Formulae
Circumference and Area of a Circle
For a circle of radius r, we have
(i) Circumference of the circle \[=2\pi r\]
(ii) Area of the circle \[=\pi {{r}^{2}}\]
(iii) Area of the semicircle \[=\frac{1}{2}\pi {{r}^{2}}\]
(iv) Perimetre of the semicircle \[=\pi r+2r\]
Area of Ring
Let R & r be the outer and inner radii of ring. Here area of the ring \[=\pi ({{R}^{2}}-{{r}^{2}})\]
Length of Arc, Area of Sector and Segment
Let an arc AB makes an angle\[\theta <180{}^\circ \]at the center of a circle of radius r then we have:
(i) Length of the arc AB\[=\frac{2\pi r\theta }{360}\]
(ii) Area of the sector OACB \[=\frac{\pi {{r}^{2}}\theta }{360}\]
(iii) Perimetre of the sector\[\text{OACB}=\text{OA}+\text{OB}+\]length of\[\overset\frown{AB}+2r+\frac{2\pi \theta }{360}\]
(iv) Area of the minor segment ACBA = (Area of the sector OACB) - (area of\[\Delta OAB\])\[=\left( \frac{\pi {{r}^{2}}\theta }{360}-\frac{1}{2}{{r}^{2}}\sin \theta \right)\]
(v) Area of the major segment BDAB = (area of the circle) - (area of the minor segment ACBA)
A chord of a circle of radius 14 cm makes a right angle of the centre. Find the area of the major segments of the circle.
(a) \[\text{59}0\,\text{c}{{\text{m}}^{\text{2}}}\]
(b) \[\text{56}0\,\text{c}{{\text{m}}^{\text{2}}}\]
(c) \[\text{595}\,\text{c}{{\text{m}}^{\text{3}}}\]
(d) \[\text{995}\,\text{c}{{\text{m}}^{\text{2}}}\]
(e) None of these
Answer: (b)
Explanation
Let AB be the chord of a circle of centre 0 & radius = 14 cm
so that\[\angle \text{AOB}=\text{9}0{}^\circ \]
Area of the sector OACB
\[=\frac{\pi {{r}^{2}}\theta }{360}c{{m}^{2}}=\left( \frac{22}{7}\times 14\times 14\times \frac{90}{360} \right)c{{m}^{2}}=154c{{m}^{2}}\]
Area of \[\Delta \text{OAB}=\frac{1}{2}{{r}^{2}}\sin \theta \]
\[=\left( \frac{1}{2}\times 14\times 14\times \sin 90{}^\circ \right)=\text{98c}{{\text{m}}^{\text{2}}}=\] Area of the minor segment ACBA
= (area of the sector OACB) - (area of the \[\Delta \text{OAB}\])\[=(\text{154}-\text{98})\text{c}{{\text{m}}^{\text{2}}}=\text{56}\,\text{c}{{\text{m}}^{2}}\] Area of the major segment BDAB
= (area of the circle) - (area of the minor segment)
\[\text{=}\left\{ \left( \frac{22}{2}\times 14\times 14 \right)-56 \right\}\text{c}{{\text{m}}^{\text{2}}}=(\text{616}-\text{56})\text{c}{{\text{m}}^{\text{2}}}=\text{56}0\text{ c}{{\text{m}}^{\text{2}}}\] 
Quadrilateral
We know that a geometrical figure bounded by four lines segment is called quadrilateral. In this section we will study about area and perimetre of different quadrilaterals
Perimetre and Area of Rectangle
Let ABCD be a rectangle in which length \[AB=\ell \] units, breadth BC = b units then we have:
(i) Area\[=(l+b)\]square units
(ii) \[\text{Length}=\frac{area}{breadth},\]\[breadth=\frac{area}{length}\]
(iii) Diagonal\[=\sqrt{{{l}^{2}}+{{b}^{2}}}\]units
(iv) Perimetre\[=2(l+b)\]units
Area of Walls of Room
Let I, b and h are length, breadth and height of the room respectively then area of walls of room\[~=\{\text{2(}l+\text{b)}\times \text{h}\}\] sq units.
Perimetre and Area of Square
Let ABCD be a square with each side equal to 'a' units then we have (i) Area\[={{a}^{2}}\]sq units
(ii) Area\[=\left\{ \frac{1}{2}{{\text{(diagonal)}}^{\text{2}}} \right\}\]sq units
(iii) Diagonal\[=a\sqrt{2}\]units
(iv) Perimetre = 4a units
Perimetre and Area of Quadrilateral
(1) When one diagonal and length of the perpendiculars from opposite vertices on it are given
\[\therefore \] Let ABCD be a quadrilateral with diagonal AC
Let \[BL\bot AC\]& \[DM\bot AC\]
Then area of the quadrilateral with diagonal AC
= [(area of\[\Delta ABC\]) + (area of\[\Delta ACD\])]sq. units
\[=\left\{ \left( \frac{1}{2}\times AC\times BL \right)+\left( \frac{1}{2}\times AC\times DM \right) \right\}\]sq units
\[\therefore \]area of the quadrilateral ABCD \[=\left\{ \frac{1}{2}\times AC\times ({{h}_{1}}+{{h}_{2}}) \right\}\]sq. units
where\[{{h}_{1}}\]&\[{{h}_{2}}\] are the lengths of perpendiculars from opposite vertices to the diagonal AC.
(2) When diagonals intersect at right angles
Let ABCD be a quadrilateral whose diagonals AC and BD intersect at O at right angles then area of the quadrilateral ABCD = [(area of\[\Delta ABC\]) + (area of\[\Delta ACD\])] \[=\left( \frac{1}{2}\times AC\times BO \right)+\left( \frac{1}{2}\times AC\times OD \right)\] \[=\frac{1}{2}\times AC\times (BO+OD)\]
\[=\frac{1}{2}\times AC\times BD=\frac{1}{2}\times \] (product of diagonals)
Area of parallelogram = base x height
Let ABCD be a parallelogram and let \[CL\bot AB\]
Then area of parallelogram ABCD
\[\text{=2}\times \text{(area of }\Delta \text{ABC})\] \[\text{=}\left( \frac{1}{2}\times AB\times CL \right)\text{=(AB}\times \text{CL})=(\text{base}\times \text{height})\]
similarly if \[CM\bot AD\] then area of parallelogram\[\text{ABCD}=(\text{AD}\times \text{CM})\]\[=(\text{base}\times \] \[\text{bright})\]
Area of a Rhombus
A parallelogram with all sides equal is called rhombus. The diagonal of a rhombus bisect each other at right angles. Area of rhombus \[=\frac{1}{2}\times \] (product of diagonals)
Area of a Trapezium
Let ABCD be a trapezium in which \[~\text{AB}\,\,\text{ }\!\!|\!\!\text{ }\,\,\text{ }\!\!|\!\!\text{ }\,\,\text{DC}\]
& Let\[CL\bot AB\]. Draw\[AM\bot CD\]
Let \[\text{CL}=\text{AM}=\text{h}\]
Area of trapezium ABCD = [Area of\[\Delta ABC\]+ area of\[\Delta ACD\]]
\[=\left[ \left( \frac{1}{2}\times AB\times h \right)\times \left( \frac{1}{2}\times CD\times more... 
Solids
The objects having definite shape and size are called solids. A solid occupies a definite space.
Cuboid
Solids like matchbox, chalk box, a tile, a book an almirah, a room etc. are in the
Shape of a cuboid
Formulae
For cuboid of length = I, breath = b and height = h, we have:
(i) Volume \[=(l\times b\times h)\]
(ii) Total surface area \[=2(lb\times bh\times lh)\]
(iii) Lateral surface area \[=[2(l+b)\times h]\]
Cube
Solids like ice cubes, sugar cubes, dice etc. are the shape of cube Formula for a cube having each edge = a units, we have:
(i) Volume\[={{a}^{3}}\]
(ii) Total surface area \[=6{{a}^{2}}\]
(iii) Lateral surface area \[=4{{a}^{2}}\]
Cylinder
Solids like measuring jar, circular pencils, circular pipes, road rollers, gas cylinder, are said to have a cylindrical shape. Formula for a cylinder of base radius = r & height (or length) = h, we have
(i) Volume\[=\pi {{r}^{2}}h\]
(ii) Curved surface area \[=2\pi rh\]
(iii) Total surface area \[=(2\pi rh+2\pi {{r}^{2}})=2\pi r(h+r)\]
Hollow Cylinder
Solids like a hollow cylinder having external radius = R, internal radius = r & height = h then , we have
(i) Volume of material = (external volume) - (internal volume) \[=(\pi {{R}^{2}}h-\pi {{r}^{2}}h)=\pi h({{R}^{2}}-{{r}^{2}})\]
(ii) Curved surface area of hollow cylinder = (external surface area) - (internal surface area) \[=(2\pi Rh-2\pi rh)=2\pi h(R-r)\]
(iii) Total surface area of hollow cylinder = (curved surface area) + 2\[\times \](area of the base ring) \[=2\pi h(R-r)+2(\pi {{R}^{2}}-\pi {{r}^{2}})\] \[=2\pi (R-r)(R+r+h)\]
Cone
Consider a cone in which base radius = r, height = h & slant height\[l=\sqrt{{{h}^{2}}+{{r}^{2}}}\] then we have
(i) Volume of the cone \[=\frac{1}{3}\pi {{r}^{2}}h\]
(ii) Curved surface area of the cone \[=\pi rl\]
(iii) Total surface area of the cone = (curved surface area) + (area of the base) \[\pi rl+\pi {{r}^{2}}=\] \[\pi r(l+r)\]
Sphere
Objects like a football, a cricket ball, etc. are said to have the shape of the a sphere. For a sphere of radius r, we have
(i) Volume of the sphere \[=\left( \frac{4}{3}\pi {{r}^{3}} \right)\]
(ii) Surface area of the sphere \[(4\pi {{r}^{2}})\]
Hemisphere
A plane through the centre of a sphere cuts it into two equal parts, each part is called hemisphere. For a hemisphere of radius r, we have:
(i) Volume more... 
Centroid and Incenter of a Triangle
The coordinate of centroid of a triangle whose vertices are \[({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}})\] and \[({{x}_{3}},{{y}_{3}})\] is \[\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\]
The coordinate of the in centre of a triangle ABC whose vertices are \[A({{x}_{1}},{{y}_{1}}),B({{x}_{2}},{{y}_{2}})\] and \[C({{x}_{3}},{{y}_{3}})\] is \[\left( \frac{a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{a+b+c},\frac{a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{a+b+c} \right)\]
The ratio in which line \[y=x-2\] divides the line segment joining (8, 9) and (3,-1) is_____.
(a) 3:2
(b) \[\frac{3}{4}:\frac{4}{2}\]
(c) 2:3
(d) 3:3
(e) None of these
Answer: (c)
Explanation:
Given line is \[y=x-2\] ..... (i)
Let P=(8,9) and Q=(3,-1)
let line (i) divides PQ, in the ratio K : 1 at point R;
then \[R=\left( \frac{8k+3}{k+1},\frac{9k-1}{k+1} \right)\]
Since R lies on \[y=x-2\], therefore,
\[\frac{9k-1}{k+1}=\frac{8k+3}{k+1}-2\]
\[\Rightarrow \]\[\frac{9k-1}{k+1}-\frac{8k+3}{k+1}=-2\] \[\Rightarrow \]\[\frac{9k-1-8k-3}{k+1}=-2\]
\[\Rightarrow \]\[\frac{k-4}{k+1}=-2\] \[\Rightarrow \]\[k-4=-2(k+1)\] \[\Rightarrow \]\[k-4=-2k-2\] \[\Rightarrow \]\[3k=4-2\] \[\Rightarrow \]\[k=\frac{2}{3}\]
Then the ratio is k : 1 \[\Rightarrow \]\[\frac{2}{3}:1\] \[\Rightarrow \]\[2:3\]
Find the coordinate of point which trisect the line segment joining the points (1, 3) and (3, 9).
(a) (5, 5), (3, 3)
(b) (5, 5), (7, 7)
(c) \[\left( \frac{5}{3},5 \right)\left( \frac{7}{3},7 \right)\]
(d) \[\left( \frac{5}{3},\frac{7}{3} \right)\left( 5,7 \right)\]
(e) None of these
Answer: (c)
Explanation:
The coordinate of point P is
\[x=\frac{3\times 1+2\times 1}{3}=\frac{5}{3}\]
\[y=\frac{3\times 2+9\times 1}{3}=\frac{{{\bcancel{15}}^{3}}}{3}=5\]
The coordinate of point Q is
\[x=\frac{1\times 1+3\times 2}{3}=\frac{7}{3}\]
\[y=\frac{3\times 1+9\times 2}{3}=\frac{21}{3}=7\]
Thus the coordinate of point P is \[\left( \frac{5}{3},5 \right)\] and Q is \[\left( \frac{7}{3},7 \right)\].
The coordinate of centroid of a triangle whose vertices are (3, 2), (-3,-1) and (0, -1) is.....
(a) (0, 0)
(b) (0, 3)
(c) (3, 0)
(d) (0,-5)
(e) None of these
Answer: (a)
more... 
Section Formula
Let the coordinate of end points of a line segment be \[({{x}_{1}},{{y}_{1}})\]and\[({{x}_{2}},{{y}_{2}})\] then the coordinate of point C which divides the line segment in the ratio m : n is:
\[x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\,\,y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]
Proof:
Let point \[C(x,y)\] be the point which divides the line segment joining two given points \[A({{x}_{1}},{{y}_{1}})\] and \[B({{x}_{2}},{{y}_{2}})\] internally in the ratio m : n.
From A, B and C draw AD, BF and CE perpendicular to x - axis.
From A and C draw AG and CH perpendicular to CE and BF.
Then \[AG={{x}_{1}}-{{x}_{1}}\]
\[CH={{x}_{2}}-x\]
\[CG=y-{{y}_{1}}\]
\[BH={{y}_{2}}-y\]
\[\Delta \text{AGC}\] and \[\Delta \text{BCH}\] are similar. Therefore,
\[\frac{AG}{CH}=\frac{CG}{BH}=\frac{AC}{BC}\]
\[\Rightarrow \] \[\frac{x-{{x}_{1}}}{{{x}_{2}}-x}=\frac{y-{{y}_{1}}}{{{y}_{2}}-y}=\frac{m}{n}\] \[\Rightarrow \] \[\frac{x-{{x}_{1}}}{{{x}_{2}}-x}=\frac{m}{n}\] \[\Rightarrow \] \[nx-n{{x}_{1}}=m{{x}_{2}}-mx\] \[\Rightarrow \]
\[(m+n)x=m{{x}_{2}}+n{{x}_{1}}\] \[x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n}\]
Similarly \[\frac{y-{{y}_{1}}}{{{y}_{2}}-y}=\frac{m}{n}\] \[\Rightarrow \] \[y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}\]
Thus the coordinate of point C is\[\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)\]
Note: If point C is the midpoint of AB then the coordinate of C is \[\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\] 
Distance Formula
Let the coordinate of two points A and B be \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] respectively then the distance between two points A and B can be found in the following way:
Take points A and B draws AE perpendicular to BD.
Thus \[OC={{x}_{1}},AC={{y}_{1}}\]
\[OD={{x}_{2}},BD={{y}_{2}}\]
Now \[AE={{x}_{2}}-{{x}_{1}}\] and \[BE={{y}_{2}}-{{y}_{1}}\]
In right \[\Delta \text{AEB}\]
\[\text{A}{{\text{B}}^{\text{2}}}~~=\text{A}{{\text{E}}^{\text{2}}}+\text{B}{{\text{E}}^{\text{2}}}\]
\[\Rightarrow \]\[A{{B}^{2}}={{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}\] \[\Rightarrow \]\[AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\]
Important Points to Use Distance Formula
In various geometrical problems in which we need to use distance formula. Before using this formula the following points must keep in your mind:
1. If \[A=({{x}_{1}}-{{y}_{1}})\]and\[B=({{x}_{2}}-{{y}_{2}})\] then \[AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}\] or \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{1}})}^{2}}}\]
2. The distance of point \[C(x,y)\] from origin is \[\sqrt{{{x}^{2}}+{{y}^{2}}}\].
3. To show three points P, Q. and R are collinear find PQ, QR and PR and then show that the greatest of these three is equal to the sum of other two.
4. Use the following geometrical facts:
(i) A triangle is equilateral if all sides are equal.
(ii) A quadrilateral is parallelogram if opposite sides are equal.
(iii) A quadrilateral is rectangle if opposite sides are equal and diagonals are equal.
(iv) A quadrilateral is rhombus if all four sides are equal.
The circumcentre of a circle whose vertices are (-1, 0), (7, -6) and (-2, 3) is.... .
(a) (5, 0)
(b) (3, -3)
(c) (-3, 3)
(d) (4, 3)
(e) None of these
Answer: (b)
Explanation:
Let the coordinate of circumcentre be \[p(x,y)\] and A = (-1, 0), B = (7, -6) and C = (-2, 3)
Since P is the circumcentre
\[\text{P}{{\text{A}}^{\text{2}}}=\text{P}{{\text{B}}^{\text{2}}}\]
\[\Rightarrow \]\[{{(x+1)}^{2}}+{{y}^{2}}={{(x-7)}^{2}}+{{(y+6)}^{2}}\]
\[\Rightarrow \]\[{{(x+1)}^{2}}-{{(x-7)}^{2}}={{(y+6)}^{2}}-{{y}^{2}}\]
\[\Rightarrow \]\[(x+1+x-7)(\bcancel{y}+1-\bcancel{y}+7)=(y+6+y)(\bcancel{y}+6-\bcancel{y})\]
\[\Rightarrow \]\[(2x-6).8=(2y+6).6\] \[\Rightarrow \]\[4(x-3)=12(y+3)\]
\[\Rightarrow \]\[4x-12=3y+9\] \[\Rightarrow \]\[4x-3y=21\] ...(i)
\[P{{A}^{2}}=P{{C}^{2}}\]
\[\Rightarrow \]\[{{(x+1)}^{2}}+{{y}^{2}}={{(x+2)}^{2}}+{{y}^{2}}\] \[\Rightarrow \]\[4x+6y+13=2x+1\]
\[\Rightarrow \]\[x+3y=-6\] ...(ii)
adding (i) and (ii), we get
\[\begin{align} & 4x-3y=21 \\ & \,\,\underline{x+3y=-6} \\ & \,\,5x\,\,\,\,\,\,\,\,\,\,=15 \\ \end{align}\]
\[x=3\]
Now put the value of x in (ii), we get
\[3+3y=-6\] \[\Rightarrow \] \[3y=-9\] \[y=-3\]
the coordinate of circumcentre is (3, - 3). 
Introduction
In Euclidean geometry/we have studied about point lines etc. It is studied as independently. In 1637 Rene Wescartes firstly introduce the use of algebra in the study of geometry. This new type of geometry was further named as coordinate geometry. In this chapter we will study about coordinate geometry.
Co-ordinate Geometry
It is the branch of mathematics in which we solve geometrical problems with the help of algebra. To solve geometrical problem by this method need coordinate system.
Cartesian coordinate system
Suppose X'OX and Y'OY be two mutually perpendicular lines through any point "O". The point O is known as origin. Taking a convenient unit of length and starting from the origin mark off a number scale on horizontal as well vertical line. The line X'OX is known as X - axis and the line Y'OY is known as Y - axis.
If we are taking two lines together then it is known as coordinate axis.
The coordinate of a point in Cartesian coordinate system is always written as ordered pair.\[(x,y)\]
i.e. \[(a,b)\ne (b,a)\]
The x - coordinate of a point is known as abscissa and y - coordinate is known as ordinate.
For a point (2, 4), 2 is abscissa and 4 is the ordinate.
Quadrants
Let X'OX and Y'OY are the coordinate axis and it divides Euclidean plane into four plane region. Every region is known as quadrants. So there are four quadrant for certesian plane which is shown below.
In I quadrant, \[x>0\] and \[\text{y}>0\]
In II quadrant, \[x>0\] and \[\text{y}>0\]
In III quadrant, \[x>0\] and \[\text{y}<0\]
In IV quadrant, \[x>0\] and \[\text{y}<0\]
Write the quadrants of points in which they lie.
(a) (2, 3)
(b) (-4, 3)
(c) (-5, -2)
(d) (5, 2)
(e) None of these
Solution:
(a) (2, 3), Here, \[x>0\] and Y > 0, therefore, it lies I quadrant
(b) (- 4, 3), Here, \[x<0\] and y > 0, therefore, it lies II quadrant
(c) (- 5, - 2), Here, \[x<0\] and y < 0, therefore, it lies in III quadrant
(d) (5,-2), Here, \[x>0\] and y < 0, therefore, it lies in IV quadrant
Important Terms
(i) Point lies in the first quadrant if \[x>0\], y > 0
(ii) Point lies in the second quadrant if \[x<0\], y > 0
(iii) Point lies in the third quadrant if \[x<0\], y < 0
(iv) Point lies in the fourth quadrant if \[x>0\], y < 0
(v) Point lies on x - axis if y = 0
(vi) Point lies on y - axis if \[x=0\]
(vii) The coordinate of origin is (0, more... 
Theorems Related to Tangent of a Circle
Theorem 1
A tangent of a circle is perpendicular to the radius through the point of contact
Given: I is the tangent to the circle c (o, r) at point A
To prove: \[OA\bot \ell \]
Construction:
Take a point B on line I other than A and join OB. Let OB cut the circle at the point C.
Proof:
Since O does not lies l. Therefore, the shortest distance from 0 to I will be the perpendicular so it is sufficient to prove OA is the shortest distance.
From figure
\[\text{OA}=\text{OC}\] \[\Rightarrow \] \[~\text{OA}<\text{PN}\]
OA is the shorter than any other segment joining 0 to any point on I. Hence, \[~\text{OA}\bot \ell \]
Theorem 2
The length of tangent drawn from an external point to a circle are equal.
Given: PA and PB are the tangents from point P to circle C (o, r).
To Prove: PA = PB
Construction:
Join PO, OA and BO
Proof:
In \[\Delta \text{POA}\] and \[\Delta \text{POB}\]
\[\text{PO}=\text{OP}\] (common)
\[\text{OA}=\text{OB}\] (radius of circle)
\[\angle \text{OAP}=\angle \text{OBP}=\text{9}0{}^\circ \]
By R.H.S Criteria
\[\Delta \text{POA}\cong \Delta \text{POB}\]
\[\frac{\text{By C}\text{.P}.\text{C}\text{.T}}{\text{PA}=\text{PB}}\]
Hence, proved
Another method:
In\[\Delta \text{POA}\]
\[P{{A}^{2}}+O{{A}^{2}}=P{{O}^{2}}\] ...(i)
In \[\Delta \text{POB}\]
\[\text{P}{{\text{B}}^{\text{2}}}+\text{O}{{\text{B}}^{\text{2}}}=\text{P}{{\text{O}}^{\text{2}}}\] ...(ii)
From (i) and (ii) we get
\[\text{P}{{\text{A}}^{\text{2}}}+\text{O}{{\text{A}}^{\text{2}}}=\text{P}{{\text{B}}^{\text{2}}}+\text{O}{{\text{B}}^{\text{2}}}\]
\[\Rightarrow \]\[\text{P}{{\text{A}}^{\text{2}}}=\text{P}{{\text{B}}^{\text{2}}}\](Since\[\text{O}{{\text{A}}^{\text{2}}}=\text{O}{{\text{B}}^{\text{2}}}\], because they are radius) \[\Rightarrow \] \[\text{PA}=\text{PB}\]
Hence, proved.
In an isosceles A PQR in which PQ = QR. A circle which passes through P and R intersect at point L and M to the side more... 
Introduction
In this chapter we will study about a non-rectilinear figure circle about which we have studied many things in previous classes.
Circle
It is a locus of a point which moves in a plane in such a way that its distance from a fixed point is always constant.
We know that the fixed point is called centre and the fixed distance is called its radius \[\text{D}=\text{2r},\text{ C}=\text{2pr}\] Where D is diameter, C is circumference of circle r is radius.
Terms Related to Circle
Secant
When a line intersect a circle at two distinct points is called a secant of the circle.
Here, line m is the secant line for circle C(0, r)
Tangent
A line which touches the circle at exactly one point is called a tangent to the circle. The point at which line touches the circle is called point of contact. Q is said to be point of contact of tangent.
Concentric Circles
Circles are said to be concentric if and only if they have a same centre and different radius.
Arc
A continuous piece of circumference of a circle is called arc.
Here, PQ is arc of C(o, r).
Concurrent Arc
Two arc are said to be concurrent if they subtend same angle at the centre.
Minor and Major Arcs
If the length of an arc is less then the arc of a semicircle then it is called minor arc and which is greater than the semicircles is called major arc. arc \[\overset\frown{PRQ}\] is minor arc and \[\overset\frown{PSQ}\] is major arc
Segment
The region bounded by chord and an arc is called segment. The segment which contains minor arc is called minor segment and which contains major arc is called major segment.
Congruent Circles
Two circles are said to be congruent if they have same radii.
Important Properties
Results on Area of Quadrilateral
(i) Any diagonal of a parallelogram divides it into two triangles of equal area.
In the above given figure,
\[\text{ar(}\Delta \text{ABC})=\text{ar}(\Delta \text{ACD)}\]
Similarly
\[\text{ar(}\Delta \text{ABD})=\text{ar}(\Delta \text{BCD)}\]
(ii) Parallelograms which are on the same base and between the same parallel lines are equal in area.
In the above given figure,
\[ar(l{{l}^{gm}}ABCD)=ar(l{{l}^{gm}}ABEF)\]
(iii) Triangles which are on the same base and between the same parallel lines are equal in area.
In the above given figure \[QR|\,\,|AB\]
Then \[~\text{ar(}\Delta \text{PQR)}=\text{ar(}\Delta \text{QRS)}\]
(iv) Area of trapezium= \[\frac{1}{2}\] (sum of parallel sides) \[\times \] (distance between them)
In the figure given below shows a pentagon in which TU drawn parallel to SP meet at QP produced at U and RV parallel to SQ, meet PQ produced at V then:
(a) ar(pentagon PQRST) \[=\text{ar(}\Delta \text{STU)}+\text{ar(}\Delta \text{QRV)}\]
(b) ar(pentagon PQRST) \[~=\text{ar(}\Delta \text{SUV)}\]
(c) ar(pentagon PQRST) \[~=\text{ar(}\Delta \text{SUV)}+\text{ar(}\Delta \text{TUP)}\]
(d) ar(pentagon PQRST) \[=\text{ar}(\Delta \text{SUV})+\text{ar(}\Delta \text{QRV)}\]
(e) None of these
Answer: (b)
Explanation:
Since \[RV|\,|SQ\] \[\Rightarrow \] \[~\text{ar(}\Delta \text{SPT)}=\text{ar}(\Delta \text{TUP)}\]
\[\text{ar(}\Delta \text{SQR)}+\text{ar(}\Delta \text{SPT)}+\text{ar(}\Delta \text{SPQ)}=\text{ar(}\Delta \text{SRV)}+\]\[\text{ar(}\Delta \text{SUP)}+\text{ar(}\Delta \text{BPQ)}\]
\[\Rightarrow \] \[\text{ar(pentagon PQRST)}=\text{ar(}\Delta \text{SUV)}\]
In a trapezium non-parallel sides are equal. When we join the midpoint of diagonals and parallel sides a quadrilateral is formed, the quadrilateral formed is.....
(a) Rhombus
(b) Rectangle
(c) Trapezium
(d) Square
(e) None of these
Answer: (b)
For a trapezium which one of the following statements is correct?
(a) Adjacent sides are parallel
(b) Vertically opposite sides are parallel
(c) The line joining the mid points of the diagonal is parallel to the parallel side and equal more... 
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