Selina Solutions Concise Mathematics Class 6 Chapter 27: Quadrilateral

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

Exercise 27(A) Solutions

Exercise 27(B) Solutions

Access Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral

Exercise 27(A)

1. Two angles of a quadrilateral are 89⁰ and 113⁰. If the other two angles are equal; find the equal angles.

Solution:

Let us consider the other angle as x⁰

As per the question, we have

89⁰ + 113⁰ + x⁰ + x⁰ = 360⁰

2x⁰ = 360⁰ – 202⁰

2x⁰ = 158

x⁰ = 158 / 2

We get,

x = 79⁰

Therefore, the other two equal angles are 79⁰ each.

2. Two angles of a quadrilateral are 68⁰ and 76⁰. If the other two angles are in the ratio 5: 7; find the measure of each of them.

Solution:

Given

Two angles are 68⁰ and 76⁰

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

Hence,

68⁰ + 76⁰ + 5x + 7x = 360⁰

12x + 144⁰ = 360⁰

12x = 360⁰ – 144⁰

12x = 216⁰

x = 216⁰ / 12

We get,

x = 18⁰

Now, the other angles is calculated as below

5x = 5 × 18⁰ = 90⁰

7x = 7 × 18⁰ = 126⁰

Therefore, the values of the other angles are 90⁰ and 126⁰

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 360⁰

Hence,

(4x)⁰ + 5(x + 2)⁰ + (7x – 20)⁰ + 6(x + 3)⁰ = 360⁰

4x + 5x + 10⁰ + 7x – 20⁰ + 6x + 18⁰ = 360⁰

22x + 8⁰ = 360⁰

22x = 360⁰ – 8⁰

22x = 352⁰

x = 352⁰ / 22

We get,

x = 16⁰

Hence, the value of x is 16⁰

Therefore, the angles are,

(4x)⁰ = (4 × 16)⁰

= 64⁰

5(x + 2)⁰ = 5(16 + 2)⁰

= 90⁰

6(x + 3)⁰ = 6(16 + 3)⁰

= 114⁰

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 = 90⁰

∠B = (2x + 4)⁰

∠C = (3x – 5)⁰

∠D = (8x – 15)⁰

We know that,

All the angles in a quadrilateral is 360⁰

So,

∠A + ∠B + ∠C + ∠D = 360⁰

90⁰ + (2x + 4)⁰ + (3x – 5)⁰ + (8x – 15)⁰ = 360⁰

On further calculation, we get

90⁰ + 2x + 4⁰ + 3x – 5⁰ + 8x – 15⁰ = 360⁰

74⁰ + 13x = 360⁰

13x = 360⁰ – 74⁰

13x = 286⁰

x = 286⁰ / 13

We get,

x = 22⁰

The value of x is 22⁰

Now,

∠B = 2x + 4 = 2 × 22⁰ + 4

= 48⁰

∠C = 3x – 5 = 3 × 22⁰ – 5

= 61⁰

Therefore, ∠B = 48⁰ and ∠C = 61⁰

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 180⁰

So,

∠A + ∠D = 180⁰

x + 2x = 180⁰

3x = 180⁰

We get,

x = 60⁰

Therefore, ∠A = 60⁰

∠D = 2x

= 2 × 60⁰

= 120⁰

Therefore, ∠D = 120⁰

Now,

∠B + ∠C = 180⁰

5y + 4y = 180⁰

9y = 180⁰

We get,

y = 20⁰

Now,

∠B = 5y = 5 × 20⁰

= 100⁰

∠C = 4y = 4 × 20⁰

= 80⁰

Therefore, ∠A = 60⁰; ∠B = 100⁰; ∠C = 80⁰ and ∠D = 120⁰

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 360⁰

Hence,

x + 4x + 3x + 4x + 48⁰ = 360⁰

12x = 360⁰ – 48⁰

12x = 312

We get,

x = 26⁰

Hence, the value of x is 26⁰

(ii) ∠ABC = 4x

4 × 26⁰ = 104⁰

Therefore, ∠ABC = 104⁰

(iii) ∠ACD = 180⁰ – 4x – 48⁰

= 180⁰ – 4 × 26⁰ – 48⁰

= 180⁰ – 104⁰ – 48⁰

We get,

= 28⁰

Therefore, ∠ACD = 28⁰

7. Given: In quadrilateral ABCD; ∠C = 64⁰, ∠D = ∠C – 8⁰; ∠A = 5(a + 2)⁰ and ∠B = 2(2a + 7)⁰.

Solution:

Given

∠C = 64⁰

∠D = ∠C – 8⁰

= 64⁰ – 8⁰

We get,

∠D = 56⁰

∠A = 5 (a + 2)⁰

∠B = 2(2a + 7)⁰

We know that, sum of all the angles in a quadrilateral = 360⁰

So,

∠A + ∠B + ∠C + ∠D = 360⁰

5(a + 2)⁰ + 2(2a + 7)⁰ + 64⁰ + 56⁰ = 360⁰

On further calculation, we get

5a + 10⁰ + 4a + 14⁰ + 64⁰ + 56⁰ = 360⁰

9a + 144⁰ = 360⁰

9a = 360⁰ – 144⁰

9a = 216⁰

We get,

a = 24⁰

∠A = 5(a + 2)

= 5(24 + 2)

We get,

= 130⁰

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 = 360⁰

70⁰ + ∠a + ∠b + ∠c = 360⁰

70⁰ + a + (2a + 15) + (3a + 5) = 360⁰

70⁰ + a + 2a + 15 + 3a + 5 = 360⁰

6a + 90⁰ = 360⁰

6a = 360⁰ – 90⁰

6a = 270⁰

We get,

a = 45⁰

Hence, ∠a = 45⁰

b = 2a + 15 = 2 × 45⁰ + 15

= 90⁰ + 15

= 105⁰

c = 3a + 5 = 3 × 45⁰ + 5

= 135⁰ + 5

= 140⁰

Therefore, ∠a = 45⁰; ∠ b = 105⁰ and ∠c = 140⁰

9. Three angles of a quadrilateral are equal. If the fourth angle is 69⁰; find the measure of equal angles.

Solution:

Given that,

Three angles of a quadrilateral are equal

Let us consider each angle as x⁰

Hence,

x⁰ + x⁰ + x⁰ + 69⁰ = 360⁰

3x = 360⁰ – 69⁰

3x = 291⁰

x = 291⁰ / 3

We get,

x = 97⁰

Therefore, the measure of all the equal angles is 97⁰

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 = 360⁰

3x + 4x + 6x + 7x = 360⁰

20x = 360⁰

x = 360⁰ / 20

We get,

x = 18⁰

So,

∠P = 3x = 3 × 18⁰

= 54⁰

∠Q = 4x = 4 × 18⁰

= 72⁰

∠R = 6x = 6 × 18⁰

= 108⁰

∠S = 7x = 7 × 18⁰

= 126⁰

Now, adding two adjacent angles, we get

∠Q + ∠R = 72⁰ + 108⁰

= 180⁰ and

∠P + ∠S = 54⁰ + 126⁰

= 180⁰

Therefore, PQ || RS

Since,

∠P + ∠Q = 54⁰ + 72⁰

= 126⁰

Which is not equal to180⁰

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 = 70⁰

Hence,

∠DAB = 180⁰ – 70⁰ [By straight line]

∠DAB = 110⁰

Hence,

∠EAD + ∠DAB = 180⁰

The sum of angles of a quadrilateral is 360⁰

110⁰ + 80⁰ + 56⁰ + 3x – 6⁰ = 360⁰

3x = 360⁰ – 110⁰ – 80⁰ – 56⁰ + 6⁰

3x = 360⁰– 240⁰

3x = 120⁰

x = 120⁰ / 3

We get,

x = 40⁰

Therefore, the value of x is 40⁰

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 = 180⁰

Substituting the value of angle A and D, we get

4x + 5x = 180⁰

9x = 180⁰

x = 20⁰

Now,

∠A = 4x = 4 × 20⁰

= 80⁰

∠D = 5x = 5 × 20⁰

= 100⁰

Similarly since, AB || DC

∠B + ∠C = 180⁰

3x – 15⁰ + 4x + 20⁰ = 180⁰

7x + 5⁰ = 180⁰

7x = 180⁰ – 5⁰

7x = 175⁰

We get,

x = 25⁰

∠B = 3x – 15⁰ = 3 × 25⁰ – 15⁰

= 75⁰ – 15⁰

= 60⁰ and

∠C = 4x + 20⁰ = 4 × 25⁰ + 20⁰

= 100⁰ + 20⁰

= 120⁰

Exercise 27(B)

1. In a trapezium ABCD, side AB is parallel to side DC. If ∠A = 78⁰ and ∠C = 120⁰, 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 = 180⁰

Hence,

∠B + ∠C = 180⁰

∠B + 120⁰ = 180⁰

∠B = 180⁰ – 120⁰

We get,

∠B = 60⁰

Also,

∠A + ∠D = 180⁰

78⁰ + ∠D = 180⁰

∠D = 180⁰ – 78⁰

We get,

∠D = 102⁰

Therefore, ∠B = 60⁰ and ∠D = 102⁰

2. In a trapezium ABCD, side AB is parallel to side DC. If ∠A = x⁰ 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 = 180⁰

Hence,

∠A + ∠D = 180⁰

x⁰ + (3x – 20)⁰ = 180⁰

x⁰ + 3x – 20⁰ = 180⁰

4x⁰ = 180⁰ + 20⁰

4x⁰ = 200⁰

x⁰ = 200⁰ / 4

We get,

x⁰ = 50⁰

Hence, the value of x is 50⁰

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 = 360⁰

Hence,

∠A + ∠B + ∠C + ∠D = 360⁰

3x + 4x + 5x + 6x = 360⁰

18x = 360⁰

x = 360⁰ / 18

We get,

x = 20⁰

Now, the angles are,

∠A = 3x = 3 × 20⁰

∠A = 60⁰

∠B = 4x = 4 × 20⁰

∠B = 80⁰

∠C = 5x = 5 × 20⁰

∠C = 100⁰

∠D = 6x = 6 × 20⁰

∠D = 120⁰

Here,

The sum of ∠A and ∠D = 180⁰

Therefore, AB is parallel to DC and the angles are co-interior angles whose sum = 180⁰

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 115⁰; find the angles B, C and D.

Solution:

Since, the base angles of an isosceles trapezium are equal

Hence,

∠A = ∠B = 115⁰

Also,

∠A and ∠D are co-interior angles

The sum of co-interior angles of a quadrilateral is 180⁰

So,

∠A + ∠D = 180⁰

115⁰ + ∠D = 180⁰

∠D = 180⁰ – 115⁰

We get,

∠D = 65⁰

Hence,

∠D = ∠C = 65⁰

Therefore, the values of angles B, C and D are 115⁰, 65⁰ and 65⁰

7. Two opposite angles of a parallelogram are 100⁰ each. Find each of the other two opposite angles.

Solution:

Given

Two opposite angles of a parallelogram are 100⁰ each

The sum of adjacent angles of a parallelogram = 180⁰

Hence,

∠A + ∠B = 180⁰

100⁰ + ∠B = 180⁰

∠B = 180⁰ – 100⁰

We get,

∠B = 80⁰

We know that,

The opposite angles of a parallelogram are equal

∠D = ∠B = 80⁰

Therefore, the other two opposite angles ∠D = ∠B = 80⁰

8. Two adjacent angles of a parallelogram are 70⁰ and 110⁰ respectively. Find the other two angles of it.

Solution:

Given

Two adjacent angles of a parallelogram are 70⁰ and 110⁰ respectively

We know that,

Opposite angles of a parallelogram are equal.

Hence, ∠C = ∠A = 70⁰ and ∠D = ∠B = 110⁰

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 = 360⁰

So,

∠A + ∠B + ∠C + ∠D = 360⁰

2x + 3x + 2x + 3x = 360⁰

10x = 360⁰

x = 360⁰ / 10

We get,

x = 36⁰

Hence, the measure of each angle is as follows

∠A = ∠C = 2x = 2 × 36⁰

∠A = ∠C = 72⁰

∠B = ∠D = 3x = 3 × 36⁰

∠B = ∠D = 108⁰

Since the opposite angles are equal and

The adjacent angles are supplementary

i.e ∠A + ∠B = 180⁰

72⁰ + 108⁰ = 180⁰

180⁰ = 180⁰ and

∠C + ∠D = 180⁰

72⁰ + 108⁰ = 180⁰

180⁰= 180⁰

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 = 90⁰

(i) What is the measure of angle B.

(ii) Write the special name of the parallelogram.

Solution:

Given

In a parallelogram ABCD, ∠A = 90⁰

(i) We know that,

In a parallelogram, adjacent angles are supplementary

Hence,

∠A + ∠B = 180⁰

90⁰ + ∠B = 180⁰

∠B = 180⁰ – 90⁰

We get,

∠B = 90⁰

Therefore, the measure of ∠B = 90⁰

(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 + 5⁰. 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 360⁰

Hence,

∠A + ∠B + ∠C + ∠D = 360⁰

x + 5⁰ + x + 5⁰ + x + 5⁰ + x + 5⁰ = 360⁰

4x + 20⁰ = 360⁰

4x = 360⁰ – 20⁰

4x = 340⁰

x = 340⁰ / 4

We get,

x = 85⁰

Hence, the value of x is 85⁰

(ii) Each angle of the quadrilateral ABCD = x + 5⁰

= 85⁰ + 5⁰

We get,

= 90⁰

Therefore, each angle of the quadrilateral = 90⁰

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

15. If three angles of a quadrilateral are 90⁰ each, show that the given quadrilateral is a rectangle.

Solution:

If each angle of quadrilateral is 90⁰, then the given quadrilateral will be a rectangle

We know that,

The sum of interior angles of a quadrilateral is 360⁰

Hence,

∠A + ∠B + ∠C + ∠D = 360⁰

90⁰ + 90⁰ + 90⁰ + ∠D = 360⁰

270⁰ + ∠D = 360⁰

∠D = 360⁰ – 270⁰

We get,

∠D = 90⁰

Since,

Each angle of the quadrilateral = 90⁰

Therefore, the given quadrilateral is a rectangle.