Class 8 SELINA Solutions Maths Chapter 16: Understanding Shapes (Including Polygons)
Understanding Shapes (Including Polygons) Exercise Ex. 16(A)
Solution 1(i)
Correct option: (c) 360°
The sum of the exterior angles of a polygon is 360°.
⇒ ∠x + ∠y + ∠z + ∠r = 360°
Solution 1(ii)
Correct option: (c) 140°
Sum of the interior angles of a polygon with n sides = (2n – 4) × 90°
Since, quadrilateral has 4 sides.
Sum of the interior angles of a quadrilateral = (2×4 – 4) × 90° = 360°
As the angles are in the ratio 2 : 5 : 7 : 4,
2x + 5x + 7x + 4x = 360°
⇒ 18x = 360°
⇒ x = 20°
Largest angle = 7x = 140°
Solution 1(iii)
Correct option: (a) concave
Sum of the interior angles of a polygon = 360°
∠A + ∠B + ∠C + ∠D = 360°
⇒ 45° + 55° + ∠C + 60° = 360°
⇒ ∠C = 200°
Since ∠C > 180° and it is an internal angle of quadrilateral ABCD, it is a concave quadrilateral.
Solution 1(iv)
Correct option: (d) 60°
In quadrilateral ABCD,
∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + ∠B = 360° – (∠C + ∠D)
⇒ ∠A + ∠B = 360° – (40° + 80°)
⇒ 2∠PAB + 2∠ABP = 360° – 120° = 240°
⇒ ∠PAB + ∠ABP = 120°
Now, ∠APB + ∠PAB + ∠ABP = 180° … (Interior angles of a triangle)
⇒ ∠APB = 180° – (∠PAB + ∠ABP)
⇒ ∠APB = 180° – 120° = 60°
Solution 1(v)
Correct option: (b) 7
The sum of interior angles of a polygons with n sides = (2n – 4) × 90°
⇒ 900° = (2n – 4) × 90°
⇒ 2n – 4 = 10
⇒ n = 7
Solution 2
The sum of interior angles of a polygon with n sides = (2n – 4) × 90°
∴ Sum of angles of a polygon with 10 sides = (2×10 – 4) × 90°
= 1440°
Solution 3
16 right angles = 16 × 90° = 1440°
The sum of interior angles of a polygon with n sides = (2n – 4) × 90°
∴ 1440° = (2n – 4) × 90°
⇒ 2n – 4 = 16
⇒ n = 10
Solution 4
The sum of interior angles of a polygon with n sides = (2n – 4) × 90°
(i)
870° = (2n – 4) × 90°
⇒ 2n – 4 = 29/3
⇒ n = 6.8333…
Which is not possible.
(ii)
2340° = (2n – 4) × 90°
⇒ 2n – 4 = 26
⇒ n = 15
So, it is possible to have a polygon whose sum of interior angles is 2340°.
(iii)
7 right angles = 7×90° = 630°
630° = (2n – 4) × 90°
⇒ 2n – 4 = 7
⇒ n = 5.5
Which is not possible.
Solution 5(i)
The sum of interior angles of a polygon with n sides = (2n – 4) × 90°
Number of sides in a hexagon = 6
⇒ Sum of interior angles of hexagon = (2×6 – 4) × 90° = 720°
Let the measure of each angle be x.
⇒ 6x = 720°
⇒ x = 120°
Solution 5(ii)
The sum of interior angles of a polygon with n sides = (2n – 4) × 90°
Number of sides = 14
⇒ Sum of interior angles of a polygon = (2×14 – 4) × 90° = 2160°
Let the measure of each angle be x.
⇒ 14x = 2160°
Solution 6
The sum of the exterior angles of a polygon is 360°.
Hence, the sum of the exterior angles of a polygon with 7 sides is 360°.
Solution 7
The sum of the exterior angles of a polygon is 360°.
(6x – 1)° + (10x + 2)° + (8x + 2)° + (9x – 3)° + (5x + 4)° + (12x + 6)° = 360°
⇒ (50x + 10)° = 360°
⇒ 50x = 350
⇒ x = 7
⇒ (6x – 1)° = 41°, (10x + 2)° = 72°, (8x + 2)° = 58°, (9x – 3)° = 60°, (5x + 4)° = 39°, (12x + 6)° = 90°
Solution 8
Sum of interior angles of a pentagon = (2×5 – 4) × 90° = 540°
⇒ 4x + 5x + 6x + 7x + 5x = 540°
⇒ 27x = 540°
⇒ x = 20°
⇒ 4x = 80°, 5x = 100°, 6x = 120°, 7x = 140°, 5x = 100°
Solution 9
Sum of interior angles of a hexagon = (2×6 – 4) × 90° = 720°
Let each equal angle measure x°.
⇒ 120° + 160° + 4x = 720°
⇒ 4x = 440°
⇒ x = 110°
Hence, the measure of each equal angle is 110°.
Solution 10
(i)
Sum of interior angles of a pentagon = (2×5 – 4) × 90° = 540°
(ii)
Since, sides AB and ED are parallel to each other.
⇒ ∠A + ∠E = 180°
(iii)
Now, ∠A + ∠B + ∠C + ∠D + ∠E = 540°
Since, ∠B : ∠C : ∠D = 5 : 6 : 7
⇒ 180° + 5x + 6x + 7x = 540°
⇒ 18x = 360°
⇒ x = 20°
⇒ ∠B = 5x = 100°, ∠C = 6x = 120°, ∠C = 7x = 140°
Solution 11
Two right angles = 2×90° = 180°
Sum of interior angles of a polygon with n sides = (2n – 4) × 90°
⇒ 180° + (n – 2)×120° = (2n – 4) × 90°
⇒ 180° + n × 120° – 240° = n × 180° – 360°
⇒ 300° = n × 60°
⇒ n = 5
Hence, the number of sides is 5.
Solution 12
Since, sides AB and FE are parallel to each other.
⇒ ∠A + ∠F = 180°
Also, ∠B : ∠C : ∠D : ∠E = 6 : 4 : 2 : 3
Let ∠B = 6x, ∠C = 4x, ∠D = 2x, ∠E = 3x
Now, sum of interior angles of a hexagon = (2×6 – 4) × 90° = 720°
⇒ ∠A + ∠F + ∠B + ∠C + ∠D + ∠E = 720°
⇒ 180° + 6x + 4x + 2x + 3x = 720°
⇒ 15x = 540°
⇒ x = 36°
⇒ ∠B = 216° and ∠D = 72°
Solution 13
Sum of interior angles of a hexagon = (2×6 – 4) × 90° = 720°
⇒ (x + 10)° + (2x + 20)° + (2x – 20)° + (3x – 50)° + (x + 40)° + (x + 20)° = 720°
⇒ (10x + 20)° = 720°
⇒ 10x = 700°
⇒ x = 70°
Solution 14
Sum of interior angles of a pentagon = (2×5 – 4) × 90° = 540°
Since, the three angles are in the ratio 1 : 3 : 7.
⇒ 40° + 60° + x + 3x + 7x = 540°
⇒ 100° + 11x = 540°
⇒ x = 40°
⇒ 7x = 280°
Hence, the biggest angle is 280°.
Understanding Shapes (Including Polygons) Exercise Ex. 16(B)
Solution 1(i)
Correct option: (c) 6
Since, each interior angle is double of its exterior angle.
∴ Sum of interior angles = 2 × Sum of all exterior angles
⇒ (2n – 4) × 90° = 2 × 360°
⇒ (2n – 4) = 8
⇒ n = 6
Solution 1(ii)
Correct option: (c) 9
Sum of the interior angles of a polygon with n sides = (2n – 4) × 90°
⇒ 1260° = (2n – 4) × 90°
⇒ 2n – 4 = 14
⇒ n = 9
Solution 1(iii)
Correct option: (a) 4
Sum of the interior angles of a polygon with n sides = (2n – 4) × 90°
Sum of the exterior angles of a polygon = 360°
⇒ (2n – 4) × 90° = 360°
⇒ 2n – 4 = 4
⇒ 2n = 8
⇒ n = 4
Solution 1(iv)
Correct option: (c) 6
Sum of all interior angles = (2n – 4) × 90°
⇒ n × 120° = (2n – 4) × 90°
⇒ 4n = 3(2n – 4)
⇒ 4n = 6n – 12
⇒ n = 6
Solution 1(v)
Correct option: (d) none of these
Let the number of sides of a regular polygon be n.
Sum of interior angle and the corresponding exterior angle is 180°.
⇒ 3x + 2x = 180°
⇒ x = 36°
Each interior angle = 3x = 108°
Each exterior angle = 2x = 72°
Sum of all exterior angles = 360°
⇒ n × 72° = 360°
⇒ n = 5
Solution 2
(i)
Number of sides = 8
(ii)
Number of sides = 12
(iii)
Number of sides = n
⇒ n = 5
(iv)
Number of sides = n
⇒ n = 8
(v)
Number of sides = n
⇒ n = 12
(vi)
Number of sides = n
⇒ n = 9
Solution 3(i)
Sum of all interior angles = (2n – 4) × 90°
⇒ n × 160° = (2n – 4) × 90°
⇒ 16n = 18n – 36
⇒ n = 18
Solution 3(ii)
Sum of all interior angles = (2n – 4) × 90°
⇒ n × 135° = (2n – 4) × 90°
⇒ 3n = 4n – 8
⇒ n = 8
Solution 3(iii)
Sum of all interior angles = (2n – 4) × 90°
⇒ n × 108° = (2n – 4) × 90°
⇒ 108n = 180n – 360
⇒ 72n = 360
⇒ n = 5
Solution 4(i)
Sum of all exterior angles = 360°
⇒ n × 30° = 360°
⇒ n = 12
Solution 4(ii)
Sum of all exterior angles = 360°
⇒ n × 36° = 360°
⇒ n = 10
Solution 5(i)
The sum of interior angles of a polygon with n sides = (2n – 4) × 90°
⇒ n × 170° = (2n – 4) × 90°
⇒ 17n = 18n – 36
⇒ n = 36
Hence, it is possible to have a regular polygon with each interior angle 170°.
Solution 5(ii)
The sum of interior angles of a polygon with n sides = (2n – 4) × 90°
⇒ n × 138° = (2n – 4) × 90°
⇒ 138n = 180n – 360
⇒ 42n = 360
⇒ n = 8.57
Which is not possible.
Solution 6(i)
The sum of exterior angles of a polygon with n sides =360°
⇒ n × 80° = 360°
⇒ 8n = 36
⇒ n = 4.5
Which is not possible.
Solution 6(ii)
Each exterior angle = 40% of a right angle = 36°
The sum of exterior angles of a polygon with n sides =360°
⇒ n × 36° = 360°
⇒ n = 10
Hence, it is possible to have a regular polygon whose each exterior angle is 40% of a right angle.
Solution 7
Since, interior angle is equal to its exterior angle.
⇒ Sum of all the interior angles of a polygon = Sum of all the exterior angles of a polygon
⇒ (2n – 4) × 90° = 360°
⇒ 2n – 4 = 4
⇒ n = 4
Hence, the number of sides is 4.
Solution 8
Exterior angle of a regular polygon = one-third of its interior angle
Sum of all the exterior angles of a polygon = one-third of the sum of its interior angles
⇒ 360° = (2n – 4) × 30°
⇒ 2n – 4 = 12
⇒ n = 8
Hence, the number of sides in the polygon is 8.
Solution 9
Interior angle of a regular polygon = 5 × Exterior angle
∴ Sum of all the interior angles of a polygon = 5 × Sum of all the exterior angles
(i)
∴ (2n – 4) × 90° = 5 × 360°
∴ 2n – 4 = 20
∴ n = 12
Measure of each interior angle 
(ii)
Measure of each exterior angle 
(iii)
Number of sides in the polygon = 12
Solution 10
Interior angle : Exterior angle = 2 : 1
⇒ Interior angle = 2x and Exterior angle = x
(i)
Sum of an interior angle and its corresponding exterior angle is 180°.
⇒ 2x + x = 180°
⇒ x = 60°
(ii)
Sum of all the exterior angles = 360°
⇒ Number of sides 
Solution 11
Exterior angle : Interior angle = 1 : 4
⇒ Exterior angle = x and Interior angle = 4x
The sum of an interior angle and its corresponding exterior angle is 180°.
⇒ x + 4x = 180°
⇒ x = 36°
Now, sum of all the exterior angles = 360°
⇒ Number of sides 
Solution 12
Sum of interior angles of a regular polygon = 2 × Sum of its exterior angles
⇒ (2n – 4) × 90° = 2 × 360°
⇒ 2n – 4 = 8
⇒ n = 6
Solution 13
AB, BC and CD are three consecutive sides.
∠BAC = 20°
(i)
In ΔABC, AB = BC
⇒∠ACB = ∠BAC = 20°
Now, ∠ACB + ∠ABC + ∠BAC = 180°
⇒∠ABC = 180°– (∠ACB + ∠BAC) = 140°
∴ each interior angle = 140°
(ii)
Since, an exterior angle + interior angle = 180°
⇒ Each exterior angle = 40°
(iii)
Each exterior angle 
⇒ n = 9
Hence, the number of sides is 9.
Solution 14
Let the alternate sides AB produced and CD produced meet at O.
⇒∠OBC, ∠OCB and ∠BOC act as the interior angles of triangle BOC.
⇒∠OBC + ∠OCB + ∠BOC = 180°
⇒∠OBC + ∠OCB = 180° – ∠BOC
In a regular polygon, all the angles are equal.
⇒∠ABC = ∠BCD
⇒ 180°–∠ABC = 180°–∠BCD
⇒∠OBC = ∠OCB
∴ 2∠OCB = 180°– 90° = 90°
∴∠OCB = ∠OBC = 45°
⇒∠ABC = 180°– 45° = 135°
Each interior angle 
⇒ 3n = 4n – 8
⇒ n = 8
Solution 15
(i)
Sum of all interior angles of a pentagon = (2×5 – 4) × 90° = 540°
⇒ Each interior angle of a regular pentagon 
⇒∠BAE = 108°
(ii)
In ΔABE,
∠ABE + ∠AEB + ∠BAE = 180°
Since, ∠AEB = ∠ABE … Angles opposite to equal sides
⇒ 2∠ABE = 72°
⇒∠ABE = ∠AEB = 36°
(iii)
∠BED = 108°–∠AEB = 72°
Solution 16
Since the ratio between the number of sides of two regular polygons is 3 : 4.
Let the number of sides be 3x and 4x.
Sum of the interior angles of the 1st polygon = (2×3x – 4) × 90°
Sum of the interior angles of the 2nd polygon = (2×4x – 4) × 90°
Since, the ratio between the sum of their interior angles is 2 : 3.
⇒ 3(6x – 4) = 2(8x – 4)
⇒ 18x – 12 = 16x – 8
⇒ 2x = 4
⇒ x = 2
Number of sides in 1st polygon = 3x = 6
Number of sides in the 2nd polygon = 4x = 8
Solution 17
Sum of the exterior angles of a polygon = 360°
Since, the three of the exterior angles are 40°, 51° and 86°, and rest three are x° each.
⇒ 40° + 51° + 86° + x° + x° + x° = 360°
⇒ 3x° = 360° – (40° + 51° + 86°)
⇒ 3x = 183
⇒ x = 61
Solution 18(i)
Interior angle = 5 × exterior angle
Since, an interior angle + exterior angle = 180°
⇒ 6 × exterior angle = 180°
⇒ Exterior angle = 30°
∵ Each exterior angle
⇒ 30° 
⇒ n = 12
Hence, the number of sides is 12.
Solution 18(ii)
Since, the ratio between an exterior angle and its interior angle is 2 : 7.
Let exterior angle be 2x and interior angle be 7x.
Since, an interior angle + exterior angle = 180°
⇒ 7x + 2x = 180°
⇒ 9x = 180°
⇒ x = 20°
∴ Each exterior angle = 2x = 40°
∵ each exterior angle 
⇒ 40° 
⇒ n = 9
Hence, the number of sides is 9.
Solution 18(iii)
Le the interior angle be x.
∵ Exterior angle = Interior angle + 60°
⇒ Exterior angle = x + 60°
Since, an interior angle + exterior angle = 180°
⇒ x + (x + 60°) = 180°
⇒ x = 60°
⇒ Exterior angle = 120°
∵ Each exterior angle
⇒ 120°
⇒ n = 3
Hence, the number of sides is 3.
Understanding Shapes (Including Polygons) Exercise TEST YOURSELF
Solution 1(i)
Correct option: (b) 24
Each interior angle of a regular polygon is 165°.
∴ Sum of all the interior angles of a polygon = (2n – 4) × 90°
⇒ n × 165° = (2n – 4) × 90°
⇒ 33n = 36n – 72
⇒ n = 24
Solution 1(ii)
Correct option: (a) 130°
Two angles of a quadrilateral are 50° each.
Let the measure of the remaining two angles be x° each.
Sum of the interior angles of a polygon with n sides = (2n – 4) × 90°
⇒ 50° + 50° + x° + x° = 4 × 90°
⇒ 50 + x = 180
⇒ x = 130
Hence, the measure of each equal angle is 130°.
Solution 1(iii)
Correct option: (c) 
A polygon has n sides.
Number of diagonals 
Solution 1(iv)
Correct option: (b) 10
Interior angle = 4 × Exterior angle
Let the number of sides be n.
Sum of all the interior angles = (2n – 4) × 90°
Sum of all the exterior angles = 360°
⇒ (2n – 4) × 90° = 4 × 360°
⇒ 2n – 4 = 16
⇒ n = 10
Solution 1(v)
Correct option: (a) 89°
Two angles of a quadrilateral are 69° and 113°.
Let the measure of the remaining two angles be x° each.
Sum of the interior angles of a polygon with n sides = (2n – 4) × 90°
⇒ 69° + 113° + x° + x° = 4 × 90°
⇒ 91 + x = 180
⇒ x = 89°
Solution 2
Since, the angles are in the ratio 3 : 5 : 7.
Let the angles be 3x, 5x and 7x.
⇒ 7x – 3x = 76°
⇒ x = 19°
⇒ 3x = 57°, 5x = 95°, 7x = 133°
Fourth angle = 360° – (57° + 95° + 133°) = 75°
Solution 3(i)
Since, the angles are in the ratio 4 : 5 : 6.
Let the angles be 4x, 5x and 6x.
⇒ 4x + 6x = 160°
⇒ x = 16°
⇒ 4x = 64°, 5x = 80°, 6x = 96°
Fourth angle = 360° – (64° + 80° + 96°) = 120°
Thus, the angles of a quadrilateral are 64°, 80°, 96° and 120°.
Solution 4
Let the fourth angle of the quadrilateral be x.
Sum of three angles of a quadrilateral is 5 times the fourth angle.
∴ Sum of three angles = 5x
Since, sum of all the interior angles of a quadrilateral = (2 × 4 – 4) × 90°
⇒ Sum of all three angles + fourth angle = 360°
⇒ 5x + x = 360°
⇒ x = 60°
Hence, the fourth angle is 60°.
Solution 5
Four angles of a heptagon measures 132° each.
Let the measure of the remaining three angles be x° each.
Since, the sum of all the interior angles of a polygon = (2n – 4) × 90°
⇒ Sum of interior angles of a heptagon = (2×7 – 4) × 90° = 900°
⇒ 4 × 132° + 3 × x = 900°
⇒ 528° + 3x = 900°
⇒ 3x = 372°
⇒ x = 124°
Hence, each of the three angles measures 124°.
Solution 6
Each interior angle of a regular polygon is 144°.
Let the number of sides be n.
Since, sum of all the interior angles of a polygon = (2n – 4) × 90°
⇒ n × 144° = (2n – 4) × 90°
⇒ 8n = (2n – 4) × 5
⇒ 8n = (n – 2) × 10
⇒ 8n = 10n – 20
⇒ 2n = 20
Each interior angle of a regular polygon with 20 sides
Solution 7
The exterior angles of a pentagon are in the ratio 1 : 2 : 3 : 4 : 5.
Let the angles be x, 2x, 3x, 4x and 5x.
Since, the sum of all the exterior angles of a polygon = 360°
⇒ x + 2x + 3x + 4x + 5x = 360°
⇒ 15x = 360°
⇒ x = 24°
⇒ 2x = 48°, 3x = 72°, 4x = 96°, 5x = 120°
⇒ Interior angles are (180° – 24°), (180° – 48°), (180° – 72°), (180° – 96°) and (180° – 120°).
That is, the interior angles are 156°, 132°, 108°, 84° and 60°.
Solution 8
The sum of interior angles of a pentagon = (2×5 – 4) × 90° = 540°
⇒ x° + (x – 10)° + (x + 20)° + (2x – 44)° + (2x – 70)° = 540°
⇒ 7x – 104 = 540
⇒ 7x = 644
⇒ x = 92
Solution 9
Since, sides AB and CD are parallel.
⇒∠A + ∠D = 180° and ∠B + ∠C = 180°
Since, ∠A : ∠D = 2 : 3 and ∠B : ∠C = 7 : 8
Let ∠A = 2x, ∠D = 3x, ∠B = 7y and ∠C = 8y.
⇒ 2x + 3x = 180° and 7y + 8y = 180°
⇒ x = 36° and y = 12°
∴∠A = 2x = 72°, ∠B = 7y = 84°, ∠C = 8y = 96° and ∠D = 3x = 108°
Solution 10
Let ∠D = x
⇒∠C = 3x, ∠B = 2x, ∠A = 3x
Sum of all the interior angles of a quadrilateral = (2×4 – 4) × 90° = 360°
∴∠A + ∠B + ∠C + ∠D = 360°
⇒ 3x + 2x + 3x + x = 360°
⇒ x = 40°
∴∠A = 120°, ∠B = 80°, ∠C = 120° and ∠D = 40°
Solution 11
Sum of all the interior angles of a polygon = (2n – 4) × 90°
Sum of exterior angles of a polygon = 360°
Since, sum of interior angles of a regular polygon is thrice the sum of its exterior angles.
⇒ (2n – 4) × 90° = 3 × 360°
⇒ 2n – 4 = 12
⇒ n = 8
Solution 12
Sum of all the interior angles of a quadrilateral = (2×4 – 4) × 90° = 360°
(i)
⇒ (4x)° + 5(x + 2)° + (7x – 20)° + 6(x + 3)° = 360°
⇒ 4x + 5x + 10 + 7x – 20 + 6x + 18 = 360
⇒ 22x = 352
⇒ x = 16
(ii)
∴ (4x)° = 64°, 5(x + 2)° = 90°, (7x – 20)° = 92°, 6(x + 3)° = 114°
Solution 13
The given figure is a quadrilateral.
(i)
Sum of all the interior angles of a quadrilateral = 360°
⇒∠A + ∠B + ∠C + ∠D = 360°
⇒ 90° + (2x + 4)° + (3x – 5)° + (8x – 15)° = 360°
⇒ 90 + 2x + 4 + 3x – 5 + 8x – 15 = 360
⇒ 13x = 286
⇒ x = 22
(ii)
∠B = (2x + 4)° = 48° and ∠C = (3x – 5)° = 61°