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Class- 10
Term -1 COMPLETE MATHS
IN ONE SHOT REVISION
SUBSCRIBE FOR
MORE CONTENTS
CHAPTER - 1 REALNUMBERS
FORMULAS---
HCF(a,b) XLCM(a,b) =ax b
2n x 5m in adenominatorno
other number then it is
terminating
CHAPTER - 1 REALNUMBERS
Q1. Given that HCF(306,657)=9, findLCM (306,657).
CHAPTER - 1 REALNUMBERS
Findwhich is terminating or non-terminating :
CHAPTER - 1 REALNUMBERS
CHAPTER - 2 POLYNOMIALS
GRAPH ANALYSIS FORPARABOLAS:
 If parabola is “U” shapethen a>0but
if parabola is “∩” then a<0.
 Y-axis intersection always gives ‘C’ if
“ +ve” then Cis also “+ve” and if ‘-ve’
then Cis also ‘-ve’.
 Wherethe X-axis intersect that are
the roots of that equation.
y=x^{2} -3x-4
General Quadratic equation =
ax2+bx+c
CHAPTER–2POLYNOMIALS
General Quadratic equation = ax2+bx+c
Relationshipbetween roots and coefficient of
quadratic equation---
α +β = -b/a
α .β =c/a
Formation of
Quadraticequation
when the
roots/zeroes are
given ---
X2–(a-b)x +a.b =0
CHAPTER–2POLYNOMIALS
Relationship between roots and coefficient of
cubic equation ---
α +β +γ =-b/a
αβ+ βγ +γα=c/a
αβγ =-d/a
General Cubic equation ---
ax3 + bx2 + cx + d = 0
CHAPTER–2POLYNOMIALS
Findthe number of zeroes of p(x) .
CHAPTER–2POLYNOMIALS
Q22.If 5is a zero of the quadraticpolynomial, x2 –kx –15,
then the valueof k is
(a) 2
(b)-2
(c)4
(d) –4
CHAPTER– 3PAIR OF LINEAR
EQUATION IN TWO VARIABLE
General equation --
a1 x + b1 y + c1= 0,and
A2 x + b2 y +c2=0
CHAPTER– 3PAIR OF LINEAR
EQUATION IN TWO VARIABLE
CHAPTER– 3PAIR OF LINEAR
EQUATION IN TWO VARIABLE
Algebraic methods ---
1. Substitution method
2. Cross multiplicationmethod
3. Elimination method
CHAPTER– 3PAIR OF LINEAR
EQUATION IN TWO VARIABLE
Examples--- how tosolvebyelimination
and bysubstitution methods:
1. 2x+ 3y =8
4x +3y=7
CHAPTER– 3PAIR OF LINEAR
EQUATION IN TWO VARIABLE
CHAPTER – 6 TRIANGLES
Twotriangles are similarif :
1. Theircorrespondingangles are equal and
2. Theircorrespondingsides are in the same
ration (orproportion).
Similarity criterion intriangles 
1. AAA/ AA
2. SAS
3. SSS
CHAPTER – 6 TRIANGLES
Thales theorem :
Statement : ifa lineis drawn parallel to
one side of a triangle to intersect the
other two sides in distinct points, the
other two sides are divided in the same
ratio.
CHAPTER – 6 TRIANGLES
Pythagoras Theorem
If triangle is right angled triangle , then :
(hypotenuse)2= (base)2 + (perpendicular)2
CHAPTER – 6 TRIANGLES
Important results:
In right angled triangle , if we make a
perpendicular on the hypotenuse from
the opposite vertex , then all three
triangles are similar.
CHAPTER – 7 COORDINATE GEOMETRY
SIGN CONVERSIONS:
CHAPTER – 7 COORDINATE GEOMETRY
Distance formula :
CHAPTER – 7 COORDINATE GEOMETRY
Find the distance betweenthepoints
(-2,1)and(3,2)
CHAPTER – 7 COORDINATE GEOMETRY
If we want to find the distancefrom
the origin (0,0), then :
CHAPTER – 7 COORDINATE GEOMETRY
Find thedistance between the
points : (0,0)and (3,3)
CHAPTER – 7 COORDINATE GEOMETRY
Mid point formula :
CHAPTER – 7 COORDINATE GEOMETRY
Findthe mid point of this line
whose point are(1,1)and ( 2,4).
CHAPTER – 7 COORDINATE GEOMETRY
Section formula :
CHAPTER – 7 COORDINATE GEOMETRY
Centroid formula
Thepoint of intersection of the medians
of the triangle.
Here ,Centroid is point M
CHAPTER–8TRIGONOMETRY
Sine (sin)
Cosine(cos)
Tangent (tan)
Cosecant(cosec)
Secant (sec)
Cotangent (cot)
Opposite side / Hypotenuse
Adjacent side / Hypotenuse
Opposite side / Adjacent side
Hypotenuse / Opposite side
Hypotenuse / Adjacent side
Adjacent side / Opposite side
TRIGNOMETRYRATIOS
CHAPTER–8TRIGONOMETRY
TRIGNOMETRYRATIOSTRICK
TOMEMORISE
PANDIT (P) BADRI (B) PRASAAD (P)
HAR (H) HAR (H) BOLE (B)
CHAPTER–8TRIGONOMETRY
Changing trigonometryratios
CHAPTER–8TRIGONOMETRY
Trigonometry ratiosComplementary angles
sin (90 –θ) =cos θ
cos(90–θ) =sinθ
tan( 90 –θ) = cot θ
cot( 90 –θ) = tan θ
sec(90–θ) =cosecθ
cosec(90–θ) =secθ
CHAPTER–8TRIGONOMETRY
CHAPTER–8TRIGONOMETRY
CHAPTER– 12AREASRELATEDTO CIRCLES
BASICFORMULASOFCIRCLE
Area of circle = ∏r2
Circumference of circle = 2∏r
Circumference of semi-circle= ∏r
NEWFORMULASOFCIRCLE
Area of sector = θ/360 x ∏r2
Length of the arc= θ/360x 2∏r
Area of segment =area of sector –
area of ∆AOB
= θ/360 x ∏r2 - ½ sin θxr2
Area of equilateral ∆=
CHAPTER –15 PROBABILITY
PRPBABILITYmainFORMULA
P(E)=number of favorable outcomes to E
Total possibleoutcomes
Complementary events P(C):
P(E)+P(C) =1
CHAPTER–15PROBABILITY
Important terms
Prime numbers = 2,3,5,7,11…….etc.
Composite numbers =
4,6,8,9,12,….etc.
Tossingacoin =head or tail (total
outcomes =2 =21)
Tossingtwo coins=(total outcome =
4 =22)
Tossing three coins =( totaloutcomes
=8 =23)
Formula tofind total
outcome of dice =(6)n
;where ‘n’ isthe no. of
times thedice thrown
CHAPTER–15PROBABILITY
TERM –1MATHS समाप्त
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