Question-5431

Question Number 5431 by Rasheed Soomro last updated on 14/May/16

Commented by Yozzii last updated on 14/May/16

$${What}\:{does}\:\boldsymbol{\mathrm{d}}\:{measure}? \\ $$

Commented by Rasheed Soomro last updated on 14/May/16

d is a name of semi-circle. It has no effect on problem. I wanted to delete it but couldn′t.

$$\mathrm{d}\:\mathrm{is}\:\mathrm{a}\:\mathrm{name}\:\mathrm{of}\:\mathrm{semi}-\mathrm{circle}. \\ $$$$\mathrm{It}\:\mathrm{has}\:\mathrm{no}\:\mathrm{effect}\:\mathrm{on}\:\mathrm{problem}. \\ $$$$\mathrm{I}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{delete}\:\mathrm{it}\:\mathrm{but}\:\mathrm{couldn}'\mathrm{t}. \\ $$

Commented by Yozzii last updated on 14/May/16

$${No}\:{problem}.\: \\ $$

Commented by Yozzii last updated on 15/May/16

$${I}'{ll}\:{try}\:{it}\:{soon}.\:{I}'{ve}\:{been}\:{busy}. \\ $$

Commented by Rasheed Soomro last updated on 16/May/16

$$\underset{\boldsymbol{\mathrm{D}}} {\overset{\boldsymbol{\mathrm{G}}} {\mathcal{OO}}} \\ $$

Commented by Yozzii last updated on 16/May/16

Let the square be of side length 2r. Hence, the radius of the semicircle is r. Since the quadrilateral in the figure is a square, its diagonals are mutually orthogonal. So, if O is the centre of the semicircle, X is the vertex of the square where angle x arises, and J is one of the end points of the arc of the semicircle, then △OJX is right−angled. By symmetry of the figure ∠OXJ=(x/2) and ∠JOX=90°. Sincr OJ=r and OX=r(√2) ⇒cos(x/2)=((r(√2))/( (√(r^2 +2r^2 ))))=((√2)/( (√3))) ⇒x=2cos^(−1) (√(2/3))≈70.53°

$${Let}\:{the}\:{square}\:{be}\:{of}\:{side}\:{length}\:\mathrm{2}{r}. \\ $$$${Hence},\:{the}\:{radius}\:{of}\:{the}\:{semicircle}\:{is}\:{r}. \\ $$$${Since}\:{the}\:{quadrilateral}\:{in}\:{the}\:{figure} \\ $$$${is}\:{a}\:{square},\:{its}\:{diagonals}\:{are}\:{mutually} \\ $$$${orthogonal}.\:{So},\:{if}\:{O}\:{is}\:{the}\:{centre} \\ $$$${of}\:{the}\:{semicircle},\:{X}\:{is}\:{the}\:{vertex}\:{of} \\ $$$${the}\:{square}\:{where}\:{angle}\:{x}\:{arises},\:{and} \\ $$$${J}\:{is}\:{one}\:{of}\:{the}\:{end}\:{points}\:{of}\:{the}\:{arc} \\ $$$${of}\:{the}\:{semicircle},\:{then}\:\bigtriangleup{OJX}\:{is} \\ $$$${right}−{angled}.\:{By}\:{symmetry}\:{of}\:{the} \\ $$$${figure}\:\angle{OXJ}=\frac{{x}}{\mathrm{2}}\:{and}\:\angle{JOX}=\mathrm{90}°.\:{Sincr}\:{OJ}={r}\:{and} \\ $$$${OX}={r}\sqrt{\mathrm{2}}\:\Rightarrow{cos}\frac{{x}}{\mathrm{2}}=\frac{{r}\sqrt{\mathrm{2}}}{\:\sqrt{{r}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} }}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{x}=\mathrm{2}{cos}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\approx\mathrm{70}.\mathrm{53}° \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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