Understanding Shapes-iii (special Types Of Quadrilaterales)

(i) AD = BC [In a parallelogram diagonals bisect each other]

(ii) ∠DCB = ∠BAD [alternate interior angles are equal]

(iii) OC = OA [In a parallelogram diagonals bisect each other]

(iv) ∠DAB+ ∠CDA = 180° [Sum of adjacent angles in a parallelogram is 180°]

(i) ∠ABC = ∠Y = 100° [In a parallelogram opposite angles are equal]

∠x + ∠Y = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠x + 100° = 180°

∠x = 180°-100°

∠x = 80°

∠x = ∠z = 80° [In a parallelogram opposite angles are equal]

(ii) ∠PSR + ∠Y = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠Y + 50° = 180°

∠Y = 180°-50°

∠Y = 130°

∠x = ∠Y = 130° [In a parallelogram opposite angles are equal]

∠PSR = ∠PQR = 50° [In a parallelogram opposite angles are equal]

∠PQR + ∠Z = 180° [Linear pair]

50° + ∠Z = 180°

∠Z = 180°-50°

∠Z = 130°

(iii) In ΔPMN

∠MPN + ∠PMN + ∠PNM = 180° [Sum of all the angles of a triangle is 180°]

30° + 90° + ∠z = 180°

∠z = 180°-120°

∠z = 60°

∠y = ∠z = 60° [In a parallelogram opposite angles are equal]

∠z = 180°-120° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠z = 60°

∠z + ∠NML = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

60° + 90°+ ∠x = 180°

∠x = 180°-150°

∠x = 30°

(iv) ∠x = 90° [vertically opposite angles are equal]

In ΔDOC

∠x + ∠y + 30° = 180° [Sum of all the angles of a triangle is 180°]

90° + 30° + ∠y = 180°

∠y = 180°-120°

∠y = 60°

∠y = ∠z = 60° [alternate interior angles are equal]

(v) ∠x + ∠POR = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠x + 80° = 180°

∠x = 180°-80°

∠x = 100°

∠y = 80° [In a parallelogram opposite angles are equal]

∠QRS =∠x = 100°

∠QRS + ∠Z = 180° [Linear pair]

100° + ∠Z = 180°

∠Z = 180°-100°

∠Z = 80°

(vi) ∠y = 112° [In a parallelogram opposite angles are equal]

∠y + ∠TUV = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

∠z + 40° + 112° = 180°

∠z = 180°-152°

∠z = 28°

∠z =∠x = 28° [alternate interior angles are equal]

(i) No, In a parallelogram opposite angles are equal.

(ii) Yes, In a parallelogram opposite sides are equal and parallel.

(iii) No, In a parallelogram diagonals bisect each other.

∠HOP + 70° = 180° [Linear pair]

∠HOP = 180°-70°

∠HOP = 110°

∠HOP = ∠x = 110° [In a parallelogram opposite angles are equal]

∠x + ∠z + 40° = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

110° + ∠z + 40° = 180°

∠z = 180° - 150°

∠z = 30°

∠z +∠y = 70°

∠y + 30° = 70°

∠y = 70°- 30°

∠y = 40°

(i) 3y – 1 = 26 [In a parallelogram opposite sides are of equal length]

3y = 26 + 1

Y =

Y = 9 units

3x = 18 [In a parallelogram opposite sides are of equal length]

x =

x = 6 units

(ii) y – 7 = 20 [In a parallelogram diagonals bisect each other]

y = 20 + 7

Y = units

x-y = 16 [In a parallelogram diagonals bisect each other]

x -27 = 16

x = 16 + 27 = 43 units

In parallelogram RISK

∠SKR + ∠ISK = 180° [In a parallelogram sum of the adjacent angles is equal to 180°]

120° + ∠ISK = 180°

∠ISK = 180° - 120°

∠z = 60°

In parallelogram CLUE

∠UEC = ∠z = 70° [In a parallelogram opposite angles are equal]

In ΔEOS

70° + ∠x + 60° = 180° [Sum of angles of a triangles is 180°]

∠x = 180° - 130°

∠x = 50°

We know that opposite angles of a parallelogram are equal.

Therefore (3x - 2)° = (50 - x)°

3x - 2° = 50° - x

3x° + x = 50° + 2°

4x = 52°

x =

Measure of opposite angles are: 3x - 2 = 3 × 13°-2 = 37°

(50 - x)° = 50 - 13 = 37°

Sum of adjacent angles = 180°

Other two angles are 180° - 37° = 143° each

Let one of the adjacent angle is x°

Therfore other adjacent angle =

Sum of adjacent angles = 180°

Other angle = 180° - 108° = 72°

Therefore angles of the parallelograms are 72°, 72°, 108° and 108°

Let one of the adjacent angle is x°

Therfore other adjacent angle = 70°

Sum of adjacent angles = 180°

Therefore angles of the parallelograms are 70°, 70°, 110° and 110°

Let one of the adjacent angles are x°

Therfore adjacent angles are = x° and 2x°

Sum of adjacent angles = 180°

Other angle = 180° - 60° = 120°

Therefore angles of the parallelograms are 60°, 60°, 120° and 120°

Let one of the adjacent angle ∠D = 135°

Therfore other adjacent angle ∠A = x°

Sum of adjacent angles = 180°

∠A =

∠A = ∠C =45° and ∠D = ∠B =135° [Measure of opposite angles of a parallelogram are equal]

Let one of the adjacent angle ∠A = 70°

Therfore other adjacent angle ∠B = x°

Sum of adjacent angles = 180°

∠B =

∠A = ∠C = 70° and ∠D = ∠B =110° [Measure of opposite angles of a parallelogram are equal]

Let one of the adjacent angle ∠A = 130°

Therfore other adjacent angle ∠B = x°

Sum of adjacent angles = 180°

∠B =

∠A = ∠C = 130° and ∠D = ∠B =70° [Measure of opposite angles of a parallelogram are equal]

Let each angle of parallelogram ABCD = x°

Sum of all the angles = 360°

Therfore each angle of the parallelogram is equal to 90°

Yes, this quadrilateral is a parallelogram. Aparallelogram with each angle equal to 90° is a rectangle.

We know that opposite sides of a parallelogram are equal and parallel.

Perimeter = Sum of all sides

Perimeter = 4 + 3 + 4 + 3 = 14 cm

Perimeter of the parallelogram = 150 cm

Let one of the sides = x cm

Other side = (x + 25) cm

We know that opposite sides of a parallelogram are equal and parallel.

Perimeter = Sum of all sides

x + x + 25 + x + x + 25 = 150

4x + 50 = 150

4x = 150 – 50

x =

Therefore sides of the parallelogram are 25 cm and 50 cm.

Shorter side of the parallelogram = 4.8 cm

Longer side of the parallelogram =

=

We know that opposite sides of a parallelogram are equal and parallel.

Perimeter = Sum of all sides

Perimeter of the parallelogram = 4.8 + 7.2 + 4.8 + 7.2 = 24cm

Therefore perimeter of the parallelogram 24 cm.

We know that sum of the adjacent angles of a parallelogram = 180°

(3x-4)° + (3x+10)° = 180°

3x°- 4° + 3x° + 10° = 180°

6x° = 180°- 6°

x =

Measure of one angle: 3x-4 = 3 × 29°-4 = 83°

Measure of other angle = (3x + 10)°= 3 × 29 + 10 = 97°

Therefore angles of the parallelogram are 83° 83°, 97° and 97°

∠ABC = ∠ADC = 30° [Measure of opposite angles is equal in a parallelogram]

∠BDC = 10°………….. given

∠BDA = 30° - 10° = 20°

∠DAB = 180° - 30° = 150°

∠BCD = ∠DAB = 150° [Measure of opposite angles is equal in a parallelogram]

∠DBA = ∠BDC = 10° [Alternate interior angles are equal]

In ΔDOC

∠BDC + ∠ACD + ∠DOC = 180° [Sum of all angles og a triangle is 180°]

10° + 70° + ∠DOC = 180°

∠DOC = 180°- 80°

∠DOC = 100°

∠DOC = ∠AOB = 100° [Vertically opposite angles are equal]

∠DOC + ∠AOD = 180° [Linear pair]

100° + ∠AOD = 180°

∠AOD = 180°- 100°

∠AOD = 80°

∠AOD = ∠BOC = 80° [Vertically opposite angles are equal]

∠ABC + ∠BCD = 180° [In a parallelogram sum of adjacent angles is 180°]

30° + ∠ACB + ∠ACD = 180°

30° + ∠ACB + 70° = 180°

∠ACB = 180° - 100°

∠ACB = 80°

∠ACB = ∠ACB = 80° [Alternate interior angles are equal]

In ΔBEC

∠BEC + ∠ECB +∠CBC = 180° [Sum of angles of a triangle is 180°]

90° + 40° + ∠CBC = 180°

∠CBC = 180°-130°

∠CBC =50°

∠B = ∠D = 50° [Opposite angles of a parallelogram are equal]

∠A + ∠B = 180° [Sum of adjacent angles of a triangle is 180°]

∠A + 50° = 180°

∠A = 180°-50°

∠A = 130°

In ΔDFC

∠DFC + ∠FCD +∠CDF = 180° [Sum of angles of a triangle is 180°]

90° + ∠FCD + 50° = 180°

∠FCD = 180°-140°

∠FCD =40°

∠A = ∠C = 130° [Opposite angles of a parallelogram are equal]

∠C = ∠FCE +∠BCE + ∠FCD

∠DCF + 40° + 40° = 130°

∠DCF = 130° - 80°

∠DCF = 50°

Given ABCD is a parallelogram in which DP⊥AB and AQ ⊥BC.
Given ∠PDQ = 60°
In quad. DPBQ
∠PDQ + ∠DPB + ∠B + ∠BQD = 360° [Sum of all the angles of a Quad is 360°]
60° + 90° + ∠B + 90° = 360°
∠B = 360° – 240°
Therefore, ∠B = 120°
But ∠B = ∠D = 120° [Opposite angles of parallelogram are equal]
∠B + ∠C = 180° [Sum of adjacent interior angles in a parallelogram is 180°]
120° + ∠C = 180°
∠C = 180° – 120° = 60°
Therefore, ∠A = ∠C = 70° (Opposite angles of parallelogram are equal)

In parallelogram ABCD

∠C =∠A = 55° [In a parallelogram opposite angles are equal]

In parallelogram AEFG

∠A =∠F = 55° [In a parallelogram opposite angles are equal]

In parallelogram BDEF

BD = EF ………..(i) [In a parallelogram opposite sides are equal]

In parallelogram DCEF

DC = EF ………..(ii) [In a parallelogram opposite sides are equal]

From equations (i) and (ii), we get

BD = EF = DC

Hence, BD = DC Proved

In parallelogram BDEF

BD = EF and BF = DE ……..(i) [In a parallelogram opposite sides are equal]

In parallelogram DCEF

DC = EF and DF = CE ……..(ii) [In a parallelogram opposite sides are equal]

In parallelogram AFDE

AF = DE and DF = AE ……..(ii) [In a parallelogram opposite sides are equal]

Therefore DE = AF =BF ……..(iv)

Similarly: DF = CE = AE …….(v)

But, DE = DF ……given

From equations (iv) and (v), we get

AF + BF = AE + EC

AB = AC

Therefore ΔABC is an isosceles triangle.

(i) OB = OD

OB = OD [In a parallelogram diagonals bisect each other]

(ii) ∠OBY =∠ODX [Alternate interior angles are equal]

(iii) ∠BOY= ∠DOX[Vertically opposite angles are equal]

(iv) ΔBOY ≅ ΔDOX

In ΔBOY and ΔDOX

OB = OD [In a parallelogram diagonals bisect each other]

∠OBY =∠ODX [Alternate interior angles are equal]

∠BOY= ∠DOX[Vertically opposite angles are equal]

ΔBOY ≅ΔDOX [ASA rule]

Now, state if XY is bisected at O.

Hence OX = OY [Corresponding parts of congruent triangles]

(i) ∠A = ∠C

True,

∠C =∠A = 55° [In a parallelogram opposite angles are equal]

(ii) ∠FAB=∠A

True,

AF is the angle bisectoe of angle A

(iii) ∠DCE=∠C

True,

CE is the angle bisectoe of angle A

(iv) ∠CEB= ∠FAB

True,

∠C = ∠A [In a parallelogram opposite angles are equal]

[AF and CE are angle bisectors]

(v) CE || AF

True, Since one pair of opposite angles are equal, therefore Quad AEFC is aparallelogram.

AL and CM are perpendiculars on diagonal BD.

AL = CM [In a parallelogram length of perpendiculars drawn on diagonal from opposite vertices are equal]

In parallelogram ABCD:

AB = CD……………..(i) [In aparallelogram opposite sides are equal and parallel]

AE = CF…….. (ii) given

On subtracting (ii) from (i)

AB-AE = CD-CF

BE = DF

BE parallel to DF

Therefore quad BFDE is aparallelogram, since one pair of opposite sides are equal and parallel.

In a parallelogram ABCD

AB = 10 cm, AD = 6 cm
⇒ DC = AB = 10 cm and AD = BC = 6 cm [In a parallelogram opposite sides are equal]
Given that bisector of ∠A intersects DE at E and BC produced at F.
Draw PF || CD
From the figure, CD || FP and CF || DP
Hence PDCF is a parallelogram. [Since one pair of opposite sides are equal and parallel]
AB || FP and AP || BF
⇒ ABFP is also a parallelogram
Consider ΔAPF and ΔABF
∠APF = ∠ABF [Since opposite angles of a parallelogram are equal]
AF = AF (Common side)
∠PAF = ∠AFB (Alternate angles)
ΔAPF ≅ ΔABF (By ASA congruence criterion)
⇒ AB = AP (CPCT)
⇒ AB = AD + DP
= AD + CF [Since DCFP is a parallelogram]
∴ CF = AB – AD
CF = (10 – 6) cm = 4 cm

(i) True, Rhombus is a parallelogram.

(ii) True, Rhombus has all four sides equal.

(iii) False, Rhombus has all four sides equal.

(iv) False, In rhombus no angle is right angle.

(v) True, in rhombus diagonals bisect each other at right angles.

(vi) False, in rhombus diagonals are of unequal length.

(vii) True, Rhombus has all four sides equal.

(viii) True, Rhombus is a parallelogram since opposite sides equal and parallel.

(ix) True, Rhombus is a quadrilateral since it has four sides.

(x) True, Rhombus becomes square when any one angle is 90°.

(xi) False, Rhombus is never a square. Since in a square each angle is 90°.

(i) A rhombus is a parallelogram in which opposite sides are equal and parallel.

(ii) A square is a rhombus in which all four sides are equal.

(iii) A rhombus has all its sides of equal length.

(iv) The diagonals of a rhombus bisect each other at right angles.

(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a rhombus.

No, Diagonals of a rhombus bisect each other at 90°.

A parallelogram is rhombus only when its diagonals bisect each other at right angles.

No it is not always a rhombus.

In rhombus ABCD

∠ACB = 40° given

∠ACB = ∠CAD = 40° [Altermate interior angles are equal]

In ΔAOD

∠AOD = 90° [In rhombus diagonals bisect eact other at right angles]

∠AOD + ∠CAD +∠ADB = 180° [Angle sum property of a triangle]

90° + 40° + ∠ADB = 180°

∠ADB = 180°-130°

∠ADB = 50°

We know in rhombus diagonals bisect each other at right angle.

In ΔAOB

AO = , BO =

Using pythagorous theorem in ΔAOB

AB² = AO² + BO²

AB² = 6² + 8²

AB² = 36 + 64

AB² = 100

AB = = 10cm

Therefore each side of a rhombus is 10cm.

Steps of Construction:

(i) Draw diagonal AC of length 10 cm.

(ii) Draw perpendicular bisector of AC at point O.

(iii) From point ‘O’ out two arcs of length 3cm to get points B and D.

(iv) Join AD and DC to get rhombus ABCD.

Steps of construction:

(i) Draw a line segment AB of length 3.5 cm

(ii) From point A and B draw angles of 40 and 140 respectively.

(iii) From points A and B draw two arcs of length 3.5 cm each is get points D and C.

(iv) Join ABCD to get rhombus ABCD.

Steps of construction:

(i) Draw a line segment of 4 cm

(ii) From point A draw a perpendicular from point A and cut a length of 3.2 cm to get point E.

(iii) From point E and a line parallel to AB.

(iv) From points A and B cut two arcs of length 4 cm on the drawn parallel line to get points D and C.

(v) Join AD and BC to get rhombus ABCD.

Steps of construction:

(i) Draw a line segment AB of length 6 cm.

(ii) From point ‘A’ draw an arc of length 5 cm and from point B draw an arc of length 6 cm. Such that both the arcs intersect at ‘C’.

(iii) Join AC and BC.

(iv) From point A draw an arc of length 6 cm and from point C draw an arc of 6cm, so that both the arcs intersect at point D.

(v) Joint AD and DC to get rhombus ABCD.

(i) In ΔBOC and ΔDOC

BO = DO [In a rhombus diagonals bisect each other]

CO = CO Common

BC = CD [All sides of a rhombus are equal]

ΔBOC≅ΔDOC [SSS Congurency]

(ii) ∠BCO =∠DCO from above [corresponding parts of congruent triangles]

(i) In ΔBOC and ΔDOC

BO = DO [In a rhombus diagonals bisect each other]

CO = CO Common

BC = CD [All sides of a rhombus are equal]

ΔBOC≅ΔDOC [SSS Congurency]

∠BCO =∠DCO from above [corresponding parts of congruent triangles]

Hence, each diagonal of a rhombus bisect the angle through which it passes.

We know in rhombus diagonals bisect each other at right angle.

In ΔAOB

BO =

Using pythagorous theorem in ΔAOB

AB² = AO² + BO²

10² = AO² + 8²

100-64 = AO²

AO² = 36

AO = = 6cm

Therefore length of diagonal AC of rhombus ABCD is 6 × 2 = 12cm.

We know in rhombus diagonals bisect each other at right angle.

In ΔAOB

BO =

AO =

Using pythagorous theorem in ΔAOB

AB² = AO² + BO²

AB² = 4² + 3²

AB² = 16 + 9

AB² = 25

AB = = 6cm

Therefore length of each side of a rhombus ABCD is 5cm.

(i) True, In a rectangle two pairs of sides are equal.

(ii) False, In a rectangle two pairs of sides are equal.

(iii) True, In a rectangle diagonals are of equal length.

(iv) True, In a rectangle diagonals bisect each other.

(v) False, Diagonals of a rectangle need not be perpendicular.

(vi) False, Diagonals of a rectangle need not be perpendicular. Diagonals only bisect each other.

(vii) True, Diagonals are of equal length and bisect each other.

(viii) False, Diagonals are of equal length and bisect each other. Diagonals of a rectangle need not be perpendicular

(ix) False, In a square all sides are of equal length.

(x) True, All rhombuses are parallelograms, since opposite sides are equal and parallel.

(xi) True, All squares are rhombuses, since all sides are equal in a square and rhombus. All squares are rectangles, since opposite sides are equal and parallel.

(xii) False, All squares are parallelograms, since opposite sides are parallel and equal.

(i) True, square is a rectangle, since opposite sides are equal and parallel and each angle is right angle.

(ii) True, In a square all sides are of equal length.

(iii) True, in a square diagonals bisect each other at right angle.

(v) False, in a square diagonals are of equal length. Length of diagonals is not equal to the length of sides

(i) A rectangle is a parallelogram in which opposite sides are parallel and equal.

(ii) A square is a rhombus in which all the sides are of equal length.

(iii) A square is a rectangle in which opposite sides are equal and parallel and each angle is a right angle.

No, diagonals of a rectangle are of equal length equal.

In ΔACB and ΔCAD

AB = CD [Opposite sides of a rectangle are equal]

BC = DA

AC = CA Common

ΔACB ≅ΔCAD (SSS Congurency)

ABCD is a rectangle

Let the side is x

Length of rectangle = 3x

Breadth of the rectangle = 2x

Given perimeter of rectangle = 20 cm

Perimeter of the rectangle = 2(length + breadth)

20 = 2(3x + 2x)

10x = 20

x = 2

Therefore Length of the rectangle = 3×2 = 6cm

Therefore breadth of the rectangle = 2×2 = 4cm

ABCD is a rectangle

Let the side is x

Length of rectangle = 5x

Breadth of the rectangle = 4x

Given perimeter of rectangle = 90 cm

Perimeter of the rectangle = 2(length + breadth)

90 = 2(5x + 4x)

18x = 90

x = 5

Therefore Length of the rectangle = 5×5 = 25cm

Therefore breadth of the rectangle = 4×5 = 20cm

ABCD is a rectangle

In ΔABC using pythagorous theorem,

AC² = AB² + BC²

AC² = 12² + 5²

AC² = 144 + 25

AC² = 169

AC =

AC = 13cm

Therefore length of diagonal is 13cm.

Steps of construction:

(i) Draw a lien segment AB of length 8 cm

(ii) From point ‘A’ draw an arc of length 10 cm.

(iii) From point B draw an angle of 90°, and the arc from point A cuts it at point C.

(iv) Join Ac

(v) From point A draw an angle of 90° and point C drawn an arc of length 8 cm to get point D.

(vi) Join CD and AD to get required rectangle.

Steps of construction:

(i) Draw a line segment AB of length 4.8 cm.

(ii) From points A and B draw perpendiculars.

(iii) Cut and arc of 4.8 cm from point A and B on the perpendiculars to get point D and C.

(iv) Join DC and AD to get required rectangle.

(i) Four sides of equal length

Rhombus and square are the quadrilaterals that have all four sides of equal length.

(ii) Four right angles

Rectangle and square have all four angles right angles.

(i) a quadrilateral?

A square is a quadrilateral because it has four equal sides.

(ii) a parallelogram?

A square is a parallelogram since it has opposite sides equal and parallel.

(iii) a rhombus?

A square is a rhombus because it has all four sides of equal length.

(iv) a rectangle?

A square is a rectangle because its opposite sides are equal and parallel and each angle is right angle.

(i) bisect each other

In a Parallelogram, rectangle, rhombus and square diagonals bisect each other.

(ii) are perpendicular bisector of each other

In a Rhombus and square diagonals are perpendicular bisector of each other

(iii) are equal.

In a square and rectangle diagonals are of equal length.

ABC is a right angled triangle. O is the mid point of hypotenuse AC, such that OA = OC

Draw CD||AB and join AD, such that AB = CD and AD = BC

Now quad ABCD is a rectangle, since each angle is a right angle and opposite sides are equal and

parallel.

We know in a rectangle diagonals are of equal length and they bisect each other.

Therefore, AC = BD

And also, AO = OC =BO =OD

Hence, O is equidistant from A, B and C.

a. By measuring each angle, because in a rectangle each angle is a right ange.

b. By measuring opposite sides. Since in a rectangle opposite sides are of equal length.

c. By measuring the lengths of diagonals. Since in a rectangle diagonals are of equal length.