Let $ABCD$ be a square. The area of the blue square is $\dfrac12$ of the area of $ABCD$ .

Find the angle $x$ in degrees.

The answer is 15.

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Let $a$ be the side of the big square. There are total 4 congruent wedges of brown color. Let the smaller side of the wedge be $x$ & $y$ be the larger side, so that $\tan X=\frac{x}{y}$ Now, applying Pythagoras to one of the wedge, $x^2+y^2=a^2…(1)$

Since the area of blue square is half the area of big square that means the sum of areas of the four wedges is also half the area of big square . $\Rightarrow 4•\frac{xy}{2}=\frac{a^2}{2}$

$\Rightarrow xy=\frac{a^2}{4}...(2)$

Dividing $(1)$ by $(2)$ we get, $\frac{x}{y}+ \frac{y}{x}=4$ Further solving this gives $\frac{x}{y}= \tan X=2\pm \sqrt 3$ Dropping the larger value, $\tan X=2-\sqrt 3 \Rightarrow X=\boxed{15°}$

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this is construction of bhaskara's proof of pythagoras!

Nihar Mahajan
- 6 years, 7 months ago

It is to solve cos(x) - sin(x) = 1/sqrt(2). Squaring this, we get sin(2x) = 1/2, i.e. x = 15(deg)

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i don't understand how "cos(x) - sin(x) = 1/sqrt(2)"? can anyone explain?

Shadman Wasif
- 6 years, 6 months ago

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Let, 'a' be the side of larger square.

The triangles are all congruent. So, Side ( Smaller Square ) = Larger Side of Triangle - Smaller Side of Triangle or, Side ( Smaller Square ) = a cosx - a sinx or, a/ sqrt(2) = a ( cosx - sinx ) or, cosx - sinx = 1/ sqrt(2).

Sankalp Ranjan
- 6 years, 6 months ago

May you please explain how did you arrive at it?

Sanjeet Raria
- 6 years, 6 months ago

Beautiful!

Ujjwal Rane
- 6 years, 7 months ago

HOW BY SOLVING FURTHER COS(X)-SIN(X) WE GET X =15

tushar ahooja
- 6 years, 6 months ago

Let $M$ be the midpoint of $AB$ , $O$ the common center of both squares and $P$ the vertex of the blue square such that $\angle APB$ is right angled. Let $a$ be the side length of the original square.

As $ABCD$ is a square and $M$ is the circumcenter of $\triangle APB$ , we have $OM = \frac{a}{2} = MB = MP$ . Also, as the blue square's area is one half of the original square's area, its diagonal is equal to $\frac{1}{\sqrt{2}}$ times the original square's diagonal. By the Pythagorean theorem, the original diagonal has length equal to $a\sqrt{2}$ and hence the diagonal of the blue square measures $a$ . Then, as $OP$ is half of the diagonal, $\frac{a}{2} = OP = OM = MP$ and hence $\triangle OPM$ is equilateral and $\angle OMP = 60^{\circ}$ .

Now:

$\angle AMP = \angle AMO + \angle PMO = 90^{\circ} + 60^{\circ} = 150^{\circ}$

And since $MP = MA$ , this imples $\angle MAP = 15^{\circ}$ , which is the desired angle.

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Well done Ariel! Maybe you should explain why O is common center of both squares.

Veljko Grcak
- 6 years, 5 months ago

*
(y/p)^2 + 2
*
(y/p) -1=0
=> (y/p) = {-1 +- root(3) }/2
Thus y/(y+p) = {root(3)-1}/{1+root(3)}
Thus tan (x) = {tan A - tan B }/ {1+tanA*tanB} = tan (A-B)
tan 45 =1 , tan 60 = root(3)
Thus y/y+p = tan (60 -45) = tan 15
Thus x= 15.

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*
ar(triangle) + ar(blue square) = ar(ABCD) => ar(triangle) = 1/8 * ar(ABCD) = a^2 /8 or, 1/2 * a^2 * sinx
*
cosx = a^2 /8 or, sinx*cosx = 1/4 => sin2x = 1/2 => x = 15 degrees

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$\rightarrow \quad { sin }^{ 2 }x+{ cos }^{ 2 }x+2sinxcosx=\frac { 3 }{ 2 } \quad \rightarrow \quad sin2x=\frac { 1 }{ 2 } \quad \rightarrow \quad x={ 15 }^{ o }$

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*
cos(x)-Lsin(x)=L
*
(cos(x)-sin(x))
As l^2/L^2=1/2 (areas)
we get
(cos(x)-sin(x))^2=1/2 or
1 -sin(2x)=1/2
From here 2x=30º or x=15º

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*
asin(x)
*
acos(x)=1/4
*
a^2
*
sin(2x)...
so, sin(2x)=1/2=>2x=30 deg
so, x=15 deg

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well done dear

Ali Elshaip
- 6 years, 6 months ago

*
j^2. Also the total area of the four red triangles is k^2-j^2 =j^2. The area of one triangle is (1/2)
*
(k
*
cos(x))
*
(k
*
sin(x))=(1/2)
*
k^2
*
cos(x)
*
sin(x)=j^2
*
cos(x)
*
sin(x). The total area of the four triangles is j^2=4
*
j^2
*
cos(x)
*
sin(x) or cos(x)
*
sin(x)=1/4. The solution is x=15 or 75 degrees which are complementary angles. Since x is illustrated as the smaller angle in the figure, therefore x=15 degrees.

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Let the side of the LARGER square =

a.Since the four right triangles are congruent,

the LONGER leg of the triangle =

a cos x°while the SHORTER leg of the triangle =

a sin x°.So it would follow that the

side of the SMALLER square =

a cos x° - a sin x°.Therefore,

Area of LARGER square =

a ².Area of SMALLER square =

( a cos x° - a sin x° ) ².Given that the area of the SMALLER square is 1/2 of the area of the LARGER square, we have this ;

a ² / 2 = ( a cos x° - a sin x° ) ²a / √2 = ( a cos x° - a sin x° )a / √2 = a ( cos x° - sin x° )1 / √2 = cos x° - sin x°½ = cos ² x° - 2 cos x° sin x° + sin ² x½ = cos ² x° + sin ² x° - 2 cos x° sin x°Pythagorean identity states that;

cos ² x° + sin ² x° = 1½ = 1 - 2 cos x° sin x°2 cos x° sin x° = 1 - ½2 cos x° sin x° = ½Double angle identity states that;

sin ( 2 x ) = 2 cos x° sin x°sin ( 2 x ) = ½2 x = 30°x = 15°