Two corners of a triangle have angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{2}$ $pi / 2$ . If one side of the triangle has a length of $6$ $6$ , what is the longest possible perimeter of the triangle?

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Two corners of a triangle have angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{2}$ $pi / 2$ . If one side of the triangle has a length of $6$ $6$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2018-05-20T23:10:50

$12 + 6 \sqrt{2}$ $12 +6sqrt2$
or
$\approx 20.49$ $~~20.49$

Explanation:

okay the total angles in triangle are $π$ $pi$

$π - \frac{π}{4} - \frac{π}{2}$ $pi - pi/4 - pi/2$

$\frac{4 π}{4} - \frac{π}{4} - \frac{2 π}{4} = \frac{π}{4}$ $(4pi)/4 - pi/4 - (2pi)/4 = pi/4$

so we have a triangle with angles: $\frac{π}{4}, \frac{π}{4}, \frac{π}{2}$ $pi/4,pi/4,pi/2$ so 2 sides have the same length and the other is the hypotenuse.

using the Pythagorean theorem:

$a^{2} + b^{2} = c^{2}$ $a^2 + b^2 = c^2$

we know that the hypotenuse is longer than the other 2 sides:

$c = \sqrt{a^{2} + b^{2}}$ $c = sqrt(a^2 + b^2)$

$c = \sqrt{6^{2} + 6^{2}}$ $c = sqrt(6^2 + 6^2)$

$c = \sqrt{36 + 36} = 6 \sqrt{2} \approx 8.49$ $c = sqrt(36 + 36) = 6sqrt2~~8.49$

so the permitter is:

$6 + 6 + 6 \sqrt{2} = 12 + 6 \sqrt{2} \approx 20.49$ $6+6+6sqrt2 = 12 +6sqrt2 ~~20.49$

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Two corners of a triangle have angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{2}$ $pi / 2$ . If one side of the triangle has a length of $6$ $6$ , what is the longest possible perimeter of the triangle?

1 Answer

Explanation:

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Two corners of a triangle have angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{2}$ $pi / 2$ . If one side of the triangle has a length of $6$ $6$ , what is the longest possible perimeter of the triangle?