Question

Find the area of the square

Answer 1

Area of Square = 2(b - a)² where a² + b² = 7b   and b² + (2b-a)² = 7²

Step-by-step explanation:

I am too close to solve it :

Equation of circles

x² + y ² = 7x

x² + y² = 7y    

x² + y² = 7²   ( bigger circle)

its a symmetrical figure

if  S = (a , b)   Then P  = ( b , a)    Q would be (h , b)   & R would be (b , k)

Solving for h using multiplication of slope of perpendicular lines = -1

(b - a)/(h - b)  *  (a-b)/(b - a) = -1

=> a - b = -h + b

=> h = 2b -a

similarly k = 2 b  - a

so points are

S = (a , b)  , P ( b , a)   Q = (2b - a  , a)  & R = (b , 2b-a)

Area of Square = 2(b - a)²

now putting points in circle equation we have

a² + b² = 7b      

and b² + (2b-a)² = 7²

By solving these we can get get a & b

then 2(b - a)²

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Answer 2

$\displaystyle\large\boxed {\sf {Area\ of\ square\ PQRS\approx 0.791324197\ cm^2}}$

Let,

• $\displaystyle\sf {PQ=QR=RS=PS=x\implies PR=x\sqrt2}$

• $\displaystyle\sf {OM=a\implies MA=7-a}$

• $\displaystyle\sf {\angle AOP=\theta}$

• $\displaystyle\sf {\angle POS=2\alpha}$

Here $\displaystyle\sf {OP=OS}$ and $\displaystyle\sf {\angle AOP=\angle SOB=\theta}$ due to symmetry in the quadrant.

Then,

$\displaystyle\longrightarrow\sf {\angle AOP+\angle POS=\angle SOB=\angle AOB}$

$\displaystyle\longrightarrow\sf {\alpha=45^{\circ}-\theta}$

We know that if two chords in a circle intersect each other, then the product of the two parts formed due to the point of intersection in the two individual chords must be equal.

In this case if one chord is nothing but a diameter, then the other chord will be bisected, and hence the product of the two parts of the diameter formed due to the intersection is equal to the square of half the other chord.

Therefore, in the semicircle OPA,

$\displaystyle\longrightarrow\sf {MP^2=OM\cdot MA}$

$\displaystyle\longrightarrow\sf {MP=\sqrt{a(7-a)}=\sqrt{7a-a^2}}$

And also, if we construct a semicircle ABC by joining the same large quadrant to its left along the side OB with OB, OC = 7 cm as its radii, where AO is produced leftwards to meet C, then the diameter will be AC, and hence,

$\displaystyle\longrightarrow\sf {MR^2=CM\cdot MA}$

$\displaystyle\longrightarrow\sf {MR^2=(OC+OM)\cdot MA}$

$\displaystyle\longrightarrow\sf {MR=\sqrt{(7+a)(7-a)}}$

Therefore,

$\displaystyle\longrightarrow\sf {PR=MR-MP}$

$\displaystyle\longrightarrow\sf {x=\dfrac {\sqrt{(7+a)(7-a)}-\sqrt{a(7-a)}}{\sqrt2}\quad\quad\dots (1)}$

In $\displaystyle\sf {\triangle OMP,}$

$\displaystyle\longrightarrow\sf {OP^2=OM^2+MP^2}$

$\displaystyle\longrightarrow\sf {OP=OS=\sqrt{7a}}$

And,

$\displaystyle\longrightarrow\sf {\theta=\sin^{-1}\left (\dfrac {\sqrt{a(7-a)}}{\sqrt{7a}}\right)=\cos^{-1}\left (\dfrac {a}{\sqrt{7a}}\right)}$

Consider $\displaystyle\sf {\triangle OPS.}$

$\displaystyle\sf {OP=OS=\sqrt{7a}\quad;\quad\angle POS=2\alpha}$

Then, by law of cosines,

$\displaystyle\longrightarrow\sf {PS^2=OP^2+OS^2-2\cdot OP\cdot OS\cos\angle POS}$

$\displaystyle\longrightarrow\sf {x=\sqrt{2\cdot7a(1-\cos(2\alpha)}}$

$\displaystyle\longrightarrow\sf {x=\sqrt{2\cdot7a\cdot 2\sin^2\alpha}}$

$\displaystyle\longrightarrow\sf {x=2\sin\alpha\sqrt{7a}}$

$\displaystyle\longrightarrow\sf {x=2\sin(45^{\circ}-\theta)\sqrt{7a}}$

$\displaystyle\longrightarrow\sf {x=2\left [\dfrac {1}{\sqrt2}\cos\theta-\dfrac {1}{\sqrt2}\sin\theta\right]\sqrt{7a}}$

$\displaystyle\longrightarrow\sf {x=\sqrt2\left [\dfrac {a}{\sqrt{7a}}-\dfrac {\sqrt{a(7-a)}}{\sqrt{7a}}\right]\sqrt{7a}}$

$\displaystyle\longrightarrow\sf {x=\sqrt2\left (a-\sqrt{a(7-a)}\right)\quad\quad\dots (2)}$

From (1) and (2),

$\displaystyle\longrightarrow\sf {\dfrac {\sqrt{(7+a)(7-a)}-\sqrt{a(7-a)}}{\sqrt2}=\sqrt2\left (a-\sqrt{a(7-a)}\right)}$

$\displaystyle\longrightarrow\sf {2a=\sqrt{(7+a)(7-a)}+\sqrt{a(7-a)}}$

$\displaystyle\longrightarrow\sf {4a^2=49-a^2+7a-a^2+2(7-a)\sqrt{a(7+a)}}$

$\displaystyle\longrightarrow\sf {\sqrt{a(7+a)}=\dfrac {6a^2-7a-49}{2(7-a)}}$

$\displaystyle\longrightarrow\sf {a(7+a)=\dfrac {36a^4-84a^3-539a^2+686a+2401}{4(7-a)^2}}$

$\displaystyle\longrightarrow\sf {36a^4-56a^3-347a^2-686a+2401=0}$

On solving this equation I got approx. two positive real values for $\displaystyle\sf {a:}$

• $\displaystyle\sf {a\approx1.856987688}$

• $\displaystyle\sf {a\approx4.080535131}$

But, since $\displaystyle\sf {x}$ is the length of the side of the square,

$\displaystyle\longrightarrow\sf {x\ \textgreater\ 0}$

From (2),

$\displaystyle\longrightarrow\sf {a\ \textgreater\ \sqrt{a(7-a)}}$

$\displaystyle\longrightarrow\sf {a^2\ \textgreater\ 7a-a^2}$

and since $\displaystyle\sf {a\neq0,}$

$\displaystyle\longrightarrow\sf {a\ \textgreater\ 3.5}$

Therefore,

$\displaystyle\longrightarrow\sf {a\approx4.080535131\ cm}$

Then, using (2),

$\displaystyle\longrightarrow\sf {x=0.889564049\ cm}$

Hence, the area of the square PQRS is,

$\displaystyle\longrightarrow\sf {\underline {\underline {x^2\approx 0.791324197\ cm^2}}}$