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Help with finding x 

 

 

 Jun 2, 2019
 #1
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Let the distance between the center of the base and the northwest corner of the base be  y m

Distance between the center of the base and the west side of the base  =  3/2 m  =  1.5 m

Distance between the center of the base and the north side of the base  =  1.5/2 m  =  0.75 m

 

 

By the Pythagorean Theorem,

 

\((1.5)^2+(0.75)^2\ =\ y^2\\~\\ (\frac32)^2+(\frac34)^2\ =\ y^2\\~\\ \frac94+\frac{9}{16}\ =\ y^2\\~\\ \frac{45}{16}\ =\ y^2\\~\\ y^2\ =\ \frac{45}{16}\)

 

Again by the Pythagorean Theorem,

 

\(y^2+2^2\ =\ x^2\\~\\ \frac{45}{16}+4\ =\ x^2\\~\\ \frac{109}{16}\ =\ x^2\\~\\ x\ =\ \sqrt{\frac{109}{16}}\\~\\ x\ =\ \frac{\sqrt{109}}{4}\)___

 Jun 2, 2019

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