+0

# geometry

0
40
2

ABCD is a square with side length x and length DE is 1. Find the value of x for which the area of the shaded region is 1. Jan 11, 2021

#1
+1

Call the side of the square , S

The area of triangle BDE =  (1/2)DE * height=  (1/2)(1) (S)  = S/2

Call the intersection of AC and BD =  F

Call the intersection of BE  and AC  = G

Angle BFG  =90      BF = S/sqrt (2)

Triangle BFG is right  and we have that

(1/2) BF * FG  =1

(1/2) S/sqrt (2) * FG  =1

S/sqrt(8)  *FG  =1

FG =  Sqrt (8)/ S

Area of triangle DFC =  (1/2)S  * height   =  (1/2)S (S/2)  =  S^2/4

CG =   S/sqrt (2)  - FG    =

S/sqrt (2)  - sqrt(8) / S  =

[S^2  - sqrt (2)sqrt (8)]  / [Ssqrt (2)] =

[S^2 - 4] / [ Ssqrt (2)]

And angle GCE  = 45....sin GCE = 1/sqrt (2)

CE = S - 1

So....the area of triangle  GCE  = (1/2) CG * CE sin  GCE  =

(1/2) [ S^2 - 4 ]  / [  S sqrt (2) ]  * (S - 1) (1/sqrt (2))  =

(S-1)(S^2-4)/ [ 4S]

So

Area of  BDE   -  (  white area  of BDE  )   = 1

Area of BDE  - ( area of DFC   - area of CGE )   =   1

S/2  - ( S^2/4   -  (S-1)(S^2 - 4) / [4S]  )     = 1

Simplifying this we get

S +  4/S  = 8      multiply through by S

S^2  + 4  = 8S     rearrange

S^2  - 8S  =  -4      complete the square on S

S^2 -8S + 16  = -4 + 16

(S - 4)^2 =  12   take the positive root

S - 4 = sqrt (12)

S = 4 + sqrt (12)

S = 4 + 2sqrt (3)

S = 2 ( 2 + sqrt (3))   ≈  7.464  =   x   Jan 11, 2021
edited by CPhill  Jan 12, 2021
edited by CPhill  Jan 12, 2021
#2
+2

ABCD is a square with side length x and length DE is 1.
Find the value of x for which the area of the shaded region is 1. $$\text{Let D=(0,0) } \\ \text{Let A=(0,x) } \\ \text{Let C=(x,0) } \\ \text{Let E=(1,0) } \\ \text{Let B=(x,x) }$$

line AC (Y = mX+b): $$Y =-X+x$$
line EB (Y = mX+b): $$Y =\dfrac{(X-1)x}{x-1}$$

intersection AC-EB: $$Y=-X-x=\dfrac{(X-1)x}{x-1}$$
intersection ($$x_i,y_i$$): $$x_i = \dfrac{x^2}{2x-1}$$
$$y_i = -x_i+x$$

area:

$$\begin{array}{|rcll|} \hline A &=& [ABC] - \dfrac{x}{2}*x*\dfrac12-\dfrac{1}{2}*x*(x-x_i ) \\ &=& \dfrac{x^2}{2} - \dfrac{x}{2}*x*\dfrac12-\dfrac{1}{2}*x*(x-\dfrac{x^2}{2x-1} ) \\ && \ldots \\ A &=& \dfrac{x^2}{2}*\left( \dfrac{x}{2x-1} - \dfrac12 \right) \\ && \ldots \\ A &=& \dfrac{x^2}{4}*\dfrac{1}{(2x-1)} \quad | \quad \mathbf{A=1} \\ 1 &=& \dfrac{x^2}{4}*\dfrac{1}{(2x-1)} \\ 4(2x-1) &=& x^2 \\ x^2 -8x+4 &=& 0 \\\\ x &=& \dfrac{8\pm\sqrt{64-4*4}}{2} \\ && \ldots \\ \mathbf{x} &=& \mathbf{4+2\sqrt{3}} \qquad x=7.464101615 \\ \hline \end{array}$$ Jan 12, 2021