Tag: construction of tangent to a circle

Circum radius formula

Consider the given angles.

${{y}^{\circ }},\left( {{y}^{\circ }}+{{20}^{\circ }} \right),\left( {{y}^{\circ }}+{{40}^{\circ }} \right),\left( {{y}^{\circ }}+{{60}^{\circ }} \right),\left( {{y}^{\circ }}+{{80}^{\circ }} \right)$

We know that the sum of all angles of pentagon

$ {{y}^{\circ }}+\left( {{y}^{\circ }}+{{20}^{\circ }} \right)+\left( {{y}^{\circ }}+{{40}^{\circ }} \right)+\left( {{y}^{\circ }}+{{60}^{\circ }} \right)+\left( {{y}^{\circ }}+{{80}^{\circ }} \right)={{540}^{\circ }} $

$ 5{{y}^{\circ }}+{{200}^{\circ }}={{540}^{\circ }} $

$ 5{{y}^{\circ }}={{340}^{\circ }} $

Hence, the smallest angle of the pentagon is ${{68}^{\circ }}$.

Each side of the pentagon makes an angle x at the center

$\implies 5x= 360 $

$x = 72$

Now lets consider side AB which is a chord to the circle

Let OP be a perpendicular to AB

$\implies AP = BP \implies AB = 2AP$

IN $\triangle OAP$

$\angle OPA = 90$

$\angle POA = \dfrac{x}{2} = \dfrac{72}{2} = 36$

$\sin 36 = \dfrac{AP}{OA}$

$AP = 0.6 \times 6 = 3.6$

$AB = 2 \times 3.6 = 7cm$

As we know that all angles in an equilateral triangle measures $60^o$. Hence we need only the length of the side to construct an equilateral triangle.

We can construct a rectangle when:

Suppose such a triangle is possible Then the sum of the lengths of any two side would be greater than the length of the third side  Let us check this
Is 4.5+5.8>10.2  Yes 
Is 5.8+10.2>4.5  Yes
Is 10.2+4.5>5.8  Yes
Therefore the triangle is possible

We have three measurements to construct a $\Delta$ le,

Pentagon is a five sided polygon.

We know that internal angle of regular pentagon is $\cfrac{(n-2)}{n}180^{\circ}$ where n = number of sides.