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.

SChand Composite Mathematics Class 7 Chapter 14 Perimeter and Area Exercise 14A

 Exercise 14 A

Question 1 

Copy and complete the following table: 

$\begin{array}{c|c|c|c|}\hline \text { Length } & \text { Breadth } & \text { Area } & \text { Perimeter } \\\hline \text { (P) } 5 \mathrm{~m} & (3 \mathrm{~m}) & 15 \mathrm{~m}^{2} & 116 \mathrm{~m})\\\text { (ii) } 5 \mathrm{~cm} & (7 \mathrm{~m}) & \left(35 \mathrm{~m}^{2}\right) & 24\mathrm{~cm} \\\text { (iii) }(15 \mathrm{~cm}) & 9.5 \mathrm{~cm} & 142.5 \mathrm{~cm}^{2} & 49\mathrm{~cm} \\\text { (iv) }(16 \mathrm{~cm}) & 30 \mathrm{~cm} & 480 \mathrm{~cm}^{2} & (92\mathrm{~cm}) \\\hline\end{array}$

Question 2

Find (a) the area (b) the perimeter of the square PQRS if. 

 (i) $P Q=Q R=20 \mathrm{~cm} \quad$ Area $=P Q \times Q R=20 \times 20=400 \mathrm{~cm}^{2}$

Perimeter $=4 \times$ side $=4 \times 20=80 \mathrm{~cm}$

(ii) $P Q=Q R=1.2 \mathrm{~m} \quad$ Area $=P Q \times Q R=1.2 \times 1.2=1.14 \mathrm{~m}^{2}$

Perimeter $=4 \times$ side $=4 \times 1.2=4.8 \mathrm{~m}$

Question 3

Find the perimeter of the following squares: 

(i) Area = $=64 \mathrm{~cm}^{2}$

Sol:

$\begin{aligned} S^{2} &=64 \\ \text { Side } &=8 \mathrm{~cm} \end{aligned}$

Perimeter $=4 \times$ side

$=4 \times 8$

$=32 \mathrm{~cm}$

(ii) area $=196 \mathrm{~cm}^{2}$

Sol: $s^{2}=196$

Taking square root 

Side = 14 cm

Perimeter = $4 \times$ side

$=4 \times 14$

$=56 \text { Cm}$

(iii) Area = $2.25 \mathrm{m}^{2}$

Sol: $s^{2}=$ 2.25 

Taking square root side = 1.5m

P= $4 \times 1.5$

=6m

Question 4

(i) The perimeter of the square is 20cm . find its area . 

Sol: Side of square = $\frac{1}{4} \times $ perimeter of square

$=\frac{1}{4} \times 20=5 \mathrm{~cm}$

Area of square = Side $\times$ side 

$=5 \times 5=25$ $\operatorname{cm}^{2}$

(ii) The perimeter of a square field in 3km. Find its area in hectares. 

Sol: Side = $\frac{1}{4} \times $ perimeter

$=\frac{1}{4} \times 8=2 \mathrm{~km}$ or 2000m

Area of square = $side \times side$

$=2,000 \times 2000=40,00,000 \mathrm{~m}^{2}$

1 hectare = 10000 $m^{2}$

$1 m^{2}=\frac{1}{10000}$ Hectare 

$\frac{4000000}{10000}$= 400 hectare

S Chand Class 10 CHAPTER 14 Circle Exercise 14A

  Exercise 14A

Question 1

Ans: (i) $A B$ is diameter of the circle

So $\angle A C B=90^{\circ}$

So $\angle B A C+\angle A B C=90^{\circ}$

$\Rightarrow 46^{\circ}+\angle A D C=90^{\circ}$

$\Rightarrow \angle A B C=90^{\circ}-46^{\circ}=44^{\circ}$

 

(ii) $A C$ is diameter of the circle

So $\angle A D C=\angle A B C=90^{\circ}$

So $\angle B A C+\angle B C A=90^{\circ}$.

$\begin{aligned}\Rightarrow & \angle B C A=90^{\circ}-70^{\circ}=20^{\circ} \\\text { and } & \angle D A C+\angle D C A=90^{\circ} \\\Rightarrow & \text { 40 }+\angle D C A=90^{\circ} \\\Rightarrow & \angle D C A=90^{\circ}-40^{\circ}=50^{\circ} \\& \text { So } \angle B C D=\angle B C A+\angle A C D=20^{\circ}+50^{\circ}=70^{\circ}\end{aligned}$

(iii) Arc $A B$ subtends $\angle A O B$ at the centre and $\angle A C B$ at 

He remaining part of the circle So $\angle A O B=2 \angle A C B$

$\Rightarrow \angle x=2 \angle y \Rightarrow 2 \angle y=120^{\circ}$

$\Rightarrow \quad \angle y=\frac{120^{\circ}}{2}=60^{\circ}$

So $\quad \angle y=60^{\circ}$

(iv) Bc is diameter of the circle 

So $\angle B A C=90^{\circ}$

$\Rightarrow \angle B A D+\angle O A C=90^{\circ}$

$\Rightarrow \angle B A D+65^{\circ}=90^{\circ}$

$\Rightarrow \angle B A D=90^{\circ}-65^{\circ}=25^{\circ}$

But $\angle B C D=\angle B A D$

Question 2

Ans: In the figure $O$ is the centre af the circle $\angle O A B$ $=30^{\circ}$ and $\angle O C B=40^{\circ}$ Join $O B$

(IMAGE TO BE ADDED)

In $\triangle O A B, O A=O B$

So $\angle O B A=\angle O A B=30^{\circ}$

similarly in $\triangle O B C, O B=O C$

So $\angle O B C=\angle O C B=$ to

So $\angle A B C=\angle O B A+\angle O B C=30^{\circ}+40^{\circ}=70^{\circ}$

Now are $A B$ subtends $\angle A O C$ at the center

and $\angle A B C$ at the remaining part of the circle So $\angle A O C=2 \angle A B C=2 \times 70^{\circ}=140^{\circ}$.

So $\angle A O C=140$

Question 3

Ans: In the figure $O$ is the center and $\mathrm{BOC}$ is the diameter of the circle $\angle A O B=80^{\circ}$

 $A C$ and $A B$ are joined

(IMAGE TO BE ADDED)

Now arc AB subtends 

$\angle A O B$ at the center and $\angle A C B$ at the remaining part of the circle 

So $\angle A O B=2 \angle A C B$

$\Rightarrow 80^{\circ}=2 \angle A C B \Rightarrow \angle A C B=\frac{80^{\circ}}{2}=40^{\circ}$

But in $\triangle O A C \cdot O A=O C$

So $\angle O A C=\angle O C A=\angle A C B=40^{\circ}$

But $\angle B A C=90^{\circ}$

$\begin{aligned}&\Rightarrow \angle O A C+\angle O A B=90^{\circ} \\&\Rightarrow 40^{\circ}+\angle O A B=90^{\circ} \\&\Rightarrow \angle O A B=90^{\circ}-40^{\circ}=50^{\circ}\end{aligned}$

Hence $\angle O A B=50^{\circ}$ and $\angle O A C=40^{\circ}$

Question 4

Ans: In the figure 0 is the center of the circle $\angle A O B=140^{\circ}, \angle O A C=50^{\circ}$

Join $A B$

(IMAGE TO BE ADDED)

So  Reflex $\angle A O B=360^{\circ}-140^{\circ}=220^{\circ}$

Now major arc AB subtends 

Ref $\angle A O B$ at the center $\angle A C B$ at the

Remaining part of the circle

(i) So ref $\angle A O B=2 \angle A C B$

$=220^{\circ}=2 \angle A C B \Rightarrow \angle A C B=\frac{1}{2} \times 2200$

$\Rightarrow \angle A C B=110^{\circ}$

(ii)In quad. OACB

$\angle B O A+\angle O A C+\angle A C B+\angle O B C=360^{\circ}$

$=140^{\circ}+50^{\circ}+110^{\circ}+\angle O B C=360^{\circ}$

$\Rightarrow 300^{\circ}+\angle O B C=360^{\circ}$

$\Rightarrow \angle O B C=360^{\circ}-300^{\circ}=60$

(iii)

$\begin{aligned} & \angle O A B+\angle O B A+\angle A O B=180^{\circ} \\ \Rightarrow & \angle O A B+\angle O B A+140^{\circ}=180^{\circ} \\ \Rightarrow & \angle \angle O A B=180^{\circ}-140^{\circ}=40^{\circ} \\ \text { So } \angle O A B=\frac{40}{2}=20^{\circ} \end{aligned}$

(iv) In $\triangle A B C$

$\angle B A C+\angle C B A+\angle A C B=180^{\circ}$

$\begin{aligned}&\Rightarrow 30^{\circ}+\angle C B A+110^{\circ}=180^{\circ} \\&\Rightarrow 140^{\circ}+\angle C B A=180^{\circ} \\&\Rightarrow \angle C B A=180^{\circ}-140^{\circ}=40^{\circ}\end{aligned}$

Question 5

Ans: (i) In the figure $\angle A D C=50^{\circ} \angle B D C=40^{\circ}$

AB is the diameter of the circle 

If $\angle B D A=90^{\circ}$

So $\angle C O A=\angle B D A-\angle B D C=90^{\circ}-40^{\circ}=50^{\circ}$

(ii)  $\angle B C A=90^{\circ}$

So $\angle B A C+\angle A B C=90^{\circ}$

$\Rightarrow \angle D A C+50^{\circ}=90^{\circ}$

$\Rightarrow \angle B A C=90^{\circ}-50^{\circ}$

$\Rightarrow \angle B A C=40$

Hence (i) $\angle C D A=50^{\circ}$ (ii) BAC = $40^{\circ}$ and (iii) $\angle B C A=90^{\circ}$

Question 6

Ans: In the figure AC is the diameter of the circle and $\angle P C D=75^{\circ}$

ii $\begin{aligned} & \angle A B C=90^{\circ} \\ \text { ii } & \angle B C D+\angle B A D=180^{\circ} \\ \Rightarrow & 75^{\circ}+\angle B A D=180^{\circ} \\ \Rightarrow & \angle B A D=180^{\circ}-75^{\circ}=105^{\circ} \\ & \text { But } \angle E A F=\angle B A D=105^{\circ} \\ & \text { SO } \angle E A F=105^{\circ} \end{aligned}$

Question 7

Ans: A quadrilateral ABCD which is inscribed in a circle with center O. CD is produced to E $\angle A D E=$70^{\circ}$ \text { and } \angle O B A=45^{\circ}$

Join OA, OB OCI AC

$A B C D$ is a cyclic. quadrilateral

So Ext. $\angle A D E=$ interior opposite $\angle A B C$

So $\angle A B C=70^{\circ}$

(IMAGE TO BE ADDED)

Arc AC subtends $\angle A O C$ at the center and $\angle A B C$ on the

remaining part of the circle So $\angle A O C=2 \angle A B C$

$=2 \times 70^{\circ}=140^{\circ}$

Now in $\triangle O A C, O A=O C$

So $\angle O C A=\angle O A C$

But $\angle O C A+\angle O A C+\angle A O C=180^{\circ}$

$\begin{aligned}&\Rightarrow \angle O C A+\angle O C A+140^{\circ}=180^{\circ} \\&\Rightarrow 2 \angle O C A=180^{\circ}-140^{\circ}=40^{\circ}\end{aligned}$

So $\angle O C A=\frac{40^{\circ}}{2}=20^{\circ}$

and $\angle O A C=20^{\circ}$

Now $\angle B A C=\angle B A O+\angle O A C$

$=45^{\circ}+20^{\circ}=65^{\circ}$

Hence (i) $\angle B A C=65^{\circ}$ and (ii $\angle O C A=20^{\circ}$

Question 8

Ans:  In the figure ABCD is a cyclic quadrilateral in Which O is the center of the circle

$\angle A O C=120^{\circ}$

But $\angle A O C+ Ref  \cdot \angle A O C=360^{\circ}$

(IMAGE TO BE ADDED)

So Ref $\angle A O C=360^{\circ}-\angle A O C$

$=360^{\circ}-120^{\circ}=240^{\circ}$

Now arc ADC subtends $\angle A O C$ at the center and  \angle A B C at the remaining part of the circle 

So 

$\begin{aligned} & \angle A O C=2 \angle A B C \\ \Rightarrow & 120^{\circ}=2 x \end{aligned}$

$\begin{aligned} \Rightarrow & x=\frac{120^{\circ}}{2}=60^{\circ} \\ & \text { But } \angle B+\angle D=180^{\circ} \\ \Rightarrow & x+y=180^{\circ} \\ \Rightarrow & 60^{\circ}+y=180^{\circ} \\ \Rightarrow & y=180^{\circ}-60^{\circ} \\ \Rightarrow & y=120^{\circ} \end{aligned}$

Hence $x=60^{\circ}$ and $y=120^{\circ}$

Question 9

Ans: In the figure, $A B \| C D$

0 is the center of the circle and $\angle A D C=25^{\circ}$

 Join $O A$ and $O B$

(IMAGE TO BE ADDED)

if $A B \| C D$

So $\angle B A D=\angle A D C=25^{\circ}$

In $\triangle O R D \mid O A=O D$

So $\angle O A D=\angle A D O$ or $\angle A D C=25^{\circ}$

So $\angle O A B=\angle O A D+\angle B A D=25^{\circ}+25^{\circ}=50^{\circ}$

But in $\triangle O A B$

$O A=O B$

So $\angle O A B=\angle O B A=50^{\circ}$

But $\angle O A B+\angle O B A+\angle A O B=180^{\circ}$

$\begin{aligned}&\Rightarrow 50^{\circ}+50^{\circ}+\angle A O B=180^{\circ} \\&\Rightarrow \angle A O B+100^{\circ}=180^{\circ} \\&\Rightarrow \angle A O B=180^{\circ}-100^{\circ}=80^{\circ}\end{aligned}$

Now arc AB subtends  $\angle A O B$ at the centre and $\angle A E D$ at the remaining part of the circle 

So $\begin{aligned} & \angle A O B=2 \angle A E B \\ \Rightarrow & \angle A E B=\frac{1}{2}\left(A O B=\frac{1}{2} \times 80^{\circ}=40^{\circ} \right.\\ & \text { Hence } \angle A E B=40^{\circ} \end{aligned}$

Question 10

Ans: In the figure $A B C D$ is a cycric quadrilatoral In which $A D ||D C$

$\angle B C D=100^{\circ} \text { and } \angle B A C=40^{\circ}$

(IMAGE TO BE ADDED)

if $A B \| D C$ and $A C$ is its transversal

So $\angle B A C=\angle A C D=40^{\circ}$

But $\angle B C D=100^{\circ}$

So

$\angle B C A=\angle B C D-\angle A C D=100^{\circ}-40^{\circ}=60^{\circ}$

if $A B C D$ is a cyclic quadrilateral

So $\angle B A D+\angle B C D=180^{\circ}$

$\begin{aligned}&\Rightarrow \angle B A D+100^{\circ}=180^{\circ} \\

&\Rightarrow \angle B A D=180^{\circ}-100^{\circ}=80^{\circ} \\&\Rightarrow \angle B A C+\angle C A D=80^{\circ} \\&\Rightarrow 40^{\circ}+\angle C A D=80^{\circ} \\&\Rightarrow \angle C A D=80^{\circ}-40^{\circ}=40^{\circ}\end{aligned}$

But $\angle C B D=\angle C A D$

$=40^{\circ}$

Hence ii $\angle C A D=40^{\circ}$ (ii) $\angle C B D=40^{\circ}$

(iii) $\angle B C A=60^{\circ}$

Question 11

Ans: ABCD is a parallelogram and a circle pass through A and D, Intersect AB at E and DC at F

EF is joined 

$\angle B E F=80^{\circ}$

If ADFE is a cyclic quadrilateral 

So Ext. $\angle BE F=$ int $\cdot$ OPP. $\angle A B F$

So $\angle A D F=80^{\circ}$ or $\angle A D C=80^{\circ}$

But In ||gm ABCD 

$\angle A D C=\angle A B C$

So $\angle A B C=80^{\circ}$

Question 12

Ans: In the figure , AD is the diameter of the circle and ABCD is a cyclic quadrilateral $\angle B C D=125^{\circ}$

Join BD 

(IMAGE TO BE ADDED)

 

(i) If ABCD is a cyclic quadrilateral 

So  $\angle B A D+\angle B C D=180^{\circ}$

$\angle B A D+125^{\circ}=180^{\circ}$

$\angle 13 A D=180^{\circ}-125^{\circ}=55^{\circ}$

(ii) Now in $\triangle A B D$.

$\angle A B D=90^{\circ}$

So $\angle B A D+\angle A D B=90^{\circ}$

$\Rightarrow 55^{\circ}+\angle A O B=90^{\circ}$

$\Rightarrow \angle A D B=90^{\circ}-55^{\circ}=35^{\circ}$

So $\angle A D B=35^{\circ}$

Hence  $\angle B A D$ or $\angle D A B=55^{\circ}$ and  $\angle A D B=35^{\circ}$

Question 13

Ans: In the figure, 

ABCD is a cyclic trapezium in which AD||BC and $\angle B=70^{\circ}$

(i) If AD||BC 

So $\angle A B C+\angle B A D=180^{\circ}$

$70^{\circ}+\angle B A D=180^{\circ}$

$\Rightarrow \angle B A D=180^{\circ}-70^{\circ}$

SO $\angle B A D=110^{\circ}$

(ii) In Cyclic Trapezium 

$\angle B A D+\angle B C D=180^{\circ}$

$110^{\circ}+\angle B C D=180^{\circ}$

$\Rightarrow \angle B C D=180^{\circ}-110^{\circ}$

$\Rightarrow \angle B C D=180^{\circ}-110^{\circ}=70^{\circ}$

Hence $\angle B A D=110^{\circ}, \angle B C D=70^{\circ}$

Question 14

Ans: In the Figure $A B$ is the diameter as the circle $A P Q$ and $R B Q$ are straight lines $\angle A=35^{\circ}$ and $\angle Q=25^{\circ}$

(IMAGE TO BE ADDED)

(i) $\angle P R B$ and $\angle P A B$ are in the same segment So $\angle P R B=\angle P A B=35^{\circ}$

(ii) In $\triangle P B Q$

Ext. $\angle P B R=\angle B P Q+\angle Q$

$=90^{\circ}+25^{\circ}=115^{\circ}$

(iii) In $\triangle B R P$

$\angle B R P+\angle P B R+\angle BP R=180^{\circ}$

$\Rightarrow 35^{\circ}+115^{\circ}+\angle B P R=180^{\circ}$

$\Rightarrow 150^{\circ}+\angle B P R=180^{\circ}$

$\Rightarrow \angle B P R=180^{\circ}-150^{\circ}=30^{\circ}$

Question 15

Ans: In the figure O is the center of the circle and $\angle A O C=130^{\circ}$

So Reflex $\angle A O C=360^{\circ}-130^{\circ}=230^{\circ}$

Now major  arc AC subtends reflex $\angle A O C$ at the center and $\angle A B C$ at the remaining pant of the circle

So $\operatorname{Reflex} \angle A O C=2 \angle A B C$

$\Rightarrow 230^{\circ}=2 \angle A B C \Rightarrow \angle A B C=\frac{230^{\circ}}{2}=115^{\circ}$

Question 16

Ans: In the figures AOB is the diameter of the circle  $\angle B C D=140^{\circ}$

If ABCD is a cyclic quadrilateral 

So $\angle B C D+\angle B A D=180^{\circ}$

$140^{\circ}+\angle B A D=180^{\circ}$

$\Rightarrow \angle B A D=180^{\circ}-140^{\circ}$

$\angle B A B=40^{\circ}$

Now in $\triangle A O B$

$\angle A D B=90^{\circ}$

So $\angle BA D+\angle D B A=90^{\circ}$

$\Rightarrow 40^{\circ}+\angle D B A=90^{\circ}$

$\Rightarrow \angle D B A=90^{\circ}-40^{\circ}=50^{\circ}$

Question 17

Ans: In the figure ABCD is a cyclic Quadrilateral in Which AB is produced to D 

$\angle C B D=65^{\circ}$

If ABCD is a cyclic quadrilateral 

So Ext. $\angle C B D=$ interior opposite $\angle A E C$

$\Rightarrow \angle A E C=\angle C B D=65^{\circ}$

IF Arc $A B C$ subtends $\angle A O C$ at the center and LAEC at the remaining part of the circle So $\angle A O C=2 \angle A E C=2 \times 65^{\circ}=130^{\circ}$

But $\angle A O C+$ reflex $\angle A O C=360^{\circ}$

$\Rightarrow 130^{\circ}+x^{\circ}=360^{\circ} \Rightarrow x^{\circ}=360^{\circ}-130^{\circ}=230^{\circ}$

Question 18

Ans: In the figure, 

Two diagonals of a cyclic quadrilateral ABCD intersect Each other at P inside the circle.

AB= 8cm, CD = 5cm and Area $(\triangle A P B)=24 \mathrm{CB}^{2}$

If diagonals AC and BD intersect each other at P 

So, AP. PC = BP. PD 

$\begin{aligned}\Rightarrow \frac{A P}{B P}=\frac{D P}{C P} \Rightarrow \frac{A P}{D P}=\frac{B P}{C P} & \\\text { and } \angle A P B=& \angle C P D\end{aligned}$

So $\triangle A P B \backsim \triangle C P D$

So $\frac{\text { area } \triangle A P B}{\text { area } \triangle C P B}=\frac{A B^{2}}{C D^{2}}$

$\Rightarrow \frac{24 \mathrm{~cm}^{2}}{\text { area } \mathrm{△CPD}}=\frac{(8)^{2}}{(5)^{2}}$

$\Rightarrow \frac{24}{\text { area } \triangle C P D}=\frac{64}{25}$

Area △CPD = $\frac{24 \times 25}{64}=\frac{3 \times 25}{8}$

$=\frac{75}{8}=9 \frac{3}{8} \mathrm{~cm}^{2}$

Question 19

Ans: In the figure, 

(i) O is the center of the circle and $\angle B A D=30^{\circ}$

If ABCD is a cyclic quadrilateral 

So $\angle B A D+\angle B C D=180^{\circ}$

$\Rightarrow 30^{\circ}+p=180^{\circ}$

$\Rightarrow p=180^{\circ}-30^{\circ}=150^{\circ}$

Arc BCD subtends $\angle B O D$ at the center and $\angle B A D$ at the remaining part of the circle 

So  $\angle D O D=2 \angle B A D$

$q=2 \times 30^{\circ}=60^{\circ}$

if $\angle B A D$ and $\angle B E D$ are in the same segment

So

$\begin{aligned}& \angle B E D=\angle B A D \\\Rightarrow & r=300\end{aligned}$

Hence $p=150^{\circ}, q=60^{\circ}$  and $r=30^{\circ}$

Question 20

Ans: In the figure ABCD is a cyclic quadrilateral in which its diagonals AC and BD intersect each other at P 

$\angle B A D=65^{\circ}, \angle A B D=70^{\circ}$ $\angle B D C=45^{\circ}$

if $A B C D$ is a cyclic quadrilateral

So $\angle B A D+\angle B C D=180^{\circ}$

$\begin{aligned}\Rightarrow & 65^{\circ}+\angle B C D=180^{\circ} \\\Rightarrow & \angle BC D=180^{\circ}-65^{\circ}=115^{\circ}\end{aligned}$

(ii) In $\triangle A B D$,

$\angle B A D+\angle A B D+\angle A D B=180^{\circ}$

$\Rightarrow 65^{\circ}+70^{\circ}+\angle A D B=180^{\circ}$

$\Rightarrow 135^{\circ}+\angle A D B=180^{\circ}$

$\Rightarrow \angle A D B=1.80^{\circ}-135^{\circ}=45^{\circ}$

$\text { if } \angle A D C=\angle A D B+\angle B D C=45^{\circ}+45^{\circ}=90^{\circ}$

So $A D C$ is a semicircle

Hence is the diameter of the circle .

Question 21

Ans:  In the figure $A O B$ is the diameter as the circle with center O

$\angle E C D=\angle E D C=32^{\circ}$

(IMAGE TO BE ADDED)

(i) In $\triangle C DE$

Ext $\angle C E F=\angle E C D+\angle E D C=32^{\circ}+32^{\circ}=64^{\circ}$

(ii) Arc CF subtends $\angle C O F$ at the center and $\angle C D F$ or $\angle C F F$ at the remaining part of the circle

So $\angle C O F=2 \angle C D F=2 \times 32^{\circ}=64^{\circ}$

Question 22

Ans: In the figure ABCD is a cyclic quadrilateral 

AC and BD are joined 

$\angle D A C=27^{\circ}, \angle D B A=50^{\circ}, \angle A D B=33^{\circ}$ $\angle A D B=\angle A C B$

So $\angle A C B=33^{\circ}$

similarly $\angle A B D=\angle A C D$

So $\angle A C B=50^{\circ}$

So $\angle D C B=\angle A C B+\angle A C D$

$=33^{\circ}+50^{\circ}=83^{\circ}$

In cyclic quad. ABCD 

$\begin{aligned} \angle D C B+\angle B A D &=180^{\circ} \\ \Rightarrow 83^{\circ}+\angle B A D &=180^{\circ} \end{aligned}$

$\begin{aligned} \Rightarrow & \angle B A D=180^{\circ}-83^{\circ}=97^{\circ} \\ \Rightarrow & \angle D A C+\angle O A C=97^{\circ} \\ \Rightarrow & \angle B A C+27^{\circ}=97^{\circ} \\ \Rightarrow & \angle D A C=97^{\circ}-27^{\circ}=70^{\circ} \\ & or \angle C A B=70^{\circ} \\ & \angle D B C=\angle D A C \end{aligned}$

=$27^{\circ}$

Hence (i)  $\angle D B C=27^{\circ}$ (ii) $\angle D C B=83^{\circ}$  $\angle C A B=70^{\circ}$

Question 23

Ans: In the figure $A B C D E$ is a pentagon inscribed in the circle with centre 0

$A B=B C=C D$ and $\angle A B C=132^{\circ}$

Join $B E$ and $C E$

(IMAGE TO BE ADDED)

(i) In cyclic quad. $A E C B$

$\angle A E C+\angle A B C=180^{\circ}$

(IMAGE TO BE ADDED)

$\Rightarrow \angle A E C+132^{\circ}=180^{\circ}$

$\Rightarrow A E C=180^{\circ}-132^{\circ}=48^{\circ}$

if $A B=B C$

So $\angle A E B=\angle B E C$

$=\frac{48^{\circ}}{2}=24$

(ii) if $A B=B C=C D$

So $\quad \angle A E B=\angle B E C=\angle C E D=24^{\circ}$

So $\angle A E D=\angle A E B+\angle A E C+\angle C E D$

$=24^{\circ}+24^{\circ}+24^{\circ}=72^{\circ}$

(iii) Arc CD subtends $\angle C O D$ at the center and 

$\angle C E D$ at the remaining part of the circle

 So $\angle C O D=2 \angle C E D=2 \times \angle 4^{\circ}=48^{\circ}$

S.chand class 6 Mathematics Chapter 14 Exercise 14A

 Exercise 14A

Question 1

1. Add:

(i) $x, 5 x, 2 x, 3 x$

(ii) $5 n, 7 n, 4 n, 3 n$

(iii) $5 b^{2},-3 b^{2}$,

(iv) $2 x,-7 x, 4 x,-x$

(v) $-x_{1},-x,-x$

(vi) $-2 a x^{4} y^{2},-9 a x^{4} y^{2}$

(vii) $6 a^{4} b^{2} x^{2},-6 a^{4} b^{2} x^{2}$

(viii) $3(a+b), 6(a+b)$

Question 2

Simplify the following.

(i) $5 b^{2}+3 b^{2}-8 b^{2}$

(ii) $9 x^{3}-10 x^{3}+8 x^{3}$

(iii) $4 m+2 m-m+2 m-2 m$

Question 3

(i) $\begin{array}{r}x+5 y \\3 x+y \\\hline$\end{array}

(ii) $\begin{aligned}&4 m+6 n \\&5 m+2 n \\&\hline \end{aligned}$

(iii) $\begin{array}{r}p+q\\3p+4 q\\ \hline $

(iv) $\begin{array}{r} -6x-2 \\ -x-5 \\ \hline\end{array}

(v) $\begin{array}{r} x^{2}+2\\-6 x^{2}+3$\\ \hline \end{array}

(vi) $\begin{aligned}&-c^{2}+7 c \\ &-4 c^{2}+8 c \\&\hline \end{aligned}$

(vii) $\begin{aligned}&3 a b-9 x y \\&7 a b+4 x y \\ &\hline \end{aligned}$

(viii) $\begin{aligned}&2 x^{2}-5 x \\ &9 x^{2}+2 x \\ &\hline \end{aligned}$

Question 4

Add the following.

(i) $\begin{aligned} &x^{2}+2 x-7 \\&x^{2}-5 x+3 \\ &\hline \end{aligned}$

(ii) $\begin{aligned} &5 a^{2}-2 a b-9 b^{2} \\ &-a^{2}+4 a b-b^{2} \\ &\hline \end{aligned}$

(iii) $\begin{array}{r} x^{2}-3 x+4 \\ 4 x^{2}-2 x+7 \\ -x^{2}+5 x-2 \\ \hline \end{array}$

Question 5

(i) $5 x^{2}+5 x-6$ and $4 x^{2}-x+8$

(ii) $3 y^{2}-4 y+5,2 y^{2}-7 y-1$, and $y^{2}-3 y-5$

Question 6

Simplify:

(i) $(2 x-y)+(2 y-3 x)+(3 y-x)$

(ii) $3 y^{2}-6-2 y+y^{2}-3 y+7-y^{2}-4 y-y^{2}+5$

Multiple Choice Questions (MCQ)

Tick (✔) the correct option.

7.If $P=2 a-5 b-7 c, Q=9 b-6 a-10 c$ and $R=17 c-4 b+4 a$, then $P+Q+R$ equals

(a) $15 a-9 b$

(b) $-34 c$

(c) 0

(d) $-12 a-10 b-7 c$

RS Aggarwal solution class 8 chapter 14 Polygons Exercise 14A

Exercise 14A

Page-182

Q1 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 1:

Find the measure of each exterior angle of a regular
(i) pentagon
(ii) hexagon
(iii) heptagon
(iv) decagon
(v) polygon of 15 sides.

Answer 1:

Exterior angle of an n-sided polygon = ${\left(\frac{360}{n}\right)}^o$
(i) For a pentagon:

(ii) For a hexagon:

(iii) For a heptagon:

(iv) For a decagon:

(v) For a polygon of 15 sides:

Q2 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 2:

Is it possible to have a regular polygon each of whose exterior angles is 50°?

Answer 2:

Each exterior angle of an n-sided polygon = ${\left(\frac{360}{n}\right)}^o$
If the exterior angle is 50°, then:



Since n is not an integer, we cannot have a polygon with each exterior angle equal to 50°.

Q3 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 3:

Find the measure of each interior angle of a regular polygon having
(i) 10 sides
(ii) 15 sides.

Answer 3:

For a regular polygon with n sides:


(i) For a polygon with 10 sides:


(ii) For a polygon with 15 sides:
 

Q4 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 4:

Is it possible to have a regular polygon each of whose interior angles is 100°?

Answer 4:

Each interior angle of a regular polygon having n sides =

If each interior angle of the polygon is 100°, then:



Since n is not an integer, it is not possible to have a regular polygon with each interior angle equal to 100°.

Q5 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 5:

What is the sum of all interior angles of a regular
(i) pentagon
(ii) hexagon
(iii) nonagon
(iv) polygon of 12 sides?

Answer 5:

Sum of the interior angles of an n-sided polygon =

(i) For a pentagon:


(ii) For a hexagon:
 

(iii) For a nonagon:


(iv) For a polygon of 12 sides:

Q6 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 6:

What is the number of diagonals in a
(i) heptagon
(ii) octagon
(iii) polygon of 12 sides?

Answer 6:

Number of diagonal in an n-sided polygon = $\frac{n\left(n-3\right)}{2}$
(i) For a heptagon:

 $n=7\Rightarrow \frac{n\left(n-3\right)}{2}=\frac{7\left(7-3\right)}{2}=\frac{28}{2}=14$

(ii) For an octagon:

 $n=8\Rightarrow \frac{n\left(n-3\right)}{2}=\frac{8\left(8-3\right)}{2}=\frac{40}{2}=20$

(iii) For a 12-sided polygon:

 $n=12\Rightarrow \frac{n\left(n-3\right)}{2}=\frac{12\left(12-3\right)}{2}=\frac{108}{2}=54$

Q7 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 7:

Find the number of sides of a regular polygon whose each exterior angle measures:
(i) 40°
(ii) 36°
(iii) 72°
(iv) 30°

Answer 7:

Sum of all the exterior angles of a regular polygon is ${360}^o$​.

(i)


(ii)


(iii)


(iv)

Q8 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 8:

In the given figure, find the angle measure x.

Answer 8:

Sum of all the interior angles of an n-sided polygon =


∴ x = 105

Q9 | Ex-14A | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 9:

Find the angle measure x in the given figure.

Answer 9:

For a regular n-sided polygon:
Each interior angle = $180-\left(\frac{360}{n}\right)$
In the given figure:

∴ x = 108