Two corners of a triangle have angles of $\frac{π}{2}$ $( pi )/ 2$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $1$ $1$ , what is the longest possible perimeter of the triangle?

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Two corners of a triangle have angles of $\frac{π}{2}$ $( pi )/ 2$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $1$ $1$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2016-06-04T07:06:59

Longest possible perimeter is $3.4142$ $3.4142$ .

Explanation:

As two angles are $\frac{π}{2}$ $pi/2$ and $\frac{π}{4}$ $pi/4$ , third angle is $π - \frac{π}{2} - \frac{π}{4} = \frac{π}{4}$ $pi-pi/2-pi/4=pi/4$ .

For longest perimeter side of length $1$ $1$ , say $a$ $a$ , has to be opposite smallest angle which is $\frac{π}{4}$ $pi/4$ and then using sine formula other two sides will be

$\frac{1}{sin (\frac{π}{4})} = \frac{b}{sin (\frac{π}{2})} = \frac{c}{sin (\frac{π}{4})}$ $1/(sin(pi/4))=b/sin(pi/2)=c/(sin(pi/4))$

Hence $b = \frac{1 \times sin (\frac{π}{2})}{sin (\frac{π}{4})} = \frac{1 \times 1}{\frac{1}{\sqrt{2}}} = \sqrt{2} = 1.4142$ $b=(1xxsin(pi/2))/(sin(pi/4))=(1xx1)/(1/sqrt2)=sqrt2=1.4142$

and $c = 1$ $c=1$

Hence longest possible perimeter is $1 + 1 + 1.4142 = 3.4142$ $1+1+1.4142=3.4142$ .

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Two corners of a triangle have angles of $\frac{π}{2}$ $( pi )/ 2$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $1$ $1$ , what is the longest possible perimeter of the triangle?

1 Answer

Explanation:

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Two corners of a triangle have angles of $\frac{π}{2}$ $( pi )/ 2$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $1$ $1$ , what is the longest possible perimeter of the triangle?