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# geometry hard problem!

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In rectangle WXYZ, A is on side WX such that AX = 4, B is on side YZ such that BY = 18, and C is on side XY such that angle ACB= 90 degrees and CY = 2CX. Find AB.

Jun 20, 2020

#1
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AB works out to 4*sqrt(11).

Jun 20, 2020
#2
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In rectangle $$WXYZ$$,
$$A$$ is on side $$WX$$ such that $$AX = 4$$,
$$B$$ is on side $$YZ$$ such that $$BY = 18$$, and
$$C$$ is on side $$XY$$ such that angle $$ACB= 90^\circ$$ and
$$CY = 2CX$$.
Find $$AB$$.

$$\begin{array}{|lrcll|} \hline &AC^2 &=& 4^2+cx^2 \\ &CB^2 &=& (2cx)^2+18^2 \\ &x^2= AC^2+CB^2 &=& 4^2+cx^2+(2cx)^2+18^2 \\ &x^2 &=& 16+cx^2+ 4cx^2+ 324 \\ &x^2 &=& 5cx^2+ 340 \qquad (1) \\ \hline &x^2 &=& (18-4)^2 + (3cx)^2 \\ &x^2 &=& 14^2 + 9cx^2\\ &x^2 &=& 9cx^2 + 196 \qquad (2) \\ \hline (2)=(1): & 9cx^2 + 196 &=& 5cx^2+ 340 \\ & 9cx^2 - 5cx^2 &=& 340 - 196 \\ & 4cx^2 &=& 144 \\ & cx^2 &=& 36 \\ & \mathbf{cx} &=& \mathbf{6} \\ \hline (1): &x^2 &=& 5cx^2+ 340 \quad | \quad cx=6 \\ &x^2 &=& 5*6^2+ 340 \\ &x^2 &=& 520 \\ &x &=& 2\sqrt{130} \\ &\mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array}$$

AB is $$\approx \mathbf{22.8}$$

Jun 20, 2020