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### Access Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral Exercise 27(B)

Exercise 27(B)

**1. In a trapezium ABCD, side AB is parallel to side DC. If âˆ A = 78 ^{0} and âˆ 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,

âˆ B + âˆ C = 180^{0}

âˆ B + 120^{0} = 180^{0}

âˆ B = 180^{0} – 120^{0}

We get,

âˆ B = 60^{0}

Also,

âˆ A + âˆ D = 180^{0}

78^{0} + âˆ D = 180^{0}

âˆ D = 180^{0} – 78^{0}

We get,

âˆ D = 102^{0}

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

**2. In a trapezium ABCD, side AB is parallel to side DC. If âˆ A = x ^{0} and âˆ D = (3x â€“ 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,

âˆ A + âˆ D = 180^{0}

x^{0} + (3x â€“ 20)^{0} = 180^{0}

x^{0} + 3x â€“ 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 âˆ A: âˆ B: âˆ C: âˆ 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,

âˆ A + âˆ B + âˆ C + âˆ 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,

âˆ A = 3x = 3 Ã— 20^{0}

âˆ A = 60^{0}

âˆ B = 4x = 4 Ã— 20^{0}

âˆ B = 80^{0}

âˆ C = 5x = 5 Ã— 20^{0}

âˆ C = 100^{0}

âˆ D = 6x = 6 Ã— 20^{0}

âˆ D = 120^{0}

Here,

The sum of âˆ A and âˆ 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 â€¦â€¦â€¦â€¦ to each other and the other pair of opposite sides are â€¦â€¦â€¦â€¦â€¦ 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 â€“ 8) cm. Find the value of x.**

**Solution:**

We know that,

The diagonals of an isosceles trapezium are of equal length

Figure

Hence,

3x â€“ 8 = x

3x â€“ 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,

âˆ A = âˆ B = 115^{0}

Also,

âˆ A and âˆ D are co-interior angles

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

So,

âˆ A + âˆ D = 180^{0}

115^{0} + âˆ D = 180^{0}

âˆ D = 180^{0} – 115^{0}

We get,

âˆ D = 65^{0}

Hence,

âˆ D = âˆ 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,

âˆ A + âˆ B = 180^{0}

100^{0} + âˆ B = 180^{0}

âˆ B = 180^{0} – 100^{0}

We get,

âˆ B = 80^{0}

We know that,

The opposite angles of a parallelogram are equal

âˆ D = âˆ B = 80^{0}

Therefore, the other two opposite angles âˆ D = âˆ 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, âˆ C = âˆ A = 70^{0} and âˆ D = âˆ 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,

âˆ A + âˆ B + âˆ C + âˆ 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

âˆ A = âˆ C = 2x = 2 Ã— 36^{0}

âˆ A = âˆ C = 72^{0}

âˆ B = âˆ D = 3x = 3 Ã— 36^{0}

âˆ B = âˆ D = 108^{0}

Since the opposite angles are equal and

The adjacent angles are supplementary

i.e âˆ A + âˆ B = 180^{0}

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

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

âˆ C + âˆ 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) Ã— AC

= (1 / 2) Ã— 12

= 6 cm

OB = (1 / 2) Ã— BD

= (1 / 2) Ã— 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) Ã— AC

AC = 2 Ã— OA

AC = 2 Ã— 6

We get,

AC = 12 cm and

OB = (1 / 2) Ã— BD

BD = 2 Ã— OB

BD = 2 Ã— 7.5

We get,

BD = 15 cm

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

(i) We know that,

In a parallelogram, adjacent angles are supplementary

Hence,

âˆ A + âˆ B = 180^{0}

90^{0} + âˆ B = 180^{0}

âˆ B = 180^{0} – 90^{0}

We get,

âˆ B = 90^{0}

Therefore, the measure of âˆ 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,

âˆ A + âˆ B + âˆ C + âˆ 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,

âˆ A + âˆ B + âˆ C + âˆ D = 360^{0}

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

270^{0} + âˆ D = 360^{0}

âˆ D = 360^{0} – 270^{0}

We get,

âˆ D = 90^{0}

Since,

Each angle of the quadrilateral = 90^{0}

Therefore, the given quadrilateral is a rectangle.