solutions for Class 10 Maths Chapter 15 Probability

PROBABILITY :

          If there are n elementary events associated with a random experiment and m of them are favorable to an event A, then the probability of happening or occurrence of event A is denoted by P(A)

             Thus,    P(A) =

           And      

           If,         P(A) = 0, then A is called impossible event

             If,         P(A) = 1, then A is called sure event

                       P(A) + P = 1

             Where   P(A) = probability of occurrence of A.

                       P = probability of non - occurrence of A.