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Find the area of the shaded region if r = 1/4 and each side of the square is 1.

 Jul 27, 2020
 #1
avatar+1326 
+2

Angle BDC = 45º

1/    Find BD

2/    Find BO

3/    Find DO

4/    Find angle ODM

5/    Angle NDC = ∠BDC - ∠ODM

6/    NC = tan(∠NDC)

Area = 1/2( NC * DC )

 

 Jul 27, 2020
edited by Dragan  Jul 27, 2020
 #2
avatar+25556 
+2

Find the area of the shaded region if r = 1/4 and each side of the square is 1.

 

\(\begin{array}{|rcll|} \hline \mathbf{AC^2} &=& \mathbf{(1-r)^2+(1-r)^2} \\ AC^2 &=& 2(1-r)^2 \\ AC &=& (1-r)\sqrt{2} \\ \mathbf{AC} &=& \mathbf{\sqrt{1.125}} \\ \\ \mathbf{AB^2+r^2}&=& \mathbf{AC^2} \\ AB^2+0.0625&=& 1.125 \\ AB^2&=& 1.0625 \\ \mathbf{AB} &=& \mathbf{\sqrt{1.0625}} \\ \hline \end{array} \)

 

cos-rule:

\(\begin{array}{|rcll|} \hline \mathbf{r^2} &=& \mathbf{AC^2+AB^2-2*AC*AB*\cos(\varphi)} \\\\ \cos(\varphi) &=& \dfrac{AC^2+AB^2-r^2}{2*AC*AB} \\\\ \cos(\varphi) &=& \dfrac{1.125+1.0625-0.0625}{2*\sqrt{1.125}*\sqrt{1.0625}} \\\\ \cos(\varphi) &=& \dfrac{2.125}{2*\sqrt{1.1953125}} \\\\ \cos(\varphi) &=& 0.97182531581 \\\\ \varphi &=& 13.6330222254^\circ \\ \hline 45^\circ - \varphi &=& 31.3669777746^\circ \\ \mathbf{\tan(45^\circ - \varphi)} &=& \mathbf{\dfrac{x}{1} } \\ \tan(45^\circ - \varphi) &=& x \\ x &=& \tan(45^\circ - \varphi) \\ x &=& \tan(31.3669777746^\circ) \\ \mathbf{x} &=& \mathbf{0.60961179680} \\ \hline \text{Area of the shaded region} &=& \dfrac{x*1}{2} \\ \text{Area of the shaded region} &=& \dfrac{0.60961179680}{2} \\ \mathbf{\text{Area of the shaded region}} &=& \mathbf{0.30480589840} \\ \hline \end{array}\)

 

 

laugh

 Jul 27, 2020

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