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Quadrilaterals 1. 2. A quadrilateral ABCD is a parallelogram if (a) AB = CD (b) AB BC (c) ∠A = 600 , ∠C = 600 , ∠B = 1200 (d) AB = AD In figure, ABCD and AEFG are both parallelogram if ∠C = 800 , then∠DGF is (a) 1000 (b) 600 (c) 800 (d) 1200 3. 4. 5. In a square ABCD, the diagonals AC and BD bisects at O. Then ∆AOB is (a) acute angled (b) obtuse angled (c) equilateral (d) right angled ABCD is a rhombus. If ∠ACB = 300 , then ∠ADB is (a) 300 (b) 1200 (c) 600 (d) 450 The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral. 6. Show that each angles of a rectangle is a right angle. 7. A 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. 9. 10. In a parallelogram ABCD, bisectors of adjacent angles A and B intersect each other at P. prove that ∠APB = 900 In figure diagonal AC of parallelogram ABCD bisects ∠A show that (i) if bisects ∠C and (ii) ABCD is a rhombus 11. 12. In figure ABCD is a parallelogram. AX and CY bisects angles A and C. prove that AYCX is a parallelogram. The line segment joining the mid-points of two sides of a triangle is parallel to the third side. 13. Prove that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it is a square. ANSWERS Ans01. (c) Ans05. Suppose angles of quadrilateral ABCD are 3x, 5x, 9x, and 13x Ans02. Ans03. (c) ∠A + ∠B + ∠C + ∠D = 3600 (d) Ans04. (c) [ sum of angles of a quadrilateral is 3600 ] 3 x + 5 x + 9 x + 13 x = 3600 30 x = 3600 x = 120 ∴ ∠A = 3 x = 3 × 12 = 360 ∠B = 5 x = 5 ×12 = 600 ∠C = 9 x = 9 × 12 = 1080 ∠D = 13 x = 13 × 12 = 1560 Ans06. We know that rectangle is a parallelogram whose one angle is right angle. Let ABCD be a rectangle. ∠A = 900 To prove ∠B = ∠C = ∠D = 900 Proof: ∵ AD BC and AB is transversal ∴ ∠A + ∠B = 1800 900 + ∠B = 1800 ∠B = 1800 − 900 = 900 ∠C = ∠A Ans07. ∴ ∠C = 900 ∠D = ∠B ∴ ∠D = 900 ∵ AB CD and EF cuts them at P and R. ∴ ∠APR = ∠PRD 1 1 ∴ ∠APR = ∠PRD 2 2 i.e. ∠1 = ∠2 ∴ PQ RS [ Alternate int erior angles ] [ Alternate] Ans08. ABCD is a rectangle AC and BD is diagonals. To prove AC = BD Proof: In ∆ DAB and CBA AD = BC [In a rectangle opposite sides are equal] ∠A = ∠B [900 each] AB = AB common [common] ∴ ∆DAB ≅ ∆CAB [ By SAS ] ∴ AC = BD [ By CPCT ] Ans09. Given ABCD is a parallelogram is and bisectors of ∠A and ∠B intersect each other at P. To prove ∠APB = 900 Proof: 1 1 ∠1 + ∠2 = ∠A + ∠B 2 2 1 (i ) = ( ∠A + ∠B ) → 2 In ∆APB But ABCD is a parallelogram AD BC ∠1 + ∠2 + ∠APB = 1800 ∴ ∠A + ∠B = 1800 1 ∠1 + ∠2 = ×1800 2 = 900 ∴ Ans10. (i) AB DC and AC is transversal ∴ and ∠1 = ∠2 ∠3 = ∠4 But ∠1 = ∠3 ∴ ∠ 2 = ∠4 ∴ (ii) ∴ ∴ ∴ ( Alternate angles ) ( Alternate angles ) AC bi sec sts ∠C In ∆ABC and ∆ADC AC = AC [common] ∠1 = ∠3 [ given] ∠2 = ∠4 [ proved ] ∆ABC ≅ ∆ADC AB = AD ABCD is a r hom bus [ By CPCT ] 900 + ∠APB = 1800 ∠APB = 900 Hence proved [ By angle sum property ] Ans11. Given in a parallelogram AX and CY bisects ∠A and ∠C respectively and we have to show that AYCX in a parallelogram. In ∆ADX and ∆CBY ∠D = ∠B → (i ) [opposite angles of a paralle log ram] 1 ∠A 2 1 and ∠BCY= ∠C 2 But ∠A=∠C ∴ By (2) and (3), we get ∠DAX = → [ given] [given] (ii) → (iii) ∠DAX = ∠BCY → (iv) Also ∴ AD = BC [opposite sides of paralle log ram] → (v ) From (i ), (iv) and (v), we get ∴ But ∆ADX ≅ ∆CBY [ By ASA] DX = BY [CPCT ] AB = CD AB − BY = CD − DX [oppsite sides of paralle log ram] or AY = CX But AY XC [∵ ABCD is a gm] ∴ AYCX is a paralle log ram Ans12. Given ∆ ABC in which E and F are mid points of side AB and AC respectively. To prove: EF||BC Construction: Produce EF to D such that EF = FD. Join CD Proof: In ∆AEF and ∆CDF AF = FC [∵ F is mid − po int of AC ] ∠1 = ∠2 [vertically opposite angles ] EF = FD [ BY construction] ∴ ∆AEF ≅ ∆CDF [ By SAS ] ∴ [ By CPCT ] AE = CD and AE = BE ∴ [∵ E is the mid − po int] BE = cd and AB CD [∴∠BAC = ∠ACD] ∴ BCDE is a paralle log ram EF BC Henceproved Ans13. Given in a quadrilateral ABCD, AC = BD, AO = OC and BO = OD and ∠AOB = 900 To prove: ABCD is a square. Proof: In ∆AOB and ∆COD OA = OC [ given] OB = OD [ given] and ∠AOB = ∠COD [vertically opposite angles ] ∴ ∆AOB ≅ ∆COD [ By SAS ] ∴ AB = CD [ By CPCT ] ∠1 = ∠2 [ By CPCT ] But there are alternate angles ∴ AB CD ABCD is a parallelogram whose diagonals bisects each other at right angles ∴ ABCD is a rhombus Again in ∆ABD and ∆BCA AB = BC [ Sides of a r hom bus ] AD = AB [ Sides of a r hom bus ] and BD = CA [Given] ∴ ∆ABD ≅ ∆BCA ∴ ∠BAD = ∠CBA [ By CPCT ] There are alternate angles of these same side of transversal ∴ ∠BAD + ∠CBA = 1800 or ∠BAD = ∠CBA = 900 Hence ABCD is a square Quadrilateral 1. In fig ABCD is a parallelogram. It ( A ) 80° ( B ) 60° ( C ) 20° ( D ) 40° 2. ∠DAB = 60° and ∠DBC = 80° then ∠CDB is D C 0 0 80 60 A B If the diagonals of a quadrilateral bisect each other, then the quadrilateral must be. (a) Square (b) Parallelogram (c) Rhombus (d) Rut angle 3. The diagonal AC and BD of quadrilateral ABCD are equal and are perpendicular bisector of each other then quadrilateral ABCD 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 ABCD taken in order, is a rectangle if (a) ABCD is a parallelogram (b) ABCD is a rut angle (c) Diagonals AC and BD are perpendicular (d) AC=BD 5. If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram. D 6. C In fig ABCD is a parallelogram and X,Y are the points on the diagonal BD such that DX<BY show X that AYCX is a parallelogram. O Y B A G C D 7. Show that the line segments joining the mid points of opposite sides of a quadrilateral bisect each other. H F A E B 8. ABCD is a rhombus show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D 9. Prove that a quadrilateral is a rhombus if its diagonals bisect each other at right angles. 10. 11. Prove that the straight line joining the mid points of the diagonals of a trapezium is parallel to the parallel sides. In fig ∠B is a right angle in ∆ABC.D is the mid-point of AC.DE AB intersects BC at E. show that A (i) E is the mid-point of BC D (ii) DE ⊥ BC (iii) BD = AD C B E 12. ABC is a triangle and through vertices A, B and C lines are drawn parallel to BC, AC and AB respectively intersecting at D, E and F. prove that perimeter of ∆DEF is double the perimeter of ∆ABC . 13. Prove that in a triangle, the line segment joining the mid points of any two sides is parallel to the third side. ANSWERS Ans1. (D) Ans5. Given A quadrilateral ABCD in which AB = DC and AD = BC Ans2. Ans3. (B) (B) To prove: ABCD is a parallelogram Ans4. (A) D C Construction: Join AC Prof: In ∆ABC and ∆ADC AD = BC ( given ) A AB = DC B AC = AC [ common ] ∴∆ABC ≅ ∆ADC [ by sss ] ∴∠BAC = ∠DAC [ By CPCT ] ∴ ABCD Is a parallelogram Ans6. ABCD is a parallelogram. The diagonals of a parallelogram bisect bisect each other ∴OD = OB But DX = BY [given] ∴ OD − DX = OB − BY or OX = OY Now in quad AYCX, the diagonals AC and XY bisect each other ∴ AYCX is a parallelogram. Ans7. Given ABCD is quadrilateral E, F, G, H are mid points of the side AB, BC, CD and DA respectively To prove: EG and HF bisect each other. In ∆ABC , E is mid-point of AB and F is mid-point of BC ∴ EF AC And EF = 1 AC....... ( i ) 2 Similarly HG AC and GD = 1 AC...... ( ii ) 2 From (i) and (ii), EF HG and EF = GH ∴ EFGH is a parallelogram and EG and HF are its diagonals Diagonals of a parallelogram bisect each other Thus EG and HF bisect each other. Ans8. ABCD is a rhombus In ∆ABC and ∆ADC C D AB = AD [Sides of a rhombus] BC = DC [Sides of a rhombus] AC = AC [Common] B A ∴∆ABC ≅ ∆ADC [By SSS Congruency] ∴∠CAB = ∠CAD And ∠ACB = ∠ACD Hence AC bisects ∠A as well as ∠C Similarly, by joining B to D, we can prove that ∆ABD ≅ ∆CBD Hence BD bisects ∠B as well as ∠D Ans9. Given ABCD is a quadrilateral diagonals AC and BD bisect each other at O at right angles To Prove: ABCD is a rhombus Proof: ∵ diagonals AC and BD bisect each other at O D C ∴ OA = OC , OB = OD And ∠1 = ∠2 = ∠3 = 90° 0 3 Now In ∆BOA And ∆BOC OA = OC Given OB = OB [Common] And ∠1 = ∠2 = 90° (Given) ∴∆BOA = ∆BOC (SAS) ∴ BA = BC (C.P.C.T.) 2 1 A B Ans10. Given a trapezium ABCD in which AB DC and M,N are the mid Points of the diagonals AC and BD. We need to prove that MN AB DC Join CN and let it meet AB at E C D Now in ∆CDN and ∆EBN M ∠DCN = ∠BEN [Alternate angles] ∠CDN = ∠BEN [Alternate angles] A E N B And DN = BN [given] ∴∆CDN ≅ ∆EBN [SAA] ∴ CN = EN [By C.P.C.T] Now in ∆ACE , M and N are the mid points of the sides AC and CE respectively. ∴ MN AE Or MN AB Also AB DC ∴ MN AB DC Ans11. Proof: ∵ DE AB and D is mid points of AC Ans12. ∴ BC = AF In ∆DCE and ∆DBE CE=BE DE= DE And ∠DEC = ∠DEB = 90° ∴∆DCE = ∆DBE ∴∆DCE ≅ ∆DBE ∴ CD = BD ∵ BCAF Is a parallelogram ∵ ABCE Is a parallelogram ∴ BC = AE AF + AE = 2 BC A E F Or EF = 2 BC Similarly ED = 2AB and FD = 2AC C B ∴ Perimeter of ∆ABC = AB + BC + AC Perimeter of ∆DEF = DE + EF + DF = 2AB+2BC+2AC = 2[AB+BC+AC] = 2 Perimeter of ∆ABC Hence Proved D Ans13. Given: A ∆ABC in which D and E are mid-points of the side AB and AC respectively TO Prove: DE BC A Construction: Draw CF BA 3 Proof: In ∆ADE and ∆CFE 1 D E F 2 ∠1 = ∠2 [Vertically opposite angles] B AE=CE [Given] And ∠3 = ∠4 [Alternate interior angles] ∴∆ADE ≅ ∆CFE [By ASA] ∴ DE=FE [By C.P.C.T] But DA = DB ∴ DB = FC Now DB FC ∴ DBCF is a parallelogram ∴ DE BC Also DE = EF = 4 1 BC 2 C