Maths M.C.Q. [1 Marks Each]

Number of sides $\frac{360^\circ}{24^\circ}=15$

Each exterior angle of a regular n-sided polygon $=\frac{160}{\text{n}}=40$
$\Rightarrow\text{n}=\frac{360}{\text{40}}=9$

We know that, the measures of each exterior angle of a polygon having n sides is given by $\frac{360^\circ}{\text{n}}$
$\therefore$ The number of sides, $\text{n}=\frac{360^\circ}{\text{Exterior angle}}=\frac{360^\circ}{180^\circ}=\frac{360^\circ}{45^\circ}=8$

 In $\triangle\text{BDC}=90^\circ$
Using Pythagoras Theorem, We have,
$\text{BC}^2=\text{BD}^2+\text{CD}^2$
$\Rightarrow\text{BC}^2=10^2+24^2=100+576$
$\Rightarrow\text{BC}^2=676$
$\Rightarrow\text{BC}=\sqrt{676}$
$\Rightarrow\text{BC}=256\text{cm}$

 Fourth angle $= 360^\circ - (70^\circ + 70^\circ + 70^\circ )$
$= 150^\circ .$

 The diagonals of a rectangle are equal in length.
$2x + 1 = 3x - 1$
$1 + 1 = 3x - 2x$
$2 = x$
Thus, the value of $x$ is $2.$

 For an n-sided polygon:
Number of diagonals $=\frac{\text{n}(\text{n}-3)}{2}$
$\therefore\text{n}=12$
$\Rightarrow\frac{12(12-3)}{2}=54$

$n = 8$
$(n - 2) 180^\circ = 1080^\circ .$

$\because ABCD$ is a Quadrilateral
To Proof $\angle\text{1}+ \angle\text{2}+\angle\text{3}+\angle\text{4}=360^\circ$
$\angle\text{A}+ \angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$\angle\text{A}+ \angle\text{1}=180^\circ$
$\angle\text{B}+ \angle\text{2}=180^\circ$
$\angle\text{C}+ \angle\text{3}=180^\circ$
$\angle\text{D}+ \angle\text{4}=180^\circ$
$\angle\text{A}+ \angle\text{B}+\angle\text{C}+\angle\text{D}+ \angle\text{1}+ \angle\text{2}+\angle\text{3}+\angle\text{4}$
$= 180^\circ+180^\circ+180^\circ+180^\circ$
$=720^\circ$
$360^\circ+\angle1+\angle2+\angle3+\angle4=720^\circ$
$\angle1+\angle2+\angle3+\angle4=360^\circ$
Each polygon have $360^\circ $ as it's exterior angel but it's Interior angle is depend on it's sides, so the answer is $D.$

Since of increasing size of regular polygon $\Rightarrow $ its exterior angle decreases.
$\therefore$ Mass exterior angle is for a regular triangle (equilateral triangle) and $= 180 -$ interior angle
$= 180^\circ - 60^\circ = 120^\circ $

Given, length of rectangle is $4\ cm$
Breadth of rectangle $= 2\ cm$
By the formula of perimeter of rectangle, we know that;
Perimeter $= 2 ($Length $+$ Breadth$)$
$P = 2(4 + 2)$
$P = 2 × 6$
$P = 12\ cm.$

 For a regular polygon with $n$ sides:
Each exterior angle $=\frac{ 360}{ \text{n}}$
Each interior angle $=180-\frac{360}{\text{n}}$
$\therefore180-\frac{360}{\text{n}}=3\Big(\frac{360}{\text{n}}\Big)$
$\Rightarrow180=4\Big(\frac{360}{\text{n}}\Big)$
$\Rightarrow\text{n}=\frac{4\times360}{180}=8$

Perimeter of parallelogram $= 2$ (Sum of Parallel sides)
$P = 2 (12 + 7)$
$P = 2 (19)$
$P = 38\ cm$

$(n - 2) 180^\circ = 10 \times 90^\circ $
$\Rightarrow n = 7.$

For a pentagon:
$n = 5$
Number of diagonals $=\frac{\text{n}(\text{n}-3)}{2}=\frac{5(5-3)}{2}=5$

$n = 7$
$(n - 2) 180^\circ = 900^\circ$

Measure of each angle
$= \frac{360^\circ}{4}=90^\circ$

Each interior angle of a regular decagon $=180-\frac{360}{10}=180-36=144^\circ$

A convex quadrilateral is a four sided figure with interior angles of less than $180$ degrees each and both of its diagonals contained within the shape. It has got two Diagonals.

Required measure $= \frac{360^\circ}{9}=40^\circ$

We know that the sum of exterior angles, taken in order of any polygon is $360^\circ $ and triangle is also a polygon. Hence, the sum of all exterior angles of a triangle is $360^\circ .$

Perimeter $= 2 (4 + 2)cm = 12cm.$

We know that, the sum of all the angles of a quadrilateral is $360^\circ $. Also, an obtuse angle is more than 90^\circ and less than $180^\circ $.

$\therefore (2x + 25) + (3x - 5) = 180$
$\Rightarrow 2x + 25 + 3x - 5 = 180$
$\Rightarrow 5x = 180 - 20$
$\Rightarrow 5x = 160$
$\Rightarrow x = 32$

Perimeter $= 4$ side
$= 4 \times 6 = 24cm$

Required measure $= \frac{360^\circ}{15}=24^\circ$

Let the fourth angle be $x.$
$80^\circ+80^\circ+80^\circ\ \text{x}^\circ=360^\circ$
$\Rightarrow240^\circ+\text{x}=360^\circ$
$\Rightarrow\text{x}=360^\circ-240^\circ$
$\Rightarrow\text{x}=120^\circ$

Number of diagonals in an n-sided polygon $=\frac{\text{n}(\text{n}-3)}{2}$
$\text{n}=6$
$\therefore\frac{\text{n}(\text{n}-3)}{2}=\frac{6(6-3}{2}$
$=\frac{18}{9}=9$

$3x - 2 = 2x - 1$
$\Rightarrow x = 1.$

The adjacent angles of a parallelogram sums up to $180^\circ .$
Thus,$45^\circ + x = 180^\circ$
$x = 180^\circ - 45^\circ$
$x = 135^\circ$

Let $ABCD$ is a parallelogram.

$AE$ and $AD$ is the bisector angles of adjacent angles of $\angle\text{A}$ and $\angle\text{D}.$
As we know that,
$\angle\text{A}+\angle\text{D}=180^\circ$ (Sum of interior angles on the same side of traversal is $180^\circ )$
$\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{D}=\frac{1}{2}\times180^\circ$
$=90^\circ\ ...(\text{i})$
Now, in triangle $AOD,$
$\angle\text{AED}+\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{D}=180^\circ$
$($AE and $AD$ is the angle bisector of $\angle\text{A}$ and $\angle\text{D}).$
$\angle\text{AED}+90^\circ=180^\circ$ (From eq $(i))$
$\angle\text{AED}=180^\circ-90^\circ=90^\circ$
So, the bisectors of any two adjacent angles of a parallelogram intersect at $90^\circ .$

Number of sides $= \frac{360^\circ}{45^\circ}=8.$

Exterior angle $= 180^\circ - 165^\circ = 15^\circ $
$\therefore$ Number of sides $\frac{360^\circ}{15^\circ}=24$

Sum of the ratios $= 1 + 2 + 3 + 4 = 10$
$\therefore$ Smallest angle $=\frac{1}{10}\times360^\circ=36^\circ$

Suppose, $A B C D$ is a quadrilateral.
Let angle $A$ is $x$
Then, $x+2 x+3 x+4 x=360^{\circ}$ [Angle sum property of quadrilateral] $10 x=360^{\circ}$ $x=36^{\circ}$
Hence, the greatest angle is $4 x=4 \times 36=144^{\circ}$

Required measure
$= \frac{360^\circ-(110^\circ+100^\circ)}{2}=75^\circ$

$n = 10$
$(n – 2) 180^\circ = 1440^\circ .$

We know that, in a trapezium, the angles on either sides of base are supplementary angle. In trapezium $ABCD,$
$\therefore\angle\text{A}+\angle\text{D}=180^\circ$
$\Rightarrow55^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-50^\circ$
$\Rightarrow\angle\text{D}=120^\circ$

$x + y = 8$
$y + 5 = 10$
$\Rightarrow y = 5$
$\therefore x + 5 = 8$
$\Rightarrow x = 3.$

For a regular n-sided polygon:
Number of diagonals $=\frac{\text{n}(\text{n}-3)}{2}$
For an actagon:
$\text{n}=8$
$\frac{8(8-3)}{2}=\frac{40}{2}=20$

In a rectangle, all angles are equal, i.e. all equal to $90^\circ .$

We know that, in a parallelogram, opposite sides are equal and parallel. Also, opposite angles are equal. So, to construct a parallelogram uniquely, we require the measure of any two nonparallel sides and the measure of an angle. Hence, the minimum number of measurements required to draw a unique parallelogram is $3.$

$\angle{\text{C}} = \angle{\text{A}}= 60^\circ$

We know that, the sum of exterior angles taken in an order of a polygon is $360^\circ $ Since, each exterior angle measures $45^\circ $, therefore the number of sides = Sum of exterior angles/ Measure of an exterior angle.
$=\frac{360^\circ}{45^\circ}=8$

Measure of the adjacent angle$= 180^\circ - 65^\circ = 115^\circ .$

The adjacent angles of parallelogram are supplementary.
$\angle\text{B}+ \angle\text{B}=180^\circ$
$70^\circ+\angle\text{B}=180^\circ$
$\angle\text{B}=180^\circ-70^\circ=110^\circ$

$(2n - 4) \times 90 = 1080$
$(2n - 4) = 12$
$2n = 16$
Or $n = 8$
Each interior angle $=180-\frac{360}{\text{n}}=180-\frac{360}{8}=180-45=135^\circ$

Fourth angle $= 360^\circ - (65^\circ + 75^\circ + 85^\circ )$
$= 135^\circ$

We know that the sum of all exterior angles of a polygon is $360$ degrees.

Let the measure of smallest anlge be $x^\circ $ and other is $(2x - 24)^\circ .$
$\therefore x + (2x - 24) = 180$
$\Rightarrow x + 2x = 180 + 24$
$\Rightarrow 3x = 204$
$\Rightarrow x = 68$
Hence, the samllest angle is $68^\circ .$
Ite adjacent is $= (180 - 68)^\circ = 112^\circ .$
Therefore, the largest angle is $112^\circ .$

As we know, the sum of adjacent angles of a parallelogram is equal to $180^\circ $ and opposite angles are equal to each other.
Thus, in parallelogram $ABCD$ angle $A$ and angle $B$ are adjacent to each other
Let angle $A = x$ and angle $B = 2x.$
So, $x + 2x = 180^\circ $
$3x = 180^\circ $
$x = 60^\circ $

Sum $= 360^\circ - (130^\circ + 40^\circ ) = 190^\circ .$

By the angle sum property of triangle, we know that;
Sum of all the angles of a triangle $= 180^\circ $
Let the unknown angle be $x$
$80^\circ + 50^\circ + x = 180^\circ $
$x = 180^\circ - 130^\circ $
$x = 50^\circ $

A rhombus is a parallelogram that has all its four sides equal. Thus, the perimeter of rhombus,
$P = 4 \times $ side-length
$P = 4 \times 5$
$P = 20cm$

$\angle{\text{A}}+\angle{\text{B}}=180^\circ$
$\angle{\text{A}}:\angle{\text{B}}=1:2$
Stun of the ratios $= 1 + 2 = 3$
$\therefore\angle\text{A}=\frac{1}{3}\times180^\circ=60^\circ$

We know, all sides are equal of a square. Then,
$\therefore AB = BC$
$\Rightarrow 2x + 3 = 3x - 5$
$\Rightarrow 3x - 2x = 3 + 5$
$\Rightarrow x = 8$