Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons Exercise 28(A) has in-depth answers for exercise problems are explained under this exercise. The solutions are prepared with the aim of improving conceptual knowledge among students. Formulas and shortcut tricks are provided to help students obtain a better academic performance in the exam. While revising the textbook problems, students can cross check their answers and also clear doubts by using solutions. For enhancing problem solving abilities, students can refer to Selina Solutions Concise Mathematics Class 6 Chapter 28 Polygons Exercise 28(A) PDF, free from the links available below.
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Exercise 28(A)
1. State, which of the following are polygons:
Solution:
(i) The given figure is not closed.
Hence, the figure is not a polygon.
(ii) The given figure is closed.
Hence, the figure is a polygon.
(iii) The given figure is closed.
Hence, the figure is a polygon.
(iv) In the given figure, one of the sides is an arc.
Hence, the figure is not polygon.
(v) The side intersects each other in the given figure.
Hence, the figure is not polygon.
2. Find the sum of interior angles of a polygon with:
(i) 9 sides
(ii) 13 sides
(iii) 16 sides
Solution:
(i) 9 sides
Number of sides n = 9
The Sum of interior angles of polygon = (2n – 4) × 900
= (2 × 9 – 4) × 900
= (18 – 4) × 900
= 14 × 900
We get,
= 12600
(ii) 13 sides
Number of sides n = 13
The sum of interior angles of polygon = (2n -4) × 900
= (2 × 13 – 4) × 900
= (26 – 4) × 900
= 22 × 900
We get,
= 19800
(iii) 16 sides
Number of sides n = 16
The sum of interior angles of polygon = (2n – 4) × 900
= (2 × 16 – 4) × 900
= (32 – 4) × 900
= 28 × 900
We get,
= 25200
3. Find the number of sides of a polygon, if the sum of its interior angles is:
(i) 14400
(ii) 16200
Solution:
(i) 14400
The sum of interior angles of polygon = 14400
Let the number of sides = n
The sum of interior angle of polygon is (2n – 4) × 900
The side of polygon can be calculated as,
(2n – 4) × 900 = 14400
2n – 4 = 14400 / 900
2(n – 2) = 14400 / 900
n – 2 = 14400 / (2 × 900)
On further calculation, we get
n – 2 = 160 / 2
We get,
n – 2 = 8
n = 8 + 2
n = 10
Hence, the side of polygon = 10
(ii) 16200
Given
The sum of interior angles of polygon = 16200
Let number of sides = n
The sum of interior angles of polygon = (2n – 4) × 900
The side of polygon can be calculated as,
(2n – 4) × 900 = 16200
2(n – 2) × 900 = 16200
n – 2 = 16200 / (2 × 900)
n – 2 = 8100 / 900
We get,
n – 2 = 9
n = 9 + 2
n = 11
Hence, the side of polygon = 11
4. Is it possible to have a polygon, whose sum of interior angles is 10300.
Solution:
Given
The sum of interior angles of polygon = 10300
Let us consider the number of sides = n
The sum of interior angle of polygon = (2n – 4) × 900
The side of polygon is calculated as,
(2n – 4) × 900 = 10300
2(n – 2)= 10300 / 900
On further calculation, we get
(n – 2) = 10300 / (2 × 900)
(n – 2) = 1030 / 180
n = 5.72 + 2
n = 7.72
which is not a whole number.
Therefore, it is not a polygon, whose sum of interior angles is 10300
5. (i) If all the angles of a hexagon arc equal, find the measure of each angle.
(ii) If all the angles of an octagon are equal, find the measure of each angle.
Solution:
(i) Number of sides of polygon n = 6
Let us consider each angle be = x0
We know,
The sum of interior angles of hexagon = 6x0
The sum of interior angle of polygon = (2n – 4) × 900
The sum of the interior angles of polygon can be calculated as,
(2n – 4) × 900 = Sum of angles
(2 × 6 – 4) × 900 = 6x0
(12 – 4) × 900 = 6x0
6x0 = 8 × 900
x0 = (8 × 900) / 6
We get,
x = 1200
Therefore, each angle of hexagon = 1200
(ii) Number of sides of octagon n = 8
Let us consider each angle be = x0
We know that,
The sum of interior angles of octagon = 8x0
The sum of interior angles of polygon = (2n – 4) × 900
The sum of interior angles of polygon can be calculated as,
(2n – 4) × 900 = Sum of angles
(2n – 4) × 900 = 8x0
(2 × 8 – 4) × 900 = 8x0
12 × 900 = 8x0
8x0 = 12 × 900
x0 = (12 × 900) / 8
We get,
x0 = 1350
Therefore, each angle of octagon = 1350
6. One angle of a quadrilateral is 900 and all other angles are equal; find each equal angle
Solution:
Let us consider all the three equal angle of a quadrilateral be x0
The sum of angles of a quadrilateral = 3600
x + x + x + 900 = 3600
3x + 900 = 3600
3x = 3600 – 900
3x = 2700
x = 2700 / 3
We get,
x = 900
The measure of each equal angle = 900
7. If angles of quadrilateral are in the ratio 4: 5: 3: 6; find each angle of the quadrilateral.
Solution:
Let us consider the angles of quadrilateral be 4x, 5x, 3x and 6x
We know,
The sum of angles of quadrilateral = 3600
4x + 5x + 3x + 6x = 3600
18x = 3600
x = 3600 / 18
We get,
x = 200
Now, all the angles are,
4x = 4 × 200
= 800
5x = 5 × 200
= 1000
3x = 3 × 200
= 600
6x = 6 × 200
= 1200
Therefore, the angles of the quadrilateral are 800, 1000, 600 and 1200
8. If one angle of a pentagon is 1200 and each of the remaining four angles is x0, find the magnitude of x.
Solution:
Given
One angle of a pentagon = 1200
Number of sides of pentagon n = 5
Let us consider all other equal angle of pentagon be x
The sum of interior angle of polygon is (2n – 4) × 900
The sum of the interior angle of pentagon can be calculated as,
(2n – 4) × 900 = (2 × 5 – 4) × 900
= 6 × 900
We get,
= 5400
Therefore, the sum of interior angles of pentagon is 5400
Now,
x + x + x + x + 1200 = 5400
4x + 1200 = 5400
4x = 5400 – 1200
4x = 4200
x = 4200 / 4
We get,
x = 1050
Hence, the value of x = 1050
9. The angles of a pentagon are in the ratio 5: 4: 5: 7: 6; find each angle of the pentagon.
Solution:
Let us consider all the angle of pentagon as 5x, 4x, 5x, 7x and 6x
The sum of the interior angle of polygon is (2n – 4) × 900
The sum of the interior angle of pentagon can be calculated as,
(2n – 4) × 900 = (2 × 5 – 4) × 900
= 6 × 900
We get,
= 5400
The sum of interior angles of pentagon = 5400
Hence,
5x + 4x + 5x + 7x + 6x = 5400
27x = 5400
x = 5400 / 27
We get,
x = 200
Thus, each angle,
5x = 5 × 200
= 1000
4x = 4 × 200
= 800
5x = 5 × 200
= 1000
7x = 7 × 200
= 1400
6x = 6 × 200
= 1200
Therefore, all the angles of a pentagon are 1000, 800, 1000, 1400 and 1200
10. Two angles of a hexagon are 900 and 1100. If the remaining four angles arc equal, find each equal angle.
Solution:
Let us consider all the angle of hexagon as x
Number of sides in hexagon n = 6
The sum of interior angle of polygon is (2n – 4) × 900
The sum of interior angle of hexagon can be calculated as,
(2n – 4) × 900 = (2 × 6 – 4) × 900
= (12 – 4) × 900
= 8 × 900
We get,
= 7200
The sum of interior angles of pentagon is 7200
Hence,
900 + 1100 + x + x + x + x = 7200
2000 + 4x = 7200
4x = 7200 – 2000
4x = 5200
We get,
x = 1300
Hence, the measure of each equal angle = 1300
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