Prove that a + b = x + y. Join BD. In ABD, we have AOB = 180º AOB = 180º ( 1 + 2) AOB = 180º A

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1 bhilasha lasses lass- IX ate: SLUTIN (hap 8,9,0) 50 ob no The sides and of a quadrilateral are produced as shown in fig. rove that a + b = x + y. Join. In, we have y a + + = 80º = 80º ( + ) b x + = bº In, we have + = aº dding (i) and (ii), we get ( + ) + ( + ) = aº + bº xº + yº = aº + bº Hence, x + y = a + b. In parallelogram, = 0 cm. The altitudes corresponding to the sides and are respectively 7 cm and 8 cm. ind. We have, rea of a gm = ase Height. ar ( gm ) = lso, ar ( gm ) = N N 7 cm 0 cm 8 cm = (0 7) cm = ( 8) cm rom (i) and (ii), we get 0 7 = 8 = cm = 8.75 cm. 3. In a quadrilateral, and are the bisectors of and respectively. rove that = ( + ). In, we have = 80º = 80º = 80 and ( + ) [360º ( + )] [ = 360º + = 360º ( + )] = 80º 80º + = ( + ) ( + ) 4. is a parallelogram and line segments X,Y are angle bisector of and respectively then show X Y. 3 Y X Since opposite angles are equal in a parallelogram. Therefore, in parallelogram, we have = = = [X and Y are bisectors of and respectively] Now, and the transversal Y intersects them. = 3 [ lternate interior angles are equal] rom (i) and (ii), we get = 3

2 Thus, transversal intersects X and Y at and Y such that = 3 i.e. corresponding angles are equal. X Y 5. If = 50, find. 50 Given : In a circle with centre at = 50. To find: =? rocedure. = 50 = = (50 ) = 00 [rc subtends at the centre and at remaining part of c] In, = = radius = (pposite angles of equal sides of a ) Now, + + = 80 [Sum of angles of a triangle] º = 80 + = = 80 = is the centre of the circle. = 0, = 55. ind and Given. = 0, = 55 To find. =? and =? rocedure. Let = y and = x = [s = = radius] = 0 In and, = = = 0 =...[Radii of circle]...[roved]...[ommon] y x (SS theorem of congruence) x = y...(...t) lso, =...(...T) + = 80 = 80 = 90...[Linear pair] = 90...[ = 90 ] So in, + + = 80 y = 80 y = 80 0 = 70 x = y = In figure is a parallelogram and = 60º. If the bisectors and of angles and respectively, meet at on, prove that is the mid-point of We have, = 60º + = 80º 60º + = 80º = 0º Now, and transversal intersects them. = = 30º [ = 30º] Thus, in, we have = [ach equal to 30º] = [ ngles opposite to equal sides are equal] Since is the bisector of. Therefore, = = 60º Now, and transversal intersects them. = = 60º [ = 60º]

3 Thus, in, we have = [ach equal to 60º] = [Sides opp, to equal angles are equal] = rom (i) and (ii), we get =... (ii) [ is a gm = ] is the mid point of. 8. In the adjoining figure, is a gm whose diagonals and intersect at. line segment through meets at and at. rove that are () = ar ( gm ). iagonal of gm divides it into two triangles of equal area. ar() = ar( gm ).(i) In and, we have = [diagonals of a gm bisect each other] = [vert. opp. ] = [alt. int. ] ar() = ar() ar() + ar(quad. ) = ar() + ar(quad. ) ar(quad. ) = ar() = ar() = ar( gm ) [using (i)] ar( gm ) 9. Show that the quadrilateral, formed by joining the mid-points of the sides of a square, is also a square. Given : square in which,, R, S are the mid-points of sides,,, respectively., R, RS and S are joined. To rove : RS is a square. onstruction : Join and. R S N roof : In, and are the mid-points of sides and respectively. and = In, R and S are the mid-points of and respectively. RS and RS = rom (i) and (ii), we have RS and = RS...(iii) Thus, in quadrilateral RS one pair of opposite sides are equal and parallel. Hence, RS is a parallelogram. Now, in s and R, we have = R, is a square and = [ = R and = ] and = R [ach equal to 90 ] So, by SS criterion of congruence R = R...(iv) orresponding parts of congruent s are equal rom (iii) and (iv), we have = R = RS ut, RS is a gm. R = S So, = R = RS = S Now, N...(v) [rom (i)]...(vi) Since and S are the mid-points of and respectively. S...(vii) Thus, in quadrilateral N, we have N N So, N is a parallelogram. N = N [rom (vi)] [rom (vii)]

4 N = [N = ] N = 90 z S = 90 iagonals of square are 90 Thus, RS is a quadrilateral such that = R = RS = S and S = 90. Hence, RS is a square. 0. The medians and of a triangle intersect at G. rove that area of G = area of quadrilateral G. Join. Since the line segment Joining the midpoints of two sides of a triangle is parallel to the third side. So,. G learly, s and are on the same base and between the same parallel lines. So, ar () = ar () ar () ar (G) = ar () ar (G) ar (G) = ar (G) We know that a median of a triangle divides it into two triangels of equal area. Therefore, ar () = ar () ar(g) + ar (G) = ar(quad. G) + ar (G) ar(g) + ar(g) = ar (quad. G) + ar (G) [Using (i)] ar (G) = ar (quad. G).,, G, H are respectively, the mid-points of the sides,, and of parallelogram. Show that the area of quadrilateral GH is half the area of the parallelogram. Given: quadrilateral in which,, G, H are respectively the mid-points of the sides,, and. To rove : H (i) ar ( gm GH) = onstruction : Join and H G ar ( gm ) Since HG and gm H are on the same base H and between the same parallel lines. ar(hg) = ar ( gm H) Similarly, H and gm H are on the same base H and between the same parallels. ar (H) = ar ( gm H) dding (iii) and (iv),we get ar (HG) + ar (H) = ar ( gm GH) = [ar( gm H) + ar( gm H)] ar ( gm ).. In ig. is a trapezium in which side is parallel to side and is the mid-point of side. If is a point on the side such that the segment is parallel to side. G rove that = ( + ). Given : trapezium in which, is the mid-point of and is a point on such that. To rove: = ( + ) roof: In, is the mid-point of and G (Given) G is the mid-point of Since segment joining the mid-points of two sides of a triangle is half of the third side. G = Now, is a trapezium in which. ut, G In, G is the mid-point of (proved above) and. is the mid-point of G = Segment joining the mid po ints of two sides of a is half of the third sides rom (i) and (ii), we have

5 G + G = = () + ( + ) () 3. The side of a parallelogram is produced to any point. line through parallel to meets produced in and the parallelogram R completed. Show that ar ( gm ) = ar ( gm R). onstruction: Join and. lso, s X and X are on the same base X and between the same parallels X and. ar (X) = ar (X) learly, sy and Y are on the same base Y and between the same parallels and Y. ar(y) = ar (Y) rom (i), (ii) and (iii), we get...(iii) ar(x) = ar (Y). 5. If a line intersects two concentric circles (circles with the same centre) with centre at,, and, rove that = (figure)[nrt] R To prove: ar ( gm ) = ar( gm R) roof: Since and are diagonals of parallelograms and R respectively. ar () = and, ar () = ar ( gm ) ar ( gm R) Now, s and are on the same base and between the same parallels and ar() = ar () ar() ar () = ar () ar() [Subtracting ar () from both sides] ar () = ar () ar ( gm ) = ar ( gm R) [Using (i) and (ii)] ar( gm ) = ar( gm R). 4. In ig. XY, X and Y. rove that: X ar (X) = ar (Y). Join X and Y. Since s X and Y are on the same base and between the sum parallels and XY ar(x) = ar(y) Y raw or outer circle, is chord =...() ( from centre bisect the chord) & for inner circle, is chord =...() Subrtract equation () from eqation () = = 6. Two equal chords and of a circle with centre, when produced meet at a point. rove that = and =. Given. Two equal chords and intersecting at a point. To prove. = and =. onstruction. Join, raw L and L roof. In triangles L and, ( ) L = L = [ach equal to 90 ] and = [ommon] L [y RHS criteria] L =...() [...T] Now, = [Given]

6 L =.() Subtracting () from (), we get L L = =. gain, = and = + = + = Hence, = and =. 7. arallelogram & rectangle have the same base and also have equal areas. Show that perimeter of the parallelogram is greater than that of the rectangle. Given: gm and a rectangle with the same base and equal areas. To rove: erimeter of gm > erimeter of rectangle i.e > Given : Two circles of radius r & r intersect at two different points &. and is bisector of. = & = 90º To prove : entres of circles & lie on. onstruction : Join to, and also to,. roof : = = r and = = r quadrilateral is kite. is bisector of is shorter digagonal., are on. 9. Three girls Reshma, Salma and andeep are playing a game by standing on a circle of radius 5 m drawn in a pack. Reshma throws a ball to Salma, Salma to andeep. andeep to Reshma. If the distance between Reshma and Salma and between Salma and andeep is 6 m each. What is the distance between Reshma and andeep? [NRT] Let the position of Reshma, Salma and andeep be at R, S and on the circumference of the circular park. roof: Since opposite sides of a parallelogram and a rectangle are equal. = [ is a gm ] and, = = + = + [ is a rectangle] Since, of all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest. < and < > and > + > + dding (ii) and (iii), we get...(iii) > > is the centre of the circle. 8. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. [NRT] I st ethod Reshma R S Salma RS = S = 6 m andeep Radius = R = S = 5m. In rt angled R R = R...() In rt angled SR R = RS S...() rom () and (), we get R = RS S (5) x = (6) (5 x) (Let = x) 5 x = 36 x 5 + 0x 0x = x = 4 x = 0 istance between Reshma and andeep = R = R = R 4 = 5 x = 5 0

7 = m is a diameter of the circle with centre and chord is equal to radius, and produced meet at. rove that = Given : is a diameter of the circle with centre and chord is equal to radius. and produced meet at. To rove : = 60 0 onstruction : Join. roof : In, =...(i) [Radii of the same circle] = [Given] rom (i) and (ii), = = is equilateral = 60 0 = = = (60 0 ) = 30 0 [ngle subtended by any arc of a circle at the centre is twice the angle subtended by it at any point of the reaming part of the circle] = (iii) nd, = (iv) + = 90º 30 + = 90º (external) = 60º = 60º. [ngle in a semi-circle]

Theorems on Area. Introduction Axioms of Area. Congruence area axiom. Addition area axiom

Theorems on Area. Introduction Axioms of Area. Congruence area axiom. Addition area axiom 3 Theorems on rea Introduction We know that Geometry originated from the need of measuring land or recasting/refixing its boundaries in the process of distribution of certain land or field among different