Quadrilaterals. 1. A quadrilateral ABCD is a parallelogram if. (a) AB = CD. (c) C 80, then DGF is

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1 Quadrilaterals 1. quadrilateral is a parallelogram if (a) = (b) (c) = 6, = 6, = 12 (d) = 2. In figure, and EFG are both parallelogram if = 8, then GF is (a) (b) (c) (d) 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 3, = then is (a) (c) 3 (b) 12 6 (d) The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

2 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 angles enclose a rectangle. 8. Prove that diagonals of a rectangle are equal in length.

3 9. In a parallelogram, bisectors of adjacent angles and intersect each other at P. prove that P = 9 1. In figure diagonal of parallelogram bisects show that (i) if bisects and (ii) is a rhombus

4 11. In figure is a parallelogram. X and Y bisects angles and. prove that YX is a parallelogram. 12. The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

5 13. Prove that if the diagonals of a quadrilateral are equal and bisect each other atright angles then it is a square.

6 NSWERS ns1. (c) ns2. (c) ns3. (d) ns4. (c) ns5. Suppose angles of quadrilateral are 3x, 5x, 9x, and 13x 3x + 5x + 9x + 13x = 36 3x = 36 x = 12 = 3x = 3 12 = [ 36 ] = sum of angles of a quadrilateral is = 5x = 5 12 = 6 = 9x = 9 12 = 18 = 13x = = 156 ns6. We know that rectangle is a parallelogram whose one angle is right angle. Let be a rectangle. = 9 To prove = = = 9 Proof: and is transversal + = = 18 = 9 = 9 = = 18 9 = = 9 ns7. and EF cuts them at P and R. PR = PR [ lternate int erior angles] 1 1 PR = PR 2 2 i. e. 1 = 2 PQ RS [ lternate]

7 ns8. is a rectangle and is diagonals. To prove = Proof: In and = [In a rectangle opposite sides are equal] = [9 each] = common [common] [ y SS] = [ y PT] ns9. Given is a parallelogram is and bisectors of and intersect each other at P. To prove P = 9 Proof: = = ( + ) 2 ( i) ut is a parallelogram + = = 18 2 = 9 In P + + P = y angle sum property [ ] 9 + P = 18 P = 9 Hence proved ns1. (i) and is transversal 1 = 2 ( lternate angles) and 3 = 4 ( lternate angles) ut 1 = 3 2 = 4 bi sec sts (ii) In and = [ common] 1 = 3 [ given] 2 = 4 [ proved] = [ y PT] is a r hombus

8 ns11. 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] 1 X = 2 [ given] ( ii) and 1 Y= 2 [given] (iii) ut = y (2) and (3), 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 ns12. 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 ] 1 = 2 [ 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

9 ns13. Given in a quadrilateral, =, O = O and O = O and To prove: is a square. Proof: In O and O O = 9 O = O [ given] O = O [ given] and O = O [ vertically opposite angles] O O [ y SS] = [ y PT ] 1 = 2 [ 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 + = 18 or = = 9 Hence is a square

10 1. In fig is a parallelogram. It = 6 and = 8 then ( ) ( ) ( ) ( ) If the diagonals of a quadrilateral bisect each other, then the quadrilateral must be. (a) Square (b) Parallelogram (c) Rhombus (d) Rut angle Quadrilateral 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) = is 5. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

11 6. In fig is a parallelogram and X,Y are the points on the diagonal such that X<Y show X that YX is a parallelogram. O Y 7. Show that the line segments joining the mid points of opposite sides of a quadrilateral bisect each other. H G F E

12 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 rightangles.

13 1. Prove that the straight line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides. 11. 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

14 12. 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. 13. Prove that in a triangle, the line segment joining the mid points of any two sides is parallel to the third side.

15 NSWERS ns1. () ns2. () ns3. () 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. 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 = 1... ( i) 2 Similarly HG and G = 1... ( ii) 2 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.

16 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 1 = 2 = 3 = 9 Now In Ond O O = O Given O = O[ommon] nd 1 = 2 = 9 (Given) O = O(SS) = (.P..T.) 3 1 2

17 ns1. 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] M N N = EN[lternate angles] nd N = N[given] E 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 ns11. Proof: E and is mid points of In E and E E=E E= E nd E = E = 9 E = E E E = = E ns12. FIs a parallelogram F + E = 2 Or EF = 2 = F EIs a parallelogram Similarly E = 2 and F = 2 Perimeter of = + + Perimeter of EF = E + EF + F F E = = 2[++] = 2 Perimeter of Hence Proved

18 ns13. Given: in which and E are mid-points of the side and respectively TO Prove: E onstruction: raw F Proof: In E 1 = 2 angles] E=E nd 3 = 4 and FE [Vertically opposite [Given] E FE [y S] E=FE ut = = F Now F [lternate interior angles] [y.p..t] 3 1 E 2 4 F F is a parallelogram E lso E = EF = 1 2

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