Frank Solutions Class 9 Maths Chapter 18 Rectilinear Figures help students with 100% accurate answers, along with shortcut tips and tricks. Those who find Maths difficult are advised to practise Frank Solutions thoroughly. Practising textbook problems using these solutions improves conceptual knowledge among students, undoubtedly. It also boosts confidence and skills in solving problems quickly. Access the Frank Solutions Class 9 Maths Chapter 18 Rectilinear Figures PDF, both online and offline, from the link provided below.
Chapter 18 Rectilinear Figures has perfect answers provided for the questions in the textbook. A figure, all of whose edge intersections are at right angles, is known as a rectilinear figure. Students can use Frank Solutions on a daily basis, as they provide vital questions that may arise in examinations. They can also check whether their answers are correct or not while revising the textbook problems.
Frank Solutions for Class 9 Maths Chapter 18 Rectilinear Figures – Download the PDF
Access Frank Solutions for Class 9 Maths Chapter 18 Rectilinear Figures
1. Find the sum of the interior angles of a polygon of:
(i) 7 sides
(ii) 12 sides
(iii) 9 sides
Solution:
(i) When n = 7
∴ Sum of interior angles = (n – 2) x 1800
= (7 – 2) x 1800
On further calculation, we get,
= 5 x 1800
= 9000
(ii) When n = 12
∴ Sum of interior angles = (n – 2) x 1800
= (12 – 2) x 1800
On further calculation, we get,
= 10 x 1800
= 18000
(iii) When n = 9
∴ Sum of interior angles = (n – 2) x 1800
= (9 – 2) x 1800
On further calculation, we get,
= 7 x 1800
= 12600
2. Find the measure of each interior angle of a regular polygon of:
(i) 6 sides
(ii) 10 sides
(iii) 15 sides
Solution:
(i) When n = 6
∴ Each interior angle of a regular polygon = {(n – 2) x 1800} / n
= {(6 – 2) x 1800} / 6
On further calculation, we get,
= (4 x 1800) / 6
= 1200
(ii) When n = 10
∴ Each interior angle of a regular polygon = {(n – 2) x 1800} / n
= {(10 – 2) x 1800} / 10
On further calculation, we get,
= (8 x 1800) / 10
= 1440
(iii) When n = 15
∴ Each interior angle of a regular polygon = {(n-2) x 1800} / n
= {(15 – 2) x 1800} / 15
On further calculation, we get,
= (13 x 1800) / 15
= 1560
3. Find each exterior angle of a regular polygon of:
(i) 9 sides
(ii) 15 sides
(iii) 18 sides
Solution:
(i) When n = 9
∴ Each exterior angle of a regular polygon = (3600 / n)
= (3600 / 9)
We get,
= 400
(ii) When n = 15
∴ Each exterior angle of a regular polygon = (3600 / n)
= (3600 / 15)
We get,
= 240
(iii) When n = 18
∴ Each exterior angle of a regular polygon = (3600 / n)
= (3600 / 18)
We get,
= 200
4. Find the number of sides in a regular polygon, when each interior angle is:
(i) 1200
(ii) 1400
(iii) 1350
Solution:
(i) Each interior angle of a regular polygon = {(n-2) x 1800} / n
{(n-2) x 1800} / n = 1200
On calculating further, we get,
1800 (n – 2) = 1200 (n)
3 (n – 2) = 2n
3n – 6 = 2n
3n – 2n = 6
We get,
n = 6
(ii) Each interior angle of a regular polygon = {(n-2) x 1800} / n
{(n – 2) x 1800} / n = 1400
On calculating further, we get,
1800 (n – 2) = 1400 (n)
9 (n – 2) = 7n
9n – 18 = 7n
9n – 7n = 18
2n = 18
We get,
n = 9
(iii) Each interior angle of a regular polygon = {(n – 2) x 1800} / n
{(n – 2) x 1800} / n = 1350
On calculating further, we get,
1800 (n – 2) = 1350 (n)
4 (n – 2) = 3n
4n – 8 = 3n
4n – 3n = 8
We get,
n = 8
5. Find the number of sides in a regular polygon, when each exterior angle is:
(i) 200
(ii) 600
(iii) 720
Solution:
(i) Each exterior angle = (3600 / n)
(3600 / n) = 200
We get,
n = 18
(ii) Each exterior angle = (3600 / n)
(3600 / n) = 600
We get,
n = 6
(iii) Each exterior angle = (3600 / n)
(3600 / n) = 720
We get,
n = 5
6. The angles of a pentagon are 1000, 960, 740, 2x0 and 3x0. Find the measures of the two angles 2x0 and 3x0.
Solution:
We know that,
A pentagon has 5 sides
∴ Sum of interior angles = (n – 2) x 1800
= (5 – 2) x 1800
On calculating further, we get,
= 3 x 1800
We get,
= 5400
Given angles are 1000, 960, 740, 2x0 and 3x0
∴ 1000 + 960 + 740 + 2x0 + 3x0 = 5400
5x0 + 2700 = 5400
5x0 = 5400 – 2700
x0 = (5400 – 2700) / 5
We get,
x0 = 540
Hence,
2x0 = 2 x 54 = 1080 and 3x0 = 3 x 54 = 1620
Therefore, the two angles are 1080 and 1620 respectively
7. The three angles of a quadrilateral are 710, 1100, 950. Find its fourth angle.
Solution:
A quadrilateral is a polygon with four sides
∴ Sum of interior angles = (n – 2) x 1800
= (4 – 2) x 1800
= 2 x 1800
We get,
= 3600
Given, the three interior angles are 710, 1100, 950
Let x be the fourth angle
∴ 710 + 1100 + 950 + x = 3600
x + 2760 = 3600
We get,
x0 = 3600 – 2760
We get,
x0 = 840
Hence, the fourth angle is 840
8. Find the angles of a pentagon which are in the ratio 4: 4: 6: 7: 6.
Solution:
A pentagon has 5 sides
∴ Sum of interior angles = (n-2) x 1800
= (5 – 2) x 1800
= 3 x 1800
We get,
= 5400
Ratio of the angles are 4x0, 4x0, 6x0, 7x0 and 6x0
∴ 4x0 + 4x0 + 6x0 + 7x0 + 6x0 = 5400
27x0 = 5400
We get,
x0 = 200
Therefore, the interior angles of a pentagon are 800, 800, 1200, 1400 and 1200
9. Find the angles of a quadrilateral whose angles are in the ratio 1: 4: 5: 2.
Solution:
A quadrilateral is a polygon with four sides
∴ Sum of interior angles = (n – 2) x 1800
= (4 – 2) x 1800
= 2 x 1800
We get,
= 3600
Given ratio of the angles = 1: 4: 5: 2
Hence,
The interior angles are x0, 4x0, 5x0 and 2x0
x0 + 4x0 + 5x0 + 2x0 = 3600
12x0 = 3600
We get,
x0 = 300
Therefore, the interior angles of the quadrilateral are 300, 1200, 1500 and 600
10. The angles of a pentagon are x0, (x – 10)0, (x + 200), (2x – 44)0 and (2x – 70)0. Find the angles.
Solution:
A pentagon has 5 sides
∴ Sum of interior angles = (n – 2) x 1800
= (5 – 2) x 1800
= 3 x 1800
We get,
= 5400
Given angles are x0, (x – 10)0, (x + 20)0, (2x – 44)0 and (2x – 70)0
∴ x0 + (x – 10)0 + (x + 20)0 + (2x – 44)0 + (2x – 700) = 5400
7x0 – 1040 = 5400
7x0 = 5400 + 1040
7x0 = 6440
We get,
x0 = 920
Therefore, the interior angles of the pentagon are 920, 820, 1120, 1400 and 1140
11. The angles of a hexagon are (2x + 5)0, (3x – 5)0, (x + 40)0, (2x + 20)0, (2x + 25)0 and (2x + 35)0. Find the value of x.
Solution:
A hexagon has 6 sides
∴ Sum of interior angles = (n – 2) x 1800
= (6 – 2) x 1800
= 4 x 1800
We get,
= 7200
Given angles of a hexagon are (2x + 5)0, (3x – 5)0, (x + 40)0, (2x + 20)0, (2x + 25)0 and (2x + 35)0
∴ (2x + 5)0 + (3x – 5)0 + (x + 40)0 + (2x + 20)0 + (2x + 25)0 + (2x + 35)0 = 7200
12x + 1200 = 7200
We get,
x = 500
12. One angle of a hexagon is 1400 and the remaining angles are in the ratio 4: 3: 4: 5: 4. Calculate the measures of the smallest and the largest angles.
Solution:
A hexagon has 6 sides
∴ Sum of interior angles = (n – 2) x 1800
= (6 – 2) x 1800
= 4 x 1800
We get,
= 7200
One angle of a hexagon is given to be 1400
Ratio of the remaining five angles = 4: 3: 4: 5: 4
∴ The interior angles are 4x0, 3x0, 4x0, 5x0 and 4x0
∴ 1400 + 4x0 + 3x0 + 4x0 + 5x0 + 4x0 = 7200
20x0 + 1400 = 7200
x0 = 5800 / 20
We get,
x0 = 290
Smallest angle is 3x0 = 3 x 290 = 870 and
Largest angle is 5x0 = 5 x 290 = 1450
Therefore, the smallest angle is 870 and the largest angle is 1450
13. One angle of a pentagon is 1600 and the rest are all equal angles. Find the measure of the equal angles.
Solution:
A pentagon has 5 sides
∴ Sum of interior angles = (n – 2) x 1800
= (5 – 2) x 1800
= 3 x 1800
We get,
= 5400
One angle of a pentagon is given to be 1600
Ratio of the remaining four angles = 1: 1: 1: 1
∴ The interior angles are x0, x0, x0 and x0
∴ 1600 + x0 + x0 + x0 + x0 = 5400
4x0 = 5400 – 1600
4x0 = 3800
We get,
x0 = 950
Therefore, each equal angle is 950
14. Calculate the measure of each angle of a nonagon.
Solution:
A nonagon has 9 sides
∴ Each interior angle of a regular polygon = {(n – 2) x 1800} / n
= {(9 – 2) x 1800} / 9
= (7 x 1800) / 9
We get,
= 1400
15. Calculate the measure of each angle of a regular polygon of 20 sides.
Solution:
Given
n = 20
∴ Each interior angle of the regular polygon = {(n – 2) x 1800} / n
= {(20 – 2) x 1800} / 20
= (18 x 1800)/ 20
We get,
= 1620
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