Geometric Constructions - 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 $6\left ( \dfrac {\sqrt{3}}{4}a^{2} \right )=6*\dfrac {\sqrt{3}}{4}\dfrac {20\sqrt{3}}{3}\dfrac {20\sqrt{3}}{3}=200\sqrt{3}$

Least count of normal scale is $.1\ \ cm$ that is minimum length that can be measured using a normal scale is $.1\ \ cm$.

Least count of normal scale is $.1\ \ cm$ that is minimum length that can be measured using a normal scale is $.1\ \ cm$

Least count of normal scale is $.1\ \ cm$ that is minimum length that can be measured using a normal scale is $.1\ \ cm$

For constructing a line segment of given length steps are following:

A line can be constructed using a Ruler and a Compass.

Let length of each part $=x$

Correct sequence is :

Correct sequence is :

Step 1 $:$ Draw a line .

Yes, we can construct a copy of line segment of length $2.345$ using scale/compass. By using scale/compass we measure the length and by stretching the compass & cut the same length of line.

Correct sequence is :

Line segment of length $3.153\ \ cm$ can not be drawn using a compass as the least count of the compass is $1\ \ mm$ or $.1\ \ cm$

$1.$ Draw a line segment of length $12.4\ \ cm$ using a ruler.

Correct sequence is :

Draw a line segment $AB$ of $7.8\ \ cm$ using ruler 

Draw a line segment $AB$ of length $6.5\ \ cm$

Draw a line segment $XY$ of length $13.6\ \ cm$

Draw a line segment $PQ$ is length $9.6\ \ cm$ using a ruler.

Correct sequence is :

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

Let circle C have center (1, 1) and radius 1. Circle T has center (0, y), passes through the origin (0, 0), so its radius is r = y. Circle T touches C externally, meaning the distance between their centers equals the sum of their radii: sqrt((0 - 1)^2 + (y - 1)^2) = r + 1 = y + 1. Squaring both sides: 1 + y^2 - 2y + 1 = y^2 + 2y + 1, which simplifies to 2 - 2y = 2y + 1, giving 4y = 1, so y = 1/4. Thus, the radius is 1/4.

Circum radius formula

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,