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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 ∠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.
