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

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,
$\therefore AB = BC$
$\Rightarrow 2x + 3 = 3x - 5$
$\Rightarrow 3x - 2x = 3 + 5$
$\Rightarrow x = 8$