Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral Exercise 27(A) are available in PDF format to boost the exam preparation of students. The method of finding the angles of a quadrilateral is the main concept talked about under this exercise. Multiple solved examples are present to help students grasp the various tricks of solving complex problems with ease. Students can find Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral Exercise 27(A) PDF, from the links provided here.

## Selina Solutions Concise Mathematics Class 6 Chapter 27: Quadrilateral Exercise 27(A) Download PDF

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Exercise 27(A)

**1. Two angles of a quadrilateral are 89 ^{0} and 113^{0}. If the other two angles are equal; find the equal angles.**

**Solution:**

Let us consider the other angle as x^{0}

As per the question, we have

89^{0} + 113^{0} + x^{0} + x^{0} = 360^{0}

2x^{0} = 360^{0} – 202^{0}

2x^{0} = 158

x^{0} = 158 / 2

We get,

x = 79^{0}

Therefore, the other two equal angles are 79^{0} each.

**2. Two angles of a quadrilateral are 68 ^{0} and 76^{0}. If the other two angles are in the ratio 5: 7; find the measure of each of them.**

**Solution:**

Given

Two angles are 68^{0} and 76^{0}

Let us consider the other two angles as 5x and 7x

Hence,

68^{0} + 76^{0} + 5x + 7x = 360^{0}

12x + 144^{0} = 360^{0}

12x = 360^{0} – 144^{0}

12x = 216^{0}

x = 216^{0} / 12

We get,

x = 18^{0}

Now, the other angles is calculated as below

5x = 5 Ã— 18^{0} = 90^{0}

7x = 7 Ã— 18^{0} = 126^{0}

Therefore, the values of the other angles are 90^{0} and 126^{0}

**3. Angles of a quadrilateral are (4x) ^{0}, 5(x+2)^{0}, (7x â€“ 20)^{0} and 6(x + 3)^{0}. Find**

**(i) the value of x.**

**(ii) each angle of the quadrilateral.**

**Solution:**

Given

The angles of quadrilateral are,

(4x)^{0}, 5(x + 2)^{0}, (7x â€“ 20)^{0} and 6(x + 3)^{0}

We know that the sum of angles in a quadrilateral is 360^{0}

Hence,

(4x)^{0} + 5(x + 2)^{0} + (7x â€“ 20)^{0} + 6(x + 3)^{0} = 360^{0}

4x + 5x + 10^{0} + 7x â€“ 20^{0} + 6x + 18^{0} = 360^{0}

22x + 8^{0} = 360^{0}

22x = 360^{0} – 8^{0}

22x = 352^{0}

x = 352^{0} / 22

We get,

x = 16^{0}

Hence, the value of x is 16^{0}

Therefore, the angles are,

(4x)^{0} = (4 Ã— 16)^{0}

= 64^{0}

5(x + 2)^{0} = 5(16 + 2)^{0}

= 90^{0}

6(x + 3)^{0} = 6(16 + 3)^{0}

= 114^{0}

And,

(7x – 20)Â° = (7Ã—16 – 20)Â°

= 92Â°

**4. Use the information given in the following figure to find:**

**(i) x**

**(ii) âˆ B and âˆ C**

**Solution:**

Here, given that,

âˆ A = 90^{0}

âˆ B = (2x + 4)^{0}

âˆ C = (3x â€“ 5)^{0}

âˆ D = (8x â€“ 15)^{0}

We know that,

All the angles in a quadrilateral is 360^{0}

So,

âˆ A + âˆ B + âˆ C + âˆ D = 360^{0}

90^{0} + (2x + 4)^{0} + (3x â€“ 5)^{0} + (8x â€“ 15)^{0} = 360^{0}

On further calculation, we get

90^{0} + 2x + 4^{0} + 3x â€“ 5^{0} + 8x â€“ 15^{0} = 360^{0}

74^{0} + 13x = 360^{0}

13x = 360^{0} – 74^{0}

13x = 286^{0}

x = 286^{0} / 13

We get,

x = 22^{0}

The value of x is 22^{0}

Now,

âˆ B = 2x + 4 = 2 Ã— 22^{0} + 4

= 48^{0}

âˆ C = 3x â€“ 5 = 3 Ã— 22^{0} â€“ 5

= 61^{0}

Therefore, âˆ B = 48^{0} and âˆ C = 61^{0}

**5. In quadrilateral ABCD, side AB is parallel to side DC. If âˆ A: âˆ D = 1: 2 and âˆ C: âˆ B = 4: 5**

**(i) Calculate each angle of the quadrilateral.**

**(ii) Assign a special name to quadrilateral ABCD.**

**Solution:**

Given

âˆ A: âˆ D = 1: 2

Let us consider âˆ A = x and âˆ D = 2x

âˆ C: âˆ B = 4: 5

Let us consider âˆ C = 4y and âˆ B = 5y

Also, given

AB || DC and the sum of opposite angles of quadrilateral is 180^{0}

So,

âˆ A + âˆ D = 180^{0}

x + 2x = 180^{0}

3x = 180^{0}

We get,

x = 60^{0}

Therefore, âˆ A = 60^{0}

âˆ D = 2x

= 2 Ã— 60^{0}

= 120^{0}

Therefore, âˆ D = 120^{0}

Now,

âˆ B + âˆ C = 180^{0}

5y + 4y = 180^{0}

9y = 180^{0}

We get,

y = 20^{0}

Now,

âˆ B = 5y = 5 Ã— 20^{0}

= 100^{0}

âˆ C = 4y = 4 Ã— 20^{0}

= 80^{0}

Therefore, âˆ A = 60^{0}; âˆ B = 100^{0}; âˆ C = 80^{0} and âˆ D = 120^{0}

**6. From the following figure find:**

**(i) x,**

**(ii) âˆ ABC,**

**(iii) âˆ ACD.**

**Solution:**

(i)We know that,

In quadrilateral the sum of angles is equal to 360^{0}

Hence,

x + 4x + 3x + 4x + 48^{0} = 360^{0}

12x = 360^{0} â€“ 48^{0}

12x = 312

We get,

x = 26^{0}

Hence, the value of x is 26^{0}

(ii) âˆ ABC = 4x

4 Ã— 26^{0} = 104^{0}

Therefore, âˆ ABC = 104^{0}

(iii) âˆ ACD = 180^{0} â€“ 4x – 48^{0}

= 180^{0} â€“ 4 Ã— 26^{0} – 48^{0}

= 180^{0} â€“ 104^{0} – 48^{0}

We get,

= 28^{0}

Therefore, âˆ ACD = 28^{0}

**7. Given: In quadrilateral ABCD; âˆ C = 64 ^{0}, âˆ D = âˆ C â€“ 8^{0}; âˆ A = 5(a + 2)^{0} and âˆ B = 2(2a + 7)^{0}.**

**Solution:**

Given

âˆ C = 64^{0}

âˆ D = âˆ C – 8^{0}

= 64^{0} – 8^{0}

We get,

âˆ D = 56^{0}

âˆ A = 5 (a + 2)^{0}

âˆ B = 2(2a + 7)^{0}

We know that, sum of all the angles in a quadrilateral = 360^{0}

So,

âˆ A + âˆ B + âˆ C + âˆ D = 360^{0}

5(a + 2)^{0} + 2(2a + 7)^{0} + 64^{0} + 56^{0} = 360^{0}

On further calculation, we get

5a + 10^{0} + 4a + 14^{0} + 64^{0} + 56^{0} = 360^{0}

9a + 144^{0} = 360^{0}

9a = 360^{0} – 144^{0}

9a = 216^{0}

We get,

a = 24^{0}

âˆ A = 5(a + 2)

= 5(24 + 2)

We get,

= 130^{0}

**8. In the given figure**

**âˆ b = 2a + 15**

**And âˆ c = 3a + 5; find the values of b and c**

**Solution:**

âˆ b = 2a + 15 and

âˆ c = 3a + 5

Sum of angles of a quadrilateral = 360^{0}

70^{0} + âˆ a + âˆ b + âˆ c = 360^{0}

70^{0} + a + (2a + 15) + (3a + 5) = 360^{0}

70^{0} + a + 2a + 15 + 3a + 5 = 360^{0}

6a + 90^{0} = 360^{0}

6a = 360^{0} – 90^{0}

6a = 270^{0}

We get,

a = 45^{0}

Hence, âˆ a = 45^{0}

b = 2a + 15 = 2 Ã— 45^{0} + 15

= 90^{0} + 15

= 105^{0}

c = 3a + 5 = 3 Ã— 45^{0} + 5

= 135^{0} + 5

= 140^{0}

Therefore, âˆ a = 45^{0}; âˆ b = 105^{0} and âˆ c = 140^{0}

**9. Three angles of a quadrilateral are equal. If the fourth angle is 69 ^{0}; find the measure of equal angles.**

**Solution:**

Given that,

Three angles of a quadrilateral are equal

Let us consider each angle as x^{0}

Hence,

x^{0} + x^{0} + x^{0} + 69^{0} = 360^{0}

3x = 360^{0} – 69^{0}

3x = 291^{0}

x = 291^{0} / 3

We get,

x = 97^{0}

Therefore, the measure of all the equal angles is 97^{0}

**10. In quadrilateral PQRS, âˆ P: âˆ Q : âˆ R: âˆ S = 3: 4: 6: 7.**

**Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other. Is PS also parallel to QR?**

**Solution:**

Given

âˆ P: âˆ Q: âˆ R: âˆ S = 3: 4: 6: 7

Let âˆ P = 3x

âˆ Q = 4x

âˆ R = 6x and

âˆ S = 7x

Hence,

âˆ P + âˆ Q + âˆ R + âˆ S = 360^{0}

3x + 4x + 6x + 7x = 360^{0}

20x = 360^{0}

x = 360^{0} / 20

We get,

x = 18^{0}

So,

âˆ P = 3x = 3 Ã— 18^{0}

= 54^{0}

âˆ Q = 4x = 4 Ã— 18^{0}

= 72^{0}

âˆ R = 6x = 6 Ã— 18^{0}

= 108^{0}

âˆ S = 7x = 7 Ã— 18^{0}

= 126^{0}

Now, adding two adjacent angles, we get

âˆ Q + âˆ R = 72^{0} + 108^{0}

= 180^{0} and

âˆ P + âˆ S = 54^{0} + 126^{0}

= 180^{0}

Therefore, PQ || RS

Since,

âˆ P + âˆ Q = 54^{0} + 72^{0}

= 126^{0}

Which is not equal to180^{0}

Therefore, PS and QR are not parallel

**11. Use the information given in the following figure to find the value of x.**

**Solution:**

Given

A, B, C and D are the vertices of quadrilateral and BA is produced to E

Here,

âˆ EAD = 70^{0}

Hence,

âˆ DAB = 180^{0} – 70^{0} [By straight line]

âˆ DAB = 110^{0}

Hence,

âˆ EAD + âˆ DAB = 180^{0}

The sum of angles of a quadrilateral is 360^{0}

110^{0} + 80^{0} + 56^{0} + 3x – 6^{0} = 360^{0}

3x = 360^{0} â€“ 110^{0} â€“ 80^{0} â€“ 56^{0} + 6^{0}

3x = 360^{0}– 240^{0}

3x = 120^{0}

x = 120^{0} / 3

We get,

x = 40^{0}

Therefore, the value of x is 40^{0}

**12. The following figure shows a quadrilateral in which sides AB and DC are parallel. If âˆ A: âˆ D = 4: 5, âˆ B = (3x â€“ 15) ^{0} and âˆ C = (4x + 20)^{0}, find each angle of the quadrilateral ABCD.**

**Solution:**

Let us consider âˆ A = 4x and

âˆ D = 5x

Since, AB || DC

So,

âˆ A + âˆ D = 180^{0}

Substituting the value of angle A and D, we get

4x + 5x = 180^{0}

9x = 180^{0}

x = 20^{0}

Now,

âˆ A = 4x = 4 Ã— 20^{0}

= 80^{0}

âˆ D = 5x = 5 Ã— 20^{0}

= 100^{0}

Similarly since, AB || DC

âˆ B + âˆ C = 180^{0}

3x â€“ 15^{0} + 4x + 20^{0} = 180^{0}

7x + 5^{0} = 180^{0}

7x = 180^{0} – 5^{0}

7x = 175^{0}

We get,

x = 25^{0}

âˆ B = 3x â€“ 15^{0} = 3 Ã— 25^{0} – 15^{0}

= 75^{0} â€“ 15^{0}

= 60^{0} and

âˆ C = 4x + 20^{0} = 4 Ã— 25^{0} + 20^{0}

= 100^{0} + 20^{0}

= 120^{0}