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RS Aggarwal Solutions For Class 8 Maths also provides the foundation to higher studies. RS Aggarwal Solutions For Class 8 Maths Chapter 16 Parallelograms provides the concepts in an effective way.

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## Exercise 16A Page No: 193

**1. ABCD is a parallelogram in which âˆ A=110 ^{o}. Find the measure of each of the angles âˆ B, âˆ C and âˆ D.**

**Solution: **

Given âˆ A=110^{o}

But we know that sum of adjacent angles of a parallelogram is 180^{o}

âˆ A + âˆ B = 180^{o}

110^{o}+ âˆ B = 180^{o}

âˆ B = 180^{o}– 110^{o}= 70^{o.}

Also âˆ B + âˆ C = 180^{o }[Since âˆ B and âˆ C are adjacent angles]

70^{o}+ âˆ C = 180^{o}

âˆ C = 180^{o}– 70^{o}= 110^{o.}

Now âˆ C + âˆ D = 180^{o }[Since âˆ C and âˆ D are adjacent angles]

110^{o}+ âˆ D = 180^{o}

âˆ D = 180^{o}– 110^{o}= 70^{o}

**2. Two adjacent angles of a parallelogram are equal. What is the measure of each of these angles?**

**Solution: **

Let âˆ A and âˆ B are two adjacent angles.

But we know that sum of adjacent angles of a parallelogram is 180^{o}

âˆ A + âˆ B = 180^{o}

But given that âˆ A = âˆ B

Now substituting we get

âˆ A + âˆ A = 180^{o}

2âˆ A = 180^{o}

âˆ A=180/2 = 90

**3. Two adjacent angles of a parallelogram are in the ratio 4:5. Find the measure of each of its angles.**

**Solution: **

Let âˆ A and âˆ B are two adjacent angles.

But we know that sum of adjacent angles of a parallelogram is 180^{o}

âˆ A + âˆ B = 180^{o}

Given that adjacent angles of a parallelogram are in the ratio 4:5 and let that ratio be multiple of x

âˆ A + âˆ B = 180^{o}

4x + 5x = 180^{o}

9x = 180^{o}

x = 180/9

x= 20^{o}

âˆ A = 4x = 4 Ã— 20 = 80^{o}

âˆ B = 5x = 5 Ã— 20 = 100^{o}

Also âˆ B + âˆ C = 180^{o }[Since âˆ B and âˆ C are adjacent angles]

100^{o}+ âˆ C = 180^{o}

âˆ C = 180^{o}– 100^{o}= 80^{o.}

Now âˆ C + âˆ D = 180^{o }[Since âˆ C and âˆ D are adjacent angles]

80^{o}+ âˆ D = 180^{o}

âˆ D = 180^{o}– 80^{o}= 100^{o}

**4. Two adjacent angles of a parallelogram are (3x-4) ^{o }and (3x+16)^{o}. Find the value of x and hence find the measure of each of its angle.**

**Solution: **

Let âˆ A = (3x-4)^{o }and âˆ B = (3x+16)^{o}

Let âˆ A and âˆ B are two adjacent angles.

But we know that sum of adjacent angles of a parallelogram is 180^{o}

âˆ A + âˆ B = 180^{o}

(3x-4)^{o} + (3x+16)^{o}= 180^{o}

6x + 12 = 180^{o}

6x = 180^{o}-12^{o}

6x = 168^{o}

X= 168^{o}/6 = 28^{0}

âˆ A = (3x-4)^{o}=3 Ã— 28 â€“ 4 = 80^{o}

âˆ B = (3x+16)^{o}= 3 Ã— 28 + 16 = 100^{o}

Also âˆ B + âˆ C = 180^{o }[Since âˆ B and âˆ C are adjacent angles]

100^{o}+ âˆ C = 180^{o}

âˆ C = 180^{o}– 100^{o}= 80^{o.}

Now âˆ C + âˆ D = 180^{o }[Since âˆ C and âˆ D are adjacent angles]

80^{o}+ âˆ D = 180^{o}

âˆ D = 180^{o}– 80^{o}= 100^{o}

**5. The sum of two opposite angles of a parallelogram is 130 ^{o}. Find the measure of each of its angle.**

**Solution: **

Let ABCD be parallelogram.

Given that sum of two opposite angles of a parallelogram is 130^{o}

Given âˆ B + âˆ D = 130^{o}

âˆ B + âˆ B = 130^{o} [since opposite angles of a parallelogram are equal]

2âˆ B = 130^{o}

âˆ B= 130/2 = 65^{o}

Therefore âˆ D = 65^{o}

Also âˆ B + âˆ C = 180^{o }[Since âˆ B and âˆ C are adjacent angles]

65^{o}+ âˆ C = 180^{o}

âˆ C = 180^{o}– 65^{o}= 115^{o}

âˆ A = âˆ C = 115^{o }[since opposite angles of a parallelogram are equal]

**6. Two sides of a parallelogram are in the ratio 5:3. If its perimeter is 64cm, find the lengths of its sides.**

**Solution: **

Given that two sides of a parallelogram are in the ratio 5:3 so let x as common multiple.

Also given that its perimeter is 64cm

But we know that perimeter = 2 (length + width) = 2(5x + 3x)

64 = 2(5x + 3x)

64 = 2(8x)

64 = 16x

x = 64/16 = 4

5x=20

3x=12

Therefore the sides are 20 and 12

**7. The perimeter of a parallelogram is 140cm. If one of the sides is longer than the other by 10cm, find the length of each of its sides.**

**Solution: **

Given that one of the sides is longer than the other by 10cm and let it be x and x+10

Also given that its perimeter is 140cm

But we know that perimeter = 2 (length + width) = 2(x + (x + 10))

140 = 2(2x + 10)

140 = 4x + 20

140 – 20 = 4x

x = 120/4 = 30

x=30cm

x + 10 = 40cm

Therefore the sides are 30cm and 40cm.

## Exercise 16B

**Select the correct answer in each of the following:**

**1. The two diagonals are not necessarily equal in a **

**(a)rectangle (b)square (c) rhombus (d) isosceles trapezium**

**Solution:**

(c) Rhombus

**Explanation: **

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 becomes square. Thatâ€™s why the diagonals of rhombus not necessarily equal

**2. The length of the diagonals of a rhombus are 16cm and 12cm. The length of each side of the rhombus is**

**(a)8cm (b)9cm (c)10cm (d)12cm**

**Solution:**

(c)10cm

**Explanation: **

Let ABCD be rhombus with AC and BD diagonals.

Let AC and BD bisect at O.

Therefore AO = 16/2=8

BO = 12/2=6cm

In right angled triangle AOB, by Pythagoras theorem we can write as

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

AB^{2}=8^{2}+6^{2}

AB^{2}= 64 + 36

AB= âˆšÂ 100

AB =10

Length of rhombus is 10cm

**3. Two adjacent angles of a parallelogram are (2x+25) ^{0} and (3x-5)^{0}. The value of x is**

**(a)28 (b)32 (c)36 (d)42**

**Solution:**

(b)32

**Explanation: **

We know that sum of adjacent angles of a parallelogram is 180^{o}

(2x+25)^{o} + (3x-5)^{o}= 180^{o}

5x + 20 = 180^{o}

5x = 180^{o}-20^{o}

5x = 160^{o}

x= 160^{o}/5 = 32^{0}

**4. The diagonals do not necessarily intersect at right angles in a**

**(a)parallelogram (b)rectangle (c)rhombus (d)kite**

**Solution: **

(a)parallelogram

**Explanation: **

The diagonals do not necessarily intersect at right angles in parallelogram. But opposite sides and opposite angles should be equal.

**5. The length and breadth of a rectangle are in the ratio 4:3. If the diagonal measures 25cm then the perimeter of the rectangle is**

**(a)56cm (b)60cm (c)70cm (d) 80cm**

**Solution:**

(c)70cm

**Explanation:**

Given length and breadth of a rectangle are in the ratio 4:3

And diagonal= 25cm

Let ratio in multiple of x i.e.4x and 3x

According to the Pythagoras theorem we can write as

(4x)^{2}+(3x)^{2}=25^{2}

16x^{2}+9x^{2}=625

25x^{2}=625

X^{2}=625/25=25

x=5

Length=20cm

Breadth=15cm

Perimeter= 2(l +b) = 2(20+15) = 2(35) = 70cm

## RS Aggarwal Solutions for Class 8 Maths Chapter 16- Parallelograms

Chapter 16, Parallelograms, contains 2 Exercises. RS Aggarwal Solutions given here contains the answers to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.

- Definition of parallelogram
- Definition of rhombus
- Definition of rectangle
- Definition of square
- Definition of trapezium
- Definition of isosceles trapezium
- Properties of parallelogram
- Diagonal properties of a rectangle
- Diagonal properties of a rhombus
- Diagonal properties of a square

### Also, Access RS Aggarwal Solutions for Class 8 Maths Chapter 16 Parallelograms Exercises

### Chapter Brief of RS Aggarwal Solutions for Class 8 Maths Chapter 16 – Parallelograms

The RS Aggarwal Solutions for Class 8 Maths Chapter 15 â€“ Parallelograms deals with the definition of parallelogram, definition of some geometrical shapes like rhombus, rectangle, square, trapezium, isosceles trapezium. It also deals with properties of parallelograms, diagonal properties of a rectangle, diagonal properties of rhombus and diagonal properties of a square.