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Work along with the answer included is much appreciated! Thx CPhill you’re the best

 Aug 6, 2018
 #1
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+1

In triangle HAC, using Pythagoras's Theorem, the hypotenuse HC =Sqrt(8)

In triangle AHX, using Pythagoras's Theorem, HX=Sqrt(2)

In triangle BCY, using Pythagoras's Theorem, CY=1/sqrt(2)

 

Therefore, XY =HC - HX - CY =XY =sqrt(8) - sqrt(2) - 1/sqrt(2) =1/sqrt(2) =WZ

By similar reasoning, XW=YZ = 1/sqrt(2)

 

Therefore, the area of the quadrilateral WXYZ =[1/sqrt(2)]^2 =1/2

 Aug 6, 2018
 #2
avatar+22358 
+2

Please help with this geometry question

 

In rectangle \(ADEH\), points B and C trisect \(\overline{AD}\),
and points G and F trisect \(\overline{HE}\).
In addition, \(\overline{AH}=\overline{AC}=2\).
What is the area of quadrilateral \(WXYZ\) ?

 

\(\text{Let $\overline{AH}=\overline{AC}=2$ } \\ \text{Let $\overline{AB}=\overline{BC}=\dfrac{\overline{AC}}{2} = 1$ } \\ \text{Let $\overline{AB}=\overline{XZ} = 1$ } \\ \text{Let $\overline{BX}=\dfrac{\overline{AH}}{2} = 1$ } \)

 

\(\text{Let triangle $XWZ = $ triangle $BYC$ } \)

 

\(\begin{array}{|rcll|} \hline \mathbf{A_{\text{quadrilateral } WXYZ}} &\mathbf{=} & \mathbf{ A_{XWZ} + A_{XYZ} } \quad | \quad A_{XWZ} = A_{BYC} \\\\ &=& A_{BYC} + A_{XYZ} \quad | \quad A_{BYC} =\dfrac{\overline{BC}\cdot x}{2} \quad A_{XYZ} =\dfrac{\overline{XZ}\cdot (1-x)}{2} \\\\ &=& \dfrac{\overline{BC}\cdot x}{2} + \dfrac{\overline{XZ}\cdot (1-x)}{2} \quad | \quad \overline{BC}=\overline{XZ}=1 \quad \\\\ &=& \dfrac{x}{2} + \dfrac{1-x}{2} \\\\ &=& \dfrac{x+1-x}{2} \\\\ &\mathbf{=} & \mathbf{ \dfrac{1}{2} } \\ \hline \end{array}\)

 

laugh

 Aug 6, 2018
edited by heureka  Aug 6, 2018
edited by heureka  Aug 6, 2018
edited by heureka  Aug 6, 2018

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