Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral has accurate answers on fundamental concepts covered under this chapter. A quadrilateral is defined as a plane closed figure with four sides. Experts prepare the solutions with the intention to boost exam preparation of students. Those who find difficulty in solving questions can use these solutions, to speed up the problem solving abilities. For a better academic score, Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral, PDF links are presented below with a free download option.

Chapter 27 deals with the method of finding the angle of a given set of quadrilaterals. The important formula and the method of solving problems are given in a descriptive manner, for a better conceptual knowledge among students.

## Selina Solutions Concise Mathematics Class 6 Chapter 27: Quadrilateral Download PDF

### Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 27: Quadrilateral

## Access Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral

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 × 18^{0} = 90^{0}

7x = 7 × 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 – 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 – 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 – 20)^{0} + 6(x + 3)^{0} = 360^{0}

4x + 5x + 10^{0} + 7x – 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 × 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)° = (7×16 – 20)°

= 92°

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

**(i) x**

**(ii) ∠B and ∠C**

**Solution:**

Here, given that,

∠A = 90^{0}

∠B = (2x + 4)^{0}

∠C = (3x – 5)^{0}

∠D = (8x – 15)^{0}

We know that,

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

So,

∠A + ∠B + ∠C + ∠D = 360^{0}

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

On further calculation, we get

90^{0} + 2x + 4^{0} + 3x – 5^{0} + 8x – 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,

∠B = 2x + 4 = 2 × 22^{0} + 4

= 48^{0}

∠C = 3x – 5 = 3 × 22^{0} – 5

= 61^{0}

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

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

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

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

**Solution:**

Given

∠A: ∠D = 1: 2

Let us consider ∠A = x and ∠D = 2x

∠C: ∠B = 4: 5

Let us consider ∠C = 4y and ∠B = 5y

Also, given

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

So,

∠A + ∠D = 180^{0}

x + 2x = 180^{0}

3x = 180^{0}

We get,

x = 60^{0}

Therefore, ∠A = 60^{0}

∠D = 2x

= 2 × 60^{0}

= 120^{0}

Therefore, ∠D = 120^{0}

Now,

∠B + ∠C = 180^{0}

5y + 4y = 180^{0}

9y = 180^{0}

We get,

y = 20^{0}

Now,

∠B = 5y = 5 × 20^{0}

= 100^{0}

∠C = 4y = 4 × 20^{0}

= 80^{0}

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

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

**(i) x,**

**(ii) ∠ABC,**

**(iii) ∠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} – 48^{0}

12x = 312

We get,

x = 26^{0}

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

(ii) ∠ABC = 4x

4 × 26^{0} = 104^{0}

Therefore, ∠ABC = 104^{0}

(iii) ∠ACD = 180^{0} – 4x – 48^{0}

= 180^{0} – 4 × 26^{0} – 48^{0}

= 180^{0} – 104^{0} – 48^{0}

We get,

= 28^{0}

Therefore, ∠ACD = 28^{0}

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

**Solution:**

Given

∠C = 64^{0}

∠D = ∠C – 8^{0}

= 64^{0} – 8^{0}

We get,

∠D = 56^{0}

∠A = 5 (a + 2)^{0}

∠B = 2(2a + 7)^{0}

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

So,

∠A + ∠B + ∠C + ∠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}

∠A = 5(a + 2)

= 5(24 + 2)

We get,

= 130^{0}

**8. In the given figure**

**∠b = 2a + 15**

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

**Solution:**

∠b = 2a + 15 and

∠c = 3a + 5

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

70^{0} + ∠a + ∠b + ∠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, ∠a = 45^{0}

b = 2a + 15 = 2 × 45^{0} + 15

= 90^{0} + 15

= 105^{0}

c = 3a + 5 = 3 × 45^{0} + 5

= 135^{0} + 5

= 140^{0}

Therefore, ∠a = 45^{0}; ∠ b = 105^{0} and ∠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, ∠P: ∠Q : ∠R: ∠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

∠P: ∠Q: ∠R: ∠S = 3: 4: 6: 7

Let ∠P = 3x

∠Q = 4x

∠R = 6x and

∠S = 7x

Hence,

∠P + ∠Q + ∠R + ∠S = 360^{0}

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

20x = 360^{0}

x = 360^{0} / 20

We get,

x = 18^{0}

So,

∠P = 3x = 3 × 18^{0}

= 54^{0}

∠Q = 4x = 4 × 18^{0}

= 72^{0}

∠R = 6x = 6 × 18^{0}

= 108^{0}

∠S = 7x = 7 × 18^{0}

= 126^{0}

Now, adding two adjacent angles, we get

∠Q + ∠R = 72^{0} + 108^{0}

= 180^{0} and

∠P + ∠S = 54^{0} + 126^{0}

= 180^{0}

Therefore, PQ || RS

Since,

∠P + ∠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,

∠EAD = 70^{0}

Hence,

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

∠DAB = 110^{0}

Hence,

∠EAD + ∠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} – 110^{0} – 80^{0} – 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 ∠A: ∠D = 4: 5, ∠B = (3x – 15) ^{0} and ∠C = (4x + 20)^{0}, find each angle of the quadrilateral ABCD.**

**Solution:**

Let us consider ∠A = 4x and

∠D = 5x

Since, AB || DC

So,

∠A + ∠D = 180^{0}

Substituting the value of angle A and D, we get

4x + 5x = 180^{0}

9x = 180^{0}

x = 20^{0}

Now,

∠A = 4x = 4 × 20^{0}

= 80^{0}

∠D = 5x = 5 × 20^{0}

= 100^{0}

Similarly since, AB || DC

∠B + ∠C = 180^{0}

3x – 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}

∠B = 3x – 15^{0} = 3 × 25^{0} – 15^{0}

= 75^{0} – 15^{0}

= 60^{0} and

∠C = 4x + 20^{0} = 4 × 25^{0} + 20^{0}

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

= 120^{0}

Exercise 27(B)

**1. In a trapezium ABCD, side AB is parallel to side DC. If ∠A = 78 ^{0} and ∠C = 120^{0}, find angles B and D.**

**Solution:**

Given

AB || DC and BC is transversal

We know that,

The sum of co-interior angles of a parallelogram = 180^{0}

Hence,

∠B + ∠C = 180^{0}

∠B + 120^{0} = 180^{0}

∠B = 180^{0} – 120^{0}

We get,

∠B = 60^{0}

Also,

∠A + ∠D = 180^{0}

78^{0} + ∠D = 180^{0}

∠D = 180^{0} – 78^{0}

We get,

∠D = 102^{0}

Therefore, ∠B = 60^{0} and ∠D = 102^{0}

**2. In a trapezium ABCD, side AB is parallel to side DC. If ∠A = x ^{0} and ∠D = (3x – 20)^{0}; find the value of x.**

**Solution:**

Given

AB || DC and BC is transversal

The sum of co-interior angles of a parallelogram = 180^{0}

Hence,

∠A + ∠D = 180^{0}

x^{0} + (3x – 20)^{0} = 180^{0}

x^{0} + 3x – 20^{0} = 180^{0}

4x^{0} = 180^{0} + 20^{0}

4x^{0} = 200^{0}

x^{0} = 200^{0} / 4

We get,

x^{0} = 50^{0}

Hence, the value of x is 50^{0}

**3. The angles A, B, C and D of a trapezium ABCD are in the ratio 3: 4: 5: 6. Let ∠A: ∠B: ∠C: ∠D = 3: 4: 5: 6. Find all the angles of the trapezium. Also, name the two sides of this trapezium which are parallel to each other. Give reason for your answer**

**Solution:**

Let us consider the angles of a parallelogram ABCD be 3x, 4x, 5x and 6x

We know that,

The sum of angles of a parallelogram = 360^{0}

Hence,

∠A + ∠B + ∠C + ∠D = 360^{0}

3x + 4x + 5x + 6x = 360^{0}

18x = 360^{0}

x = 360^{0} / 18

We get,

x = 20^{0}

Now, the angles are,

∠A = 3x = 3 × 20^{0}

∠A = 60^{0}

∠B = 4x = 4 × 20^{0}

∠B = 80^{0}

∠C = 5x = 5 × 20^{0}

∠C = 100^{0}

∠D = 6x = 6 × 20^{0}

∠D = 120^{0}

Here,

The sum of ∠A and ∠D = 180^{0}

Therefore, AB is parallel to DC and the angles are co-interior angles whose sum = 180^{0}

**4. In a isosceles trapezium one pair of opposite sides are ………… to each other and the other pair of opposite sides are …………… to each other.**

**Solution:**

In an isosceles trapezium one pair of opposite sides are **parallel** to each other and the other pair of opposite sides are **equal** to each other.

**5. Two diagonals of an isosceles trapezium are x cm and (3x – 8) cm. Find the value of x.**

**Solution:**

We know that,

The diagonals of an isosceles trapezium are of equal length

Figure

Hence,

3x – 8 = x

3x – x = 8

2x = 8

x = 8 / 2

We get,

x = 4

Therefore, the value of x is 4 cm

**6. Angle A of an isosceles trapezium is 115 ^{0}; find the angles B, C and D.**

**Solution:**

Since, the base angles of an isosceles trapezium are equal

Hence,

∠A = ∠B = 115^{0}

Also,

∠A and ∠D are co-interior angles

The sum of co-interior angles of a quadrilateral is 180^{0}

So,

∠A + ∠D = 180^{0}

115^{0} + ∠D = 180^{0}

∠D = 180^{0} – 115^{0}

We get,

∠D = 65^{0}

Hence,

∠D = ∠C = 65^{0}

Therefore, the values of angles B, C and D are 115^{0}, 65^{0} and 65^{0}

**7. Two opposite angles of a parallelogram are 100 ^{0} each. Find each of the other two opposite angles.**

**Solution:**

Given

Two opposite angles of a parallelogram are 100^{0} each

The sum of adjacent angles of a parallelogram = 180^{0}

Hence,

∠A + ∠B = 180^{0}

100^{0} + ∠B = 180^{0}

∠B = 180^{0} – 100^{0}

We get,

∠B = 80^{0}

We know that,

The opposite angles of a parallelogram are equal

∠D = ∠B = 80^{0}

Therefore, the other two opposite angles ∠D = ∠B = 80^{0}

**8. Two adjacent angles of a parallelogram are 70 ^{0} and 110^{0} respectively. Find the other two angles of it.**

**Solution:**

Given

Two adjacent angles of a parallelogram are 70^{0} and 110^{0} respectively

We know that,

Opposite angles of a parallelogram are equal.

Hence, ∠C = ∠A = 70^{0} and ∠D = ∠B = 110^{0}

**9. The angles A, B, C and D of a quadrilateral are in the ratio 2: 3: 2: 3. Show this quadrilateral is a parallelogram.**

**Solution:**

Given

Angles of a quadrilateral are in the ratio 2: 3: 2: 3

Let us consider the angles A, B, C and D be 2x, 3x, 2x and 3x

We know that,

The sum of interior angles of a quadrilateral = 360^{0}

So,

∠A + ∠B + ∠C + ∠D = 360^{0}

2x + 3x + 2x + 3x = 360^{0}

10x = 360^{0}

x = 360^{0} / 10

We get,

x = 36^{0}

Hence, the measure of each angle is as follows

∠A = ∠C = 2x = 2 × 36^{0}

∠A = ∠C = 72^{0}

∠B = ∠D = 3x = 3 × 36^{0}

∠B = ∠D = 108^{0}

Since the opposite angles are equal and

The adjacent angles are supplementary

i.e ∠A + ∠B = 180^{0}

72^{0} + 108^{0} = 180^{0}

180^{0} = 180^{0} and

∠C + ∠D = 180^{0}

72^{0} + 108^{0} = 180^{0 }

180^{0}= 180^{0}

Quadrilateral ABCD fulfills the condition

Therefore, a quadrilateral ABCD is a parallelogram

**10. In a parallelogram ABCD, its diagonals AC and BD intersect each other at point O.**

**If AC = 12 cm and BD = 9 cm; find; lengths of OA and OB**

**Solution:**

Given

AC and BD intersect each other at point O

So,

OA = OC = (1 / 2) AC and

Similarly,

OB = OD = (1 / 2) BD

Hence,

OA = (1 / 2) × AC

= (1 / 2) × 12

= 6 cm

OB = (1 / 2) × BD

= (1 / 2) × 9

= 4. 5 cm

**11. In a parallelogram ABCD, its diagonals intersect at point O. If OA = 6 cm and OB = 7.5 cm, find the lengths of AC and BD.**

**Solution:**

The diagonals AC and BD intersect each other at point O

So,

OA = OC = (1 / 2) AC and

OB = OD = (1 / 2) BD

So,

OA = (1 / 2) × AC

AC = 2 × OA

AC = 2 × 6

We get,

AC = 12 cm and

OB = (1 / 2) × BD

BD = 2 × OB

BD = 2 × 7.5

We get,

BD = 15 cm

**12. In a parallelogram ABCD, ∠A = 90 ^{0}**

**(i) What is the measure of angle B.**

**(ii) Write the special name of the parallelogram.**

**Solution:**

Given

In a parallelogram ABCD, ∠A = 90^{0}

(i) We know that,

In a parallelogram, adjacent angles are supplementary

Hence,

∠A + ∠B = 180^{0}

90^{0} + ∠B = 180^{0}

∠B = 180^{0} – 90^{0}

We get,

∠B = 90^{0}

Therefore, the measure of ∠B = 90^{0}

(ii) Since all the angles of a given parallelogram is right angle.

Hence the given parallelogram is a rectangle

**13. One diagonal of a rectangle is 18 cm. What is the length of its other diagonal?**

**Solution:**

We know that,

In a rectangle, the diagonal are equal

Hence,

AC = BD

Given that one diagonal of a rectangle is 18 cm

Hence, the other diagonal of a rectangle will be = 18 cm

Therefore, the length of the other diagonal is 18 cm

**14. Each angle of a quadrilateral is x + 5 ^{0}. Find:**

**(i) the value of x**

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

**(iii) Give the special name of the quadrilateral taken.**

**Solution:**

(i) We know that,

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

Hence,

∠A + ∠B + ∠C + ∠D = 360^{0}

x + 5^{0} + x + 5^{0} + x + 5^{0} + x + 5^{0} = 360^{0}

4x + 20^{0} = 360^{0}

4x = 360^{0} – 20^{0}

4x = 340^{0}

x = 340^{0} / 4

We get,

x = 85^{0}

Hence, the value of x is 85^{0}

(ii) Each angle of the quadrilateral ABCD = x + 5^{0}

= 85^{0} + 5^{0}

We get,

= 90^{0}

Therefore, each angle of the quadrilateral = 90^{0}

(iii) The name of the taken quadrilateral is a rectangle

**15. If three angles of a quadrilateral are 90 ^{0} each, show that the given quadrilateral is a rectangle.**

**Solution:**

If each angle of quadrilateral is 90^{0}, then the given quadrilateral will be a rectangle

We know that,

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

Hence,

∠A + ∠B + ∠C + ∠D = 360^{0}

90^{0} + 90^{0} + 90^{0} + ∠D = 360^{0}

270^{0} + ∠D = 360^{0}

∠D = 360^{0} – 270^{0}

We get,

∠D = 90^{0}

Since,

Each angle of the quadrilateral = 90^{0}

Therefore, the given quadrilateral is a rectangle.