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

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

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 $ABCD,$
$\therefore\angle\text{A}+\angle\text{D}=180^\circ$
$\Rightarrow55^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-50^\circ$
$\Rightarrow\angle\text{D}=120^\circ$

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