Question 1: Three angles of a quadrilateral are respectively equal to 110^o, 50^o and 40^o . Find its fourth angle.

Answer:

n = 4

Sum of internal angles = (2n-4) \times 90^o = 4 \times 90^o = 360^o

Therefore 110^o + 50^o + 40^o + x = 360^o

\Rightarrow x = 360^o - 200^o = 160^o

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Question 2: In a quadrilateral ABCD , the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5 . Find the measure of each angles of the quadrilateral.

Answer:

The ratio of angles is 1:2:4:5

Sum of internal angles = 360^o

Therefore x + 2x + 4x+ 5x = 360^o

\Rightarrow 12x = 360^o

\Rightarrow x = 30^o

Therefore angles are 30^o, 60^o, 120^o and 150^o

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Question 3: In a quadrilateral ABCD, CO and DO are the bisectors of \angle C and \angle D respectively. Prove that \angle COD = \frac{1}{2} (\angle A + \angle B) .

Answer:2018-12-21_17-53-19

To prove: \angle COD = \frac{1}{2} (\angle A + \angle B)

In \triangle COD

\angle COD = 180^o - (\angle ODC + \angle OCD)

\Rightarrow \angle COD= 180^o - ( \frac{1}{2} \angle D + \frac{1}{2} \angle C)

Now, \angle A + \angle B + \angle C + \angle D = 360^o

\Rightarrow \angle C + \angle D = 360^o- (\angle A + \angle B)

Therefore \angle COD = 180^o - \frac{1}{2} [360 - (\angle A + \angle B)]

\Rightarrow \angle COD = \frac{1}{2} (\angle A + \angle B)

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Question 4: The angles of a quadrilateral are in the ratio 3: 5: 9: 13 . Find all the angles of the quadrilateral.

Answer:

Angles are in ratio of 3:5:9:13

Sum of internal angles = 360^o

Therefore 3x + 5x + 9x + 13x = 360^o

\Rightarrow 30x = 360^o

x = \frac{360}{30} = 12

Therefore angles are 36, 60, 108 and 156

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Question 5: Two opposite angles of a parallelogram are (3x -2)^o and (50 - x)^o . Find the measure of each angle of the parallelogram.

Answer:2018-12-22_8-19-14

In a parallelogram, the sum of diagonally opposite angles are equal.

Therefore 3x - 2 = 50 - x

\Rightarrow 4x = 52

\Rightarrow x = 13

Therefore angles A and C are 3 \times 13 - 4 = 37

We know that A + B = 180

\Rightarrow B = 180 - 37 = 143

Therefore angles are 37, 143, 37, 143

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Question 6: If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.

Answer:2018-12-22_8-20-12

Given A = \frac{2}{3} B

Therefore x + \frac{2}{3} x = 180

\Rightarrow \frac{5}{3} x = 180

\Rightarrow x = 3 \times 36 = 108

Therefore \angle D = 180 - 108 = 72

Hence the angles are 108, 72, 108, 72

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Question 7: Find the measure of all the angles of a parallelogram, if one angle is 24^o less than twice the smallest angle.

Answer:2018-12-22_8-27-50

Let the smallest angle = x

Therefore the other angle would be = 2x - 24^o

We know that the A + B = 180^o

\Rightarrow 3x = 204^o

\Rightarrow x = 68^o

Hence A = 2 \times 68^o - 24^o = 136^o - 24^o = 112^o

Hence the angles are 68^o, 112^o, 68^o, 112^o

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Question 8: The perimeter of a parallelogram is 22 \ cm . If the longer side measures 6.5 \ cm what is the measure of the shorter side?

Answer:2018-12-22_8-28-12

Perimeter = 22 \ cm

Let the shorter side = x

Therefore 2 \times 6.5 + 2x = 22

\Rightarrow 2x = 22 - 13

\Rightarrow 2x = 9

\Rightarrow x = 4.5 \ cm

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Question 9: In a parallelogram ABCD, \angle D = 135^o , determine the measures of \angle A  and \angle B .

Answer:2018-12-22_8-28-30

Given \angle D = 135^o

Therefore \angle A = 180^o - 135^o = 45^o

\angle C = 180^o -135^o= 45^o

\angle B = 180^o - 45^o = 135^o

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Question 10: ABCD is a parallelogram in which \angle A=70^o . Compute \angle B, \angle C and \angle D .

Answer:2018-12-22_8-28-50

Given \angle A = 70^o

Therefore \angle B = 180^o - 70^o = 110^o

\angle C = 180^o - 110^o = 70^o

\angle D = 180^o - 70^o = 110^o

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Question 11: In the adjoining figure, ABCD is a parallelogram in which \angle A = 60^o . If the bisectors of \angle A and \angle B meet at P , prove that A D = DP, PC = BC and DC = 2AD .

Answer:2018-12-22_8-36-33

To prove i) AD = DP    ii) PC = BC    iii) DC = 2AD

Given \angle A = 60^o

AP bisects \angle A   Therefore \angle PAB = \angle PAD = 30^o

Also \angle B = 180^o - \angle A = 120^o

PB Bisects \angle B    Therefore \angle PBA = \angle PBC = 60^o

Also \angle C = 60^o

\Rightarrow \angle BPC = 180^o - 60^o - 60^o = 60^o

\Rightarrow \angle DPA = 180^o - 90^o - 60^o = 30^o

\angle D = 180^o - 60^o = 120^o

i) Consider \triangle ADP ,

Since \angle DAP = \angle DPA = 30^o

(sides opposite to equal angles in a triangle are equal)

\Rightarrow AD = DP … … … … … i)

ii) Similarly in \triangle PCB (all angles are equal to 60^o as it is an equilateral triangle)

Hence PC = PB … … … … … ii)

iii) Since AD = BC … … … … … iii) (opposite sides of a parallelogram)

Adding i) and ii)

DP + PC = AD + BC

\Rightarrow DC = AD + BC

From iii) DC = 2AD

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Question 12: In the adjoining figure, ABCD is a parallelogram in which \angle DAB = 75^o and \angle DBC = 60^o . Compute \angle CDB and \angle ADB .

Answer:2018-12-22_8-35-14

Given \angle A = 75^o \Rightarrow \angle B = 180^o - 75^o = 105^o

\Rightarrow \angle DBA = 105^o - 60^o = 45^o

\Rightarrow \angle D = 180^o - 75^o = 105^o

In \triangle DAB, \angle ADB = 180^o - 75^o - 45^o  = 60^o

\Rightarrow \angle CDB = \angle D - \angle ADB

\Rightarrow \angle CDB = 105^o - 60^o = 45^o

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Question 13: In the adjoining figure, 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 .

Answer:

To prove: AF = 2AB 2018-12-22_8-39-58

Given: ABCD is a parallelogram, E is the mid point of BC

Consider \triangle EBF and \triangle ECD

BE = EC (E is mid point of BC)

\angle DEC = \angle BEF (vertically opposite angles)

\angle DCE = \angle EBF (alternate angles)

Therefore \triangle EBF \cong \triangle ECD (By A.A.S criterion)

Hence DC = BF

Since DC = AB

\Rightarrow AB = BF

Adding BF on both sides

\Rightarrow AB + BF = BF + BF

\Rightarrow AF = 2 BF . Hence Proved.

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Question 14: Which of the following statements are true (T) and which are false (F) ?

  1. In a parallelogram, the diagonals are equal – FALSE
  2. In a parallelogram, the diagonals bisect each other – TRUE
  3. In a parallelogram, the diagonals intersect each other at right angles– FALSE
  4. In any quadrilateral, if a pair of opposite sides is equal, it is parallelogram– TRUE
  5. If all the angles of a quadrilateral are equal, it is a parallelogram– TRUE
  6. If three sides of a quadrilateral are equal, it is a parallelogram– FALSE
  7. If three angles of a quadrilateral are equal, it is a parallelogram– FALSE
  8. If all the sides of a quadrilateral are equal it is a parallelogram– TRUE
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