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The exterior angle of a quadrilateral have measures $72{}^\circ ,58{}^\circ ,2x{}^\circ $ and $3x{}^\circ .$ How do I find the value of $x?$

Last updated date: 24th Jul 2024
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Hint: Here the measure of quadrilateral angles are given in the question.
As we know, that the sum of all quadrilateral angles is equal to $360{}^\circ .$
So, add up all the angles given and simplify it to get the value of $x.$
Always remember, that the sum of all angles should be equal to $360{}^\circ .$ For that we can verify the calculated answer.

Complete step by step solution:
We know that, the sum of all exterior angle of any regular polygon is always $360{}^\circ .$ Whether that quadrilateral polygon has $3$ sides or $16$ sides, its sum will always be $360{}^\circ .$
Given that, there are some exterior angles of quadrilateral is, $72{}^\circ ,58{}^\circ ,2x{}^\circ ,3x{}^\circ $
As the sum of all angles of any polygon quadrilateral are $360{}^\circ .$
The equation becomes,
$72{}^\circ +58{}^\circ +2x{}^\circ +3x{}^\circ =360{}^\circ $
For getting the value of $x$ we have to simplify the above equation.
$72{}^\circ +58{}^\circ +2x{}^\circ +3x{}^\circ =360{}^\circ $
Add the like terms on the left side of the equation.
$130{}^\circ +5x{}^\circ =360{}^\circ $
Transpose the $130{}^\circ $ to the other side of the equation
$5x{}^\circ =360{}^\circ -130{}^\circ $
$\Rightarrow 5x{}^\circ =230{}^\circ $
$\Rightarrow x{}^\circ =\dfrac{230{}^\circ }{5{}^\circ }$

Therefore the value of $x$ from given angles of quadrilateral is $46.$

Additional Information:
The sum of all interior angles of a quadrilateral is equal to \[360{}^\circ \]
The quadrilateral word is derived from the Latin language of two words ‘quadri’ and ‘latus’ which means four and side in English respectively, Quadrilateral’s have total five types according to the shape of it which is as follows:
(1) Rectangle
(2) Square
(3) Parallelogram
(4) Rhombus
(5) Trapezium
Where, a rectangle has four sides and angles of $90{}^\circ .$ and its opposite sides are equal and parallel. A square quadrilateral has $4$ equal and parallel sides and $4$ angles of $90{}^\circ .$
A parallelogram quadrilateral has $4$ angles and sides whose opposite angles and side are equal and parallel. It has the total $180{}^\circ .$ sum of any two of adjacent angles.
A quadrilateral of rhombus has four equal sides.
But equal opposite angles and parallel opposite sides. It has the total of $180{}^\circ .$ sum of any two of adjacent angles.
A quadrilateral of trapezium has four sides and angles but it has parallel opposite sides only of one pair.

Note: As given that, here are angles of a quadrilateral $72{}^\circ ,58{}^\circ ,2x{}^\circ ,3x{}^\circ $
But the sum of any quadrilateral polygon’s angle is equal to $360{}^\circ .$ And, hence the calculated value of $x=46{}^\circ $
As the sum of all quadrilateral angles are $360{}^\circ $. We can verify the answer.
$\therefore 72{}^\circ +58{}^\circ +2x{}^\circ +3x{}^\circ =360{}^\circ $
Now put $x=46$, as calculated above in the solution,
$72{}^\circ +58{}^\circ +2\times 46+3\times 46=360{}^\circ $
$\Rightarrow72{}^\circ +58{}^\circ +92{}^\circ +138{}^\circ =360{}^\circ $
$\Rightarrow130{}^\circ +230{}^\circ =360{}^\circ $
$\Rightarrow 360{}^\circ =360{}^\circ $
From above it is clear that the calculated value of $x=46$ is correct.