10th class maths important formulae
Answers
Real number
Euclid’s division algorithm (lemma): Given positive integers ‘a’ and ‘b’, there exists unique integers q and r such that a= b.q+r, where 0 <=r< b ( where a= dividend, b= divisor, q= quotient, and r= remainder.
Polynomials
In step 1 : Factorize the given polynomials,
a) Either by splitting the terms, ( OR )
b) Using these identities :
(a+b)^2 = a*a + 2ab+b*b
(a-b)^2 = a*a – 2ab +b*b
a*a – b*b = (a+b)( a-b)
a^4 – b^4 = (a^2)^2 –( b^2)^2= (a*a+ b*b)(a*a – b*b) = (a*a + b*b)(a-b)(a+b)
(a+b)^3 = a^3 + b^3 + 3ab(a+b)
a^3 + b^3 =( a+b)(a*a +ab + b*b)
(a-b)^3 = a^3- b^3 – 3ab (a-b)
a^3- b^3= (a-b) (a*a+ ab +b*b)
(a+b+c)^2 = a*a + b*b +c*c + 2ab+2bc+2ca
a^3+b^3+c^3- 3abc = (a+b+c)(a*a + b*b + c*c –ab – bc- ac)
Remainder theorem
If (x-2) is a factor of the given expression, then take x-2 = 0 , therefore x = 2 , then substitute this value in p(x) = 5 x*x + 3 x-6
P(2) : 5(2*2) + 3(2)-6 =0 (here taking = 0 is very important. If not taken, answer can’t be found)
If (x-2) leaves a remainder of 4
P(2) : 5(2*2) + 3 (2)-6 =4 ( Here taking = 4 is very important. If not taken, answer can’t be found)
Linear equations in two variables
If pair of linear equation is : a1 + b1y +c1 =0 and a2x + b2 y + c2=0
Then nature of roots/zeroes/solutions :
i. If a1/a2 is not equal to b1/b2 then, system has unique solution, is consistent OR graph is two intersecting lines
ii. If a1/a2 = b1/b2 is not equal to c1/c2 , then system has no solution, is inconsistent OR graph is parallel lines.
iii. If a1/a2=b1/b2=c1/c2, then system has infinite solution, is consistent OR graph are coincident lines.
Quadratic Equations
Note: To find the value of ‘x’ you may adopt either ‘splitting the middle term’ or ‘formula method’.
X=( -b +-(D)^.5)/2a (where D= b*b – 4ac) Hence x= (-b+- (b*b- 4ac)^.5)/2a
Sum of the roots = -b/a & Product of roots= c/a
If roots of an equation are given, then :
Quadratic equation : x*x – (Sum of roots).x + (product of the roots) =0
If Discriminant > 0, then the roots are Real & unequal or unique, lines are intersecting.
Discriminant = 0, then the roots are real & equal , lines are coincident.
Discriminant< 0 , then the roots are imaginary (not real), parallel lines.
Ratio & proportion
Duplicate ratio of a : b is a*a : b*b (Incase of Sub-duplicate ratio you have to take ‘Square root’)
Triplicate ratio of a: b is a^3 : b^3 (Incase of Sub-triplicate ratio you have to take ‘cube root’)
Proportion a:b =c:d, continued proportion a :b = b : c, (Middle value is repeated)
Product of ‘Means’( Middle values)= Product of ‘Extremes’ (Either end values)
If a/b= c/d is given then Componendo & dividend is (a+b)/ (a-b) = (c+d) / (c-d)
Note : “Where to take “K” method ? “ You may adopt it in the following situations.
If a/b = c/d = e/f are given , then you may assume as a/b= c/d= e/f =k
Therefore a = b.k, c= d.k, e= f.k, then substitute the values of ‘a’ ‘b’ and c’c’ in the given problem.
Incase of continued proportion : a/b = b/c = k , hence , a=bk, b= ck therefore putting the value of b we can get a= c k*k & b= ck.( putting these values equation can be solved)
Similarity
If two triangles are similar then, ratio of their sides are equal.
i.e if triangle ABC~ triangle PQR then AB/PQ = BC/QR = AC/PR
If triangle ABC ~ triangle PQR then (Area of triangle ABC)/ Area of triangle PQR) = (side*side) /(side*side) = ( AB*AB) /( PQ*PQ) =( BC*BC)/( QR*QR) = (AC*AC)/ (PR*PR)
Distance and section formulae
Distance =(( x2- x1 )^2 + y2-y1)^2))^.5 ( The same formula is to be used to find the length of line segment, sides of a triangle , square , rectangle, parallelogram etc.)
To prove co-linearity of the given three points A,B and C, you have to find length of AB, BC, AC then use the condition AB + BC = AC .OR use this condition to solve the question easily :
Area of triangle formed by these points : 0.5 [x1 ( y2-y3)+ x2(y3 – y1) + x3(y1- y2)]=0
Section formula : point (x,y)=[ ( m1x2 + m2x1)/( m1 + m2) , (m1y2 + m2y1)/ (m1 + m2)]
Mid-point =[( x1 + x2 )/2 ,( y1 +y2 )/2 ]
Centroid of a triangle=[( x1+x2+x3)/3 ,( y1+y2+y3)/3]
If line is trisected then take m:n ratio as 1:2 and find co- ordinates of point p(x,y).
Equation of a line
If two points are given, then Slope (m) = (y2-y1)/(x2-x1)
If a point , and slope are given , then Slope (m)= (y-y1)/(x-x1)
If two lines are ‘Parallel’ to each other then their slopes are equal i.e m1=m2.
If two lines are ‘Perpendicular’ to each other then product of their slopes is -1 i.e m1 *m2 = -1
Depending upon the question You may have to use equation of straight line as
a) Y=mx + c, where ‘c’ is the y- intercept. OR b) (y-y1)= m.( x-x1)
Answer:
Real number
Euclid’s division algorithm (lemma): Given positive integers ‘a’ and ‘b’, there exists unique integers q and r such that a= b.q+r, where 0 <=r< b ( where a= dividend, b= divisor, q= quotient, and r= remainder.
Polynomials
In step 1 : Factorize the given polynomials,
a) Either by splitting the terms, ( OR )
b) Using these identities :
(a+b)^2 = a*a + 2ab+b*b
(a-b)^2 = a*a – 2ab +b*b
a*a – b*b = (a+b)( a-b)
a^4 – b^4 = (a^2)^2 –( b^2)^2= (a*a+ b*b)(a*a – b*b) = (a*a + b*b)(a-b)(a+b)
(a+b)^3 = a^3 + b^3 + 3ab(a+b)
a^3 + b^3 =( a+b)(a*a +ab + b*b)
(a-b)^3 = a^3- b^3 – 3ab (a-b)
a^3- b^3= (a-b) (a*a+ ab +b*b)
(a+b+c)^2 = a*a + b*b +c*c + 2ab+2bc+2ca
a^3+b^3+c^3- 3abc = (a+b+c)(a*a + b*b + c*c –ab – bc- ac)
Remainder theorem
If (x-2) is a factor of the given expression, then take x-2 = 0 , therefore x = 2 , then substitute this value in p(x) = 5 x*x + 3 x-6
P(2) : 5(2*2) + 3(2)-6 =0 (here taking = 0 is very important. If not taken, answer can’t be found)
If (x-2) leaves a remainder of 4
P(2) : 5(2*2) + 3 (2)-6 =4 ( Here taking = 4 is very important. If not taken, answer can’t be found)
Linear equations in two variables
If pair of linear equation is : a1 + b1y +c1 =0 and a2x + b2 y + c2=0
Then nature of roots/zeroes/solutions :
i. If a1/a2 is not equal to b1/b2 then, system has unique solution, is consistent OR graph is two intersecting lines
ii. If a1/a2 = b1/b2 is not equal to c1/c2 , then system has no solution, is inconsistent OR graph is parallel lines.
iii. If a1/a2=b1/b2=c1/c2, then system has infinite solution, is consistent OR graph are coincident lines.
Quadratic Equations
Note: To find the value of ‘x’ you may adopt either ‘splitting the middle term’ or ‘formula method’.
X=( -b +-(D)^.5)/2a (where D= b*b – 4ac) Hence x= (-b+- (b*b- 4ac)^.5)/2a
Sum of the roots = -b/a & Product of roots= c/a
If roots of an equation are given, then :
Quadratic equation : x*x – (Sum of roots).x + (product of the roots) =0
If Discriminant > 0, then the roots are Real & unequal or unique, lines are intersecting.
Discriminant = 0, then the roots are real & equal , lines are coincident.
Discriminant< 0 , then the roots are imaginary (not real), parallel lines.
Ratio & proportion
Duplicate ratio of a : b is a*a : b*b (Incase of Sub-duplicate ratio you have to take ‘Square root’)
Triplicate ratio of a: b is a^3 : b^3 (Incase of Sub-triplicate ratio you have to take ‘cube root’)
Proportion a:b =c:d, continued proportion a :b = b : c, (Middle value is repeated)
Product of ‘Means’( Middle values)= Product of ‘Extremes’ (Either end values)
If a/b= c/d is given then Componendo & dividend is (a+b)/ (a-b) = (c+d) / (c-d)
Note : “Where to take “K” method ? “ You may adopt it in the following situations.
If a/b = c/d = e/f are given , then you may assume as a/b= c/d= e/f =k
Therefore a = b.k, c= d.k, e= f.k, then substitute the values of ‘a’ ‘b’ and c’c’ in the given problem.
Incase of continued proportion : a/b = b/c = k , hence , a=bk, b= ck therefore putting the value of b we can get a= c k*k & b= ck.( putting these values equation can be solved)
Similarity
If two triangles are similar then, ratio of their sides are equal.
i.e if triangle ABC~ triangle PQR then AB/PQ = BC/QR = AC/PR
If triangle ABC ~ triangle PQR then (Area of triangle ABC)/ Area of triangle PQR) = (side*side) /(side*side) = ( AB*AB) /( PQ*PQ) =( BC*BC)/( QR*QR) = (AC*AC)/ (PR*PR)
Distance and section formulae
Distance =(( x2- x1 )^2 + y2-y1)^2))^.5 ( The same formula is to be used to find the length of line segment, sides of a triangle , square , rectangle, parallelogram etc.)
To prove co-linearity of the given three points A,B and C, you have to find length of AB, BC, AC then use the condition AB + BC = AC .OR use this condition to solve the question easily :
Area of triangle formed by these points : 0.5 [x1 ( y2-y3)+ x2(y3 – y1) + x3(y1- y2)]=0
Section formula : point (x,y)=[ ( m1x2 + m2x1)/( m1 + m2) , (m1y2 + m2y1)/ (m1 + m2)]
Mid-point =[( x1 + x2 )/2 ,( y1 +y2 )/2 ]
Centroid of a triangle=[( x1+x2+x3)/3 ,( y1+y2+y3)/3]
If line is trisected then take m:n ratio as 1:2 and find co- ordinates of point p(x,y).
Equation of a line
If two points are given, then Slope (m) = (y2-y1)/(x2-x1)
If a point , and slope are given , then Slope (m)= (y-y1)/(x-x1)
If two lines are ‘Parallel’ to each other then their slopes are equal i.e m1=m2.
If two lines are ‘Perpendicular’ to each other then product of their slopes is -1 i.e m1 *m2 = -1
Depending upon the question You may have to use equation of straight line as
a) Y=mx + c, where ‘c’ is the y- intercept. OR b) (y-y1)= m.( x-x1)