SELINA Solution Class 9 Chapter 14 Rectilinear Figures (Quadrilaterals: Parallelogram , Rectangle , Rhombus, Square and Trapezium) Exercise 14A

Question 1

The sum of the interior angles of a polygon is four times the sum of its exterior angles.
Find the number of sides in the polygon.

Sol:

The sum of the interior angle=4 times the sum of the exterior angles.

Therefore the sum of the interior angles = 4 × 360° =1440°.

Now we have
( 2n - 4 ) x 90° = 1440°
2n - 4 = 16
2n = 20
n = 10
Thus the number of sides in the polygon is 10.

Question 2

The angles of a pentagon are in the ratio 4: 8: 6: 4: 5.
Find each angle of the pentagon.

Sol:

Let the angles of the pentagon are 4x, 8x, 6x, 4x and 5x.
Thus we can write 
4x + 8x+ 6x + 4x +5x = 540°
                             27x =540°
                                 x = 20°

Hence the angles of the pentagon are:
4 × 20° = 80° ,
8 × 20°= 160° ,
6 × 20°= 120° ,
4 × 20°= 80° ,
5 × 20°= 100° 

Question 3

One angle of a six-sided polygon is 140^o and the other angles are equal.
Find the measure of each equal angle.

Sol:

Let the measure of each equal angles are x.
Then we can write
140° + 5x = ( 2 x 6 - 4 ) x 90°
140° + 5x = 720°
           5x = 580°
             x = 116°
Therefore the measure of each equal angles are 116°

Question 4

In a polygon, there are 5 right angles and the remaining angles are equal to 195^o each. Find the number of sides in the polygon.

Sol:

Let the number of sides of the polygon is n and there are k angles with measure 195°.

Therefore we can write :
5 x 90° + k  x 195° = ( 2n - 4 )90°
180°n - 195° k = 450 - 360°
180°n - 195° k = 90°
   12n - 13k = 6

In this linear equation n and k must be an integer. Therefore to satisfy this equation the minimum value of k must be 6 to get n as an integer.
Hence the number of sides are: 5 + 6 = 11.

Question 5

Three angles of a seven-sided polygon are 132^o each and the remaining four angles are equal. Find the value of each equal angle.

Sol:

Let the measure of each equal angles are x.

Then we can write:
3 x 132° + 4 x = ( 2 x 7 - 4 ) 90°
4 x = 900° - 369°
4 x = 504
x = 126°
Thus the measure of each equal angles are 126°.

Question 6

Two angles of an eight-sided polygon are 142^o and 176^o. If the remaining angles are equal to each other; find the magnitude of each of the equal angles. 

Sol:

Let the measure of each equal sides of the polygon is x.
Then we can write:
142° + 176° + 6 x = ( 2 x 8 - 4 ) 90°
6 x = 1080° - 318°
6 x = 762°
x = 127°
Thus the measure of each equal angles are 127°.

Question 7

In a pentagon ABCDE, AB is parallel to DC and ∠A: ∠E : ∠D = 3: 4: 5. Find angle E.

Sol:

Let the measure of the angles are 3x, 4x and 5x.
Thus
∠A + ∠B + ∠C + ∠D + ∠E =540°
3 x + ( ∠B + ∠C ) + 4 x + 5 x = 540°
                           12x + 180° = 540°
                                      12x = 360°
                                          x = 30°
Thus the measure of angle E will be 4 × 30° = 120°

Question 8

AB, BC, and CD are the three consecutive sides of a regular polygon. If BAC = 15°;
find,
(i) Each interior angle of the polygon.
(ii) Each exterior angle of the polygon.
(iii) The number of sides of the polygon.

Sol:

(i) Let each angle of measure x degree.
Therefore the measure of each angle will be :
x = 180° - 2 x 15° = 150°

(ii) Let each angle of measure x degree.
Therefore the measure of each exterior angle will be :
x = 180° - 150° = 30°

(iii) Let the number of each side is n.
Now we can write
n.150° = ( 2n - 4 ) x 90°
180°n - 150°n = 360°
30°n = 360°
n = 12.
Thus the number of sides is 12.

Question 9

The ratio between an exterior angle and an interior angle of a regular polygon is 2 : 3. Find the number of sides in the polygon.

Sol:

Let the measure each interior and exterior angles are 3k and 2k.
Let the number of sides of the polygon is n.
Now we can write:
n.3k = ( 2n - 4 ) x 90°
3nk = ( 2n - 4 ) 90°                  ....(1)
Again
n.2k = 360°
nk = 180°
From (1)
3.180° = ( 2n - 4 ) 90°
3 = n - 2 
n = 5
Thus the number of sides of the polygon is 5.

Question 10

The difference between an exterior angle of (n - 1) sided regular polygon and an exterior angle of (n + 2) sided regular polygon is 6° find the value of n.

Sol:

For (n-1) sided regular polygon:
Let the measure of each angle is x.
Therefore,
( n - 1 )x = [ 2( n - 1 ) - 4 ] 90°

x = ${\displaystyle \frac{n-3}{n-1}}$ 180°

For (n+1) sided regular polygon:
Let measure of each angle is y.
Therefore
( n + 2 )y = [ 2( n + 2 ) - 4 ] 90°

y = ${\displaystyle \frac{n}{n+2}}$ 180°

Now we have
y - x = 6°

${\displaystyle \frac{n}{n+2}180°-\frac{n-3}{n-1}180°}$ = 6°

${\displaystyle \frac{n}{n+2}-\frac{n-3}{n-1}=\frac{1}{30}}$

30n( n - 1 ) - 30( n - 3 )( n + 2 ) = ( n + 2 )( n - 1 )
- 30n + 30n + 180 = n² + n - 2
n² + n - 182 = 0
( n - 13 )( n + 14 ) = 0
n = 13, - 14
Thus the value of n is 13.

Question 11

Two alternate sides of a regular polygon, when produced, meet at the right angle.
Find:
(i)The value of each exterior angle of the polygon;
(ii) The number of sides in the polygon.

Sol:

(i) Let the measure of each exterior angle is x and the number of sides is n.
Therefore we can write :
n = ${\displaystyle \frac{360°}{x}}$
Now We have
x + x + 90° = 180°
2x = 90°
x = 45°

(ii) Thus the number of sides in the polygon is :
n = ${\displaystyle \frac{360°}{45°}}$ 
n = 8.