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.
+128474 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
+26367 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:
\(\begin{array}{|rcll|} \hline A &=& [ABC] - \dfrac{x}{2}*x*\dfrac12-\dfrac{1}{2}*x*(x-x_i ) \\ &=& \dfrac{x^2}{2} - \dfrac{x}{2}*x*\dfrac12-\dfrac{1}{2}*x*(x-\dfrac{x^2}{2x-1} ) \\ && \ldots \\ A &=& \dfrac{x^2}{2}*\left( \dfrac{x}{2x-1} - \dfrac12 \right) \\ && \ldots \\ A &=& \dfrac{x^2}{4}*\dfrac{1}{(2x-1)} \quad | \quad \mathbf{A=1} \\ 1 &=& \dfrac{x^2}{4}*\dfrac{1}{(2x-1)} \\ 4(2x-1) &=& x^2 \\ x^2 -8x+4 &=& 0 \\\\ x &=& \dfrac{8\pm\sqrt{64-4*4}}{2} \\ && \ldots \\ \mathbf{x} &=& \mathbf{4+2\sqrt{3}} \qquad x=7.464101615 \\ \hline \end{array}\)
