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

Question 0.1

State, 'true' or 'false'
The diagonals of a rectangle bisect each other.

Sol: True

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Question 0.2

State, 'true' or 'false'
The diagonals of a quadrilateral bisect each other.

Sol: False

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Question 0.3

State, 'true' or 'false'
The diagonals of a parallelogram bisect each other at right angle.

Sol: False

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Question 0.4

State, 'true' or 'false'
Each diagonal of a rhombus bisects it.

Sol: True

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Question 0.5

State, 'true' or 'false'
The quadrilateral, whose four sides are equal, is a square.

Sol: False

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Question 0.6

State, 'true' or 'false'
Every rhombus is a parallelogram.

Sol: True

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Question 0.7

State, 'true' or 'false' 
Every parallelogram is a rhombus.

Sol: False

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Question 0.8

State, 'true' or 'false'
 Diagonals of a rhombus are equal.

Sol: False 

 Question 0.9

State, 'true' or 'false' 

If two adjacent sides of a parallelogram are equal, it is a rhombus.

Sol: True

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Question 1.1

State, 'true' or 'false'
 If the diagonals of a quadrilateral bisect each other at right angle, the quadrilateral is a square.

Sol: False

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Question 2

In the figure, given below, AM bisects angle A and DM bisects angle D of parallelogram ABCD. Prove that: ∠AMD = 90°^.

Sol:

From the given figure we conclude that
∠A + ∠D = 180°  ...[Since consecutive angles are supplementary ]

${\displaystyle \frac{\angle A}{2}+\frac{\angle D}{2}}$= 90°

Again from the ΔADM

${\displaystyle \frac{\angle A}{2}+\frac{\angle D}{2}}$ + ∠M = 180°   
⇒ 90° + ∠M = 180°     ...${\displaystyle [\sin ce\frac{\angle A}{2}+\frac{\angle D}{2}=90°]}$
⇒ ∠M = 90°
Hence ∠AMD = 90°

Question 3

In the following figure, AE and BC are equal and parallel and the three sides AB, CD, and DE are equal to one another. If angle A is 102^o. Find angles AEC and BCD.

Sol:

In the given figure

Given that AE =BC
We have to find ∠AEC  ∠BCD
Let us Join EC and BD.

In the quadrilateral AECB
AE = BC and AB = EC
also AE || BC
⇒ AB || EC
So quadrilateral is a parallelogram

In parallelogram consecutive angles are supplementary
⇒  ∠A + ∠B = 180°
⇒  102° + ∠B = 180°
⇒  ∠B = 78°
In parallelogram  opposite angles are equal
⇒ ∠A = ∠BEC and ∠B = ∠AEC
⇒ ∠BEC = 102° and ∠AEC = 78°

Now consider ΔECD
EC = ED = CD [SInce AB = EC ]

Therefore ΔEDC is an equilateral triangle.
⇒ ∠ECD = 60°
∠BCD = ∠BEC + ∠ECD
⇒ ∠BCD = 102° + 60°
⇒ ∠BCD = 162°
Therefore ∠AEC = 78° and ∠BCD = 162°

Question 4

In a square ABCD, diagonals meet at O. P is a point on BC such that OB = BP.

Show that: 

(i) ∠POC =  ${\displaystyle \left[22\frac{1°}{2}\right]}$

(ii) ∠BDC = 2 ∠POC

(iii) ∠BOP = 3 ∠CPO

SOl:

Let ∠POC = x°
Diagonals of a square bisect the angles.
∴  ∠OCP = ∠OBP = ${\displaystyle \frac{90°}{2}}$ = 45°
Using exterior angle property for ΔOPC,
∠OPB = ∠OCP + ∠POC
⇒ ∠OPB = 45° + x°           ....(i)

In ΔOBP,
OB = BP                ...(Given)
∴ ∠OPB = ∠BOP    ..( angles opposite to equal sides are equal )
⇒ ∠BOP = 45° + x  ...(ii) ...[ From (i) ]

Diagonals of a Square are perpendicular to  each other.
∴ ∠BOP + ∠POC = 90°
⇒ 45 + x + x = 90°
⇒ X = 22.5°
⇒ ∠POC = x° = ${\displaystyle \left[22\frac{1°}{2}\right]}$

Now ⇒ ∠BDC = 45°   .....( Diagonals of a square bisects the angles )
⇒ ∠BCD = 2 x 22.5° = 2∠POC

And ,∠BOP = 45° + x°      ...[ From (ii) ]
⇒ ∠BOP = 45° + 22.5° = 3 x 22.5° = 3∠POC

Question 5

The given figure shows a square ABCD and an equilateral triangle ABP.

Calculate: (i) ∠AOB
                (ii) ∠BPC
                (iii) ∠PCD
                (iv) Reflex ∠APC

Sol:

In the given figure ΔAPB is an equilateral triangle.
Therefore all its angles are 60°
Again in the
ΔADB,
∠ABD = 45°
∠AOB = 180° - 60° - 45° = 75°

Again
ΔBPC
⇒ ∠BPC = 75°              ....[ Since BP = CB ]
Now,
∠C = ∠BCP + ∠PCD 
⇒ ∠PCD = 90° - 75°
⇒ ∠PCD = 15°
Therefore,
∠APC = 60° + 75°
⇒ ∠APC = 135°
⇒ Reflex ∠APD = 360° - 135° = 225°

(i) ∠AOB = 75°
(ii) ∠BPC = 75°
(iii) ∠PCD = 15°
(iv) Reflex ∠APC = 135°.  Reflex ∠APD = 225°

Question 6

In the given figure ABCD is a rhombus with angle A = 67°

If DEC is an equilateral triangle, calculate:
(i) ∠CBE
(ii) ∠DBE.

Sol:

Given that the figure ABCD is a rhombus with angle A = 67^o

In the rhombus we have
∠A = 67° = ∠C                 ....[ Opposite angles ]
∠A + ∠D = 180°               ....[ Consecutive angles are supplementary. ]
⇒ ∠D = 113°
⇒ ∠ABC = 113°

Consider ΔDBC,
DC = CB                          ....[ Sides of rhombous ]
So, ΔDBC is an isoscales triangle,
⇒ ∠CDB = ∠CBD
Also,
∠CDB + ∠CBD + ∠BCD = 180°
⇒ 2∠CBD = 113°
⇒ ∠CDB = ∠CBD = 56.5°    ....(i)

Consider ΔDCE,
EC = CB
So ΔDCE is an isoscales triangle
⇒  ∠CBE = ∠CEB
Also,
∠CBE + ∠CEB + ∠BCE = 180°
⇒ 2∠CBE = 53°
⇒ ∠CDE = 26.5°

From (i)
∠CBD = 56.5°
⇒ ∠CBE + ∠DBE = 56.5°
⇒ 26.5° + ∠DBE = 56.5°
⇒ ∠DBE = 56.5° - 26.5° = 30°

Question 7.1

In the following figures, ABCD is a parallelogram.

find the values of x and y.

SoL:

ABCD is a parallelogram.
Therefore
AD = BC
AB = DC
Thus
4y = 3x - 3                ....[ Since AD = BC ]
⇒ 3x - 4y = 3            ....(i)
6y + 2 = 4x               ....[ Since AB = DC ]
⇒  4x - 6y = 2           ....(ii)

Solving equations (i) and (ii) we have
x = 5
y = 3.

Question 7.2

In the following figures, ABCD is a parallelogram.

find the values of x and y.

SOl:

In the figure, ABCD is a parallelogram
∠A = ∠C 
∠B = ∠D              .....[ Since opposite angles are equal. ]
Therefore,
7y = 6y + 3y - 8°       ...(i)( Since ∠A = ∠C )
4x + 20° = 0              ...(ii)

Solving (i), (ii) we have
x = 12° 
Y = 16°

Question 8

The angles of a quadrilateral are in the ratio 3: 4: 5: 6. Show that the quadrilateral is a trapezium.

Sol:

Given that the angles of a quadrilateral are in the ratio 3:4:5:6 
Let the angles be 3x, 4x, 5x, 6x.
3x + 4x + 5x + 6x = 360°
⇒ x = ${\displaystyle \frac{360°}{18}}$

⇒ x =  20°
Therefore the angles are
3 x 20 = 60°
4 x 20 = 80°
5 x 20 = 100°
6 x 20 = 120°
Since all the angles are of different degrees thus forms a trapezium.

Question 9

In a parallelogram ABCD, AB = 20 cm and AD = 12 cm. The bisector of angle A meets DC at E and BC produced at F.
Find the length of CF.

Sol:

Given AB = 20 cm and AD = 12 cm.
From the above figure, it's evident that ABF is an isosceles triangle with angle BAF = angle BFA = x
So AB = BF = 20
BF = 20
BC + CF = 20
CF = 20 - 12 = 8 cm

Question 10

In parallelogram ABCD, AP and AQ are perpendiculars from the vertex of obtuse angle A as shown.
If  ∠x: ∠y = 2: 1.
find angles of the parallelogram.

Sol:

We know that AQCP is a quadrilateral. So sum of all angles must be 360.
∴ x + y + 90 + 90 = 360
x + y = 180
Given x : y = 2 : 1
So substitute x = 2y
3y = 180
y = 60
x = 120

We know that angle C = angle A = x = 120
Angle D = Angle B = 180 - x = 180 - 120 = 60
Hence, angles of a parallelogram are 120, 60, 120 and 60.

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

 Exercise 14 B

Question 1 

Find the area of the shaded portion in each case. 

(i) (Image to be added)

Sol: Area of shaded portion = Area of outer square -Area of Inner square 

$=8^{2}-4^{2}$

$=64-16$

$=48 m^{2}$

(ii)  (Image to be added) 

Sol:  : Area of shaded portion = Area of outer square -Area of Inner square

$=(24 \times 20)-(22 \times 18)$

$=480-396$

$=84 \mathrm{~m}^{2}$

Question 2

A photograph of sides 35 cm by 22 cm is mounted into a frame of external dimension 45 cm by 30 cm . find the area of the border surrounding the photograph . 

Sol: Area of border = External area of frame - Area of photograph 

$=(45 \times 30)-(35 \times 22)$

$=1350-770$

$=580 \mathrm{~cm}^{2}$

Question 3

A Verandah 1.25 m wide is constructed all along the outside of  a room 5.5m long and 4m wide. find the cost of cementing the floor of this verandah at the rate of Rs 15 per sqm.

Sol: (IMAGE TO BE ADDED)

To find the cost cementing the floor of verandah we have to find the area. 

Outer width = $\begin{aligned} & 5.5+1.25+1.21 \\=& 8 m \end{aligned}$

Outer length = $4+1.25+1.25=6.5 \mathrm{~m}$ 

Area of verandah = Area of outer - Area of inner 

$=(8 \times 6.5)-(5.5 \times 4)$

$=52-22$

$=30 \mathrm{~m}^{2}$

Cost of cementing floor = $15 \times 30$

= Rs 450 Answer 

Question 4

A sheet of paper measuring 30 cm by 20 cm. A strip 4cm wide is cut from it all around . Find the area of remaining sheet and also the area of the strip cut out . 

Sol: Width after cutting 4cm= 30-4+4

$=30-8=22 \mathrm{~cm}$

Length = 20 - 4+4= $20-8=12 \mathrm{~cm}$

Area of remaining sheet =  L$ \times $ B

$=22 \times 12=264 \mathrm{~cm}^{2}$

Area of strip = $(30 \times 20)-(264)$

$600-264=336 \mathrm{~cm}^{2}$

Question 5

Find the area of the crossroads at right angles to each other through the center of the field . 

Sol: (IMAGE TO BE ADDED)

Area of cross roads 

=Area of ABCD+ EFGH -IJKL 

$75 \times 2+62 \times 2-2 \times 2$

$=150+124-4$

$274-4=270 \mathrm{~m}^{2}$ Answer 

 

S Chand Class 10 CHAPTER 14 Circle Exercise 14B

 Exercise 14B

Question 1

Ans: In the figure in $\triangle A B C, A B=A C \quad \times XY|| B C$

To prove : $B C Y X$ is a cyclic quadrilateral proof: In $\triangle A B C$,

$\begin{aligned}&\text { XY } || B C \\&\text { So } \angle A X Y=\angle A B C \\&\text { But } \angle A B C=\angle A C B \\&\text { So } \angle A XY=\angle A C B\end{aligned}$

But Ext. $\angle A x y$ is equal to its interior opp. $\angle A C B$ So $B C YX$ is a cyclic quadrilateral.

Question 2

Ans: In the figure in quad. ABCD BC is produced to X Ext. $\angle X C D=$ int OPP. $\angle A$

To prove: Quad. $A B C D$ is a cyclic

Proof: $\angle X C D+\angle D C B=180^{\circ}$

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

So Quad $A B C D$ is a cyclic. Hence proved

Question 3

Ans: In $\triangle P Q R, P Q=P R$

A circle passing through $Q$ and $R$, intersects $P Q$ and $P R$ at $S$ and T respectively. S is Joined

To prove: ST||QR 

Proof : In $\triangle P Q R$

$P Q=P R$

So $\angle Q=\angle R$........(i)

If SQRT is a cyclic quadrilateral 

So Ext $\angle S=$ int $\cdot O P P \cdot \angle R$......(ii)

from ( i) and (ii)

$\angle Q=\angle S$

But these are corresponding angle 

So QR||ST 

Question 4

Ans: Two circles with center $\mathrm{O}_{1}$ and $\mathrm{O}_{2}$ Intersect each other at A and B 

$A O_{1} C$ and $A O_{2} D$ are diameters 

To prove: D, B and C are in a straight line or D, B and C are Collinear

Proof: If AC is the diameter of circle of center $\mathrm{O}_{1}$ 

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

similarly

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

Adding we get 

$\angle A B C+\angle A B D=90^{\circ}+90^{\circ}=180^{\circ}$

So CBD is a straight line

Hence $D, B$ and $C$ are in the same straight line.

Hence proved 

Question 5

Ans: In circle with center O, AB is its diameter AD $\perp XY$ and $B C+ XY$ Which intersect the circle at E 

To prove: CE =AD 

Construction : Join A, E.

(IMAGE TO BE ADDED)

Proof: $\angle A E B=90^{\circ}$

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

But $\angle C=\angle D=90^{\circ}$

so $A E C D$ is a rectangle

if Opposite side of a rectangle are equal

So $\quad \angle E=A D$

Question 6

Ans: In a circle with center O, AC is its Diameter and Chord AB||CD

To prove: AB= CD 

Construction : 

(IMAGE TO BE ADDED)

Proof: In $\triangle A O B$ and $\triangle C O D$.

$\begin{aligned}&O A=O C \\&O B=O D \\&\angle B A O=\angle O C D \\&\text { SO } \triangle A O B \cong \triangle C O D \\&\text { SO } A B=C D\end{aligned}$

Question 7

Ans: (IMAGE TO BE ADDED)

Two circles circles intersect each other at $P$ and $Q$ Lines $A P B$ and $C Q D$ are drawn from the point of intersection Respectively P Q, A C, BD and, $A Q, O B$ CP and $P D$ are Joined.

To prove: 

(i) AC||BD (ii) $\angle C P D=\angle A Q B$

Proof: (i) If APQC is a cyclic quad 

So Ext. $\angle B P Q=$ int OPP $\angle C$ ..............(i)

If PBDQ is a cyclic quad.

So $\quad \angle B P Q+\angle D=180^{\circ}$

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

But These are Co-interior angles 

So AC||BQ

(ii) In $\triangle A Q B$ and $\triangle C P D$,

$\angle P A Q=P C Q$

$\angle P B Q=\angle P D Q$

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

So $\triangle A Q B \sim \triangle A P D$

So Third angler $=$ Third angle

$\Rightarrow \angle A Q B=\angle C P D$

Hence proved 

Question 8

Ans: (a) ABCD is a rhombus whose diagonals AC and BD intersects each other at O. A circle With AB as diameter is drawn 

(IMAGE TO BE ADDED)

To prove: The circle passes through O 

Proof: Let the circle drawn on AB as diameter does not passes through O, Let it intersect AC at P 

Join PB 

The  $\angle A P B=90^{\circ}$

 But $\angle A O B=90^{\circ}$

So $\angle A P B=\angle A O B$

But it is not possible because  $\angle A P B$ is the exterior angle of Triangle OPB and an Exterior angle of a triangle is always greater than its interior opposite angle So our supposition is wrong 

Hence the circle will pass through O 

Question 9

Ans: $A n$ isosceles trapezium $A B C D$ in which $A D \| B C$ and

A B=D C

To prove: $A B C D$ is cyclic

Construction: Draw AE and DF perpendicular on BC 

Proof: In right $\triangle A B E$ and $\triangle D C F$

Hyp. $A B=D C$ 

side $A E=D F$

$\text { So } \triangle A B E \cong \triangle D C F$

So $\angle B=\angle C$

Now AD $\| B C$

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

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

But here are sum of opposite angles of a quad 

So ABCD is a circle 

Question 10

Ans: PQRS is a cyclic quadrilateral

$\angle A, \angle B, \angle C$ and $\angle D$ are angles in the four segment so formed exterior to the cyclic quadrilateral

To prove: $\angle A+\angle B+\angle C+\angle D=6$ right angles

constructions: Join $A S$ and $A R$

Proof: In cyclic quad .$AS D P$

$\angle P A S+\angle D=2 \mathrm{rt}$ angles...........(i)

Similarly in cyclic quad ARBQ

$\angle R A Q+\angle B=2 \mathrm{rt}$. angles.......(ii)

and In cyclic quad .ARCS 

$\angle S A R+\angle C=2 rt$. angles.........(iii)

Adding (i), (ii)and (iii)

$\angle P A S+\angle D+\angle R A Q+\angle B+\angle S A R+\angle C$

$=2+2+2=6 \mathrm{rt}$. angles

$\Rightarrow \angle P A S+\angle S A R+\angle R A Q+\angle B+\angle C+\angle D$

=6 rt-angles

$\Rightarrow \angle A+\angle B+\angle C+\angle D=6 \mathrm{rt.}$ angles

Question 11

Ans: O is the circumcenter of $\triangle A B C O D \perp B C \cdot O B$ and $O C$ are Joined

To prove: $\angle B O D=\angle A$

Proof: arc $B C$ subtends $\angle B O C$ at the center

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

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

In $\triangle O B C, O D \perp B C$

O B=O C

So OD bisects $\angle B O C$

$\Rightarrow \angle B O D=\frac{1}{2} \angle B O C$...........(ii)

From in and (ii)

$\angle B O D=\frac{1}{2} \times 2 \angle B A C=\angle B A C$

Hence proved.

Question 12

Ans: ABCD is a cyclic quadrilateral 

A circle passing through A and B meet AD and BC at E and F respectively EF is joined 

To prove: EF||DC 

Proof : IF ABCD is a cyclic quad.

So $\angle 1+\angle 3=180^{\circ}$.....(i)

Similarly ABFE is a cyclic quadrilateral 

So $\angle 1+\angle 2=180^{\circ}$........(ii)

from (i) and (ii)

$\begin{aligned}& \angle 1+\angle 3=\angle L+\angle 2 \\\Rightarrow & \angle 3=\angle 2\end{aligned}$

But these are corresponding angles 

So EF || DC Hence proved

S.chand class 6 Mathematics Chapter 14 Exercise 14B

 Exercise 14B

Question 1

1. Subtract:

(i) $3 x$ from $5 x$

(ii) $-9 x y z$ from $7 x y z$

(iii) $-7 a$ from $-3 a$

(iv) $4 b^{2}$ from $-6 b^{2}$

(v) $\frac{-2}{3} p^{3}$ from $\frac{1}{3} p^{3}$

(vi) $3(a-b-c)$ from $-4(a-b-c)$

2. Subtract:

(i) $-5 x^{2} y$ from $9 x^{2} y$

(ii) $-9 y^{2}$ from $3 x^{2}$

(iii) $3 m-8 n+5 p$ from $7 m+5 n-3 p$

3. Subtract as indicated.

(i) $\left(x^{2}\right)-(-x)^{2}$

(ii) $\left(-8 a^{2} b\right)-\left(-8 a^{2} b\right)$

(iii) $(4 p+q)-(p-q)$

(iv) $\left(7-x+x^{2}\right)-\left(x^{2}+6-3 x\right)$

4.Subtract the following.

(i) $\begin{aligned}&6 x+8 y \\&2 x+6 y \\&\hline\end{aligned}$

(ii) $\begin{array}{r}p-3 q \\4 p-5 q \\\hline\end{array}$

(iii) $\begin{array}{r} a^{2}+b^{2}\\2 a^{2}-3 b^{2}\\ \hline$

(iv) $\begin{aligned}&8 x^{2}+x y-4 y^{2} \\&3 x^{2}-3 x y+y^{2}\\ \hline \end{aligned}$

5.Take away:

(i) $4 x^{2} y-8$ from $7 x^{2} y-3$

(ii) $2 y^{2}+y z-2 z^{2}$ from $9 y^{2}-3 z^{2}$

6. Subtract:

(i) $x^{2}-3 x y+y^{2}$ from $4 x y-3 x^{2}-2 y^{2}$

(ii) $1-p+p^{2}$ from $p^{2}+p-1$

Multiple Choice Question (MCQ)

Tick (✔) the correct option.

7.What should be added to $4 c^{2}-3 b^{2}+1$ to get $a^{2}+b^{2}+c^{2}$ ?

(a) $-a^{2}-4 b^{2}+3 c^{2}+1$

(b) $a^{2}+4 b^{2}-3 c^{2}-1$

(c) $-a^{2}-4 b^{2}-3 c^{2}-1$

(d) $a^{2}-4 b^{2}+3 c^{2}-1$

RS Aggarwal solution class 8 chapter 14 Polygons Exercise 14B

Exercise 14B

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

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Question 1:

Tick (✓) the correct answer:
How many diagonals are there in a pentagon?
(a) 5
(b) 7
(c) 6
(d) 10

Answer 1:

(a) 5

For a pentagon:
$n=5$

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

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Question 2:

Tick (✓) the correct answer:
How many diagonals are there in a hexagon?
(a) 6
(b) 8
(c) 9
(d) 10

Answer 2:

(c) 9
Number of diagonals in an n-sided polygon = $\frac{n\left(n-3\right)}{2}$
For a hexagon:

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

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Question 3:

Tick (✓) the correct answer:
How many diagonals are there in an octagon?
(a) 8
(b) 16
(c) 18
(d) 20

Answer 3:

(d) 20

​For a regular n-sided polygon:
Number of diagonals =: $\frac{n\left(n-3\right)}{2}$
For an octagon:

 

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

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Question 4:

Tick (✓) the correct answer:
How many diagonals are there in a polygon having 12 sides?
(a) 12
(b) 24
(c) 36
(d) 54

Answer 4:

(d) 54
For an n-sided polygon:
Number of diagonals = $\frac{n\left(n-3\right)}{2}$

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

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Question 5:

Tick (✓) the correct answer:
A polygon has 27 diagonals. How many sides does it have?
(a) 7
(b) 8
(c) 9
(d) 12

Answer 5:

(c) 9

Page-183

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

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Question 6:

Tick (✓) the correct answer:
The angles of a pentagon are x°, (x + 20)°, (x + 40)°, (x + 60)° and (x + 80)°. The smallest angle of the pentagon is
(a) 75°
(b) 68°
(c) 78°
(d) 85°

Answer 6:

(b) 68°
​Sum of all the interior angles of a polygon with n sides =

°

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

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Question 7:

Tick (✓) the correct answer:
The measure of each exterior angle of a regular polygon is 40°. How many sides does it have?
(a) 8
(b) 9
(c) 6
(d) 10

Answer 7:

(b) 9

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

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Question 8:

Tick (✓) the correct answer:
Each interior angle of a polygon is 108°. How many sides does it have?
(a) 8
(b) 6
(c) 5
(d) 7

Answer 8:

(c) 5
​Each interior angle for a regular n-sided polygon = $180-\left(\frac{360}{n}\right)$

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

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Question 9:

Tick (✓) the correct answer:
Each interior angle of a polygon is 135°. How many sides does it have?
(a) 8
(b) 7
(c) 6
(d) 10

Answer 9:

(a) 8
​

Q10 | Ex-14B | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 10:

Tick (✓) the correct answer:
In a regular polygon, each interior angle is thrice the exterior angle. The number os sides of the polygon is
(a) 6
(b) 8
(c) 10
(d) 12

Answer 10:

(b) 8
For a regular polygon with n sides:
Each exterior angle = $\frac{360}{n}$
Each interior angle = $180-\frac{360}{n}$

Q11 | Ex-14B | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 11:

Tick (✓) the correct answer:
Each interior angle of a regular decagon is
(a) 60°
(b) 120°
(c) 144°
(d) 180°

Answer 11:

(c) 144°
Each interior angle of a regular decagon =

Q12 | Ex-14B | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 12:

Tick (✓) the correct answer:
The sum of all interior angles of a hexagon is
(a) 6 right ∠s
(b) 8 right ∠s
(c) 9 right ∠s
(d) 12 right ∠s

Answer 12:

(b) 
Sum of all the interior angles of a hexagon is right angles.
For a hexagon:

Q13 | Ex-14B | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 13:

Tick (✓) the correct answer:
The sum of all interior angles of a regular polygon is 1080°. What is the measure of each of its interior angles?
(a) 135°
(b) 120°
(c) 156°
(d) 144°

Answer 13:

(a) 135°

Q14 | Ex-14B | Class 8 | RS AGGARWAL | chapter 14 | Polygons 

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Question 14:

Tick (✓) the correct answer:
The interior angle of a regular polygon exceeds its exterior angle by 108°. How many sides does the polygon have?
(a) 16
(b) 14
(c) 12
(d) 10

Answer 14:

(d) 10

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