## RS Aggarwal Class 8 Solutions Chapter 16 Parallelograms Ex 16A

These Solutions are part of RS Aggarwal Solutions Class 8. RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16A.

**Other Exercises**

- RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16A
- RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16B

**Question 1.**

**Solution:**

In ||gm ABCD,

∠A = 110°

But ∠ C = ∠ A {Opposite angles of a ||gm are equal}

∴ ∠C = 110°

But ∠A + ∠B = 180°

(Sum of adjacent angles)

=> 110° + ∠B = 180°

=> ∠B – 180° – 110° = 70°

But ∠ D = ∠ B (opposite angles)

∴ ∠ D = 70°

Hence ∠B = 70°, ∠C = 110° and ∠D = 70° Ans.

**Question 2.**

**Solution:**

In a parallelogram, sum of two adjacent angles is 180°

But these are equal to each other

∴ Each angle will be \(\frac { 180^{ o } }{ 2 } \)

= 90° Ans.

**Question 3.**

**Solution:**

The ratio between two adjacent angles of a ||gm ABCD are in the ratio 4 : 5

Let ∠ A = 4x and ∠ B = 5x

But ∠A + ∠B = 180°

=> 4x + 5x = 180°

=> 9x = 180°

∴ x = \(\frac { 180^{ o } }{ 9 } \)

= 20°

∴ ∠A = Ax = 4 x 20° = 80°

∠B = 5x = 5 x 20 = 100° Ans.

**Question 4.**

**Solution:**

In || gm ABCD, ∠ A and ∠ B are two adjacent angles

Let ∠ A = (3x – 4)° and ∠ B = (3x + 16)°

But ∠A + ∠B = 180°

=> (3x – 4)° + (3x + 16) = 180°

=> 3x – 4° + 3x + 16° = 180°

=> 6x + 12° = 180°

=> 6x= 180° – 12°

=> 6x = 168

=> x = \(\\ \frac { 168 }{ 6 } \) = 28°

∴ x = 28°

Now ∠A = 3x – 4 = 3 x 28° – 4° = 84° – 4° = 80°

∠B = 3x + 16

= 3 x 28 + 16

= 84°+ 16° = 100°

But ∠C = ∠A (opposite angles of ||gm)

∴ ∠ C = 80°

Similarly ∠ D = ∠ B = 100°

Hence ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠D= 100° Ans.

**Question 5.**

**Solution:**

In ||gm ABCD, ∠A and ∠C are opposite angles.

∴ ∠A = ∠C= 130°

But ∠A = ∠C (opposite angles)

∴ ∠A = ∠C

= \(\frac { 130^{ o } }{ 2 } \)

= 65°

But ∠A + ∠B = 180°

(sum of adjacent angles)

=> 65° + ∠B = 180°

=> ∠B = 180° – 65° = 115°

But ∠ D = ∠ B (opposite angles)

∴ ∠D = 115°

Hence ∠A = 65°, ∠B = 115°, ∠C = 65° and ∠ D = 115° Ans.

**Question 6.**

**Solution:**

Let ABCD is a parallelogram in which AB : BC = 5:3

Let AB = 5x: and BC = 3x.

But perimeter = 64 cm.

∴ 2(5x + 3x) = 64

=> 2 x 8x = 64

=> 16x = 64

x = \(\\ \frac { 64 }{ 16 } \)

= 4

∴ AB = 5x = 5 x 4 = 20 cm

BC = 3x = 3 x 4=12 cm

But CD = AB and AD = BC

(opposite sides of ||gm)

∴ CD = 20 cm and AD = 12 cm Ans.

**Question 7.**

**Solution:**

Perimeter of parallelogram ABCD = 140 cm.

=> ∴ 2 (AB + BC) = 140 cm.

=> AB + BC = \(\\ \frac { 140 }{ 2 } \) = 70 cm.

Let BC = x

then AB = x + 10

∴ x + x + 10 = 70

=> 2x + 10 = 70

=> 2x = 70 – 10 = 60

=>x = \(\\ \frac { 60 }{ 2 } \) = 30

∴ BC = 30 cm. and

AB = 30 + 10 = 40 cm.

But AD = BC and CD = AB

(Opposite sides of parallelogram)

∴ AD = 30 cm. and CD = 40 cm.

**Question 8.**

**Solution:**

In rectangle ABCD, AC is diagonal BM ⊥ AC and DN ⊥ AC.

Now, we have to prove that

∆BMC ≅ ∆DNA

In ∆BMC and ∆DNA,

BC = AD (opposite sides of the rectangle)

∠M = ∠N (each = 90°)

∠BCM = ∠D AN (Alternate angles)

∴ ∆BMC ≅ ∆DNA

(S.A.A. axiom of congruency)

∴ BM = DN (c.p .c.t.)

**Question 9.**

**Solution:**

ABCD is a parallelogram.

AE and CF are the bisectors of ∠A and ∠C respectively.

In ∆ADE and ∆CBF,

AD = BC

(Opposite sides of the parallelogram)

∠D = ∠B

(Opposite angles of the parallelogram)

∠DAE = ∠FCB (\(\\ \frac { 1 }{ 2 } \) of equal angles)

∴ ∆ADE ≅ ∆CBF

(S.A.A. axiom of congruency)

∴ DE = BF (c.p.c.t.)

But CD = AB

(Opposite sides of the parallelogram)

∴ CD – DE = AB – BF

=> EC = AF

and AB || CD

∴ AFCE is a parallelogram

∴ AE || CF.

**Question 10.**

**Solution:**

Let ABCD is a rhombus AC and BD are its diagonals which bisect each other at right angles at O.

AC = 16cm and BD = 12cm

∴ AO = \(\\ \frac { 16 }{ 2 } \) = 8cm

BO = \(\\ \frac { 12 }{ 2 } \) = 6 cm

Now, in right ∆AOB

AB² = AO² + BO²

(Pythagorus Theorem)

= (8)² + (6)²

= 64 + 36 = 100 = (10)²

∴ AB = 10 cm

But all the sides of a rhombus are equal

∴ Each side will be 10 cm Ans.

**Question 11.**

**Solution:**

In square ABCD, AC is its diagonal

∴ Diagonals of a square bisect each angle at the vertex

∴ ∠ CAD = ∠ CAB

But ∠ DAB = 90° (Angle of a square)

∴ ∠ CAD = ∠ CAB = \(\\ \frac { 1 }{ 2 } \) ∠ DAB

= \(\\ \frac { 1 }{ 2 } \) x 90° = 45°

Hence ∠ CAD = 45° Ans.

**Question 12.**

**Solution:**

Let ABCD is a rectangle

AB : BC = 5 : 4

Let AB = 5x and BC = 4x.

But perimeter = 90cm

2(AB + BC) = 90

=> 2(5x + 4x) = 90

=> 2 x 9x = 90

=> 18x = 90

x = \(\\ \frac { 90 }{ 18 } \) = 5

∴ Length (AB) = 5x = 5 x 5 = 25 cm

Breadth (BC) = 4x = 4 x 5 = 20 cm Ans.

**Question 13.**

**Solution:**

(i) It is a rectangle

(ii) Square

(iii) rhombus

(iv) rhombus

(v) square

(vi) rectangle.

**Question 14.**

**Solution:**

(i) False

(ii) False

(iii) False

(iv) False

(v) False

(vi) True

(vii) True

(viii) True

(ix) False

(x) True

Hope given RS Aggarwal Solutions Class 8 Chapter 16 Parallelograms Ex 16A are helpful to complete your math homework.

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