Special Parallelograms

EXERCISE 1. ABCD is a rectangle in which ∠ACB = 62o , then ∠AOB is a. 46o b.

62o

c. 124o d. 56o Sol:

Correct option is (c)

We know that the diagonals of a rectangle are equal and bisect each other.

Therefore AC = DB or

1 1 AC = DB 2 2

or OC=OB

(Halves of equal are equal) … (1)

Now, in Triangle BOC OB=OC

[from (1)]

Therefore Triangle BOC is an isosceles triangle. In an isosceles triangle, angles opposite equal sides are equal. So, ∠OBC = ∠OCB or ∠OBC = 62o

[ ∠OCB = 62o given]

Since ∠AOB is the exterior angle of ∠BOC ,

Therefore ∠BOC = ∠OBC + ∠OCB [Exterior angle= sum of the remote interior angles] or ∠BOC = 62o + 62o = 124o 2. In figure 7-42, ABCD is a rectangle. The measure of ∠ODC is a. 68o b. 58o c. 22o d. 44o Sol:

Correct option is (c)

Since ABCD is a rectangle, and in rectangle, the diagonals bisect each other and are equal in length.

Therefore OD = OC

… (1)

Now, in DOC , OD=OC

[from (1)]

So, it is an isosceles triangle

Therefore ∠ODC = ∠OCD

(In an isosceles triangle, angles opposite equal side are equal)

and ∠DOC = 136o

(Vertically opposite angles are equal)

Now, in Triangle ODC

∠ODC + ∠OCD + ∠DOC = 180o

(Angle sum property of triangle)

or ∠ODC + ∠ODC + 136o = 180o

( ∠OCD = ∠ODC and ∠DOC = 136o )

or 2∠ODC = 180o − 136o

or 2∠ODC = 44o

44o or ∠ODC = 2 or ∠ODC = 22o 3. In figure 7-43, ABCD is parallelogram and DB is the diagonal. The values of x and y respectively are a. 4o and 3o b. 5o and 4o c. 3o and 4o d. 4o and 5o Sol:

Correct option is (a)

We know that the opposite sides of a parallelogram are parallel to each other. Therefore ∠CDB = ∠DBA (Alternate interior angles are equal) or 9 y = 27o or y =

27 o 9

or y = 3o

Again, AD parallel to BC and DB is the transversal.

Therefore ∠ADB = ∠DBC or 8 x = 32o or x =

32o 8

or x = 4o Therefore the values of x and y are 4o and 3o . 4. In figure 7-44, ABCD is a square; AC is the diagonal then ∠ACD is a. 90o b. 45o c. 135o d. 60o Sol:

Correct option is (b)

We know that the diagonals of a square bisects the vertex,

1 Therefore ∠DCA = × 90o = 45o 2

(Each angles of square = 90o )

5. In figure 7-45, A, B and C are respectively, the mid point of sides QR, RP and PQ respectively. The perimeter of the quadrilateral ABCQ is a. 40 cm b. 80 cm c. 104 cm d. 54 cm Sol:

Correct option is (d)

Since C is the mid point of PQ and B is the mid point of PR. Therefore PC=CQ and PB=BR Now,

PC = 1 and CQ

PB =1 BR

(Since PC=CQ and PB=BR)

Therefore In Triangle PQR

PC PB = CQ BR

(Since Each fraction =1)

By converse of basic proportionality theorem, we find that … (1)

CB parallel to QR

Again, since A is the mid point of QR, Therefore QA = AR and B is the mid point of PR, Therefore PB = BR Now,

QA PB = 1 and =1 AR BR

Therefore In Triangle QRP,

[ Since QA = AR and PB = BR ]

RA RB = AQ BP

(each fraction =1)

By converse of basic proportionality theorem, we have

BA parallel to PQ

… (2)

From (1) and (2), we find that BCQA is a parallelogram As opposite sides of parallelogram are equal,

Therefore BC = QA and BA=CQ 1 Now QA = QR 2

(Since A is the mid point of QR)

1 Therefore QA = × 24 = 12 cm 2 1 and CQ = × PQ 2

(Since C is the mid point of PQ)

1 Therefore CQ = × 30 = 15 cm 2 Now, perimeter of the quadrilateral ABCQ is AB+BC+CQ+QA=15+12+15+12=54 cm 6. In figure 7-46, PQRS is a trapezium in which PQSR , M and N are the mid points of SP and RQ respectively. The length of SR is a. 31 cm b. 20 cm c. 24 cm d. 34 cm Sol:

Correct option is (b)

Since M and N are the mid points of the non-parallel sides SP and RQ of the trapezium PQRS, ∴ MN is the median of the trapezium. We know that the length of the median is the mean of the sum of the length of the parallel sides

Therefore

or

1 ( PQ + SR) = MN 2

1 (14 + SR) = 17 2

or (14 + SR ) = 17 × 2 or (14 + SR) = 34 or SR = 34 − 14 or SR=20 cm 7. In figure 7-47, ABCD is a rhombus. If ∠B = 68o , then ∠C is a. 112o

b. 68o c. 44o d. 68o Sol:

Correct option is (a)

Since ABCD is a rhombus

Therefore AB parallel to DC each other)

(Opposite sides of a rhombus are parallel to

Now AB D and CB is the transversal,

Therefore ∠B + ∠C = 180o or 68o + ∠C = 180o or ∠C = 180o − 68o or ∠C = 112o

(Consecutive interior angles are supplementary)

8. In figure 7-48, ABCD is a parallelogram, diagonals AC and BD intersect at O. AC=8cm and BD=12 cm. Then OA+OB is a. 10 cm b. 20 cm c. 4 cm d. none of the above Sol:

Correct option is (a)

We know that the diagonals of a parallelogram bisect each other and their point of intersection is the point of bisection.

Therefore OA =

1 AC 2

1 or OA = × 8 2 or OA = 4 cm

… (1)

and OB =

1 BD 2

1 or OB = ×12 2 or OB = 6 cm

… (2)

From (1) and (2), we find that OA+OB=4cm + 6 cm = 10 cm 9. In figure 7-49, A, B and C are the mid points of the sides PQ, QR and RP of Triangle PQR respectively. The perimeter of the Triangle PQR is a. 9 cm b. 27 cm c. 18 cm d. none of the above Sol:

Correct option is (c)

Since A is the mid point of PQ and C is the mid point of PR of Triangle PQR .

Therefore

PA PC = AQ CR

(Q Each fraction =1)

By converse of Basic Proportionality Theorem, we have

AC parallel to QR

… (1)

Also C is the mid point of RP and B is the mid point of RQ in PQR

Therefore

RC RB = CP BQ

(Q Each fraction =1)

By converse of Basic Proportionality Theorem, we have

CB parallel to PQ

… (2)

From (1) and (2), we have

AC parallel to QB and CB parallel to AQ Therefore ACBQ is a parallelogram (By definition) Hence QB=AC and AQ=CB

(Opposite sides of a parallelogram are equal)

or QB=2cm, and AQ=3 cm Similarly CR= 4cm Since A is the mid point of PQ, Therefore PQ = 2 AQ or PQ=6 cm B is the mid point of QR, Therefore QR = 2QB or QR=4 cm C is the mid point of RP, Therefore RP = 2CR or RP=8 cm So, perimeter of Triangle PQR = PQ + QR + RP = 6 + 4 + 8 = 18 cm

10. In figure 7-50, ABCD is a rhombus. Its diagonals AC and BD intersect at O. If AC=6 cm and BD=8 cm. then the perimeter of the rhombus is a. 28 cm b. 40 cm c. 14 cm d. 20 cm Sol:

Correct option is (d)

We know that the diagonals of a rhombus bisect each other at right angles,

Therefore OA =

and OB =

1 1 AC = × 6 = 3 cm 2 2

1 1 BD = × 8 = 4 cm 2 2

Now, in Triangle AOB, ∠AOB = 90o (Diagonals of a rhombus bisect each other at right angles)

Therefore Triangle AOB is a right triangle, By Pythagoras theorem, we have

AO 2 + BO 2 = AB 2 or 32 + 42 = AB 2 or 9 + 16 = AB 2 or 25 = AB 2 or AB = 25 = 5 cm Now, perimeter of the rhombus = 4 × side (Since all the sides of rhombus are equal) = 4 × 5 = 20 cm 11. In figure 7-52, ABCD and PQWRS are rectangle, P is the mid point of DC and Q is the mid point of AC. If DC=8 cm and AD=6 cm, the length PR is a. 5 cm b. 7 cm c. 6 cm d. 8 cm Sol:

Correct option is (a)

Since ABCD is a rectangle, Therefore ∠D = 90o Now, in right Triangle ADC Using Pythagoras Theorem, we have

DC 2 + AD 2 = AC 2 or 82 + 62 = AC 2 or 64 + 36 = AC 2 or 100 = AC 2 or AC = 100 = 10 cm Since, Q is the mid point of AC

Therefore QC =

1 1 AC = ×10 = 5 cm 2 2

… (1)

Now, it is given that PQRC is a rectangle Therefore QC and PR are the diagonals of the rectangle PQRC

Therefore PR = QC or PR=5 cm

(Diagonals of a rectangle are equal_ [from (1)]

Hence PR= 5 cm 12. ABCD is a rhombus. AB is produced to both sides to E and F such that AB=AE=BF. Then ∠G is a. 91o b. 89o c. 90o d. 105o Sol:

Correct option is (c)

In Triangle EAD, since EA=AD (Given)

Therefore Triangle EAD is an isosceles triangle with AE=AD Therefore ∠AED = ∠ADE (In an isosceles triangle angles opposite equal sides are equal) Now ∠DAB is the exterior of angle of EAD

Therefore ∠DAB = ∠AED + ∠ADE (Exterior angle of a triangle is equal to sum of the remote interior angles) or ∠DAB = ∠AED + ∠AED ( Since ∠AED = ∠ADE )

or ∠DAB = 2∠AED or 2∠OAB = 2∠AED

(The diagonals of rhombus bisects the vertex)

or ∠OAB = ∠AED Thus, AC EG

… (1) ( Since ∠OAB and ∠AED are corresponding angles)

Again, in BFC

Since BF = AB Therefore BF = BC

( Since AB and BC are the sides of a rhombus)

Hence, Triangle BFC is an isosceles triangle with BC=BF

Therefore ∠BFC = ∠BCF (In an isosceles triangle, angle opposite equal sides are equal) Now, ∠ABC is the exterior angle of Triangle BFC

Therefore ∠ABC = ∠BFC + ∠BCF (Exterior angle of a triangle is equal to the sum of the remote interior angles)

or ∠ABC = ∠BFC + ∠BFC ( Since ∠BFC = ∠BCF ) or ∠ABC = 2∠BFC or 2∠ABD = 2∠BFC

(The diagonals of a rhombus bisects the vertex)

or ∠ABD = ∠BFC This implies that BD parallel to FG ( Since ∠ABD and ∠BFC are corresponding angles) … (2) In quadrilateral DOCG, we find that

OD parallel to GC and OC parellel to DG parallel)

(parts

of

parallel

lines

are

Therefore DOCG is a parallelogram (By definition) Now, ∠DOC = 90o (Diagonals of a rhombus are perpendicular to each other)

Therefore DOCG is a rectangle (A parallelogram whose one angle is a right angle is a rectangle) Therefore ∠G = 90o (All the angles of a rectangle = 90o ) 13. ABCD is a parallelogram in which AB=3 cm and AD=2cm. AD is produced to E so that DE=DC and EC produced meets AB produced in F. The length AF is a. 5 cm b. 8 cm c. 7 cm d. 6 cm Sol:

Correct option is (a)

In Triangle EDC , since DE=DC

(Given)

Therefore ∠E = ∠DCE ( Since Triangle EDC is isosceles with DE=DC and in an isosceles triangle angles opposite equal sides are equal) … (1) Now DC AF and EF is the transversal.

Therefore ∠DCE = ∠F

(Corresponding angles are equal)

… (2)

From (1) and (2), we find that

∠E = ∠F

(Transitive property)

Now, in Triangle AEF

∠E = ∠F

(Proved above)

Therefore Triangle AEF is an isosceles triangle Thus AE=AF (In an isosceles triangle, sides opposite equal angles are equal)

Therefore AD + DE = AF or AD + DC = AF

[ Since DE = DC , given]

or AD + AB = AF

( Since DC = AB, opposite sides of a parallelogram are equal)

or 2 + 3 = AF or AF = 5 cm 14. In figure 7-54, ABC is a triangle in which AB=2 cm and BC=4 cm. O is the mid point of AC. BO is produced to D such that BO=OD. Then length AD is a. 6 cm b. 2 cm c. 8 cm d. 4 cm Sol:

Correct option is (d)

Construction:

Join AD and CD.

Now in quadrilateral ABCD OA=OC (Since O is the mid point of AC) and BO=OD (Given)

and AC and BD are the diagonals of the quadrilateral ABCD. Therefore Quadrilateral ABCD is a parallelogram (Since a quadrilateral in which the diagonals bisect each other is a parallelogram) So, AD=BC (opposite sides of a parallelogram are equal) or AD=4 cm 15. In figure 7-55, lines AB and CD bisect each other at right angles. If AB=4 cm, then the perimeter of the quadrilateral ADBC is a. 8 cm b. 8 2 cm c. 16 cm d. 8 3 cm Sol:

Correct option is (b)

In figure 7-55 as diagonals AB and CD bisect each other at right angle. Therefore ADBC is a square. Now in Triangle OBC , since OB=OC=2cm and ∠COB = 90o

Triangle OBC is an isosceles right triangle, and in an isosceles right triangle, each acute angle = 45o . Now, side opposite 45o = 2cm Therefore Side opposite 90o = 2 2 triangle)

( 45o , 45o ,90o property of a right

Now, perimeter of a square = 4 × side = 4 × 2 2 = 8 2cm 16. The perimeter of a square is 12 cm. The sum of its diagonals is a. 12 cm b. 18 cm c. 6 2 cm d. 9 cm Sol:

Correct option is (c)

We know that perimeter of a square = 4 × side or 12 = 4 × AB or AB =

12 = 3 cm 4

BC= 3 cm (All the sides of a square are equal) Now, in right triangle ABC, using Pythagoras theorem, we have

AB 2 + BC 2 = AC 2 or 32 + 32 = AC 2 or 9 + 9 = AC 2 or AC 2 = 18 or AC = 18 = 3 2 Since diagonals of a square are equal

Therefore BD = 3 2

Now sum of the diagonals = AC+BD= 3 2 + 3 2 = 6 2 cm 17. In figure 7-57, ABCD is a rhombus. If ∠BAD = 64o then ∠BDC is a. 58o b. 116o c. 64o d. 42o Sol:

Correct option is (a)

Since rhombus is a special parallelogram in which all the sides are equal

Therefore AB parallel to DC

(Opposite sides of a parallelogram are equal)

Thus, ∠BAD + ∠ADC = 180o (Consecutive interior angles are supplementary) or, 64o + ∠ADC = 180o or ∠ADC = 180o − 64o = 116o

( Since ∠BAD = 64o , given) … (1)

Since diagonals of a rhombus bisects the vertex,

1 Therefore ∠BDC = ∠ADC 2 1 or ∠BDC = ×116 = 58o 2

[ Since ∠ADC = 116o , from (1)]

18. ABCD is a rhombus. If ∠A = 60o and AB=3cm, then length of BD is a. 3 2 cm b. 3 cm c. 2 3 cm d. 6 cm Sol:

Correct option is (b)

Since all the sides of a rhombus are equal,

Therefore AB = AD

Hence, Triangle ABD is an isosceles triangle (If two sides of a triangle are equal, it is an isosceles triangle)

Therefore ∠ABD = ∠ADB (In an isosceles triangle, angle opposite equal sides are equal) Now in

ABD

∠A + ∠ABD + ∠ADB = 180o (Angle sum property of triangles) or 60o + ∠ABD + ∠ABD = 180o

[ Since ∠ADB = ∠ABD ]

or 60o + 2∠ABD = 180o or 2∠ABD = 180o − 60o = 120o or ∠ABD =

120o = 60o 2

Now, in Triangle ABD

∠A = ∠ABD = ∠ADB = 60o As all the angles of Triangle ABD are 60o Hence, AB=BD=AD=3 cm (Since in equilateral triangle, all sides are equal) So, BD= 3cm 19. In figure 7-59, ABCD is a rhombus in which ∠DAB = 60o . If AB=5cm, then the length of the diagonal AC is a. 10 cm b. 6 cm c. 8 cm d. 5 3 cm

Sol:

Correct option is (d)

Since in a rhombus, all the sides are equal, Therefore AB = AD This implies that Triangle ABC is an isosceles triangle with AB=AD

Therefore ∠ABD = ∠ADB (In an isosceles triangle, angles opposite equal sides are equal) Now, in Triangle ABD,

∠DAB + ∠ABD + ∠ADB = 180o

(Angle sum property of triangles)

or 60o + ∠ADB + ∠ADB = 180o

( Since ∠ABD = ∠ADB )

or 60o + 2∠ADB = 180o or 2∠ADB = 180o − 60o = 120o or ∠ADB =

120o = 60o 2

Now, in Triangle ABD,

∠DAB = ∠ABD = ∠ADB = 60o

Therefore Triangle ABD is an equilateral triangle (A triangle in which each angle is 60o is an equilateral triangle) Therefore AB = BD = AD = 5 cm (In an equilateral triangle, all the sides are equal) Hence BD=5 Since, diagonals of a rhombus bisect each other at right angles

Therefore Triangle AOB is a right triangle with sides OB and OA Using Pythagoras theorem, we have

OA2 + OB 2 = AB 2 2

⎛ BD ⎞ 2 or OA + ⎜ ⎟ = AB ⎝ 2 ⎠ 2

2

⎛5⎞ or OA + ⎜ ⎟ = 52 ⎝2⎠ 2

or OA2 +

25 = 25 4

or OA2 = 25 −

25 4

or OA2 =

100 − 25 75 = 4 4

or OA =

75 5 3 = cm 4 2

Now, AC= 2 OA (Since O is the point of bisection of diagonals)

or AC= 2 ×

5 3 = 5 3 cm 2

20. In figure 7-60 ABCD is a triangle, ∠CAB = 30o . If AC=4 cm, then AB and C are respectively equal to a. 2 3 cm, 2 cm b. 3 2 cm, 3 cm c. 3 cm, 2 cm d. 2 cm, 2 3 cm Sol:

Correct option is (a)

In Triangle ABC ,

∠CAB + ∠ABC + ∠BCA = 180o

(Angle Sum Property of Triangle)

or 30o + 90o + ∠BCA = 180o

( Since ∠B = 90o , ABCD is a rectangle)

or 120o + ∠BCA = 180o

or ∠BCA = 180o − 120o = 60o Now, using 30o , 60o ,90o property of triangles, we have

1 AB = × 4 3 = 2 3 2 1 BC= × 4 = 2 cm 2

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