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Let \(\mathcal{R} \) denote the circular region bounded by x^2 + y^2 = 36. The lines x = 4 and y = 3 partition \(\mathcal{R} \) into four regions \(\mathcal{R}_1, \mathcal{R}_2, \mathcal{R}_3, and \mathcal{R}_4\). Let \([\mathcal{R}_i]\) denote the area of region \(\mathcal{R}_i\). If \([\mathcal{R}_1] > [\mathcal{R}_2] > [\mathcal{R}_3] > [\mathcal{R}_4] \), then compute \([\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]\).

 

Thanks!

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avatar+20807 
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Let \(\mathcal{R}\) denote the circular region bounded by x^2 + y^2 = 36.
The lines x = 4 and y = 3 partition \(\mathcal{R}\) into four regions \(\mathcal{R}_1, \mathcal{R}_2, \mathcal{R}_3, \text{ and } \mathcal{R}_4\).
Let \([\mathcal{R}_i]\) denote the area of region\( \mathcal{R}_i\).
If \([\mathcal{R}_1] > [\mathcal{R}_2] > [\mathcal{R}_3] > [\mathcal{R}_4]\), then compute \([\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4]\).

 

\(\text{Let $ \mathcal{R}_1 = \mathcal{R}_{11} + \mathcal{R}_{12} + \mathcal{R}_{13} + \mathcal{R}_{14} $ } \\ \text{Let $ \mathcal{R}_2 = \mathcal{R}_{21} + \mathcal{R}_{22} $ } \\ \text{Let $(1)\qquad \mathcal{R}_2 = \mathcal{R}_{13} + \mathcal{R}_{14} $ } \\ \text{Let $(2)\qquad \mathcal{R}_3 = \mathcal{R}_{12} + \mathcal{R}_{13} $ } \\ \text{Let $(3)\qquad \mathcal{R}_4 = \mathcal{R}_{21} $ } \\ \\ \text{Let $(4)\qquad \mathcal{R}_{13} = \mathcal{R}_{21} $ } \\ \text{Let $(5)\qquad \mathcal{R}_{1} -\mathcal{R}_{13}-\mathcal{R}_{14} = \mathcal{R}_{12} + \mathcal{R}_{11} $ }\)

 

\(\begin{array}{|rcll|} \hline [\mathcal{R}_1] - [\mathcal{R}_2] - [\mathcal{R}_3] + [\mathcal{R}_4] &=& ([\mathcal{R}_1] - \underbrace{[\mathcal{R}_2])}_{= [\mathcal{R}_{13}] + [\mathcal{R}_{14}]} - (\underbrace{[\mathcal{R}_3]}_{=[\mathcal{R}_{12}] + [\mathcal{R}_{13}] } - \underbrace{[\mathcal{R}_4]}_{=[\mathcal{R}_{21}] } ) \\\\ &=& ( \underbrace{[\mathcal{R}_1] - [\mathcal{R}_{13}] - [\mathcal{R}_{14}]}_{=[\mathcal{R}_{12}] + [\mathcal{R}_{11}] } ) -( [\mathcal{R}_{12}] + \underbrace{[\mathcal{R}_{13}]}_{=[\mathcal{R}_{21}]} - [\mathcal{R}_{21}] ) \\ \\ &=& ( [\mathcal{R}_{12}] + [\mathcal{R}_{11}] ) -( [\mathcal{R}_{12}] + [\mathcal{R}_{21}] - [\mathcal{R}_{21}] ) \\ \\ &=& ( [\mathcal{R}_{12}] + [\mathcal{R}_{11}] ) -[\mathcal{R}_{12}] \\ \\ &=& [\mathcal{R}_{11}] \\ \\ &=& 8 \times 6 \\ \\ &\mathbf{=}& \mathbf{ 48 } \\ \hline \end{array}\)

 

This last region is simply a rectangle of height 6 and width 8, so its area is 48.

 

Source: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwjoxuXK74ncAhVNr6QKHcfOByQQFgg3MAE&url=https%3A%2F%2Fservices.artofproblemsolving.com%2Fdownload.php%3Fid%3DYXR0YWNobWVudHMvYi84LzAwNWQ1ZDMzMmNlZmNhMmRiM2Q3YTg2YmVhZDE5NjFmYmIwYjUz%26rn%3DMjAxMGNvbnRlc3RlbnRpcmVkcmFmdHYxLjQucGRm&usg=AOvVaw3xDrFBFJINmMsnAYS_3zde

 

 

laugh

 Jul 6, 2018
edited by heureka  Jul 6, 2018

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