DAV Class 8 Maths Solution Chapter 11 Understanding Quadrilaterals
DAV School Books Class 8 Maths Solution Chapter 11 understanding quadrilaterals all Question Answer.
DAV Class 8 11th chapter understanding quadrilaterals full Chapter explanation is provided by expert teacher. DAV class 8 maths chapter wise solution can be found here.
DAV School Books Class 8 Maths Solution Chapter 11 Understanding Quadrilaterals
Worksheet 1
1) PQRS is a trapezium with PQ∥RS, If ∠P = 30°, ∠Q = 50°, find ∠R and ∠S.
Ans:
According to the problem PQ∥RS and the line PS is a transversal line.
∠P and ∠S form a pair of consecutive interior angles.
∴ ∠P + ∠S = 180°
Or, ∠S = 180° – 30° [∵We know that ∠P = 30°]
Or, ∠S = 150°
R and Q form a pair of consecutive interior angles.
∴ ∠Q + ∠R = 180°
Or, ∠R = 180° – 50° [∵We know that ∠Q = 50°]
Or, ∠R = 130°
Hence, ∠S = 150° and ∠R = 130°
2) ABCD is a quadrilateral with ∠A = 80°, ∠B = 40°, ∠C = 140°, ∠D = 100°.
i) Is ABCD is a trapezium?
ii) Is ABCD a parallelogram?
Justify your answer.
Ans:
For ABCD to be parallelogram ∠A = ∠C and ∠B = ∠D
But the given Quadrilateral ABCD does not satisfy the above properties.
∴ ABCD is not a parallelogram.
For ABCD to be a Trapezium we have to check if any two sides of the quadrilateral ABCD are parallel.
We know that if two angles of the same side of the transversal form a pair of consecutive interior angles then the lines forming the angles with the transversal are parallel to each other.
In the quadrilateral ABCD we can see that the angles ∠A and ∠D form a pair of consecutive interior angles
As ∠A+ ∠D = 80° + 100° = 180°
∴ We can say that AB∥CD
So, ABCD is a trapezium.
3) One of the angles of a parallelogram is 75°. Find the measures of the remaining angles of the parallelogram?
Ans:
From the properties of parallelogram we know that the opposite angles of a parallelogram are equal.
Which means ∠A = ∠C and ∠B = ∠D
∴ ∠A = ∠C = 75°
Now we know that the sum of all the interior angles of a quadrilateral is 360°
∴ ∠A+ ∠B+ ∠C+ ∠D = 360°
Or, 75° + ∠B+ 75° + ∠D = 360°
Or, ∠B+ ∠D = 360° – 150°= 210°
∴ ∠B = ∠D = 210°/2 = 105°
The values of ∠B, ∠C and ∠D are 105°, 75° and 105° respectively.
4) Two adjacent angles of a parallelogram are in the ratio 1:5. Find all the angles of the parallelogram.
Ans:
Let us consider the angles A and B is in the ratio 1:5
∴ ∠ A = k and ∠ B = 5k
From the properties of parallelogram we know that the opposite angles of a parallelogram are equal.
Which means ∠A = ∠C and ∠B = ∠D
∴ ∠A = ∠C = k and ∠B = ∠D = 5k
Now we know that the sum of all the interior angles of a quadrilateral is 360°
∴ ∠A+ ∠B+ ∠C+ ∠D = 360°
Or, k + 5k + k + 5k = 360°
Or, 12k = 360°
Or, k = 30°
The values of the angles ∠A, ∠B, ∠C and ∠D is 30°, 150°,30° and 150° respectively.
5) An exterior angle of a parallelogram is 110°. Find all the angles of the parallelogram.
Ans:
We know that ∠DCE + ∠BCD = 180°
∴ ∠BCD = 180° – 100° [∵ ∠DCE = 100°]
∠BCD = 80°
Now, we know that the opposite angles of a parallelogram are equal.
∴ For the parallelogram ABCD, ∠BCD = ∠DAB and ∠ABC = ∠CDA
∴ ∠BCD = ∠DAB = 80°
We know that the sum of all the internal angles of a parallelogram is 360°
So ∠BCD + ∠DAB +∠ABC + ∠CDA = 360°
Or, 80° + 80° + ∠ABC + ∠CDA = 360° [∵ ∠BCD = ∠DAB = 80°]
Or, 2∠ABC = 360° – 160° [∵ ∠ABC = ∠CDA]
Or, ∠ABC = 200°/2 = 100°
∴ ∠ABC = ∠CDA = 100°
Hence, the values of ∠ABC, ∠BCD, ∠CDA and ∠DAB are equal to 100°, 80°, 100° and 80° respectively.
6) Two adjacent sides of a parallelogram are in the ratio 3:8 and its perimeter is 110cm. Find the sides of the parallelogram.
Ans:
Let us consider that the sides AB: BC = 3:8
∴ Let AB = 3k cm and BC= 8k cm
In parallelogram the opposite sides are equal
∴ AB = CD and BC = DA
∴ AB = CD = 3k cm.
and BC = DA = 8k cm
According to the problem
AB+BC+CD+DA = 110
Or, 3k+8k+3k+8k= 110
Or, 22k = 110
Or, k = 5
∴ The sides AB = 15cm, BC = 40 cm, CD = 15 cm and DA = 40 cm
7) One side of a parallelogram is (3/4) times its adjacent side. If the perimeter of the parallelogram is 70cm, find the sides of the parallelogram.
Ans:
According to the question
AB = (3/4) × BC
Or, AB: BC = 3:4
Let AB = 3k and BC = 4k
In parallelogram the opposite sides are equal
∴ AB = CD and BC = DA
∴ AB = CD = 3k cm.
and BC = DA = 4k cm
According to the problem
AB+BC+CD+DA = 70
Or, 3k+4k+3k+4k= 70
Or, 14k = 70
Or, k = 5
∴ The sides AB = 15cm, BC = 20 cm, CD = 15 cm and DA = 20 cm
8) ABCD is a parallelogram whose diagonals intersect each other at right angles. If the length of the diagonals is 6 cm and 8 cm, find the lengths of all the sides of the parallelogram.
Ans:
According to the question the diagonals of parallelogram ABCD intersect each other at right angles.
We know that the diagonals of a rhombus intersect each other at right angles.
So we can say that ABCD is a rhombus.
Of ABCD, diagonal AC = 8 cm and BD = 6 cm
We know that the diagonals of rhombus dissect each other in equal halves.
AO = (1/2) AC = 8/2 cm = 4 cm
and BO = (1/2) BD = 6/2 cm = 3 cm
Now for △ABO
AO2+BO2= AB2
Or, 42+32 = AB2
∴ AB = √(16+9) = √25 = 5
The sides of a rhombus are equal
So AB=BC=CD=DA =5 cm
9) In figure given below, one pair of adjacent sides of a parallelogram is in the ratio 3:4. If one of its angles, ∠A is a right angle and diagonal BD = 10 cm, find the
i) Lengths of the sides of the parallelogram.
ii) Perimeter of the parallelogram.
Ans:
According to the image given in the problem
AD = 3x cm, AB = 4x cm and ∠DAB = 90°
∴ AD2 + AB2 = BD2
Or, BD2 = (3x)2+(4x)2
Or, BD = √ (9x2+16x2)
∴ BD = 5x
According to the problem BD = 10 cm
∴ 5x = 10
Or, x = 2
As ABCD is a parallelogram
i) The length of the sides
AB = CD = 4x = 20 cm and
AD = BC = 3x = 15 cm
ii) The perimeter of the parallelogram
= AB+BC+CD+DA
= (20+15+20+15) cm = 70 cm
10) ABCD is a quadrilateral in which AB = CD and AD = BC. Show that it is a parallelogram. [Hint: Draw one of the diagonals.]
Ans:
According to the problem
AB = CD and BC = DA
Draw a line joining A and C
To prove that ABCD is a parallelogram
Of the △ACD and △BCA
AB=CD (given)
BC = DA (given)
CA = AC (common)
∴ △ACD and △BCA are congruent by SSS condition.
∴∠1 = ∠2
Angle 1 and 2 form alternate interior angles
We know that if alternate interior angles are equal then the lines forming those angles are parallel
∴DA∥BC and
DA = BC
Hence, ABCD is a parallelogram.