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Q1. The diagonal of a square is 10 cm. Its area is:

• 1) 100 cm²
• 2) 20 cm²
• 3) 25 cm²
• 4) 50 cm²

Solution

Diagonal using Pythagoras Theorem: a² + a² = 10² 2a² = 100 a² = 100/2 a²=50 a=7.07 cm (where a is the length of the sides of the square.) Area = 7.07² = 50 cm²

Q2. Show that diagonals of a square are equal and bisect each other at right angles.

Solution

Let the diagonals AC and BD of the square ABCD intersect each other at a point O. We need to prove AC = BD, OA = OC, OB = OD and AOB = 90^o Now, in ABC and DCB, AB = DC (sides of square are equal) ABC = DCB (all interior angles are of 90^o) BC = BC (common side) ABC DCB (by SAS congruency) AC = DB (by CPCT) Hence, the diagonals of a square are equal in length. Now, in AOB and COD, AOB = COD (vertically opposite angles) ABO = CDO (alternate interior angles) AB = CD (sides of square are equal) AOB COD (by AAS congruence rule) AO = CO and OB = OD (by CPCT) Hence, the diagonals of a square bisect each other. Now in AOB and COB, AO = CO AB = CB (sides of square are equal) BO = BO (common) AOB COB (by SSS congruence) AOB = COB (by CPCT) But, AOB + COB = 180^o (linear pair) 2AOB = 180 ^o AOB = 90^o Hence, the diagonals of a square bisect each other at right angles.

Q3. Two parallel lines l and m are intersected by a transversal p. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

Solution

l||m and p intersects them at points A and C respectively. Bisectors of interior angles intersect at B and D. Now PAC = ACR (alternate angles) AB||DC (as alternate angles are equal) Similarly BC||AD ABCD is a parallelogram Also PAC + CAS = 180^o (Linear pair) ABCD is a ||gm with one angle 90^o ABCD is a rectangle.

Q4. The diagonals of a parallelogram PQRS intersect at O. If QOR = 90^o and QSR = 50^o, then ORS is:

• 1) 40^o
• 2) 90^o
• 3) 50^o
• 4) 70^o

Solution

Q5. Prove that the angle bisectors of a parallelogram form a rectangle.

Solution

To prove : PQRS is a rectangle Proof : Since ABCD is a parallelogram A + B = 180^o [angles on the same side of transversal] But in ABS, by angle sum property ASB = 90^o RSP = 90^o [ASB = RSP vert. opp. Angles] Similarly SRQ = 90^o , RQP = 90^o, and SPQ = 90^o Hence PQRS is a rectangle.

Q6. ABCD is a parallelogram and line segments AX and CY bisect A and C respectively. Prove that AX || CY.

Solution

A = C (opposite angles of parallelogram)

Q7. In a parallelogram ABCD, if A = 60^o, then D is equal to:

• 1) 110^o
• 2) 130^o
• 3) 120^o
• 4) 140^o

Solution

Adjacent angles of a parallelogram are supplementary.

Q8. ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersect AB at P and DC at Q. Prove that: ar(POA) = ar(QOC)

Solution

In POA and QOC, we have AOP = COQ [vertical opp. angle] AO = OC [diagonals of a ||gm bisect each other] PAC = QCA [ alt. interior angles] [ASA criteria] [congruent figures are equal in areas]

Q9. Two consecutive angles of parallelogram are in the ratio 1 : 3, then the measure of the smaller angle is:

• 1) 90^o
• 2) 45^o
• 3) 60^o
• 4) 50^o

Solution

Let the angles be x and 3x. The sum of the two consecutive angles of a parallelogram is 180⁰

Q10. ABCD is a quadrilateral and AP and DP are bisectors of A and D. The value of x is:

• 1) 95^o
• 2) 85^o
• 3) 100^o
• 4) 60^o

Solution

Given, ABCD is a quadrilateral. Therefore, A + B + C + D = 360° .... (1) Again, it is given that AP and DP are bisectors of A and D respectively. So, in triangle APD, by angle sum property of triangle, From (1), we have A + D +B + C = 360° 360° - 2APD +B + C = 360° [using (2)] - 2APD +B + C = 0 B + C = 2APD 2APD = B + C APD =

Q11. The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If DAC = 32^o, AOB = 70^o, then DBC is equal to:

• 1) 88^o
• 2) 38^o
• 3) 32^o
• 4) 24^o

Solution

Q12. In the figure, P is the mid-point of side BC of parallelogram ABCD such that BAP = DAP. Prove that AD = 2CD.

Solution

In the figure, 1 = 2 (given) But 2 = 3 (alternate interior angles) 1 = 3 AB = BP (sides opp. to equal angles are equal) But AB = CD (opposite sides of ||gm) CD = BP CD = BC (Since, P is the mid-point of BC) 2CD = BC 2CD = AD (Since, BC = AD)

Q13. All the angles of a convex quadrilateral are congruent. However, not all its sides are congruent. What type of quadrilateral is it?

• 1) Rectangle
• 2) Square
• 3) Parallelogram
• 4) Trapezium

Solution

All the angles of a rectangle are equal and the pair of opposite sides is equal. Thus, all the angles of rectangle are congruent, but all its sides are not congruent.

Q14. In a parallelogram ABCD, if A = 75^o, then B =?

• 1) 15^o
• 2) 95^o
• 3) 80^o
• 4) 105^o

Solution

In a parallelogram, adjacent angles are supplementary. Therefore,

Q15. M is the mid point of side PQ of a parallelogram PQRS. A line through Q parallel to PS meets SR at N and PS produced at L. Prove that (i) PL = 2QR (ii) QL = 2QN.

Solution

(i) In PLQ, M is the mid point of PQ and MS  QL S is the mid point of PL PL = 2PS or PL = 2QR      (ABCD is parallelogram PS = QR)    (ii) ABCD is parallelogram SR  PQ or SN  PQ In LPQ, S is the mid point of PL and SN  PQ N is the mid point of QL QL = 2QN 

Q16. In fig., PQRS is a rectangle. If RPQ = 30^o then the value of (x + y) is:

• 1) 180^o
• 2) 120^o
• 3) 90^o
• 4) 150^o

Solution

Let the diagonals bisect each other at O. Also the diagonals of a rectangle are equal. Therefore, OP = OQ Now in Hence, 180^o

Q17. The angles of a quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral.

Solution

Let the angles be 3x, 5x, 9x, and 13x. 3x + 5x + 9x + 13x = 360^o       (angle sum of a quadrilateral) 30x = 360^o  x =12^o Hence, the angles are 36^o, 60^o, 108^o and 156^o.

Q18. Which of the following is not a parallelogram?

• 1) Square
• 2) Rectangle
• 3) Trapezium
• 4) Rhombus

Solution

A parallelogram has two pairs of parallel sides. The trapezium has only one pair of parallel sides. Hence, Trapezium is not a parallelogram.

Q19. Find the measure of each angle of a parallelogram, if one of its angles is 30^o less than twice the smallest angle.

Solution

Let the smallest angle be x. Then one of the angles is 2x – 30^o (2x – 30^o) + x = 180^o               (Interior angles on same side of transversal are supplementary)  x = 70^o    One angle = (2x – 30^o) = 110^o Opposite angles of a parallelogram are equal Hence, the four angles are 70^o, 110^o, 70^o, 110^o

Q20. ABCD is a rhombus. Show that diagonal AC bisects A as well as C and diagonal BD bisects B as well as D.

Solution

Let us join AC. In ABC, BC = AB (Sides of a rhombus are equal to each other) 1 = 2 (Angles opposite to equal sides of a triangle are equal) However, 1 = 3 (Alternate interior angles for parallel lines AB and CD) 2 = 3 Therefore, AC bisects C. Also, 2 = 4 (Alternate interior angles for || lines BC and DA) 1 = 4 Therefore, AC bisects A. Similarly, it can be proved that BD bisects B and D as well.

Q21. In a quadrilateral ABCD, the line segments bisecting C and D meet at E. Prove that A + B = 2 CED.

Solution

In DEC, 1+ 2 + CED = 180^o           (angle sum property of triangle) CED = 180^o – (1+ 2) A + B+ C+D = 360^o           (angle sum property of quadrilateral)

Q22. In the given figure, ABCD is a rhombus in which diagonals AC and BD intersect at O. Then measure of AOB is.

• 1) 60^o
• 2) 45^o
• 3) 80^o
• 4) 90^o

Solution

The intersection of the diagonals of a rhombus form 90 degree (right) angles.

Q23. ABCD is a parallelogram. If DAB = 60^o and DBC = 80^o find CDB.

Solution

A = C [opp. Angles of ||gm] C = 60^o Now in BDC C + CDB + DBC = 180^o. 60^o + CDB + 80^o = 180^o CDB = 180^o - 140^o = 40^o.

Q24. ABCD is a rhombus ABC = 66^o. Determine ACD.

Solution

ABC = ADC = 66^o                      (opposite angles of a parallelogram) ODC = 33^o                                 (ODC = 1/2 ADC) In OCD, ODC + OCD + COD = 180^o 33^o + OCD +90^o = 180^o           nbsp;  (COD = 90^o, diagonals bisect at rt. angle) OCD = 57^o Hence ACD = 57^o

Q25. Show that the quadrilateral formed by joining the mid-points of the sides of a rectangle is a rhombus.

Solution

Given :- ABCD is a rectangle and E,F,G,H are the midpoints of sides AB, BC< CD and DA respectively. To Prove:- EFGH is a rhombus. Proof:-In ABC, E and F are the mid points of AB and BC respectively. EF || AC and EF = AC Similarly, GH || AC and GH = AC Hence, EF || GH and EF = GH Thus, EFGH is a parallelogram. Now, in triangles AEH and EBF, AE=BE EAH = EBF AH = BF AEH BEF and EH=EF Therefore, EFGH is a ||^gm in which adjacent sides are equal. Hence, EFGH is a rhombus

Q26. Three angles of a quadrilateral are 70^o, 120^o and 65^o. The fourth angle of the quadrilateral is :

• 1) 105^o
• 2) 95^o
• 3) 75^o
• 4) 90^o

Solution

Sum of the four angles of a quadrilateral is 360^o. Let x be the measure of the fourth angle. x + 70^o + 120^o + 65^o = 360^o x + 255^o = 360^o x = 360^o - 255^o x = 105^o So, the measure of the fourth angle is 105^o

Q27. In the given figure, LMNQ is a parallelogram in which L = 75^o and QMN = 60^o. Find NQM and LQM.

Solution

L + M = 180^o 75^o +  LMO + 60^o = 180^o LMO = 45^o LMO = NOM = 45^o                    (alternate angles) LOM = OMN= 60^o                     (alternate angles) LOM = 60^o and NOM = 45^o

Q28. ABCD is a parallelogram and E is the mid point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.

Solution

Q29. The length of the diagonal of the square is 10 cm. The area of the square is:

• 1) 70 cm²
• 2) 50 cm²
• 3) 20 cm²
• 4) 100 cm²

Solution

Let c be the diagonal, a and b are the lengths of the sides: a² + b² = c² but, a = b 2a² = 10² a² = 100 / 2 a = cm Area = a² = = 50 cm²

Q30. ABCD is a square prove that diagonal BD bisects B as well as D. 

Solution

In triangles BAD and BCD BA = CD, BC = AD and BD = BD BAD BCD                         (SSS) ABD = BCD and ADB = CDB Hence, BD bisects B as well as D.

Q31. In quadrilateral PQRS, if P ₌ 60^o and Q : R : S = 2 : 3 : 7, then S =

• 1) 150^o
• 2) 135^o
• 3) 175^o
• 4) 210^o

Solution

P = 60⁰ Sum of all angles of a quadrilateral is 360^o. Q +R +S = (360-60)^o = 300^o…..(1) It is given that Q : R : S = 2 : 3 : 7. Let angles Q, R and S be 2x, 3x and 7x respectively. From (1), we have: 2x + 3x + 7x = 300⁰ 12x = 300^o x=25^o S = 7x = 7 25^o = 175^o

Q32. The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of quadrilateral.

Solution

Let the angles be 3x, 5x, 9x, 13x 3x + 5x + 9x + 12x = 360^o. 30x = 360^o x = 12⁰ Angles are 36^o, 60^o, 108^o, 156^o.

Q33. Which of the following is not a parallelogram?

• 1) Trapezium
• 2) Square
• 3) Rhombus
• 4) Rectangle

Solution

A Trapezium is not a parallelogram because it does not have both pair of opposite sides parallel.

Q34. Show that the diagonals of a rhombus are perpendicular to each other.

Solution

Given: ABCD is a rhombus. Diagonal AC and BD intersect at O. To prove: AC and BD bisect each other at right angles. Proof: A rhombus is a parallelogram such that, AB = DC = AD = BC ---(i) Also, the diagonals of a parallelogram bisect each other. Hence, BO = DO and AO = OC ---(ii) Now, compare AOB and AOD, AB = AD (from (i) above) BO = DO (from (ii) above) AO = AO (common side) Therefore, AOB AOD (SSS Congruency Condition) Therefore, AOB = AOD (corresponding parts) BD is a straight line segment. Therefore, AOB + AOD = 180⁰ But, AOB = AOD (Proved) AOB = AOD = 180⁰/ 2 = 90⁰ i.e., the diagonals are perpendicular to each other.

Q35. Two angles of a quadrilateral are 50^o and 80^o and other two angles are in the ratio 8 : 15. Find the measure of the remaining two angles.

• 1) 80^o, 150⁰
• 2) 140^o, 90^o
• 3) 100^o, 130^o
• 4) 70^o, 160^o

Solution

Let the angles be 8x and 15x. The sum of the interior angles of a quadrilateral = 360⁰. Hence, other two angles are: 80^o, 150^o

Q36.

Solution

Q37. In fig., PQRS is a parallelogram in which PSR = 125^o, RQT is equal to

• 1) 55^o
• 2) 65^o
• 3) 75^o
• 4) 125^o

Solution

PSR = RQP = 125^o ( since PQRS is a parallelogram, opposite angles will be equal) PQT = 180^o (PQT is a st. line) PQR + RQT = 180⁰ 125⁰ + RQT = 180⁰ RQT = 55^o

Q38. In parallelogram ABCD, if A = 2x + 15^o, B = 3x - 25^o, then value of x is:

• 1) 89^o
• 2) 91^o
• 3) 38^o
• 4) 34^o

Solution

In a parallelogram, the sum of adjacent angles = 180⁰ Therefore, A + B = 180⁰ 2x + 15^o + 3x - 25^o = 180⁰ 5x -10⁰ = 180⁰ 5x = 190⁰ x = 38^o

Q39. Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.

Solution

Q40. If APB and CQD are parallel lines and a transversal PQ cut them at P and Q, then the bisectors of angles APQ, BPQ, CQP and PQD form a

• 1) Rectangle
• 2) Square
• 3) Rhombus
• 4) Any other parallelogram

Solution

Rectangle PQRS is formed by the bisectors of angles APQ, BPQ, CQP and PQD.

Q41. ABCD is a parallelogram in which P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively. AC is a diagonal, show that PQRS is a parallelogram.

Solution

ABCD is a quadrilateral; P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively. Join AC. In ABC, P and Q are the mid points of sides AB and BC respectively. PQ  AC and PQ = AC (i)         (by mid point theorem) In ADC, R and S are the mid points of sides CD and AD respectively. RS  AC and RS = AC               (ii)        (by mid point theorem) From (i) and (ii) PQ  RS and PQ = RS                                    In the quadrilateral PQRS one pair of opposite sides is equal and parallel.  PQRS is a parallelogram.

Q42. Three angles of a quadrilateral are 60^o, 110^o and 86^o, the fourth angle of the quadrilateral is:

• 1) 124^o
• 2) 94^o
• 3) 104^o
• 4) 84^o

Solution

Let fourth angle be x. Sum of the angles of a quadrilateral = 360⁰ 60⁰ + 110⁰ + 86⁰ + x = 360⁰ 256⁰ + x = 360⁰ x = 104^o

Q43. D, E and F respectively are the mid-points of the sides BC, CA and AB of ABC. Show that ar(DEF) =

Solution

D and E are mid points of BC and AC. DE||AB and DE = (Mid-point Theorem) Similarly, E and F are mid points of AC and AB. EF||BC and EF = In quadrilateral BDEF, DE||BF and FE||BD BDEF is a parallelogram Similarly, DCEF and AFDE are parallelograms. We know that diagonal of a parallelogram divides it into two triangles of equal area. Area (BFD) = Area (DEF) (For parallelogram BDEF) Area (CDE) = Area (DEF) (For parallelogram DCEF) Area (AFE) = Area (DEF) (For parallelogram AFDE) Area (AFE) = Area (BFD) = Area (CDE) = Area (DEF) Also, Area (AFE) + Area (BDF) + Area (CDE) + Area (DEF) = Area (ABC) Area (DEF) + Area (DEF) + Area (DEF) + Area (DEF) = Area (ABC) 4 Area (DEF) = Area (ABC) Area (DEF) = Area (ABC)

Q44. The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively. Find the length of each side of the rhombus.

Solution

Let ABCD be the rhombus. Diagonal AC = 24 cm and BD = 18 cm. Diagonals of a rhombus bisect each other at right angles. AO = OC = 12 cm and OB = OD = 9 cm and AOD = 90^o In right AOD, AO = 12 cm and OD = 9 cm By Pythagoras theorem, AD² = AO² + OD²  AD² = (12)² + (9)²  AD = 15 cm All sides of a rhombus are equal. Hence, each side of the rhombus = 15 cm

Q45.

Solution

Q46. In the given figure, PQRS is a parallelogram and R is 100^o then Q is

• 1) 80^o
• 2) 90^o
• 3) 100^o
• 4) 70^o

Solution

The adjacent angles of a parallelogram are supplementary. _{R + Q = 180}^o 100⁰ + _{Q = 180}^o _{Q =} 80^o

Q47. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution

In SOR and QOR, SO = QO SOR = QOR = 90^o OR = OR (common) Similarly, And, we get, PQ = QR = RS = SP Hence, PQRS is a rhombus.

Q48. Diagonals of a parallelogram ABCD intersect at point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.

Solution

Since diagonals of parallelogram divide it into two triangles of equal area. Area(ABC) = area(ACD) area (quad ABQO) + (COQ) = area(quad. CDPO) + area(AOP) ...(1) In AOP and COQ, AOP = COQ (vert. opp. Angles) OA = OC(diagonals of a||gm bisect each other) OAP = OCQ (alternate angles) (ASA congruence) ....(2) From (1) and (2), we have: area(quad. ABQO) + area() = area(quad. CDPO) + area() Hence, area(quad. ABQP) = area(quad. CDQP)

Q49. In a parallelogram PQRS, if P = (3x - 5)⁰, Q = (2x + 15)^o find the value of x.

Solution

P + Q = 180^o [Angles on the same side of a transversal are supplementary] 3x - 5 + 2x + 15 = 180⁰ 5x = (180 - 10)⁰ 5x = 170^o x = 34^o

Q50. If the three angles of a quadrilateral are 70^o, 95^o and 105^o, then the fourth angle is:

• 1) 90^o
• 2) 110^o
• 3) 85^o
• 4) 100^o

Solution

The sum of angles in a quadrilateral is 360⁰. Let fourth angle be x. Therefore, 70⁰ + 95⁰ + 105⁰ + x = 360⁰(angle sum property) x = 360⁰ - 270^o x = 90^o

Q51. If there are three or more parallel lines and the intercepts made by them on one transversal are equal, then the corresponding intercepts by any other transversal are also equal.

Solution

Given: l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p. To prove: l, m and n cut off equal intercept DE and EF on q also. Proof: Join AF, suppose it intersect line m at G In ACF, B is the mid-point AC [AB = BC] And BG||CF G is the mid-point of AF [Converse of mid point theorem] In AFD, G is the mid point of AF and GE||AD E is the mid point of DF DE = EF Hence l, m and n cut off equal intercepts DE and EF on q.

Q52. The sides ML and ON of a quadrilateral are produced as shown in the figure. Prove that x + y = a + b

Solution

L + b = 180^o                                    (linear pair) L = 180^o – b Also N + a = 180^o N = 180^o – a L + M + N + O = 360^o       (sum of the angles of a quadrilateral)   (180^o – b) + x + (180^o – a) + y = 360^o    x + y = a + b

Q53. In the given figure, PN and RM are perpendiculars to the diagonal QS of a parallelogram PQRS. Prove that PNS RMQ and PN = RM.

Solution

PS||QR and SQ is a transversal 1 = 2                            (alternate interior angles) In PNS and RMQ PNS = RMQ                   (90^o each) 1 = 2       PS = QR PNS RMQ                (AAS) PN = RM                            (c.p.c.t)

Q54. In the figure, AB||DE, AB = DE, AC||DF, AC = DF. Prove that BC||EF and BC = EF.

Solution

In quadrilateral ABED AB||DE, AB = DE (Given) ABED is a ||^gm (Quadrilateral with one pair of opposite sides equal and parallel.) AD = BE, AD||BE ....(I) In quadrilateral ACFD AC|DF, AC = DF ACFD is a ||^gm AD||CF, AD= CF ...(II) From I and II BE||CF and BE = CF BCFE is a ||^gm BC||EF and BC = EF (opposite sides of ||^gm are parallel and equal)

Q55. Points A and B are in the same sides of a line l. AD and BE are perpendiculars to l, meeting l at D and E respectively. C is the mid-point of AB. Prove that CD = CE

Solution

Draw CM l AC = CB (Given) DM = ME (by construction) (by construction) CM = CM (Common) (by SAS cong. Rule) CD = CE (by cpct)

Q56. In the given figure, ABCD is a trapezium in which AB||CD and AD = BC . Show that (i) A = B (ii) C = D (iii)

Solution

Construction : Produce AB and draw a line CE || AD. Proof : (i) Since AD || CE and transversal AE cuts them at A and E respectively. A + E = 180° ...(1) Since AB || CD and AD || CE, AECD is a parallelogram. AD = CE BC = CE (AD = BC (given)) Thus, In BCE, we have BC = CE CBE = CEB 180° -B = E B + E = 180° ...(2) from (1) and (2), we get, A + E = B + E A = B (ii) Since A = B BAD = ABD 180° - BAD = 180° - ABD ADB = BCD D = C i.e. C = D (iii) In ABC and BAD, we have BC = AD (given) AB = AB (common) A = B (proved above) (SAS Congruence condition)

Q57. The diagonals of rhombus are 12 cm and 16 cm. The length of the side of rhombus is:

• 1) 8 cm
• 2) 10 cm
• 3) 16 cm
• 4) 12 cm

Solution

Diagonals of a rhombus are perpendicular bisectors of each other, making 4 right triangles with perpendiculars and heights 6 and 8 as diagonals of rhombus bisect each other .The side of the rhombus will be hypotenuse. Using Pythagoras theorem: Side = 10 cm

Q58.

Solution

Q59. Show that the quadrilateral formed by joining the mid-points of the sides of a rhombus taken in order, form a rectangle.

Solution

Let ABCD be a rhombus and P, Q, R and S be the mid-points of sides AB, BC, CD and DA respectively. In ABD and BDC we have From (i) and (ii) we get PQRS is a ||^gm As diagonals of a rhombus bisect each other at right angles. AC BD Since SP||BD, PQ||AC and AC BD From above results, we have ||^gm PQRS is a rectangle.

Q60. Prove that the diagonals of a rhombus are perpendicular to each other.

Solution

ABCD is a rhombus. AB = BC = DC BO = OD; and AO = OC        (rhombus is a parallelogram, diagonals bisect each other) In BOC and DOC BO = DO, BC = DC and OC = OC BOCDOC                (SSS) BOC = DOC              (c.p.c.t) BOC + DOC = 180^o       (linear pair) Or BOC = DOC = 90^o    Hence, diagonals of a rhombus are perpendicular to each other.

Q61. ABCD is a parallelogram and line segments AX and CY bisect angles A and C. Show that AX||YC.

Solution

1 = 2, 3 = 4 And _{A=C AD=BC D=B} (by ASA rule) DX = BY BY = DX AB = CD AB - BY = DC - DX AY = CX AY||CX AYCX is a parallelogram AX||CY OR, A = C (Opposite angles of parallelogram ABCD) Therefore, A = C i.e., YAX = YCX …(1) Also, AYC + YCX = 180º (Because YA || CX)…. (2) Therefore, AYC + YAX = 180º [From (1) and (2)] So, AX || CY (As interior angles on the same side of the transversal are supplementary)

Q62. Prove that if the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Solution

OA = OC and OB = OD [Given] AOB = DOC [vertically opp. angles] [SAS congruence] Therefore ABO = CDO (cpct) From this, we get AB||CD ...(i) Similarly BC||AD ...(ii) From (i) and (ii) ABCD is a ||^gm.

Q63. The angle between the two altitudes of parallelogram through the vertex of an obtuse angle is 50^o. Find the angles of parallelogram.

Solution

Q64.

Solution

Q65. LMNO is a rectangle with QPR = 35^o. Determine SQR.

Solution

Let the diagonals PR and QS intersect at O. OP = OQ                                 (diagonals of a rectangle are equal and bisect each other.)  OPQ = OQP or  QPR = SQP SQP = 35^o PQR = 90^o SQP + SQR = 90^o SQR = 90^o – 35^o = 55^o Hence SQR = 55^o

Q66. Diagonals PR and QS of quadrilateral PQRS intersect at T such that PT = TR. If PS = QR, show that ar (PTS) = ar (RTQ).

Solution

Given: In quad. PQRS, diagonals PR and QS intersect at T. To prove: area(PTS) = area(RTQ) Construction : From P and R, draw perpendiculars to QS, namely PE and RF Proof : (a) In PET and RTF, TP = TR (given) PTE = RTF (vert. opp. angles) PET = RFT = 90^o (AAS congruence) area(PTE) = area(RTF) (congruent figures have equal area) ...(1) And PE = RF (CPCT) Now, in PSE and RQF, PES = RFQ (right angle) PE = RF (proved above) PS = QR (given) So, (RHS congruence) Area (PSE) = area(RQF) ...(2) Adding (1) and (2), we have: Area(PTE) + area(PSE) = area(RTF) + area(RQF) Or, area(PTS) = area(RTQ)

Q67. Two adjacent angles of a parallelogram are in the ratio 2 : 3. Find the measures of all the four angles of the parallelogram.

Solution

Let the two angles be 2x, 3x 2x + 3x = 180^o (adj. angles are supplementary) 5x = 180 ^o x = 36 ^o 2x = 72^o, 3x = 108^o The four angles are 72^o, 108^o, 72^o, 108^o

Q68. In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angle.

Solution

_{Therefore, the angle bisectors of two adjacent angles intersect at right angle.}

Q69. ABC is a triangle right angled at C. A line through the mid point M of hypotenuse AB and parallel to BC intersects AC at D. Show that: (i) D is the mid point of AC (ii) MD AC.

Solution

(i) If a line is drawn parallel to one side of a triangle through the mid point of the second side, then it bisects the third side. D is the mid-point of AC (ii) MD||BC (corresponding angles) MD AC

Q70.

Solution

Q71. In the figure, PQRS is a square. M is the midpoint of PQ and AB RM. Prove that RA = RB.

Solution

In APM and BQM APM = BQM = 90^o AMP = BMQ (vertically opp. s.) PM = QM (M is the mid point) (ASA congruence rule) AM = MB (c.p.c.t) In AMR and BMR AM = MB AMR = BMR = 90^o RM = RM (SAS congruence rule) Hence RA = RB (c.p.c.t)

Q72. D and E are the mid-point of sides AB and AC respectively of triangle ABC. If the perimeter of ABC = 35 cm, find the perimeter of ADE.

Solution

As D and E the mid-point of AB and AC. AD = AB and AE = AC DE = BC (mid-point theorem) Perimeter of ADE = AD + DE + AE = = perimeter of ABC Perimeter of ADE = 35 cm = 17.5 cm

Q73. ABCD is a parallelogram in which A = 60^o. If bisectors of A and B meets at P. Prove that AD = DP, PC = BC, DC = 2AD

Solution

(i) ABCD is a parallelogram in which A = 60°. AP and PB are the bisector of A and B respectively. PAB = PAD = 30° (AP bisects A) AB || CD and AP is the transversal. PAB = APD = 30° (Alternate angles) In APD, PAD = APD = 30° PD = AD ...(1) (Equal sides have equal angles opposite to them in a trangle) (ii) AD || BC and AB is the transversal. A + B = 180° (Sum of adjacent interior angles is 180°) 60° + B = 180° B = 120° PBA = PBC = 60° (PB is bisector B) PBA = BPC = 60° (Alternate angles) In PBC, BPC = PBC = 60° BC = PC ...(2) (Equal sides have equal angles opposite to them) (iii) CD = DP + PC CD = AD + BC [Using (1) and (2)] CD = AD + AD [AD = BC (Opposite sides of parallelogram are equal)] CD = 2AD

Q74. In a parallelogram ABCD, E and F are mid points of sides AB and CD. Show that line segment AF and CE trisect the diagonal BD.

Solution

ABCD is a parallelogram. AB || CD And hence, AE || FC Again, AB = CD (Opposite sides of parallelogram ABCD) AB =CD AE = FC (E and F are mid-points of side AB and CD) In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram. In DCQ, DF = CF (given) PF||QC DP = PQ ....(i) (by converse of mid-point them) In ABP, AE = BE (given) AP || EQ PQ = QB ….(ii) (by converse of mid-point them)So, DP = PQ = QB Hence, line segment AF and CE trisect the diagonal BD.

Q75. Prove that the triangles formed by joining in pairs, the mid points of the three sides of a triangle are congruent to each other.

Solution

ABC is a triangle in which D, E and F are the mid points of the sides BC, CA and AB respectively. F and E are the mid points of AB and AC respectively FE  BC and FE = BC                (By mid point theorem) FE = BD In AFE and FBD, AF = FB FE = BD AFE = FBD                                 (corresponding angles) AFE  FBD                            (SAS) Similarly you can prove that the other pairs of triangles are also congruent.

Q76. PQRS is a parallelogram and X and Y are points on the diagonal QS such that SX = QY. Prove that PXRY is a parallelogram.

Solution

Join PR to meet QS at O OQ = OS                     (diagonals of a parallelogram bisect each other) QY = SX  OQ - QY = OS - SX  OY = OX Thus in quadrilateral PXRY, diagonals PR and XY are such that OX = OY and OP = OR That is the diagonals bisect each other. Hence, PXRY is a parallelogram.

Q77. M, N and P are the mid points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm. Calculate BC, AB and AC.

Solution

By mid point theorem  BC = 2MN = 6 cm, AC = 2MP = 5 cm and AB = 2NP = 7 cm

Q78. In the figure PQRS is a parallelogram and X, Y are the points on the diagonal QS, such that SX = QY. Prove that quadrilateral PYRX is a parallelogram.

Solution

Join PROQ = OS         (diagonals of a parallelogram bisect each other)QY = SX OQ - QY = OS - SXOY = OXIn quadrilateral PXRYAO = OXOX =OYThus, the diagonals of the quadrilateral PXY are bisecting each other PXRY is a parallelogram.

Q79. Prove that the triangle formed by joining the mid points of the sides of an isosceles triangle is an isosceles triangle.

Solution

An isosceles triangle ABC                     In ABC, AB = AC               D, E and F are the mid points of sides AB, AC and BC respectively. By mid point theorem EF = DF In DEF, EF = DF. Hence, DEF is an isosceles triangle.

Q80. ABCD is a parallelogram. E and F are the mid points of the sides AB and CD respectively. Prove that the segments CE and AF trisect the diagonal BD.

Solution

AE||CF and AE = CF   (halves of opposite sides of a parallelogram) Therefore, AECF is a parallelogram. So, EC||AF In DPC,    FQ||CP and F is the mid point of DC So, PQ = QD     (Line through the mid point of one side of triangle and parallel to the another side, bisects the third side) Similarly in ABQ, BP = PQ Hence, BP = PQ = QD Thus, CE and AF trisect the diagonal BD.

Q81. The two opposite angles of a parallelogram are (3x - 10)^o and (2x + 35)^o. Find the measure of all the four angles of the parallelogram.

Solution

3x - 10 = 2x + 35 (opp. s of a ||^gm are equal) x = 45 (3x - 10)^o = (135 - 10)^o = 125^o Adjacent angle = 180^o - 125^o = 55^o Thus, the four angles of the parallelogram are 125^o, 55^o, 125^o, 55^o.

Q82. In a parallelogram PQRS, the bisectors of adjacent angles R and S intersect each other at the point O. Prove that ROS = 90^o.

Solution

Proof : PQRS is ||^gm R + S = 180^o (co-interior angels) 1 + 2 = 90^o In ROS 1 + 2 + ROS = 180^o 90⁰ + ROS = 180^o ROS= 90^o

Q83. If PQRS is a rhombus with PQR = 55^o, find PRS.

Solution

In triangle PQR, PQ = QR (sides of rhombus) (angles opposite to equal sides are equal) 55^o + 1 + 2 = 180^o (Angle sum property of a triangle) 21 = 180^o - 55^o = 125^o 1 = 62.5^o PRS = 3 = 1 = 62.5^o (alternate interior angles)

Q84. Prove that the bisectors of any two consecutive angles of parallelogram intersect at right angles.

Solution

As ABCD is a parallelogram AD||BC and AB is a transversal. DAB + ABC = 180^o [Angles on the same side of transversal] 1 + 2 = 90^o In AOB ,1 + 2 + AOB = 180^o 90^o + AOB = 180^o AOB = 90^o Therefore, the bisectors of any two consecutive angles of parallelogram intersect at right angles.

Q85. In fig., ABCD is a trapezium in which AB||DC. E is the mid point of AD and F is a point on BC such that EF||DC. Prove that F is the mid point of BC.

Solution

Join AC. In ADC EF||DC, and E is the mid-point of AD G is the mid-point of AC (by mid-point theorem) In CAB GF||AC G is the mid-point of CA F is the mid-point of BC (by mid-point theorem)

Q86. Prove that the diagonals of a parallelogram divide it into two congruent triangles.

Solution

Given : A ||^gm ABCD To prove : Construction : Join AC Proof : Since ABCD is a ||^gm AB||DC and AD||BC Now AB||DC and AC is a transversal DCA = BAC [alt. interior angles] Again AB||DC and AC is the transversal Hence diagonals of a ||^gm divide it into two congruent triangles.

Q87. Show that bisectors of angles of a parallelogram form a rectangle.

Solution

P, Q, R and S are the points of intersection of bisectors of the angles of the parallelogram. In ADS, DAS + ADS = (A and D are interior angles on the same side of the transversal) Also in ADS, DAS + ADS + DSA = 180^o (angle sum property) Similarly, it can be shown that APB or SPQ = 90^o Also, SRQ = RQP = 90^o Hence, PQRS is a rectangle.

Q88. ABCD is a square and on the side DC, an equilateral triangle is constructed. Prove that (i) AE = BE and (ii) DAE = 15^o

Solution

AD = BC (sides of a square) DE = CE (sides of equilateral triangle) ADE = BCE = 90^o + 60^o = 150^o _{(BY SAS congruence theorem)} AE = BE ADE = 90^o + 60^o = 150^o As AD=DC(sides of a square) DC=DE so, AD=DE _ADE=DEA DAE = = 15^o

Q89. PQRS is a parallelogram and SPQ = 60^o. If the bisectors of P and Q meet at point A on RS, prove that A is the mid-point of RS.

Solution

SPQ = 60^o = QRS (opposite angles of a parallelogram) PQR = PSR = 120^o (adjacent angles of a parallelogram) As PA and QA bisect P and Q 1 = 30^o and 4 = 60^o In PSA, 1 = 30^o , 2 = 120 ^o So, using angle sum property, we get, 3 = 30^o PS = SA ...(1) (In a , sides opp. to equal angles are equal) Similarly, QRA, 4 = 60^o and 6 = 60^o QR = AR ...(2) But PS = QR (opposite sides of a parallelogram) ...(3) From (1), (2) and (3), SA = AR A is the mid-point of RS.

Q90.

Solution

Q91. Diagonal AC of a parallelogram bisects A. Show that: (i) it bisects C also (ii) ABCD is a rhombus

Solution

_(i) 1 = 2 (given) But 1 = 4, 2 = 3 (alt. int. angles) 3 = 4 AC bisects C _{(ii) Now, 1 = 4, 2 = 3 (alt. int. angles) AC = AC (Common)} (by ASA rule) BC = DC (cpct) But, AB = DC [Given] AB = BC = CD = AD Hence, ABCD is a rhombus.

Q92. In triangle ABD, D, E and F are mid-points of sides AB, BC and CA respectively. Show that ABC is divided into four congruent triangles by joining D, E and F.

Solution

As D and E are mid-points of AB and BC of ABC, DE||AC Similarly DF||BC, EF||AB ADEF, BDFE and DFCE are all parallelograms. DE is the diagonal of the parallelogram BDFE. _{(Since, a diagonal of a parallelogram divides it into two congruent triangles)} Similarly, and Thus, all the four triangles, BDE, FED, EFC and DAF, are congruent.

Q93. In the figure, AD is the median and DE||AB, Prove that BE is also a median.

Solution

AD is a median (Given) Therefore, 'D' is the mid-point of BC (by definition of median) DE||AB (Given) E is the mid-point of AC (converse of mid-point theorem) BE is a median (by definition of median)

Q94. Characterize the following statements a TRUE or FALSE and give reason for your answer. If a pair of adjacent sides of a parallelogram is equal, it must be a rhombus or a square.

Solution

True. Because a square and a rhombus are both parallelograms having all sides equal, the difference between the two being that in a square all angles are right angles where as in a rhombus the angles are not right angles.

Q95. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Solution

Q96. In ABC, AB = 5 cm, BC = 8 cm and AC = 7 cm. If D and E are respectively mid-points of AB and AC respectively, determine the length of DE. Given reasons.

Solution

Join DE. D and E are mid-points of AB and AC respectively. DE||BC, DE = BC [Mid-point Theorem]

Q97. In the figure, it is given that BDEF and FDCE are parallelograms. If BD = 4 cm, determine CD.

Solution

BDEF and FDCE are ||^gms and BD = 4 cm (Given) BD = FE ...(I) (opposite sides of ||^gm BDEF) In ||^gm FDCE, CD = FE ...(II) (opposite sides are equal in ||^gm) From I and II BD = CD = 4 cm.

Q98. In a ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.         

Solution

In quadrilateral ABXC, AD = DX and BD = DC This shows that the diagonals of the quadrilateral ABXC bisect each other. Therefore ABXC is a parallelogram.

Q99. Prove that the line drawn through the mid point of one side of a triangle, parallel to another side, intersects the third side at its mid point.

Solution

ABC is a triangle; D is the mid point of AB. To prove that E is the mid point of AC. Let F be the mid point of the side AC and not the point E. Join DF In ABC, D is the mid point of AB and DE  BC F is the mid point of AC, by mid point theorem, DF  BC, also DE  BC Thus two intersecting lines DE and DF are parallel to the same line BC This is not possible. So F can not be the mid point of AC. Hence, E is the mid point of AC.

Q100. ABCD is a rectangle with ABD = 40^o. Determine DBC.     

Solution

ABC = 90^o ABD + DBC = 90^o  40^o + DBC = 90^o  DBC = 50^o

Q101. In the figure of PQR, PS and RT are medians and SM||RT. Prove that QM = PQ.

Solution

Proof : In PQR, PS is median S is the mid-point of QR In QRT, S is the mid-point of QR and SM||RT M is the mid-point of QT (mid-point theorem) i.e. QM = MT Since T is the mid-point of PQ i.e., QT = PT QM + MT = PT QM + QM = PT 2QM = PT But PT = PQ, 2QM = PQ QM = PQ