BRILLIANT PUBLIC SCHOOL, SITAMARHI (Affiliated up to +2 level to C.B.S.E., New Delhi) Affiliation No

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1 RILLINT PULI SHOOL, SITMRHI (ffiliated up to + level to..s.e., New elhi) ffiliation No SE oard Level IX S..- II Maths hapterwise Printable Worksheets with Solution Session : 04-5 Office: Rajopatti, umra Road, Sitamarhi(ihar), Pin-840 Website: brilliantpublic@yahoo.com Ph , Mobile: ,

2 MTHEMTIS (lass-ix) Index: S..-II SE hapter-wise Solved Test Papers. Linear Equations in two Variables 00. Quadrilaterals 0. reas of Parallelograms and Triangles ircles onstructions 0 6. Surface rea and Volume 7 7. Statistics Probability 7

3 SE TEST PPER-0 LSS IX Mathematics (Linear Equation in two Variables). The solution of a linear equation is not effected when (a) The same numbers is added to both sides of the equation. (b) We multiply or divide both the sides of the equation by the same non-zero numbers. (c) We add a number to one side and subtract the same number from the other side of the equation (d) oth (a) and (b). Which of the following is not a solution of x + y = 6 (a)(0,) (b)(4,) (c)(,) (d)(6,0). The geometrical representation of a linear equation (a) Straight line (b) urve (c) Parabola (d) Either (b) or (c) 4. The points (,0), (-,0) and (4,0) lie (a) On the origin (b) On the X-axis (c) Y-axis (d) parallel to X-axis

4 5. Examine if - is a solution of x + 4 = 5 [] 6. Solve x + = 0 [] 7. Find the value of x that satisfies the equation.4x 0.4x = 6 [] 8. Solve x + x = 56 x [] 9. Solve x x + 5x + 4 = + x + 0. Write four solutions for the following equations π x + y = 9. Find out which of the following equation have x=, y= as solution. () x + 5y = 9 () 5x + y = 4 () x + y = 7. Express Y in terms of xand x in terms of Y from the following equation 7x 8y + 5 = 0. raw the graph of x + y = 6, Find the coordinates of the point where the graph cuts the y-axis [5]

5 SE TEST PPER-0 LSS IX Mathematics (Linear Equation in two Variables) ns. (d) [NSWERS] ns. (b) ns. ns4. (a) (b) ns5. Substituting x = in the given equation + 4 = 5 ( ) L. H. S = = 0 R. H. S = 5 L. H. S R. H. S - is not a solution of the given equation ns6. x + = 0 Substituting - both the sides x + = 0 x = 8 8 x = = 4 x = 4 ns7..4x 0.4x = 6 x = 6 6 x = = The value of x = satisfies the given equation ns8. x + x = 56 x

6 5 7 x + x = 56 x 5 7 x + x = 56 x 5 7 x + x + x = x = 56 8x = 56 x = 7 ns9. x + 5x + 4 = x + x + x + 4x + x + 4 = x + x + x + x( x + 4) + x + 4 = x x + + x + ( ) ( x + )( x + ) ( x + )( x + ) 4 = x + 4 = x + x + = x + 4 ( ) ( ) x x = 8 6 x = ns0. π x + y = 9 9 y x = π 9 9 When y = 9, x = = 0 π When y = 0, x = = π π 9 8 When y =, x = = π π π When y = 9 π, x = = π 4

7 ns. () Substituting x = and y = in the given equations L.H.S= + 5 = 4 5 = 9 = R. H. S Hence x = and y = is a solution () Substituting x = and y = L. H. S = 5 + = 0 + = R. H. S Hence, x = and y = is not a solution () Substituting x = and y = in equations L. H. S = + = 7 = R. H. S x = and y = is a solution ns. 7x 8y + 5 = 0 7x = 8y 5 8y 5 x = 7 7x + 5 = 8y 8y = 7x + 5 y = 7x ns. x + y = 6 6 y x = x 0 4 y 0 - (0,). Y. X (,0). X. -. y - 5

8 SE TEST PPER-0 LSS IX Mathematics (Linear Equation in two Variables). The points (,), (,) and (4,) from line (a) Which is parallel to the x-axis (b) Which is parallel to the y-axis (c) Passing through the origin (d) None of these. linear equation in two variables (a) Has a unique solution (b) Has two solution (c) Has infinitely many solutions (d) two or four solution. The geometrical representation of a linear equations (a) Straight line (b) urve (c) Parabola (d) b or c 4. Which of the following is not a solution of x + y = 6 (a) (0,) (b) (4,) (c) (,) (d) (6,0) 5. Examine if - is a solution of 4x + 0 = 8 [] 6

9 6. 6 x x Solve + = + 9, ( x 0) [] 7. x 9 = 5 x + 8. Solve 5x 9 { 4x 5} = x ( 6x 5) [] [] 9. Solve the equation x + x + = 4 x x + 0. x + 5x + 4 = x + x +. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case () x + y = 9.5 _ y () x 0 = 0 5 () x 5y (4) x + = 0. If the point (,4) lies on the graph of the equation y = ax + 7, find the valve of a. raw the graph of the equation x + y = 6Read a solution of the equation from the [5] graph and verify the same by actual substitution. lso find the points where the line meets the two exes. 7

10 SE TEST PPER-0 LSS IX Mathematics (Linear Equation in two Variables) ns. (a) [NSWERS] ns. (c) ns. ns4. (a) (b) ns5. Substituting x = in the given equation = 8 ( ) = 8 L. H. S = R. H. S = 8 L. H. S R. H. S ns6. 6 = + 9 x + x 6 = 9 x x 4 = x 4 x = x = ns7. x 9 = 5 x + x + x ( )( ) ( x ) ( x ) = 5 x = 5 + x = 8 8

11 ns8. 5x 9 { 4x 5} = x ( 6x 5) 5x 9 + 4x 5 = x ( 6x 5) 5x 9 + 4x 5 = x 6x + 5 5x + 4x x + 6x = x = x = ns9. ns0. x + x + 4 = x x + x + x + = x + 4 x ( )( ) ( )( ) x + x + x + = x x + x 4 4 x x + x + x + x x = 4 4 4x = 6 6 x = = 4 x + 5x + 4 = x + x + x + 4x + x + 4 = x + x + x + x( x + 4) + x + 4 = x x + + x + ( ) ( x + )( x + ) ( x + )( x + ) 4 = x + 4 = x + x + 6 = x + 8 x = ns. (i) _ x + y = 9.5 Writing in the form ax + by + c = 0, we get _ x + y 9.5 = 0 Here a =, b = and c= y (ii) x 0 = 0 5 Here a =, b = _ 9.5 and c =

12 (iii) x = 5y Writing in the form ax + by + c =0 x + 5y + 0 = 0 Here a =, b = 5 and c = 0 (iv) x + = 0 Writing in the form ax + by + c = 0 x + 0. y + = 0 Here a =, b = 0 and c = ns. (,4) is a solution of y = ax + 7 Substituting x +, y = 4in y = ax + 7, we get 4 = a + 7 = a + 7 a = 7 a = 5 a = 5 ns. x + y = 6 y = 6 x x y 4 0 X Y (0,6). (,4) x + y = 6 (,)... 4 (0,4) Y 0

13 SE TEST PPER-0 LSS IX Mathematics (Linear Equation in two Variables). x = 5can be written in as equation in two variable (a). X +. Y = 5 (b). x + 0. y = 5 (c). x + 0. y + 5 = 0 (d) oth b and c. The reason that a degree one polynomial equation ax + by + c = 0is called a linear equation is that (a) It has infinitely many solution (b) The geometrical representation is a straight line (c) It has two variable (d) oth a and b. Select the equation where graph is given below (,6) Y. 4 (0,4) (a) y = x (b) y = x + (c) y = x 4 (d) x y = 0. (-,0).. X y X 4. (-4,-) lies on a line which is a graph of the equation (a) x y = 5 (b) x + y = 4 (c) 4x y = 6 (d) x 4y = 0 5. Examine whether is a solution of the equation x + = 4x []

14 6. Solve the given equation 7x + = 5 [] 7. Solve the equation x 5 = 7x 5 9 [] 8. Find the valve of K if x=, y = is a solution of the equation x + y = k [] 9. number is three times the other. Write a linear equation in two variables to represent this statement 0. Express y in terms of xand xin terms of y from the following equation 7x 8y + 5. If the point (,4) lies on the graph of the graph of the equation y = ax + 7, find the valve of a. raw the graph of x y =, from the graph check whether x =, y = is a solution of the given equation or not.. The work done by a body on application of a constant force is directly proportional to the distance travelled by the body. Express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Read the graph the work done when the distance travelled by the body is. [5] () 0.5 units () 0 units

15 SE TEST PPER-0 LSS IX Mathematics (Linear Equation in two Variables) ns. (d) [NSWERS] ns. (b) ns. ns4. (c) (d) ns5. Substituting x = in the given equation L. H. S = x + = + = R. H. S = 4x = 4 = L. H. S = R. H. S ns6. 7x + = 5 Substituting - both the sides 7x + = 5 7x = 49 x = x = ns7. x 5 = 7x 5 9 On cross-multiplication 9 x 5 = 7x 5 ( ) 7x 45 = 7x 5 7x 7x = x = 40 x = ns8. x + y = k x =, y = is a solution, it must satisfy the given equation

16 . +. = k 4 + = k k = 7 ns9. Let one number be xand the other be y now the given statement can be presented through the following liner equation x = y x y = 0 ns0. 7x 8y + 5 = 0 7x 8y = 5 7x = 8y 5 8y 5 x = 7 7x + 5 = 8y 8y = 7x + 5 y = 7x ns. (,4) is a solution of y = ax + 7 Substituting x =, y = 4in y = ax + 7, we get 4 = a + 7 = a + 7 a = 7 = 5 a = 5 ns. x y = x = + y x =, y = is not a solution of given equation ns. Let xpresent the distance travelled and y the work done by the constant force y = 5x Form the graph we find that () the work done (y) is.5 units when distance ( x) is 0.5 units and () It is 0 unit when the distance is 0 unit (0,0) x (-,-5) y y=5x x-y= (,5). (4,) x 4

17 SE TEST PPER-0 LSS - IX Mathematics (Quadrilaterals). quadrilateral is a parallelogram if (a) = (b) (c) = = = (d) = 60, 60, 0. In figure, and EFG are both parallelogram if (a) (b) (c) (d) = 0 80, then GF is. In a square, the diagonals and bisects at O. Then O is (a) acute angled (c) equilateral (b) obtuse angled (d) right angled 4. is a rhombus. If 0 0, = then is (a) (c) (b) (d) The angles of a quadrilateral are in the ratio :5:9:. Find all the angles of the [] quadrilateral. 6. Show that each angles of a rectangle is a right angle. [] 7. transversal cuts two parallel lines prove that the bisectors of the interior [] 5

18 angles enclose a rectangle. 8. Prove that diagonals of a rectangle are equal in length. [] 9. In a parallelogram, bisectors of adjacent angles and intersect each other at P. prove that P = In figure diagonal of parallelogram bisects show that (i) if bisects and (ii) is a rhombus. In figure is a parallelogram. X and Y bisects angles and. prove that YX is a parallelogram.. The line segment joining the mid-points of two sides of a triangle is parallel to the third side.. Prove that if the diagonals of a quadrilateral are equal and bisect each other at [5] right angles then it is a square. 6

19 SE TEST PPER-0 LSS - IX Mathematics (Quadrilaterals) [NSWERS] ns0. ns0. ns0. ns04. (c) (c) (d) (c) ns05. Suppose angles of quadrilateral are x, 5x, 9x, and x = sum of angles of a quadrilateral is x + 5x + 9x + x = 60 0x = 60 x = = x = = 6 = 5x = 5 = 60 = 9x = 9 = 08 = x = = [ 60 ] ns06. We know that rectangle is a parallelogram whose one angle is right angle. Let be a rectangle. = To prove = = = 90 Proof: and is transversal + = = 80 = 90 = = = = 0 =

20 ns07. and EF cuts them at P and R. PR = PR [ lternate int erior angles] PR = PR i. e. = PQ RS [ lternate] ns08. is a rectangle and is diagonals. To prove = Proof: In and = [In a rectangle opposite sides are equal] 0 = [90 each] = common [common] [ y SS] = [ y PT] ns09. Given is a parallelogram is and bisectors of and intersect each other at P. 0 To prove P = 90 Proof: + = + = ( + ) ( i) ut is a parallelogram + = 80 + = 80 0 = 90 In P + + = P 80 [ y angle sum property] 90 + P = P = 90 0 Hence proved 8

21 ns0. (i) and is transversal = ( lternate angles) and = 4 ( lternate angles) ut = = 4 bi sec sts (ii) In and = [ common] = [ given] = 4 [ proved] = [ y PT] is a r hombus ns. Given in a parallelogram X and Y bisects and respectively and we have to show that YX in a parallelogram. In X and Y = ( i) [ opposite angles of a parallelog ram] X = [ given] ( ii) and Y= [given] (iii) ut = y () and (), we get X = Y ( iv) lso = [ opposite sides of parallelog ram] ( v) From ( i), ( iv) and ( v), we get X Y [ y S] X = Y [ PT] ut = [ oppsite sides of paralle log ram] or Y = X Y = X ut Y X [ is a gm] YX is a parallelogram 9

22 ns. Given in which E and F are mid points of side and respectively. To prove: EF onstruction: Produce EF to such that EF = F. Join Proof: In EF and F F = F [ F is mid point of ] = [ vertically opposite angles] EF = F [ Y construction] EF F [ y SS] E = [ y PT ] and E = E [ E is the mid point] E = cd and [ = ] E is a parallelogram EF Henceproved ns. Given in a quadrilateral, =, O = O and O = O and To prove: is a square. Proof: In O and O O = 90 0 O = O [ given] O = O [ given] and O = O [ vertically opposite angles] O O [ y SS] = [ y PT ] = [ y PT ] ut there are alternate angles is a parallelogram whose diagonals bisects each other at right angles is a rhombus gain in and = [ Sides of a r hom bus] = [ Sides of a r hom bus] and = [ Given] = [ y PT ] There are alternate angles of these same side of transversal + = 80 or = = 90 Hence is a square 0 0 0

23 SE TEST PPER-0 LSS - IX Mathematics (Quadrilateral). In fig is a parallelogram. It = 60 and = 80 then ( ) ( ) ( ) ( ) If the diagonals of a quadrilateral bisect each other, then the quadrilateral must be. (a) Square (b) Parallelogram (c) Rhombus (d) Rut angle. The diagonal and of quadrilateral are equal and are perpendicular bisector of each other then quadrilateral is a (a) Kite (b) Square (c) Trapezium (d) Rut angle 4. The quadrilateral formed by joining the mid points of the sides of a quadrilateral taken in order, is a rectangle if (a) is a parallelogram (b) is a rut angle (c) iagonals and are perpendicular (d) = 5. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram is []

24 6. In fig is a parallelogram and X,Y are the [] points on the diagonal such that X<Y show X O that YX is a parallelogram. Y 7. Show that the line segments joining the mid points of G [] opposite sides of a quadrilateral bisect each other. H F 8. is a rhombus show that diagonal bisects as well as and diagonal bisects as well as 9. Prove that a quadrilateral is a rhombus if its diagonals bisect each other at right angles. 0. Prove that the straight line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides. E []. In fig is a right angle in. is the mid-point of. E intersects at E. show that (i) E is the mid-point of (ii) E (iii) = E. is a triangle and through vertices, and lines are drawn parallel to, and respectively intersecting at, E and F. prove that perimeter of EF is double the perimeter of.. Prove that in a triangle, the line segment joining the mid points of any two sides is parallel to the third side. [5]

25 SE TEST PPER-0 LSS - IX Mathematics (Quadrilateral) [NSWERS] ns. () ns. () ns. () ns4. () ns5. Given quadrilateral in which = and = To prove: is a parallelogram onstruction: Join Prof: In and = = = ( given) [ common] = [ by sss] [ y PT] Is a parallelogram ns6. is a parallelogram. The diagonals of a parallelogram bisect bisect each other O = O ut X = Y[given] O X = O Y or OX = OY Now in quad YX, the diagonals and XY bisect each other YX is a parallelogram.

26 ns7. Given is quadrilateral E, F, G, H are mid points of the side,, and respectively To prove: EG and HF bisect each other. In, E is mid-point of and F is mid-point of EF nd EF =... ( i) Similarly HG and G =... ( ii) From (i) and (ii), EF EFGH HGand EF = GH is a parallelogram and EG and HF are its diagonals iagonals of a parallelogram bisect each other Thus EG and HF bisect each other. ns8. is a rhombus In and = [Sides of a rhombus] = [Sides of a rhombus] = [ommon] [y SSS ongruency] = nd = Hence bisects as well as Similarly, by joining to, we can prove that Hence bisects as well as ns9. Given is a quadrilateral diagonals and bisect each other at O at right angles To Prove: is a rhombus Proof: diagonals and bisect each other at O O = O, O = Ond = = = 90 Now In Ond O O = O Given O = O[ommon] 0 4

27 nd = = 90 (Given) O = O(SS) = (.P..T.) ns0. Given a trapezium in which and M,N are the mid Points of the diagonals and. We need to prove that MN Join N and let it meet at E Now in N and EN N = EN[lternate angles] N = EN[lternate angles] nd N = N[given] N EN[S] N = EN[y.P..T] Now in E, Mand Nare the mid points of the sides and E respectively. MN EOr MN lso MN M E N ns. Proof: E and is mid points of In E and E E=E E= E nd E = E = 90 E = E E E = ns. FIs a parallelogram = F EIs a parallelogram 5

28 = E F + E = F E Or EF = Similarly E = and F = Perimeter of = + + Perimeter of EF = E + EF + F = ++ = [++] = Perimeter of Hence Proved ns. Given: in which and E are mid-points of the side and respectively TO Prove: E onstruction: raw F Proof: In E = and FE [Vertically opposite E 4 F angles] E=E [Given] nd = 4 [lternate interior angles] E FE [y S] E=FE [y.p..t] ut = = F Now F F is a parallelogram E lso E = EF = 6

29 SE TEST PPER-0 LSS - IX Mathematics (Quadrilateral). In the fig is a Parallelogram. The values of xand yare (a) 0, 5 (b) 45, 0 (c) 45, 45 (d) 55, 5. In fig if E=8cm and is the mid Point of, then the true statement is (a) = (b) E (c) E is not mid Point of (d) E. The sides of a quadrilateral extended in order to form exterior angler. The sum of these exterior angle is (a) 80 (b) 70 (c) 90 (d) is rhombus with = 40. The measure of is (a) 90 (b) 0 (c) 40 (d) In fig is a median of, E is mid-point of.e produced meet at F. Show that F = (y+5) 0 (x-0) 0 E (x+80) 0 F E [] 6. Prove that a quadrilateral is a parallelogram if the diagonals bisect each other. [] 7. In fig is a Parallelogram. P and Q are Perpendiculars from the Vertices and on diagonal. Show that (i) P Q (ii) P = Q P Q [] 7

30 8. is a Parallelogram E and F are the mid-points of and respectively. Show that the segments F and E trisect the diagonal. [] 9. In fig is a quadrilateral P,Q,R and S are the mid Points of the sides,, and, is R diagonal. Show that S Q (i) SR (ii) PQ=SR P (iii) PQRS is a parallelogram (iv) PR and SQ bisect each other 0. In,, E, Fare respectively the mid-paints of sides, and. show that is divided into four congruent triangles by Joining,E,F.. is a Parallelogram is which P and Q are mid-points of apposite sides and. If Q intersect P at S Q intersects P at R, show that (i) PQ is a Parallelogram (ii) PQ is a parallelogram (iii) PSQR is a parallelogram. l, m, nare three parallel lines intersected by transversals P and q such that l, mand ncut off equal intercepts and on P In fig Show that l, m, ncut off equal intercepts E and EF on q also. P G q E l m F n. is a rhombus and P, Q, R, and S are the mid-points of the sides,, and respectively. Show that quadrilateral PQRS is a rectangle. [5] 8

31 SE TEST PPER-0 LSS - IX Mathematics (Quadrilateral) [NSWERS] ns. () ns. () ns. () ns4. () ns5. Let M is mid Point of F Join M M F. In M, Eis mid- Point of and M EF Fis mid-point of M F = FM FM = M F = FM = M = F + FM + M = F + F + F F = F F = Hence Proved E F M ns6. is a quadrilateral in which diagonals and intersect each other at O In Oand O O + O [Given] O = O [Given] nd O = O[Vertically apposite angle O O [y SS] O = O[y.P..T] ut this is Pair of interior angles Similarly Quad is a Parallelogram. 9

32 ns7. (I) in Pand Q = [apposite sides of a Parallelogram] P = Q [each 90 ] nd P = Q P Q[S] (II) P = Q(y.P..T) ns8. F E and F=E EF Is a Parallelogram EG H and E is the mid Point of G is the mid-point of H Or HG=G..(i) Similarly H=HG.(ii) From (i) and (ii) we get H=HG=G F H G E ns9. In, P and Q are the mid-points of the sides and respectively PQ and PQ= Similarly SR and SR= PQ SR and PQ=SR Hence PQRS is a Parallelogram. ns0. and E are mid-points of sides and of E { line segment joining the mid-point of any two sides of a triangle to third side} Similarly F and EF EF, EF and FE are all Parallelograms. E is diagonal of Parallelogram FE E FE Similarly F FE nd EF FE So all triangles are congruent E F ns. () In quad PQ P Q [ ]..(i) (Given) S Q R P=, Q= lso = P 0

33 So P=Q.(ii) Therefore, PQ is a parallelogram [It any two sides of a quad equal and parallel then quad is a parallelogram] () Similarly, quadrilateral PQ is a Parallelogram because Q P and Q=P () In quad PSQR, SP QR [SP is a part of P and QR is a Part of Q] Similarly SQ PR So. PSQR is parallelogram. ns. In fig l, m, nare parallel lines intersected by two transversal P and Q. To Prove E=EF Proof: In F is mid-point of nd G F G is mid-point of F [y mid-point theorem] Now In F G is mid-point of F and GE E is mid-point of F [y mid-point theorem] E=EF Hence Proved ns. Join and which intersect at O let intersect RS at E and intersect RQ at F IN P and S are mid-points of sides and PS and PS= R Similarly RQ and RQ= RS RQ and PS = = RQ S E O F Q PS=RQ and PS=RQ PQRS is a parallelogram Now RF EO and RE FO OFRE is also a parallelogram. P gain, we know that diagonals of a rhombus bisect each other at right angles. EOF = 90 EOF = ERF[Opp. ngles of a gm] ERF = 90 Each angle of the parallelogram PQRS is 90 Hence PQRS is a rectangle.

34 SE TEST PPER-04 LSS - IX Mathematics (Quadrilateral). In fig is mid-point of and E then E is equal to (a) (b) E (c) (d). In fig and E are mid-points of and respectively. The length of E is (a) 8. cm (b) 5. cm 4.9cm 5.cm (c) 4.9 cm (d) 4. cm E E 8.cm. diagonal of a parallelogram divides it into (a) two congruent triangles (b) two similes triangles (c) two equilateral triangles (d) none of these 4. quadrilateral is a, if its apposite sides are equal: (a) Kite (b) trapezium (c) cyclic quadrilateral (d) parallelogram 5. In the adjoining Fig. =. P and P is the bisector of prove that [] (a) P = and (ii) P is a parallelogram 6. Prove that if each pair of apposite angles of a quadrilateral is equal, then it is a [] parallelogram.

35 7. In Fig. is a trapezium in which E is the mid-point of. line through E is E F l [] parallel to show that lbisects the side 8. In Fig. is a parallelogram in which X and Y X [] are the mid-points of the sides and respectively. Prove that XY is a parallelogram Y 9. is a parallelogram in which E is mid-point of EF E meeting produced in F and at L prove that E = L 0. PQRS is a rhombus if P = 65 find RSQ. is a trapezium in which and = show that (i) = (ii) = (iii). Show that diagonals of a rhombus are perpendicular to each other. [5]. In the given Fig is a parallelogram E is midpoint of and E bisects (i) E = Prove that: (ii) E bisects E (iii) E = 90

36 SE TEST PPER-04 LSS - IX Mathematics (Quadrilateral) [NSWERS] ns. () ns. () ns. () ns4. () ns5. In = = [Opposite angle of a equal sides are equal] = + [Exterior angle] L = P = ( ) ( ) P = Now P = P lso P Given) is a parallelogram ns6. Given: is a quadrilateral in which = and = To Prove: is a parallelogram Proof: = [Given] = [Given] + = +...( i) In quad = 60 ( + ) + ( + ) = 60 [y.(i)] + = 80 + = + = 80 These are sum of interior angles on the same side transversal and is a parallelogram 4

37 ns7. Join In E is mid-point of and EO O is mid point of [ line segment joining the mid point of one side of a parallel to second side bisect the third side] In O is mid point of OF F is mid point of lisect E O F l ns8. In the given fig is a parallelogram and = nd = X Ynd X = Y [X and Y are mid-point of and respectively] XY Is a parallelogram ns9. In F Eis mid-point of (Given) E F (Given) y converse of mid-point theorem is mid-point of F = F...( i) is parallelogram =...( ii) From (i) and (ii) = F onsider Land FL = F [Proved above] L = FL[lternate angles] L = FL[Vertically app. angles] L = FL[..S] L=LF F=L E L F ns0. S x x 0 P Q 5

38 R = P = 65 [opp. ngles of a parallelogram are equal] Let RSQ = x In RSQwe have RS=RQ RQS = RSQ = x [Opp. Sides of equal angles are equal] In RSQ S + Q + R = 80 [y angle sum property] x + x + 65 = 80 x = x = 5 5 x = = 57.5 RSQ = 57.5 ns. Produce and raw a line Parallel to meeting at E E + = 80.(i) [Sum of interior angles on the some side of transversal is 80 ] In E =E (given) = 4..() [in a equal side to app. ngler are equal] + 4 = 80 () y (i) and () + = + 4 = 4 = (i) = (ii) E = 80...( i) E = 80...( ii) = = = 4 = 4 = 6 { 4 = 6 = (iii) In and = [common] = [Proved above] = [given] [y SS] E 6

39 ns. Given: rhombus whose diagonals and intersect at a Point O TO Prove: O = O = O = O = 90 Proof: clearly is a Parallelogram in which === We know that diagonals of a Parallelogram bisect each other O=O and O=O Now in O and O, we have O=O = O=O O O [y SSS] O = O[y.P..T] ut O + O = 80 O = O = 90 Similarly O = O = 90 Hence diagonals of a rhombus bisect each other at 90 O ns. is a parallelogram nd E cuts them E = E [lternate interior angle] E = E [ E = E] E = E = (i) Now E= E = E E = E [ E = E lternate interior angles] (ii) E bisects (iii) Now + = 80 + = 90 E + E = 90 ut, the sum of all the angles of the triangle = E = 80 E = 90 7

40 SE TEST PPER-05 LSS - IX Mathematics (Quadrilateral). Which of the following is not a Parallelogram? (a) Rhombus (b) Square (c) Trapezium (d) Rectangle. The sum of all the four angles of a quadrilateral is (a) 80 0 (b) 60 0 (c) 70 0 (d) In Fig is a rectangle P and Q are mid- Q points of and respectively. Then length of PQ is P cm (a)5cm 4cm (b) 4cm (c).5cm (d) cm 4. In Fig is a rhombus. iagonals and intersect at O. E and F are mid points of O and O respectively. If = 6cm and = cm then EF is E O F (a)0cm (b) 5cm (c) 8cm (d) 6cm 5. The angles of quadrilateral are in the ratio :5:0: Find all the angles of the quadrilateral. [] 8

41 6. In fig is mid-points of. P is on such that [] P = Pand E P show that E = E P 7. Prove that the bisectors of the angles of a [] Parallelogram enclose a rectangle. It is given that P Q S R adjacent sides of the parallelogram are unequal. 8. Prove that a quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal [] 9. In fig is a parallelogram, X and Y are the mid- Y points of the sides and respectively show that P Q PXQY is a parallelogram X 0. Prove that the diagonals of a rhombus bisect each other at right angles. In fig is a trapezium in which and =. Show that = P. In fig and EF are Parallelogram, prove that FE is also a parallelogram F E. is a triangle right angled at. line through the mid-point M of hypotenuse [5] and parallel to intersects at. show that (i) is mid-point of (iii) M (iii) M = M = 9

42 SE TEST PPER-0 LSS - IX Mathematics (Quadrilateral) [NSWERS] ns. () ns. () ns. () ns4. () ns5. Suppose angles of quadrilaterals are x, 5 x, 0 x, and x = x, = 5 x, = 0 x, = x In a quadrilateral = 60 x + 5x + 0x + x = 60 0x = x = = 0 = = 6, = 5 = 60 = 0 = 0, = = 44 ns6. In P is mid points of and E P E is midpoint of P E = EP also P = P P = P P = E P = E E = PE = P = E + EP + P = E + E + E 40

43 E = Hence Proved ns7. is a parallelogram + = 80 or ( + ) = 90 Or P = 90 [Sum of angle of a 80 ] SPQ = P = 90 Similarly QRS = 90 and PQR = 90 P + Q + R + S = 60 PSR = 90. Thus each angle of Quad. PQRS is 90 Hence PQRS is a rectangle. ns8. Given: is a quadrilateral in which and. To Prove: is a parallelogram onstruction: Join and intersect each other at O. Proof: O O [y ecause = = 4 and 5 = 6 O=O nd O=O is a parallelogram iagonals of a gram bisect each other O ns9. is a parallelogram = and Or = i.e. X = Yand X Y XY is a parallelogram 4

44 XQ PY... ( i) similarly, we can prove that XY is a parallelogram PX YQ (ii) From (i) and (ii) we get PXQY is a Parallelogram. ns0. We are given a rhombus whose diagonals and intersect each other at O. We need to prove that O=O, O=O and O = 90 In O and O = [Sides of rhombus] O = O [vert. opp. ngles] nd O = O [lt. angles] O O [y S] O=O nd O=O [y.p..t] lso in O and O O=O [Proved] = [sides of rhombus] nd O=O [ommon] O O [y SSS] O = O[y.P..T] ut O + O = 80 [linear pair] O = O = 90 O ns. To show that =, raw p meeting at P P and P P is a parallelogram gain in P P= [ = [Given] P = VP...( i) [ngles app. Equal sides] ut P + P = 80 [y linear pair] 4

45 lso + P = 80 [ P is a gram] + P = P + P Or = P = P [from(i)] Hence = ns. is a gram = also..(i) lso EF is a gram =FE and FE.(ii) y (i) and (ii) ==FE =FE nd FE FE EF is a parallelogram Hence Proved ns. Given is a right angle at (I) M is mid-point of nd M is mid Point of [a line through midpoint of one side of a parallel to another side bisect the third side] (II). M M = [orresponding angles] M = 90 (III) In M and M = [ is mid-point of ] M=M [ommon] M M [y SS] M=M [y.p..t] M=M=M [ M is mid-point of ] M=M=. M 4

46 SE TEST PPER-0 LSS - IX Mathematics (reas of Parallelograms and Triangles). Find the area of parallelogram (a) 759foot (b) 48 square (c) 84 square foot 5 7 (d) 60 square foot E. Find the measure of angle a (a) 45 (b) (c) 40 d) triangle has an area of 45 square foot ase of the triangle is 9 foot. What is corresponding height of triangle (a) 90 foot (b) 5 foot (c) 0 foot (d) 40 square foot 4. What is area of parallelogram base=8 and corresponding altitude is 5 (a) 40 (b) 45 (c) (d) 44

47 5. Prove that is a parallelogram. If is a quadrilateral and is one of its diagonal [] In a parallelogram = 0. The altitudes to sides is 0 cm. find area of parallelogram. [] 7. If L be any Point of and the angle of rectangle is 00 [] square cm. find area of L L 8. Find the area of parallelogram. is diagonal on quadrilateral whose = 7 and is 5 9. PQRS is a quadrilateral and SQ is one of its diagonals. Show that PQRS is a Parallelogram and find its area too. [] 0. In a parallelogram PQRS. The ltitude corresponding 0 S R sides PQ and PS one respectively. 7cm and 8cm find PS and PQ=0cm. L 7cm 8cm P M Q. rea, base and corresponding altitude are area of parallelogram. x, x and x + 4respectively. Find the. Find the altitude corresponding to side EF if as = EF. If = 8 cm and altitude corresponding to is 5cm. In EF, EF = 0cm. Prove that of all the parallelograms of given sides the parallelogram which is a rectangle has the greatest area. [5] 45

48 SE TEST PPER-0 LSS - IX Mathematics (reas of Parallelograms and Triangles) ns. (d) [NSWERS] ns. (b) ns. (c) ns4. (a) ns5. Given quadrilateral in which = =, = 4 and = = 90 intersects and such that = = 90 = = is a parallelogram ns6. rea of parallelogram = M =0 0 =00 square cm. M ns7. rea of rectangle = 00 square cm re L = are rectangle = 00square cm =50 square cm 46

49 ns8. rea of parallelogram = ase orresponding ltitude =7 5 5 square cm rea of parallelogram = 5 square cm ns9. We know that, area of gram PQRS. In which PQ=SR=, SQ=4 nd S = Q = 90 PQS = QSR = 90 PQ SR PQ=SR= is a gram rea of parallelogram =ase corresponding ltitude = 4 = square units ns0. rea of gram PQRS =PQ SM =0 7 =70 square cm..(i) rea of Parallelogram PQRS =PS QL =( 8) square cm (ii) From (i) and (ii) 8=70 = 70 8 =8.75 cm ns. rea of parallelogram = ase orresponding ltitude 47

50 x x x ( ) = ( ) + 4 x x x x = + 4 x = ( )( 4) ( 9)( 6) = = = 44 square units. ns. ar ( ) = ar ( EF ) M = EF N 8 5 = 0 N 0 = 5N N = 4cm 5cm M 8cm E N 0cm F ltitude corresponding to side EF is 4 cm ns. Let PQRS be a parallelogram in which PQ = a and PS = b and h be the altitude corresponding to base PQ rea of parallelogram PQRS = ase corresponding ltitude = ah PSKis a right angled,b( PS ) being its hypotenuse. ut hypotenuse is the greatest side of rea of (ah) of gram PQRS will be greatest when h is greatest H = b, then PS PQ The gram PQRS will be a rectangle hence, the area of gram is greatest when it is a rectangle S R S R b h P K a Q P Q 48

51 SE TEST PPER-0 LSS - IX Mathematics (reas of Parallelograms and Triangles). Parallelogram on the same base and between the same parallel are equal (i) corresponding angle (ii) area (iii) congruent area (iv) same parallel. ny side of a parallelogram is called (i) ltitude (ii) base (iii) corres. ltitude (iv) area. diagonal of a parallelogram divides into triangles of equal area (i) (ii) (iii) (iv) none of these 4. Find the area if ase = and ltitude is 4 (i) 7 (ii) (iii) (iv) none of these 5. Show that ar () = ar (). and are two triangles on the same base if line segment is bisected by at O 6. Show that EF is gm. If,E and F the mid- point of the side, and of [] [] 49

52 7. Prove that ar OLP = ar MNLif MP PO [] 8. etermine the altitude corresponding to side EF if ar ( ) ar ( EF ) =, = 8and altitude is 5 cm and EF, EF = 0cm 9. Prove that the area of a trapezium is half of the product of its height and the sum of the parallel sides [] 0. Show that the area of a rhombus is half the product of the length of its diagonals. O. In parallelogram P is any point inside it. Prove that M P + ax of P = area, of P K. Show P Q (i) ar(pqrs) = ar (RS) (ii) ar( S) = ar(pqrs) X If Xis any point on side R on PQRS and RS S R. Prove that (i) ar( E)= ar( ) 4 [5] (ii) ar( E)= ar( E) and E are two equilateral triangle such that is the mid-point of E intersects at F. F E 50

53 SE TEST PPER-0 LSS - IX Mathematics (reas of Parallelograms and Triangles) [NSWERS] ns. ns. ns. ns4. ns5. ns6. (ii) (ii) (ii) (iii) O is the median of ( ) = ( ) ( ) = ( ) ( ) + ( ) = ( ) + ( ) ( ) = ( ) ar O ar O ar O ar O ar O ar O ar O ar O ar ar Join E, EF and F E and F are the mid-points of and EF EF E F EF is a gram. ns7. ar ( MPO) = ar ( MPN ) ar ( MPO) ar ( MPL) = ar ( MPN ) ar ( MPL) ar ( OLP) = ar ( MLN ) ns8. Given that ar ( ) = ar ( EF ) cm = EF N 8 5 = 0 N 0 = 5N N = 4cm 5

54 ns9. Join and. raw L (Produced) ( ) = ( ) + ( ) ar ar ar = K + L = K + K = K ( + ) K L ar ( ) = O... ( ii) dding (i) and (ii) ar + O + O ns0. ar ( ) = O... ( i) ( ) ( ) = Hence, area of rhombus = ns. ( ) ar P = PK ar ( P) = PM = PM ar ( P) + ar ( P) = PK + PM = ( PK + PM ) = MK = ar ( gram ) ns. (i) gram PQRS and RS are on the same base SR and etween the same parallel P = SR, So, ar ( PQRS ) = ar ( RS ) 5

55 (ii) ar ( S ) = ar ( RS ) ( ) ( ) ar RS ar PQRS ar S ar PQRS ( ) = ( ) ns. Join E (i) let a be the side of equilateral ar a i 4 ar ( ) =...( ) a = 4 ( E) = a...( ii) 6 From ( i) and ( ii) ar ( E) = ar ( ) 4 ar E = ar E E = 60 (ii) ( ) ( ) = 60 E = E ( ) = ( ) ar E ar E ar E ar E ( ) = ( ) 5

56 SE TEST PPER-0 LSS - IX Mathematics (reas of Parallelograms and Triangles). If a triangle and a Parallelogram are on the same base and between the same parallel, the area of the triangle is equal to that of gram. (a) (b) 4 (c) (d) none of these. Find the area of gram if base=5, altitude = 6 (a) 0 (b) (c) 6 (d). The area of a is the Product of any of its sides and corresponding altitude (a) Triangle (b) Parallelogram (c) Rhombus (d) rea xiom 4. ase = 9, corresponding altitude = 5. Find area of gram (a) 4 (b) 40 (c) 4 (d) none of these 5. In a parallelogram EFGH, EF=5cm and the corresponding altitude HM is 9cm. Find area. [] 54

57 6. Show that the median of a triangle divides it into two triangles of equal area. [] 7. The area of rectangle EFGH is 400cm. if L be any Point on EF, find area of LGH [] 8. Prove that as ( ) = or ( ) G G if [] G 9. Show that the segment joining the mid-points of a pair of opposite sides of a gram divides it into two equal gram. 0. Show that PQ divides the gram in two Part of equal area Q if diagonal of gram intersect Point O. through Point line is drawn to intersect at P and at Q O P. Show that O = are of O if O is any Point on its median. O. The triangle and are equal in area and and intersect in O. So prove that O=O O. Show that EFGH is a gram and its area of the gram. If E, F, G, H are respectively the mid points of the sides,, and. [5] 55

58 SE TEST PPER-0 LSS - IX Mathematics (reas of Parallelograms and Triangles) ns. ns. ns. ns4. (a) (a) (b) (d) [NSWERS] ns5. rea of Parallelogram EFGH = base ltitude = 5 9 (Square cm) = 45 square cm ns6. raw the ltitude E from vertex on the base. rea of = E rea of = E = E rea of =area of E ns7. s of LGH = as rectangle EFGH = 400 Square cm =00 square cm ns8. ar ( ) = ar ( ) ar ( ) ar ( G) = ar ( ) as ( G) ar ( G) = ar ( G) 56

59 ns9. Join EF is a gram and = E F = E F and E = F In quadrilateral EF, one Pair of opposite sides E and F is equal and parallel ar ( gram EF) = ar ( gram EF ) ns0. re (quad. PQ) = ar (quad PQ) = (ar gram ) OP = OQ (V O ) O = O OP = OQ OP OQ ( ) = ( )...( ) ( ) = ( ) ( quad. ) + ( ) ( quad. ) ( ) ( quad. ) = ( quad. ) ar OP R OQ i ar ar ar QO ar OQ = ar OP + ar OP ar PQ ar PQ ns. Join O and O ar ( ) = ar ( ) [Median divide].(i) ar ( O) = ar ( O)...( ii) ar ( ) ar ( O) = ar ( ) ar ( O) ar ( O) = ar ( O) ns. ar ( O) = ar ( O) [Median divides in two sof equal area].(i) is a median ( ) = ( )...( ) ar O ar O ii dding (i) and (ii) ( ) + ( ) = ( ) + ( ) ( ) = ( ) ar O ar O ar O ar O ar ar 57

60 ns. Join and HF E and F are the mid-points of and G EF= and EF.(i) GH = and GH (ii) H F GH = EF and GH EF EFGH is a gram ar HGF ar gram HF iii ar HEF ar gram HF iv ar ( gram EFGH ) are ( gram ) ( ) = ( )...( ) ( ) = ( )...( ) E ( HGF ) + ar ( HEF ) = ar ( gram HF ) + ar ( gram HF ) 58

61 SE TEST PPER-04 LSS - IX Mathematics (reas of Parallelograms and Triangles). median of a triangle divides. It in to triangles of equal areas. (a) (b) same triangle (c) (d) none. The area of a rhombus is equal to the product of its two diagonals. (a) (b) (c) 4 (d) none. Find the area of parallelogram base = 8 corresponding ltitude = 4 (a) (b) (c) 4 (d) 8 4. rea of a triangle is half the product of any of its sides and the (a) orresponding altitude (b) altitude (c) median (d) base 5. Show that ar (quad. )= (M+N) is one of the diagonals of a quadrilateral, M and N are the from and [] 59

62 6., E, F are respectively the mid-points of the sides, and of Prove [] ar (EF) = ar () 4 7. In a parallelogram PQRS, =. The altitude to side PS = cm. find area of parallelogram PQRS [] 8. line through, Parallel to meets produced in P. [] prove that area P = ar P 9. = and if median of intersect Show that ar( G) ar ( ) G. F G E 0. Show that ar = ar QRP, Q is dawn Parallel to P to intersect Produced at Q and Parallelogram QRP is completed P is any Point on produced. Q R. Show that area of PQ = area is mid point of, P is any point on. PQ is joint and line Q is dawn parallel to P to intersect at Q.. E is the mid-point of median show that 4 ar( E) = ar ( ) E. Show that ar (P) = ar (PQ) if is produced to a point Q such [5] that = Q and Q intersect at P P Q 60

63 SE TEST PPER-04 LSS - IX Mathematics (reas of Parallelograms and Triangles) ns. ns. ns. ns4. (c) (a) (b) (a) [NSWERS] ns5. ar ( quad.) = ar ( ) + ar ( ) = + = ( M + N ) ( M ) ( N ) ns6. ar ( F ) = ar ( EF ) ns7. Now, ar ( gram EF ) = ar ( EF ) = ar 4 ( ) = ar ( ) rea of parallelogram PQRS =ase orresponding ltitude = =44 square cm ns8. ar ( ) = ar ( ) ar ( P) + ar ( ) = ar ( ) + ar ( ) ( ) = ( quad. ) ar P ar ns9. is median ( ) = ( )...( ) ar ar i G is median ( ) = ( )...( ) ar G ar G ii 6

64 Subtracting (ii) and (i) ( ) ( ) = ( ) ( ) ( ) = ( )...( ) ( ) = ( )...( ) ar ar G ar ar G ar G ar G iii ar G ar G iv From (iii) and (iv) ar G = ar ( ) ( ) ns0. ns. is diagonal of gram ( ) = ( ) ( ) ( ) = ( ) ( ) ( ) = ( ) ( ) ( ) = ( ) ( ) ( ) = ( )...( ) ar ar gram... i ar PQ ar gramqrp... ii ar Q ar QP ar Q ar Q QP ar Q ar ar PQ iii From (i),(ii) and (iii) ( ) = ( ) ar gram ar gramqrp is median ar = ar... i ar PQ = ar P... ii ( ) ( ) ( ) ( ) ( ) ( ) From (i) ar ( ) = ar ( ) ar ( P) + ar ( P ) = ar ( ) ar P ar PQ ar ii ar ( PQ) = ar ( ) ( ) + ( ) = ( )...( ) ns. ar ( ) = ar ( ) ar ( ) = ar ( ) Similarly in, E is the median ar E ar ( ) = ( ) 6

65 ar E ar = ar ( ) 4 ( ) = ( ) ns. Join ar P = ar P... i ( ) ( ) ( ) = Q Q Hence, a pair of opposite side and Q of the quad Q is equal and parallel. In P and QP, P = QP P = P P = QP P QP ( ) = ( )...( ) ar P ar QP ii From (i) and (ii) ( ) = ( ) ( ) = ar ( PQ) ar P ar QP ar P P Q 6

66 SE TEST PPER-05 LSS - IX Mathematics (reas of Parallelograms and Triangles). Given below are the measurements of a parallelogram. Find the missing measurement. rea = 9 square cm, ase = 5 cm, Height =? (a) 8 (b) 450 (c) 85 (d) 5cm. How many square feet are in square yard (a) 6 (b) 9 (c) (d) 0. The perimeter of an equilateral triangle is yard. what is the length of each its sides (a) 7 yard (b) 4 yard (c) 8 yard (d) yard 4. What is the area of a triangle with base m and a height of 8m (a) 08m (b) (c) (d) 6m 08m 98m 5. In a parallelogram PQRS, PQ =. The altitude corresponding to sides PS = 5cm. find the area of parallelogram. 6. Prove that ar (O) = ar (O). iagonals and of a trapezium with intersect each other at O. [] [] 64

67 7. Prove that ar (Q) = ar (PR) P Q R P [] Q 8. Show that E if ar ( E) =ar ( ) [] R E 9. Show that the line segments joining the mid-points of parallel sides of a trapezium divides it into two parts of equal area 0. Prove that ( X ) ar ( Y ). = if and line parallel to intersects at Xand at Y. Prove that area of G= ar of quad. FGE if E and F median intersect at G.. Show that ar E area E =area E area E if diagonals of quadrilateral and intersect at a Points E. E. If area of P = Kand two point and a positive real number K. find the lows of a point p [5] 65

68 SE TEST PPER-05 LSS - IX Mathematics (reas of Parallelograms and Triangles) [NSWERS] ns. (a) ns. (d) ns. (a) ns4. (c) ns5. rea of parallelogram = base ltitude = PQ PS = 5 =65cm ns6. r ( O ) = ar ( ) ar ( ) ar ( O) = ar ( ) ar ( O ) ar ( O) = ar ( O ) O ns7. ar ( Q) = ar ( PQ) ar ( Q) = ar ( RQ) ar ( Q) + ar ( Q) = ar ( PQ) + ar ( PQ) ar ( Q) = ar ( PR) ns8. Since se and are equal in area and have a same base se and are between the same Parallel lines. E ns9. raw M Pand N P M=N=h rea of trapezium PQ = ( P Q) M + = h + = h( )...( i) 4 + rea of trapezium PQ M n Q P n N 66

69 = h( ) 4 + From (i) and () r (trap. PQ)= ar (trap. PQ) ns0. Join X ar X = ar Y ar ar ar ( ) ( ) ( X ) = ar ( X ) ( Y ) = ar ( X ) ( X ) = ar ( Y ) X Y ns. In median E of the ( E) rea ( E ) =area ( E) rea ( G ) =area ( GE ) = area ( quad.fge ) + area ( GF ) Now, median F of or F=area F area G + area GF = area quad. FGE + area GGE ( ) ( ) ( ) ( ) ( G) ( FGE) ( G ) = area ( quad. FGE) area = area quad. area ns. raw M and also N ar ( E) ar ( E ) = E M E N = EXM E N 4 = E M E N = ar E ar E ( ) ( ) ( ) ( ) = ( ) ( ) ar E ar E ar E ar E N M E ns. Let the perpendicular distance of P from be h ( ) ar P = K P P ( ) h = K K h = P P Since and K are given h is a fixed Positive real number. This means that P lies on a line Parallel to at a distance h from it. K Hence, the locus of P is a pair of lines at a distance h =, parallel to 67

70 SE TEST PPER-0 LSS - IX Mathematics (ircle). ny angle in the semicircle is (a) Right angle (b) 80 (c) 60 (d) none of these. If the angles subtended by two chords of a circle at the centre are equal the chords are (a) not equal (b) equal (c) angle equal (d) line equals. How many circle passing through three non-collinear points (a) one (b) two (c) three (d) four 4. The constant distance is called (a) diameter (b) radius (c) centre (d) circle 5. = and diagonal and intersect at P in cyclic quadrilateral Prove P P 6. Prove that =, if and are two right triangle with common [] [] hypotenuse. 7. Show that E, in isosceles triangle, = and, intersects the sides [] and at and E. 8. Prove cyclic parallelogram is a rectangle [] 9. Pair of opposite sides of a cyclic quadrilateral are equal, Prove that the other two sides are parallel. 0. Prove that the centre of the circle through,,, is the Point intersection of its diagonals.. In isosceles triangle, = E and and E are equal on side and so prove that,,e and are con cyclic. If two non parallel sides of a trapezium are equal, prove that it is asdic.. The bisector of of an isosceles triangle with = meets the circum circle of at P if P and produced meet at Q, prove that Q = [5] 68

71 SE TEST PPER-0 LSS - IX Mathematics (ircle) [NSWERS] ns. (a) ns. (b) ns. (a) ns4. (b) ns5. In s P and P = P = P [ngle in the same segment P = P [ngle in the same segment P P [S criterion] ns6. = = 90 [ is the common hypotenuse of it s + = 80 and ] Quadrilateral is cyclic Now, chord subtends and = [ngle in the same segment] ns.7 ccording to given: E forms a cyclic quadrilateral ( ) ( ) E =... i =... ii E From (i) and (ii) waged E = ut they form a Pair of corresponding angles E 69

72 ns8. Let be the given cyclic parallelogram + = 80...( i) = [Opposite angle of a parallelogram are equal]..(ii) From (i) and (ii) = = 90 is a rectangle ns9. Given: cyclic quadrilateral in which = To Prove: onstruction: Join and Proof: = = = ut these are alternate angles ns0. Given: cyclic rectangle in which diagonals and intersect at Point O To Prove: O is the centre of the circle Proof: is a rectangle = Now as the diagonals and are intersecting at O O=O, O=O O=O=O=O,,, lie on the same circle ns. Given that, = =E E = E =, 4 = 4 = = + 5 = + 5 [dding 5both side] + 5 = = 80 4 E 5 70

73 ns. In it s Kand P = K=P [istance between sides K = P ( ) =... i = + 90 = + 90 = ( ) =... ii = = 60 + = 80 K P ns. Join P and onsidered Q 4 = +... ( i) [Exterior angle is equal to the sum of two interior opposite angle] 4 =... ( ii) From (i) and (ii) + = = 5 [P is bisector of ] + = 5 + = = [ngle in the same segments] + = + = In Q, Q = 5 4 Q 7

74 SE TEST PPER-0 LSS - IX Mathematics (ircle). PS and RS are two chord s of a circle such that PQ=0cm and Rs= 4cm and PQ RS. The distance between PQ and RS is 7cm. Find the radius of circle (a) 0cm (b) cm (c) 5cm (d) none of these. circle is drawn. It divides the plane into (a) Parts (b) 4 Parts (c) 5 Parts (d) No Parts. The relation between diameter and radius of a circle is (a) r=d (b) d=r (c) d=r (d) d=πr 4. If P and Q are any two Points on a circle then PQ is called a (a) diameter (b) secant (c) chord (d) radius 5. line is Passing through the centre of a circle. If it bisects chord and of the circle. Prove that [] 6. and are two chords of circle to Prove that O bisects [] 7

75 7. If is diameter of circle with centre O and O is to chord [] so prove =0 O 8. Given a method to find the centre of a circle [] 9. In circle bisector of of Passes through the center O of the circum circle of Prove = O 0. Prove that the circle drawn with the equal sides as a diameter passes through the Point. if is the mid Point of of an isosceles triangle with =. If a Pair of opposite sides of a cyclic quadrilateral are equal, then the diagram are also equal.. = 70 and = 0 find 0 0 Q O radius equal to chord and is diameter and and produced meet at P so prove P = 60 [5] 7

76 SE TEST PPER-0 LSS - IX Mathematics (ircle) [NSWERS] ns. (b) ns. (a) ns. (c) ns4. (c) ns5. Line EF passes through the centre O and bisects chord at P and chord at Q P, S the mid-point of and Q is the mid-point of ut the line joining the mid-point of a chord to the centre of the circle is perpendicular to the chord. OP nd OQ OP = OQ = 90 OP + OQ = 80 E Q ns6. Join O and O In s Ond O O = O O = O = O = O O = O Hence, O bisects O ns7. Join Given that O is the mid-point of O is the mid-point of Now in, O is the line joining the mid points of sides and O 74

77 O = = O = O Hence proved ns8. Take three distinct points (non-collinear), and on the circle. Join and raw bisectors PQ and RS of and respectively, to intersect at O Now, P, S the centre of the circle ns9. raw OP and OQ In s OPand OQ PO = QO PO = QO O=O OP OQ OP=OQ hords and are equidistant from centre O = Q P R O P O Q S ns0. Join in s and = = = [ is mid points] = [ PT ] + = 80 = = 90 ns. Given: cyclic quadrilateral in which = To Prove: diagonal =diagonal Proof: = (ngle in same segment of circle) = 4 + = + 4 = ut these are the angles subtended by the diagonals and in the same circle = 4 75

78 ns. = = 70 (ngle in same segment) = + = = 00 + = = 80 = = 80 Q ns. Join In O, O = O (Radii of same circle) O= (Given) O=O= O is equilateral Hence, O = 60 = 0 [ngle subtended by are at centre is double the angle at any Pont of the remaining part] = 90 Exterior = P + P 90 = 0 + P P = 60 P = 60 O P 76

79 SE TEST PPER-0 LSS - IX Mathematics (ircle). What is a diameter (a) r = d (b) d = π r (c) d = r (d) d = r. Two point on a circle shows the (a) radius (b) chord (c) secant (d) diameters. The whole are of a circle is called (a) circumference (b) semi circle (c) sector (d) segment 4. One half of the whole are of a circle (a) semi circle (b) circumference (c) segment (d) sector 5. point is taken so that m =0 0 from a semi circle with as diameter. So find m and m. 0 0 [] 6. Two different circle can t interact each other at more than two points so, prove it. [] 77

80 7. O is the centre and OP so, find the length of the chord [] 8. If O is the Per perpendicular to, find the length of [] O 5 9. is chord of a circle and Produced to such that =O and O joined and produce the circle the circle and meet to if = y and O = x, prove that x = y 0. Prove that XPZ = ( zy y z) + if O id the centre of Y circle X Z P. Prove that O is the perpendicular bisector of if. Prove that the line joining the midpoint of the two parallel chords of a circle passes through the centre of the circle.. The two chords bisect each other and show that (i) and are diameter (ii) is a rectangle [5] 78

81 SE TEST PPER-0 LSS - IX Mathematics (ircle) [NSWERS] ns. (d) ns. (b) ns. (a) ns4. (a) ns5. is a diameter and is a point on the semi circle m = 90 m = 0 In, m = 80 m + 0 = 80 m = 80 0 = 60 m = 90 and m = 60 ns6. Let the two different circles intersect in three point,,. Then these points and one non-collinear We know that through three non-collinear Points, one and only one circle can pass so, it contradicts the hypothesis ns7. Perpendicular drawn from the centre to the chord bisects the chord. P = P = In rt. Triangle PO, O = OP + P ( 5) = ( ) + ( P) ( P = 6 = 4cm) = P = 4 = 8cm 79

82 ns8. O In rt. Triangle O, O = O + O O = 5 9 = 6 ab = 6 ab = 4 ns9. Proof: In O, o= O = O = y In O, is produced to, forming exterior O O = O + o = y + y = y O=O [Radii of the same circle O = O = y gain in, O O = O + O x = y + y x = y is produced to, forming exterior O X O Y ns0. Given: circle with centre P, XY and YZ are two chords To Prove: PZ = ( ZY + Y Z ) Proof: PY = ZY... ( i) Similarly arc YZ subtends YPZ at centre and YXZat remaining Part of the circle YPZ = Y Z...( ii) dding (i) and (ii) PY + YPZ = ZY + Y Z PZ = ( ZY + Y Z ) ns. Let O intersect in P produce O to meet the circle in K Now, OK is a diameter K K K = K K K = P O K 80

83 In sp and P = P>P (ommon) = P P (SS) P = P P = P P + P = 80 Each = 90 ns. Let and be the two parallel chords of the circle with centre O P and Q are the mid-points of and join OP and OQ. raw OX or OP nd OQ POX = 90 nd also QOX = 90 POX + QOX = 80 POQ is a straight line P O Q X ns. Given that the two chords, of the circle bisect each other Let these cords bisect at K In sk and K K=K [, bisect each other at K] K= K K = K [Vertically opposite ] K K = = K + = + = = lso, in quadrilateral = = = = = 90 [, is diameter so angle is semicircle) 8

84 SE TEST PPER-04 LSS - IX Mathematics (ircle). ircle having same centre are said to be (a) oncentric (b) circle (c) chord (d) secant. The line which meet a circle in two points is called a (a) chord of circle (b) diameter (c) radius (d) secant of circle. The sum of either pair of opposite angle of cyclic quadrilateral is (a) 60 (b) 90 (c)80 (d) Two circle are congruent if they have equal. (a) diameter (b) radius (c) chord (d) secant 5. Prove that E is an isosceles triangle if O and OE [] E O 6. The exterior angle formed by producing a side of a cyclic quadrilateral is equal to the interior opposite angle. Prove [] 7. Show that OMN = ONM if and are two equal chord. [] 8. From the above question. Show that MN = NM [] 9. is a quadrilateral in which = and = show,,, lie on a circle 0. iagonal is also equal when pair of opposite sides of a cyclic quadrilateral are equal. Prove.. In cyclic quadrilateral diagonal Intersect at Q. = 70 and = 0 so find. Find the value of xif,,, are concylic points X 0 0. Show that HE straight lines. and EGare supplementary. Given that EG and HE are [5] 8