Tag: maths

$AS$ and $AP$ are tangents drawn to the circle at $A$

$\implies AS = AP$

Similarly

$BP = BQ$

$QC = CR$

$RD = DS$

Given

$AD = 23$

$\implies AS + SD = 23$

$AS = 23 – 5 = 18 = AP$

$AB = 29 \implies AP + BP = 29$

$\implies 18 + BP = 29 \implies BP = 11cm$

Now consider rectangle $PBQO$

$PB – BQ , OP = OQ = radius$

$\angle PBQ = 90$    

WKT

$OP \perp BP $ and $OQ \perp BQ$

Since radius is perpendicular to tangent at point of contact

$\implies$ All the angles are 90 degree and adjacent sides are equal

So, It is a square

$\implies r = BP = 11cm$

The tools required for constructing a tangent to a circle is ruler, compass and pencil.

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: