Construction - class-VIII

Ans- Three measurements are necessarily required to construct a triangle.

When constructing an inscribed regular hexagon in a circle, we choose radius of the circle as a arc measurement.

To construct a unique quadrilateral, we will be need a minimum of $5$ dimensions.

Hexagon is a $6$-sided polygon.
So we will cut the circle into $6$ equal parts.

We will use compass tool for cutting a circle into $6$ equal parts.

While constructing a circle circumscribing and inscribing a regular hexagon, the statement true for the construction is circle outside and inside hexagon.

Above statement is false.

$\Rightarrow$  Length of side of regular hexagon $(a)=6,cm$

The circle equation is (x-1)^2 + (y+2)^2 = 93 + 1 + 4 = 98. The radius is sqrt(98) = 7*sqrt(2). For a square with sides parallel to axes, the distance from center (1, -2) to vertices is the radius.

$\Rightarrow$   Radius of a circle is $r$.

When constructing the circles circumscribing and inscribing a regular hexagon with radius $3$ m, then inscribing hexagon length of each side is $3$ m.

Area of circle $=100\pi $
$\pi r^{2}=100\pi $
$r^{2}=100$
$r=10$
Now, a regular hexagon is made up of 6 equilateral $\bigtriangleup s $ of equal areas. Now, height of equilateral $\bigtriangleup  $ is equal to radius of circle.Therefore, ar. of 1 equilateral $\bigtriangleup=\dfrac {1}{2} $ x base x height
$\Rightarrow \dfrac {\sqrt{3}}{4}a^{2}=\dfrac {1}{2}a10\Rightarrow a=\dfrac {410}{2\sqrt{3}}=\dfrac {20\sqrt{3}}{3} $
Area of hexagon $6\left ( \dfrac {\sqrt{3}}{4}a^{2} \right )=6*\dfrac {\sqrt{3}}{4}\dfrac {20\sqrt{3}}{3}\dfrac {20\sqrt{3}}{3}=200\sqrt{3}$

$AS$ 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:

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.