**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

AO^{2}+BO^{2}= AB^{2}

Or, 4^{2}+3^{2} = AB^{2}

∴ 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°

∴ AD^{2 }+ AB^{2 }= BD^{2}

Or, BD^{2 }= (3x)^{2}+(4x)^{2}

Or, BD = √ (9x^{2}+16x^{2})

∴ 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.