UNDERSTANDING SHAPES - Q&A
Test Yourself
1. In the following figure; AB, BC and CD are the sides of a regular polygon. If the angle O is the centre of the polygon and ∠AOC = 72°; find the number of sides in the polygon.
Solution:
Let the number of sides of the regular polygon be n.
The angle subtended by each side at the centre = 360°/n
In the figure, sides AB and BC subtend angles ∠AOB and ∠BOC at the centre.
Since the polygon is regular, lengths of sides are equal, so they subtend equal angles at the centre.
∠AOB = ∠BOC
Given ∠AOC = 72°
∠AOB + ∠BOC = 72°
2 × ∠AOB = 72° ⇒ ∠AOB = 36°
We know, ∠AOB = 360°/n
36° = 360°/n
n = 360/36 = 10
Number of sides = 10.
2. Find the number of sides in a regular polygon, if its interior angle is equal to its exterior angle.
Solution:
Let the exterior angle be x.
Then interior angle = x.
We know, Interior angle + Exterior angle = 180°
x + x = 180°
2x = 180° ⇒ x = 90°
Exterior angle = 90°
Number of sides = 360° / Exterior angle = 360° / 90° = 4.
Number of sides = 4.
3. The exterior angle of a regular polygon is one-third of its interior angle. Find the number of sides in the polygon.
Solution:
Let interior angle be x.
Then exterior angle = x/3.
Interior angle + Exterior angle = 180°
x + x/3 = 180°
(3x + x)/3 = 180°
4x = 540° ⇒ x = 135°
Exterior angle = 135°/3 = 45°
Number of sides = 360° / 45° = 8.
Number of sides = 8.
4. In a regular pentagon ABCDE; draw a diagonal BE and then find the measure of:
(i) ∠BAE
(ii) ∠ABE
(iii) ∠BED
Solution:
(i) For a regular pentagon, n = 5.
Each interior angle = (2n - 4) × 90° / n
= (2(5) - 4) × 90° / 5 = 6 × 90° / 5 = 540° / 5 = 108°
So, ∠BAE = 108°
(ii) In ΔABE, AB = AE (sides of regular pentagon)
So, ΔABE is isosceles, which means ∠ABE = ∠AEB.
Sum of angles in triangle = 180°
∠BAE + ∠ABE + ∠AEB = 180°
108° + 2∠ABE = 180°
2∠ABE = 72° ⇒ ∠ABE = 36°
(iii) We know ∠AED = 108° (Interior angle)
∠BED = ∠AED - ∠AEB
From (ii), ∠AEB = 36°
∠BED = 108° - 36° = 72°
5. Find the sum of interior angles of a polygon with:
(i) 9 sides
(ii) 13 sides
(iii) 22 sides
Solution:
Sum of interior angles = (2n - 4) × 90°
(i) n = 9: Sum = (2(9) - 4) × 90° = (18 - 4) × 90° = 14 × 90° = 1260°
(ii) n = 13: Sum = (2(13) - 4) × 90° = (26 - 4) × 90° = 22 × 90° = 1980°
(iii) n = 22: Sum = (2(22) - 4) × 90° = (44 - 4) × 90° = 40 × 90° = 3600°
6. Find the number of sides in a polygon if the sum of its interior angles is:
(i) 1080°
(ii) 1980°
Solution:
Formula: (2n - 4) × 90° = Sum
(i) (2n - 4) × 90 = 1080
2n - 4 = 1080 / 90 = 12
2n = 16 ⇒ n = 8
(ii) (2n - 4) × 90 = 1980
2n - 4 = 1980 / 90 = 22
2n = 26 ⇒ n = 13
7. The sum of interior angles of a polygon is 5 times the sum of its exterior angles. Find the number of sides in the polygon.
Solution:
Sum of interior angles = 5 × Sum of exterior angles
(2n - 4) × 90° = 5 × 360°
(2n - 4) × 90 = 1800
2n - 4 = 20
2n = 24 ⇒ n = 12
Number of sides = 12.
8. The angles of a pentagon are x°, (x + 10)°, (x + 20)°, (x + 30)° and (x + 40)°. Find the value of x.
Solution:
Sum of angles of a pentagon (n=5) = (2(5) - 4) × 90° = 540°
x + (x + 10) + (x + 20) + (x + 30) + (x + 40) = 540
5x + 100 = 540
5x = 440
x = 88
9. The angles of a hexagon are (x + 10)°, (2x + 20)°, (2x - 20)°, (3x - 50)°, (x + 40)° and (x + 20)°. Find x.
Solution:
Sum of angles of a hexagon (n=6) = (2(6) - 4) × 90° = 8 × 90° = 720°
(x + 10) + (2x + 20) + (2x - 20) + (3x - 50) + (x + 40) + (x + 20) = 720
10x + 20 = 720
10x = 700
x = 70
10. In a pentagon, two angles are 40° and 60°, and the rest are in the ratio 1 : 3 : 7. Find the biggest angle of the pentagon.
Solution:
Sum of angles of pentagon = 540°
Given angles: 40°, 60°. Let the others be y, 3y, 7y.
40 + 60 + y + 3y + 7y = 540
100 + 11y = 540
11y = 440 ⇒ y = 40°
The remaining angles are 40°, 3(40)°=120°, 7(40)°=280°.
The biggest angle is 280°.
11. In a regular pentagon ABCDE, diagonals AC and BE intersect at P.
(i) Find ∠BPC.
(ii) Show that AP = AE.
Solution:
Interior angle of regular pentagon = 108°.
(i) In ΔABC, AB=BC, so ∠BAC = ∠BCA = (180-108)/2 = 36°.
Similarly in ΔABE, ∠ABE = ∠AEB = 36°.
In ΔABP, ∠BAP = 36° (from ΔABC) and ∠ABP = 36°.
Exterior angle ∠BPC of ΔABP = ∠BAP + ∠ABP = 36° + 36° = 72°.
Alternatively, in ΔPBC: ∠PCB = 36°, ∠PBC = 108° - 36° = 72°. Then ∠BPC = 180 - (36+72) = 72°.
(ii) In ΔAEP:
∠PAE (which is ∠CAE) = ∠BAE - ∠BAC = 108° - 36° = 72°.
∠AEP (which is ∠AEB) = 36°.
∠APE = 180° - (72° + 36°) = 72°.
Since ∠PAE = ∠APE = 72°, the triangle is isosceles with sides opposite equal.
Therefore, AP = AE.
12. The sum of the interior angles of a regular polygon is twice the sum of its exterior angles. Find the number of sides of the polygon.
Solution:
Sum of interior angles = 2 × Sum of exterior angles
(2n - 4) × 90° = 2 × 360°
(2n - 4) × 90 = 720
2n - 4 = 8
2n = 12 ⇒ n = 6
Number of sides = 6.
13. In the following figure; AB, BC and CD are the sides of a regular polygon. If the angle BAC = 20°; find:
(i) its each interior angle.
(ii) its each exterior angle.
(iii) the number of sides in the polygon.
Solution:
(i) In ΔABC, AB = BC (sides of regular polygon).
So, ∠BCA = ∠BAC = 20°.
∠ABC (Interior angle) = 180° - (20° + 20°) = 140°.
(ii) Each exterior angle = 180° - Interior angle = 180° - 140° = 40°.
(iii) Number of sides = 360° / Exterior angle = 360° / 40° = 9.