Geometry

Use the Pythagorean theorem to find the missing height: sqrt(10^2 - 8^2) = sqrt(100-64) = 6 cm. Then area = 1/2 × base × height = 1/2 × 8 × 6 = 24 cm².

For a square of side 6cm, the inscribed circle has diameter 6cm (radius 3cm). Circle area = πr² = 9π. Square area = 6² = 36. The ratio is 9π/36 = π/4 ≈ 0.785. Among options, 11/14 ≈ 0.786 is the closest approximation to π/4.

The largest square inscribed in a circle has a diagonal equal to the circle's diameter, which is 8 cm. The area of a square with diagonal $d$ is $d^2 / 2$. Thus, the area is $8^2 / 2 = 64 / 2 = 32$ square centimeters. Distractors represent incorrect formulas or area of the circumscribed square (64).

A square inscribed in a circle has a diagonal equal to the circle's diameter. With radius 5, the diameter is 10. The area of a square is (diagonal^2)/2, so (10^2)/2 = 100/2 = 50 square units.

According to the question, a chord of length 2a makes an angle θ at the centre. So, in the triangle, side opposite to angle θ is 2a. r (r = radius of the circle) In that triangle, applying the Cosine Law, Cos θ = (r² + r² - (2a)²)/(2r²) Cos θ = (2r² - 4a²)/(2r²) Solving, we get r² = (2a²)/(1-Cosθ)------------(1) Now, Cos θ can be written as Cos 2(θ/2) (Cos 2x = (1 - tan²x)/(1 + tan²x)) Cos 2(θ/2) = (1 - tan²(θ/2))/(1 + tan²(θ/2)) Cos θ = (1 - tan²(θ/2))/(1 + tan²(θ/2)) 1 - Cos θ = 2tan²(θ/2)/(1 + tan²(θ/2)) It can be written as:1 - Cos θ = 2/(1 + Cot²(θ/2)) Divinding numerator and denominator by tan²(θ/2) and substituting the value of 1 - Cos θ in (1), we get r² = a²(1 + Cot²(θ/2)) Now, area of circle is πr². So, area = πa²(1+Cot²(θ/2)) (Correct Answer)

Of all the triangles inscribed in a semi-circle, with the diameter as base, the one with the greatest area is the one with the largest altitude (the radius), i.e. the isosceles triangle. Thus, the area = 1/2 * 2r * r = r². Hence, option (1) is correct.

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Area of right ^$\triangle$ABC = 1/2 x AB x BC= 1/2 x (6 x 8)cm² = 24 cm²     AC² = AB² + BC²                             64 cm² + 36 cm² = 100 cm². AC = 10 cm.  Also, area (^$\triangle$ABC) = ar (^$\triangle$OBC) + ar (^$\triangle$OCA) + ar (^$\triangle$OAB)= 1/2 (BC x r) + 1/2 (AC x r) + 1/2 (AB x r)= 1/2 r (6 + 10 +8) cm = 12 r cm 12 r cm = 24 cm^{2 _{$\Rightarrow$}} r = 2 cm

The radius of circumcircle  of equilateral triangle is  $\dfrac 23h=1\h=\dfrac 32$

Given,
Radius of the circle $=7:cm$
Let the side of the square be $a:cm$.
A square when inscribed in a circle then the diameter of the circle must be diagonal of the square.
Therefore,
Diagonal of square $=\sqrt {a^2+a^2}$
                                 $=a\sqrt 2$
Now,
Diameter of the circle $=2\times 7$
                                  $=14:cm$
$=>\sqrt 2 a=14$
$=>a=\dfrac{14}{\sqrt 2}$
$=>a=7\sqrt 2: cm$
Therefore,
Area of square $=a^2$
                       $=(7\sqrt 2 cm)^2$
                       $=(7\sqrt 2 cm)(7\sqrt 2 cm)$
                       $=98: cm^2$

The radius of the circle inscribed in the square
of side 2.4m
$\displaystyle r=\dfrac{2.4}{4}m=1.2m$
$\displaystyle \therefore$ Area of the circle $\displaystyle =\pi r^{2}$ square units
$\displaystyle =\pi \times 1.2 m\times 1.2 m$
$\displaystyle =1.44 \pi m^{2}$

We have,

${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{c}^{2}}$ 
${{S} _{1}}\equiv {{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{b}^{2}}-2by-{{c}^{2}}=0$               …….. (1)

${{\left( x-b \right)}^{2}}+{{\left( y-a \right)}^{2}}={{c}^{2}}$

${{S} _{2}}\equiv {{x}^{2}}+{{b}^{2}}-2bx+{{y}^{2}}+{{a}^{2}}-2ay-{{c}^{2}}=0$                        ……… (2)

Since, $a\ne b$

Centre of the circle ${{S} _{1}}=\left( a,b \right)$ and radius ${{r} _{1}}=c$.

We know that the equation of common chord is ${{S} _{1}}-{{S} _{2}}=0$

So, the equation is

$\left( b-a \right)x+\left( a-b \right)y=0$             …….. (3)

We know that the length of common chord is

$=2\sqrt{{{r} _{1}}^{2}-{{d} _{1}}^{2}}$                      ………. (4)

Where ${{r} _{1}}=$ radius and ${{d} _{1}}$ is the length of perpendicular drawn from the centre to the chord.

So,

${{d} _{1}}=\left| \dfrac{\left( b-a \right)a+\left( a-b \right)b}{\sqrt{{{\left( b-a \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right|$

$ {{d} _{1}}=\left| \dfrac{ab-{{a}^{2}}+ab-{{b}^{2}}}{\sqrt{{{\left( a-b \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{2ab-{{a}^{2}}-{{b}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-{{\left( a-b \right)}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-\left( a-b \right)}{\sqrt{2}} \right| $

$ {{d} _{1}}=\dfrac{\left( a-b \right)}{\sqrt{2}} $

From equation (4),

The length of common chord $ =2\sqrt{{{c}^{2}}-{{\left( \dfrac{a-b}{\sqrt{2}} \right)}^{2}}} $

$ =2\sqrt{{{c}^{2}}-{{\dfrac{\left( a-b \right)}{2}}^{2}}} $

$ =2\sqrt{{{\dfrac{2{{c}^{2}}-\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{{{\dfrac{8{{c}^{2}}-4\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{4{{c}^{2}}-2{{\left( a-b \right)}^{2}}} $

Hence, this is the answer.

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

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

The diagonal of the square will be equal to the diameter of the circle.
So,diagonal of the square $ = 2 \times 5 \sqrt {2} = 10 \sqrt {2} $

Diagonal of a square $ = \sqrt {2} \times side $
So, $ 10 \sqrt {2} = \sqrt {2} \times side $
$ => Side  =  10  cm $

Area of the square $ = { side }^{ 2 } = { 10 }^{ 2 } = 100 $ sq cm