Important Points
(i) A quadrilateral is a parallelogram if their opposite sides are equal.
(ii) A quadrilateral is a parallelogram if their opposite angles are equal.
(iii) If the diagonal of a quadrilateral bisect each other then it is a parallelogram.
(iv) A quadrilateral is a parallelogram if its one pair of opposite sides are equal and parallel.
(v) The diagonals of rectangle are equal.
(vi) If the two diagonals of a parallelogram is equal then the parallelogram is a rectangle.
(vii) The diagonal of the rhombus are perpendicular bisector of each other.
(viii) A parallelogram is a square if the diagonals of a parallelogram are equal and bisector at right angle.
Intercept Theorem
If a transversal intersect three and more than three parallel lines in such a way that all the intercepts are equal then the intercept on any other transversal is also equal.
Given:
Three parallel lines a, b, c intersected by a transversal \[x\] at L, M, N respectively so that LM = MN. Another transversal y cutting a, b, c at T, U, V respectively
Proof:
Since \[\text{LT }\!\!|\!\!\text{ }\,\text{ }\!\!|\!\!\text{ MU}\] and \[~\text{LM }\!\!|\!\!\text{ }\,\text{ }\!\!|\!\!\text{ TU}\]
\[\therefore \] LMUT is a parallelogram \[\therefore \] TU = LM .....(i)
Similarly
MNVU is a parallelogram
\[\therefore \] MN=UV .....(ii)
But \[\therefore \] TU = UV
Now In
\[\Delta \text{MLT}\]and \[\Delta \text{TUM}\] (Alternate)
\[\angle \text{MTL}=\angle \text{MTU}\] (Opposite angle of Parallelogram)
LT = UM (Opposite sides of Parallelogram)
MT=MT (Common)
\[\therefore \] By SOS
\[\Delta \text{MTL}\cong \Delta \text{TUM}\]
By CPCT,
LM=TU
Similarly
MN=UV But
LM=MN
\[\therefore \] TU = UV 
Introduction
We know that a plane figure which is bounded by four line segments is called quadrilateral. Previously we have studied about different types of quadrilateral like rectangle, square, parallelogram etc. In this chapter, we will discuss about different properties of quadrilateral with the help of theorems.
Quadrilateral In the figure given below:
(i) Points J, K, L and M are the vertices of quadrilateral JKLM.
(ii) The line segments J, K, L, LM and JM are the sides of this quadrilateral.
(iii) The two sides of a quadrilateral having a common point is called adjacent side.
(iv) The two sides have no common end point is called opposite side.
(v) Two angles of a quadrilateral having common arm is called adjacent angle.
(vi) Two angles of a quadrilateral having no common arm is called vertically opposite angles.
Theorem on a Quadrilateral
Theorem 1: The sum of all interior angles of a quadrilateral is 360°
Given: A quadrilateral PQSR
To Prove:
\[\angle \text{P}+\angle \text{Q}+\angle \text{R}+\angle \text{S}=\text{36}0{}^\circ \]
Construction:
Join PQ
Proof:
In \[\Delta \text{PQR}\]
\[\angle \text{Q}+\angle \text{QPR}+\angle \text{QRP}=\text{18}0{}^\circ \] .....(i)
In \[\Delta \text{PRS}\] \[\angle \text{S}+\angle \text{SRP}+\angle \text{SPR}=\text{18}0{}^\circ \] .....(ii)
Adding (i) and (ii), we get:
\[\angle \text{Q}+\angle \text{S}+(\angle \text{QPR}+\angle \text{SPR})+(\angle \text{QRP}+\angle \text{SRP})=\text{36}0{}^\circ \]
\[\Rightarrow \] \[\angle \text{Q}+\angle \text{S}+\angle \text{P}+\angle \text{R}=\text{36}0{}^\circ \]
[Because \[\angle \text{QPR}+\angle \text{SPR}=\angle \text{P},\angle \text{QRP}+\angle \text{SRP}=\angle \text{R}\]]
Hence, proved
Theorem 2:
In a parallelogram opposite sides and opposite angles are equal and diagonals bisect each other.
Given:
A parallelogram PQSR, PQ II RS and PS II QR
To Prove:
(i) \[\text{PQ}=\text{RS},~~~~\text{PS}=\text{QR}\]
(ii) \[\angle \text{P}=\angle \text{R}\] and \[\angle \text{Q}=\angle \text{S}\]
(iii) 0 is the midpoint of PR and SQ
Proof:
In\[\Delta \text{PQR}\]and\[\Delta \text{RSP}\]
\[\angle \text{RPQ}=\angle \text{PRS},\] (Alternate)
\[\angle \text{PRQ}=\angle \text{RPS}\]
and \[PR=PR\] (Common)
\[\Delta \text{PQR}=\Delta \text{PRS}\]
By C.P.C.T. (Corresponding part of congruence triangle PQ = RS, and PS = QR and\[\angle \text{Q}=\angle \text{S}\].
Similarly In
\[\Delta \text{PQS}\] and \[\Delta \text{QSR}\]
we get:
\[\Delta P\text{QS}\cong \Delta QSS\]
\[\Rightarrow \] \[\angle P=\angle R\]
so,
\[PQ=RS\] and \[PS=QR\] Proved
\[\angle P=\angle R\] and \[\angle Q=\angle S\] Proved
Now In
\[\Delta \text{POQ}\] and \[\Delta \text{ROS}\]
\[\angle \text{OPQ}=\angle \text{ORS}\] (Alternate)
\[\angle \text{OQP}=\angle \text{OSR}\]
And \[\text{PQ}=\text{QR}\] (Opposite of Parallelogram)
\[\Rightarrow \] \[\Delta POQ\cong \Delta SOR\] (By ASA)
\[\Rightarrow \] PO = OR and QO = OS (By CPCT)
\[\Rightarrow \] O is the midpoint of PR and QS (Hence, proved) 
Some Important Results
(i) The longer side of a triangle has greater angle opposite to it.
(ii) The greater angle of a triangle has longer side opposite to it.
(iii) Perpendicular line segment is the shortest in length. PM is the shortest line segment from point P to line I.
(iv) The distance between line and points lying on it is always zero.
(v) The sum of any two sides of a triangle is greater than the third side.
(vi) The difference between any two sides of a triangle is always less than its third side.
The quadrilateral which is formed by joining the mid points of a given quadrilateral is always_______.
(a) A parallelogram
(b) A trapezium
(c) A rhombus
(d) A rectangle
(e) None of these
Answer: (a)
Explanation:
Given:
A quadrilateral LMNO is a quadrilateral in which P, Q, R and S are the mid points of LM, MN, NO, OL respectively.
To Prove:
PQRS is a parallelogram
Construction:
Join LN
Proof:
In\[\Delta \text{LMN}\],
\[PQ\,|\,\,|\,LN\] and \[\text{PQ}=\frac{1}{2}\text{LN}\] ...,.(i)
[Because P and Q are the mid points of LM and MN respectively]
In \[\Delta \text{LNO}\]
\[SR\,|\,\,|LN\] and \[\text{SR}=\frac{1}{2}\text{LN}\] .....(ii)
[Because S and R are the mid - points of LO and ON respectively]
From (i) and (ii), we get
\[\text{PQ}=\text{SR}\] and \[~\text{PQ}\,\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }\,\text{SR}\] ....(iii)
Similarly we can see that
\[\text{PS}=\text{QR}\] and \[\text{PS}\,\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }\,\text{QR}\] .....(iv)
From (iii) and (iv), we conclude that quadrilateral PQRS more... 
Congruency Figures
Two geometrical figures are said to be congruent if they have same shape and size. e.g. two angles are said to be congruence if they have same measure similarly two line segments are said to be congruent if they have same length. In this section we will discuss about congruence of triangles.
Congruency of Triangles
Two triangles are said to be congruent if and only if:
(i) Their corresponding sides are equal.
(ii) Two triangles ABC and DEF are said to be congruent if:
(a) AB = DE, BC = EF and CA = FD
(b) \[\angle \text{A}=\angle \text{D,}\,\angle \text{B}=\angle \text{E}\] and\[\angle \text{C}=\angle \text{F}\]
Criteria for Congruency
S-S-S Criteria
Two triangles are said to be congruent if the three sides of one triangle are equal to the correspond three sides of other.
In the above given figure \[\Delta PQR\] and \[\Delta LMN\] in such a way that
\[\text{PQ}=\text{LM},\text{ QR}=\text{MN},\text{ PR}=\text{LN}\]
Then \[\Delta \text{PQR}\cong \Delta \text{LMN}\] (Read as\[\Delta \text{PQR}\]and\[\Delta LMN\]are congruent)
S-A-S Criteria
Two triangles are to be congruent if the two sides and included angle of one triangle is equal to the other.
In the above given figure, \[\text{PQ}=\text{LM},\text{ QR}=\text{MN}\] and\[\angle \text{Q}=\angle \text{M}\],
Then \[\Delta \text{PQR}=\Delta \text{LMN}\]
A-S-A Criteria
If the two angles and the side included by the angles are equal to the corresponding two angles and included sides of the other triangle then the two triangles are congruent.
In the above given figure \[\angle P=\angle L,\angle Q=\angle M\] and \[\text{PQ}=\text{LM}\].Therefore, \[\Delta \text{PQR}\cong \Delta \text{LMN}\]
R-H-S Criteria
This criteria is for right angle triangle. If one side and hypotenuse of a right angled triangle is equal to the corresponding side and hypotenuse of other right angled triangle then two right angled triangles are said to be congruent.
In the above given figure, LM=PQ and LN= PR, \[\angle \text{M}=\angle \text{Q}=\text{9}0{}^\circ ,\] \[\Rightarrow \] \[\Delta \text{LMN}=\Delta \text{PQR}\] 
Similar Triangles
Two triangles are said to be similar if
(i) Their corresponding angles are equal and
(ii) Their corresponding sides are in proportion.
For two triangles \[\Delta ABC\] and \[\Delta PQR\]
If \[\angle \text{A}=\angle \text{P},\text{ }\angle \text{B}=\angle Q\] and \[\angle C=\angle R,\] and \[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{PR}\]
Then \[\Delta ABC\sim \Delta PQR\]
~ : represents similarity, and the above is read as \[\Delta \]ABC is similar to \[\Delta \text{PQR}\]
Basic Proportionality Theorem or Thales Theorem
If a line is drawn parallel to any one side of a triangle intersecting the other two sides at distinct points then it divides two sides in the same ratio.
Given: In a \[\Delta ABC\] a line I which is parallel to BC and intersect AB and AC at distinct point D and E respectively.
To Prove:
\[\frac{AD}{BD}=\frac{AE}{CE}\]
Construction:
Draw \[DM\bot AB\] and \[EN\bot AB\] and join BE, CD
Proof:
Area of \[\Delta ADE=\frac{1}{2}\times AD\times EN\] and
Area of \[\Delta BDE=\frac{1}{2}\times BD\times EN\]
...(i)
Similarly 
...(ii)
But \[\Delta BDE\] and \[\Delta CDE\] have the same base DE and between same parallel lines DE and BC
\[\therefore \] \[ar(\Delta BDE)=ar(\Delta CDE)\]
\[\Rightarrow \] \[\frac{ar(\Delta ADE)}{ar(\Delta BDE)}=\frac{ar(\Delta ADE)}{ar(\Delta CDE)}\]
Hence, proved
Converse of Basic Proportionality Theorem
If any line divides any two sides of a triangle in the same ratio, the line must be parallel the third side.
Criteria for Similarity
A-A-A Criteria
When two triangles are equiangular, they are said to be similar. This criteria is known as (A-A-A).
S-S-S Criteria
Two triangles are said to be similar if their corresponding sides are in proportion.
S-A-S Criteria
If one pair of corresponding sides are proportional and the included angle are equal then the triangles are said to be similar.
Theorem
The ratio of area of two similar triangles are equal to the ratio of the square of any corresponding sides.
Given:
\[\Delta LMN\] and \[\Delta PQR\] are similar to each other.
To Prove:
\[\frac{ar(\Delta LMN)}{ar(PQR)}=\frac{L{{M}^{2}}}{P{{Q}^{2}}}=\frac{M{{N}^{2}}}{Q{{R}^{2}}}=\frac{L{{N}^{2}}}{P{{R}^{2}}}\]
Construction:
Draw \[LC\bot MN\] and \[PD\bot QR\]
Proof:
Since \[\Delta LMN\] and \[\Delta PQD\] are similar
\[\angle \text{L}=\angle \text{P},\text{ }\angle \text{M}=\angle \text{Q},\text{ }\angle \text{N}=\angle \text{R}\]
and \[\frac{LM}{PQ}=\frac{MN}{QR}=\frac{LN}{PR}\] ... (i)
In \[\Delta \text{LMC}\] and \[\Delta \text{PQD}\], we have
\[\angle \text{LMC}=\angle \text{PQD}\]
and \[\angle \text{LCM=}\angle \text{PDQ}=\text{9}0{}^\circ \]
\[\Rightarrow \] \[\Delta \text{LMC}\tilde{\ }\Delta \text{PQD}\]
\[\Rightarrow \]\[\frac{LM}{PQ}=\frac{LC}{PD}\] ...(ii)
from (i) and (ii), we get
\[\frac{LM}{PQ}=\frac{MN}{QR}=\frac{NL}{RP}=\frac{LC}{PD}\]
Now
\[\frac{Area\,of\,\Delta LMN}{Area\,of\,\Delta PQR}=\frac{\frac{1}{{}}\times LC\times MN}{\frac{1}{\bcancel{2}}\times PD\times QR}\]
\[\Rightarrow \]\[\frac{ar(\,\Delta LMN)}{ar(\Delta PQR)}=\frac{LC}{PD}\times \frac{MN}{QR}\]
\[\Rightarrow \] \[\frac{ar(\,\Delta LMN)}{ar(\Delta PQR)}=\frac{MN}{QR}\times \frac{MN}{QR}\] \[\left[ because\frac{LC}{PD}=\frac{MN}{QR} \right]\] \[\Rightarrow \] \[\frac{ar(\,\Delta LMN)}{ar(\Delta PQR)}=\frac{M{{N}^{2}}}{Q{{R}^{2}}}\] more... 
Introduction
In our day to day life, we observe different types of objects. There may be difference in size but the shapes are same. In this chapter, we will study about such types of geometrical figures which resemble same specially emphasis on triangular shape.
Similarly
When two geometrical shapes which resemble like but need to be equal in size is called similar figures. Let us observe the following examples:
(i) Any two line segments are always similar.
(ii) Two circles of different radius are always similar.
(iii) For rectilinear figure if all the corresponding angles of polygon are equal and the ratio of their corresponding sides are equal the they are said to be similar.
For the above two regular hexagon, they are similar if
\[\angle \text{A}=\angle \text{P}~~~~~\angle \text{B}=\angle \text{Q}\]
\[\angle \text{C}=\angle \text{R}~~~~~\angle \text{D}=\angle \text{S}\]
\[\angle \text{E}=\angle \text{T}~~~~~\angle \text{F}=\angle \text{U}\]
and
\[\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CD}{RS}=\frac{DE}{ST}=\frac{EF}{TU}=\frac{FA}{UP}\] 
Conditions
For pair of linear equations in two variable is......
\[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]
\[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]
(i) If \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\] then lines are intersecting.
(ii) If \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\] then lines are coincident.
(iii) If\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\] then lines are parallel.
Solve for \[\mathbf{x}\] and y; graphically.
\[x-y-1=0\]
\[2x+y-8=0\]
Solution:
We have
\[x-y-1=0\] ....(i)
\[\Rightarrow \]\[x=y+1\]
| \[x\] | 1 | 2 | 3 |
| Y | 0 | 1 | 2 |
| \[x\] | 1 | 2 | 3 |
| Y | 6 | 4 | 2 |
In the graph the coordinate of point of intersection is (3, 2). Therefore, \[x=3\] and y = 2 is the solution of given pair of linear equations.
Important Points
(i) Equation for x - axis is y = 0
(ii) Equation of y - axis\[x=0\]
(iii) Equation of a line parallel to y - axis at a distance of p from it is \[x=p\]
(iv) Equation of line parallel to x - axis at a distance of p from it is y = p
(v) Solution of pair of linear equation in two variable. Pair of linear equation is also known as a system of two simultaneous linear equations in two variable.
The general form is
\[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]
\[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\]
If a pair of linear equation has at least one solution then it is said to be consistent. If a pair of linear equation has no solution then it is said to be inconsistent. In this topic we will emphasis on the solution of pair of linear equations by graphical method.
Graphical Method
Previously we have studied about the graph of a linear equation. Here we have a pair of linear equation which represents two lines. We know that when two lines drawn in a plane the following are the possibilities:
(i) The two lines intersect at a point.
(ii) The two lines are coincident line.
(iii) The two lines are parallel to each other.
In the above given first two conditions give the consistent solution and third one gives inconsistent solution or no solution.
Introduction
Previously we have studied about a linear polynomial in two variable. The general form of linear polynomial in two variable is \[ax+by+c\], where a, \[b\ne 0\]. In this chapter we will study about linear equation in two variable.
Linear Equation in Two Variables
Suppose \[p(x)=ax+by+c\] is a linear polynomial in two variable where a, \[b\ne 0\]. Then \[p(x)=0\] is a linear equation in two variable i.e. \[ax+by+c=0\], where a, \[b\ne 0\] is a linear equation in two variable.
Solution of a Linear Equation
\[x=p\]and\[y=q\] is called the solution of a linear equation \[ax+by+c=0\] if \[ap+bq+c=0\]
Which one of the following equations is not a linear equation in two variable?
(a) \[4x+\frac{7}{2}y=4\]
(b) \[\frac{4}{x}+\frac{7}{y}=4\]
(c) \[\frac{x+3}{x-3}=4\]
(d) \[3x+7y+87=0\]
(e) None of these
Answer: (b)
Explanation:
In \[\frac{4}{x}+\frac{7}{y}=4\]
The power of variable x and y is -1. Therefore, it is not a linear equation.
Graph a Linear Equation in Two Variable
The general form of linear equation in two variable is \[ax+by+c=0\]
\[\Rightarrow \]\[by=ax-c\] \[\Rightarrow \]\[y=\left( \frac{-a}{b} \right)x-\frac{c}{a}\]
It is the form
\[y=mx+c\] Represents a line where \[m=\left( \frac{-a}{b} \right)\] and \[c=\frac{c}{a}\] and m is known as the slope of this line. That is why, we can say that the graph of a linear equation represented by a line.
The slope of line \[4x+\text{3y}-\text{4}=0\] is______
(a) \[\frac{3}{4}\]
(b) \[\frac{-4}{3}\]
(c) \[\frac{4}{3}\]
(d) \[\frac{-3}{4}\]
(e) None of these
Answer: (b)
Explanation:
\[4x+3y-4=0\]
\[\Rightarrow \]\[3y=-4x+4\] \[\Rightarrow \]\[y=\left( \frac{-4}{3} \right)x+4\]
Here, \[m=\frac{-4}{3}\]
Graph of \[\mathbf{ax+by+c=0}\]
The following steps are followed to draw a graph:
Step 1: Express \[x\] in terms of y or y in terms of \[x\].
Step 2: Select at least three values of y or \[x\] and find the corresponding values of\[x\] or y respectively, which satisfies the given equation, write these values of \[x\] and y in the form of table.
| \[x\] | |||
| \[y\] | more...
 Standard Formula
 If \[(3-2x)\]and\[(5x+8)\] are factors of \[(-10{{x}^{2}}+hx-k)\] then the value of h and k are____.
(a) (-31, 24)
(b) (31, 24)
(c) (-31, -24)
(d) (31, -24)
(e) None of these
Answer: (b)
Explanation:
Since \[(3-2x)\] is a factor of \[-10{{x}^{2}}+hx-k\]. Therefore, for \[x=\frac{3}{2}\] the value of given polynomial is zero.
i.e. \[10{{x}^{2}}+hx-k=0\]
or \[-10{{\left( \frac{3}{2} \right)}^{2}}+h\times \frac{3}{2}-k=0\]
or, \[3h-2k=45\] ..................... (i)
Similarly, \[(5x+8)\] is a factor of \[-10{{x}^{2}}+hx-k\]
That implies \[\text{8h}+\text{5k}=\text{128}\] ............. (ii)
Solve equation (i) and (ii) and get the result.
If a and b are two positive integers and \[a+b+ab+1=77\] then find the possible value of a + b.
(a) 16
(b) 18
(c) 22
(d) 23
(e) None of these
Answer: (a)
Explanation:
\[\text{a}+\text{b}+\text{ab}+\text{l}=\text{77}\] \[\Rightarrow \]\[\text{(1}+\text{a})(\text{1}+\text{b})=\text{11}\times \text{7}\] \[\Rightarrow \]If (1 + a) = 11 then (1 + b) = 7 and vice versa.
Which one of the following statements is false for the method of factorization of an algebraic expression?
(a) By using standard identities
(b) By using remainder theorem
(c) Taking out common factor from two or more than two terms
(d) Taking out common factor from a group of terms
(e) None of these
Answer: (b)
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