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# geometry

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ACEG is a rectangle. If segment BE is 30, segment CG is 40, segment DF is 15 and ∠FDE=90∘. Find CE.

Jul 10, 2020

#1
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ACEG is a rectangle. If segment BE is 30, segment CG is 40, segment DF is 15 and $$\angle FDE=90^\circ$$.
Find CE.

$$\text{Let CE=x} \\ \text{Let GE=y} \\ \text{Let AB=z} \\ \text{Let BC=y-z}$$

$$\begin{array}{|lrcll|} \hline 1): & x^2 + y^2 &=& 40^2 \\ & x &=& \sqrt{40^2-y^2} \\ \hline 2): & x^2+(y-z)^2 &=& 30^2 \quad | \quad x^2 = 40^2-y^2 \\ & 40^2-y^2+(y-z)^2 &=& 30^2 \\ & y^2 - (y-z)^2 &=& 40^2-30^2 \\ & \mathbf{y^2 - (y-z)^2} &=& \mathbf{700} \qquad (1) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline 3): & \text{area \triangle  CDE} &=& \text{area \triangle  CFE} + \text{area \triangle  FEG} \\ & \dfrac{xy}{2} &=& \dfrac{x*15}{2} + \dfrac{y*DE}{2} \\ & xy &=& x*15 + y*DE \\ & && \boxed{\dfrac{DE}{15} =\dfrac{x}{y-z}\\ DE = \dfrac{15x}{y-z} } \\ & xy &=& x*15 + y*\left( \dfrac{15x}{y-z} \right) \quad | \quad : x \\ & y &=& 15 + \left( \dfrac{15y}{y-z} \right) \\ & (y-15)(y-z) &=& 15y \\ & y-z &=& \dfrac{15y}{y-15} \\ & \mathbf{(y-z)^2} &=& \mathbf{\dfrac{225y^2}{(y-15)^2}} \qquad (2) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{y^2 - (y-z)^2} &=& \mathbf{700} \quad | \quad \mathbf{(y-z)^2=\dfrac{225y^2}{(y-15)^2}} \\ y^2 - \dfrac{225y^2}{(y-15)^2} &=& 700 \quad | \quad *(y-15)^2 \\\\ y^2(y-15)^2 - 225y^2 &=& 700(y-15)^2 \\ y^2(y^2-30y+225) - 225y^2 &=& 700(y^2-30y+225) \\ y^4-30y^3+225y^2-225y^2 &=& 700y^2-700*30y+700*225 \\ y^4-30y^3 &=& 700y^2-700*30y+700*225 \\ \mathbf{ y^4-30y^3 -700y^2+700*30y-700*225} &=& \mathbf{0} \\ \Rightarrow \mathbf{y}&=&\mathbf{36.6659934977741} \qquad \text{WolframAlpha} \\ \hline x &=& \sqrt{40^2-y^2} \\ x &=& \sqrt{40^2-36.6659934977741^2} \\ \mathbf{x}&=&\mathbf{15.9876490087} \\ \hline \end{array}$$

CE is $$\approx \mathbf{16}$$

Jul 10, 2020