ABCD is a parallelogram in which A =110°. Find the measure of each of the angles B, C and D.
Since the sum of any two adjacent angles of a parallelogram is 180°,
∠A + ∠B = 180°
110° + ∠B = 180°
∠B = (180° - 110°) = 70°
Also, ∠B + ∠C = 180° [Since, ∠B and ∠C are adjacent angles]
70° + ∠C = 180°
∠C = (180° - 70°) = 110°.
Further, ∠C + ∠D = 180° [Since, ∠C and ∠D are adjacent angles]
110° + ∠D = 180°
∠D = (180° - 110°) = 70°.
Therefore, ∠B = 70°, ∠C = 110° and ∠D = 70°.
Two adjacent angles of a parallelogram are equal. What is the measure of each of these angles?
Let ∠A, ∠B are the two adjacent angles.
According to question,
∠A = ∠B
As we know that, sum of any two adjacent angles of a parallelogram is 180°
∠A + ∠B = 180°
∠A + ∠A = 180°
2∠A = 180°
∠A = 90°
Two adjacent angles of a parallelogram are in the ratio 4 : 5. Find the measure of each of its angles.
Let x be the common multiple.
Since the sum of any two adjacent angles of a parallelogram is 180°,
∠A + ∠B = 180°
4x + 5x = 180°
9x = 180°
x = 20°
∠A = 80°
∠B = 100°
Also, ∠B + ∠C = 180° [Since, ∠B and ∠C are adjacent angles]
100° + ∠C = 180°
∠C = (180° - 100°) = 80°
Further, ∠C + ∠D = 180° [Since, ∠C and ∠D are adjacent angles]
80° + ∠D = 180°
∠D = (180° - 80°) = 100°
Therefore, ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠D = 100°.
Two adjacent angles of a parallelogram are (3x - 4)° and (3x + 16)°. Find the value of x and hence find the measure of each of its angles.
Let ∠A = (3x - 4)°, ∠B = (3x + 16)°
Since the sum of any two adjacent angles of a parallelogram is 180°,
∠A + ∠B = 180°
(3x - 4)° + (3x + 16)° = 180°
6x + 12° = 180°
6x = 180° - 12°
6x= 168°
X= 168/6 = 28°
∠A = (3x - 4)° = 80°
∠B = (3x + 16)° = 100°
Also, ∠B + ∠C = 180° [Since, ∠B and ∠C are adjacent angles]
100° + ∠C = 180°
∠C = (180° - 100°) = 80°
Further, ∠C + ∠D = 180° [Since, ∠C and ∠D are adjacent angles]
80° + ∠D = 180°
∠D = (180° - 80°) = 100°
Therefore, x= 28°, ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠D = 100°.
The sum of two opposite angles of a parallelogram is 130°. Find the measure of each of its angles.
Let ABCD is a parallelogram.
∠ B + ∠ D = 130°
∠B + ∠B = 130° (Opposite angles of parallelogram are equal)
∠B = 65°
∠D = 65°
Now, ∠B + ∠C = 180° (sum of any two adjacent angles of a parallelogram is 180°)
∠C = 180° - ∠B = 115°
Now, ∠A = ∠C = 115° (Opposite angles of parallelogram are equal)
Therefore, ∠A = 115°, ∠B = 65°, ∠C = 115° and ∠D = 65°.
Two sides of a parallelogram are in the ratio 5 : 3. If its perimeter is 64 cm, find the lengths of its sides.
Let x be the common multiple.
According to question, sides will be 5x and 3x.
Perimeter =
64 = 2 (5x + 3x)
64 = 16x
x = 4
5x = 20 cm
3x = 12 cm
So, sides will be 20 cm and 12 cm.
The perimeter of a parallelogram is 140 cm. If one of the sides is longer than the other by 10 cm, find the length of each of its sides.
Let x be the common multiple.
According to question, sides will be x and (x + 10).
Perimeter =
140 = 2 (x + (x + 10))
140 = 2 (2x + 10)
140 = 4x + 20
4x = 120
X =30 cm
X + 10 = 40 cm
So, sides will be 30 cm and 40 cm.
In the adjacent figure, ABCD is a rectangle. If BM and DN are perpendiculars from B and D on AC, prove that . Is it true that BM = DN?
In a triangles BMC and DNA
BC = DA (Opposite sides)
∠BCM = ∠DAN (alternate angles)
∠DNA = ∠BMC = 90° [DN and BM are perpendicular to AC]
So by AAS Congruency criterion
ΔBMC ≅ ΔDNA
BM = DN (By CPCT)
In the adjacent figure, ABCD is a parallelogram and line segments AE and CF bisect the angles A and C respectively. Show that .
According to question,
A=C (Opposite angles)
Line segments AE and CF bisect the A and C means,
=
∠DAE = ∠BCF ----------(i)
Now, In triangles ADE and CBF,
AD = BC (Opposite sides)
∠B = ∠D (Opposite angles)
∠DAE = ∠BCF (from (i))
Therefore, Δ ADE ≅ ΔCBF (By ASA congruency)
By CPCT, DE=BF
But, CD=AB
CD - DE = AB - BF.
So, CE = AF.
Therefore, AECF is a quadrilateral having pairs of side parallel and equal,
So, AECF is a parallelogram. Hence, AE || CF.
The lengths of the diagonals of a rhombus are 16 cm and 12 cm respectively. Find the length of each of its sides.
Let ABCD be a rhombus with diagonals AC and BD.
AC and BD bisect at O.
So, AO = = 8 cm
And BO = = 6 cm
In right angled triangle AOB,
AB2 = AO2 + OB2 (According to Pythagoras theorem)
AB2 = 82 + 62
AB2 = 64 + 36
AB =
= 10 cm.
So, Length of Rhombus of each side is 10 cm.
In the given figure ABCD is a square. Find the measure of .
Let be x.
ABCD is a square.
So, DA = DC (every side of square is equal)
Therefore,
∠ACD = ∠CAD = x°
∠ACD + ∠CAD + ∠ADC= 180° (ACD is right angled triangle)
x° + x° + 90° = 180°
2 x° = 90°
x° = 45°
The sides of a rectangle are in the ratio 5 : 4 and its perimeter is 90 cm. Find its length and breadth.
Let x be the common multiple.
Length = 5x
Breadth = 4x
Perimeter =
90 = 2 (5x + 4x)
18x = 90
X = 5
Length = 5x = 25 cm
Breadth = 4x = 20 cm
Name each of the following parallelograms.
(i) The diagonals are equal and the adjacent sides are unequal.
(ii) The diagonals are equal and the adjacent sides are equal.
(iii) The diagonals are unequal and the adjacent sides are equal.
(iv) All the sides are equal and one angle is 60°.
(v) All the sides are equal and one angle is 90°.
(vi) All the angles are equal and the adjacent sides are unequal.
(i) rectangle
(ii) square
(iii) rhombus
(iv) rhombus
(v) square
(vi) rectangle
Which of the following statements are true and which are false?
(i) The diagonals of a parallelogram are equal.
(ii) The diagonals of a rectangle are perpendicular to each other.
(iii) The diagonals of a rhombus are equal.
(iv) Every rhombus is a kite.
(v) Every rectangle is a square.
(vi) Every square is a parallelogram.
(vii) Every square is a rhombus.
(viii) Every rectangle is a parallelogram.
(ix) Every parallelogram is a rectangle.
(x) Every rhombus is a parallelogram.
(i) False
Diagonals of parallelogram bisects each other.
(ii) False
Diagonals of rectangle do not intersect at right angle. So, they are not perpendicular to each other.
(iii) False
Diagonals of rhombus bisect each other. If it’s all sides are equal then only diagonals will be equal and it will be a square.
(iv) False
All kites are rhombus but every rhombus is not a kite.
(v) False
In square all sides are equal. Length of sides may vary in case of rectangle.
(vi) True
In Parallelogram, opposite sides and opposite angles are equal. In square all sides are equal and all angles are right angles.
(vii) True
All sides are equal and diagonals bisect each other.
(viii) True
Opposite sides and opposite angles are equal.
(ix) False
Rectangle forms right angle between adjacent sides but it is not necessary for every parallelogram.
(x) True
Opposite sides and opposite angles are equal.
The two diagonals are not necessarily equal in a
A. rectangle
B. square
C. rhombus
D. isosceles trapezium
All sides of Rhombus are equal in length but in case of angle it is not necessary to be equal.
If all the angles are equal then it will become a square. That’s why diagonals of rhombus are not necessary to be equal in length.
The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is
A. 8 cm
B. 9 cm
C. 10 cm
D. 12 cm
Let ABCD be a rhombus with diagonals AC and BD.
AC and BD bisect at O.
So, AO = = 8 cm
And BO = = 6 cm
In right angled triangle AOB,
AB2 = AO2 + OB2 (According to Pythagoras theorem)
AB2 = 82 + 62
AB2 = 64 + 36
AB =
= 10 cm.
So, Length of Rhombus of each side is 10 cm.
Two adjacent angles of a parallelogram are (2x + 25)° and (3x - 5)°. The value of x is
A. 28
B. 32
C. 36
D. 42
Let ∠A = (2x + 25)°, ∠B = (3x - 5 )°
Since the sum of any two adjacent angles of a parallelogram is 180°,
∠A + ∠B = 180°
(2x + 25)° + (3x - 5)° = 180°
5x + 20° = 180°
5x = 180° - 20°
5x= 160°
X= 160/5 = 32°
So, Value of x is 32°.
The diagonals do not necessarily intersect at right angles in a
A. parallelogram
B. rectangle
C. rhombus
D. kite
The diagonals do not necessarily intersect at right angles in a parallelogram. Only opposite sides, opposite angles are equal and diagonal bisects each other in parallelogram. If diagonals intersect each other at right angle then it would be square or rhombus.
The length and breadth of a rectangle are in the ratio 4 : 3. If the diagonal measures 25 cm then the perimeter of the rectangle is
A. 56 cm
B. 60 cm
C. 70 cm
D. 80 cm
Let x be the common multiple.
Length = 4x
Breadth = 3x
According to Pythagoras theorem,
X =5
So,
Length = 4x = 20 cm
Breadth = 3x = 15 cm
Perimeter =
= 2 (20 + 15)
= 70 cm
So, perimeter of rectangle is 70 cm.
The bisectors of any two adjacent angles of a parallelogram intersect at
A. 30°
B. 45°
C. 60°
D. 90°
Let ABCD is a parallelogram.
AE and AD is the bisector angles of adjacent angles of ∠A and ∠D.
As we know that,
∠A + ∠D = 180o (Sum of interior angles on the same side of traversal is 180o)
∠A + ∠D =
= 90o ---------------(i)
Now, in triangle AOD,
∠AED + ∠A + ∠D = 180o (AE and AD is the angle bisector of ∠A and ∠D.)
∠AED + 90o = 180o (From eq (i))
∠AED = 180o - 90o = 90o
So, the bisectors of any two adjacent angles of a parallelogram intersect at 90o
If an angle of a parallelogram is two-thirds of its adjacent angle, the smallest angle of the parallelogram is
A. 54°
B. 72°
C. 81°
D. 108°
Let ∠A = x°, ∠B =
Since the sum of any two adjacent angles of a parallelogram is 180°,
∠A + ∠B = 180°
x° + = 180°
= 180°
= 540°
x° = 108°
∠A = 108°
∠B = 72°
Also, ∠B + ∠C = 180° [Since, ∠B and ∠C are adjacent angles]
72° + ∠C = 180°
∠C = (180° - 72°) = 108°
Further, ∠C + ∠D = 180° [Since, ∠C and ∠D are adjacent angles]
108° + ∠D = 180°
∠D = (180° - 108°) = 72°
Therefore, smallest angle of the parallelogram is 72°.
The diagonals do not necessarily bisect the interior angles at the vertices in a
A. rectangle
B. square
C. rhombus
D. all of these
In rectangle, only opposite sides are equal which makes diagonals are not to be perpendicular to each other. As diagonals are not perpendicular to each other, they will not bisect the interior angles.
In a square ABCD, AB = (2x + 3) cm and BC = (3x - 5) cm. Then, the value of x is
A. 4
B. 5
C. 6
D. 8
In square all sides are equal.
So, AB = BC
2x + 3 = 3x -5
3x -2x = 5 + 3
X = 8.
If one angle of a parallelogram is 24° less than twice the smallest angle then the largest angle of the parallelogram is
A. 68°
B. 102°
C. 112°
D. 176°
Let ∠A = x°,
∠B = (2x – 24)°
∠C = x° (Opposite angles are equal.)
∠D = (2x – 24)° (Opposite angles are equal.)
Since the sum of angles of a parallelogram is 360°,
∠A + ∠B + ∠C + ∠D = 360°
x° + (2x – 24)° + x° + (2x – 24)° = 360°
6x° – 48 =360°
6x° = 408°
x° = 68°
∠A = 68°
∠B = (2x – 24)° = 112°
Therefore, largest angle of the parallelogram is 112°.