The length of the hypotenuse of a right isosceles triangle is 4√3 What is the perimeter of the triangle?

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Answer 1 · 2015-06-14T00:16:58

Permeter $= 16.7$ $=16.7$

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Answer 2 · 2015-06-15T13:35:10

The perimeter of the triangle is " " $16.72616$ $color(blue)16.72616$ units.

Perimeter=AB+BC+CA

$P = A B + A C + 4 \sqrt{3}$ $P=AB+AC+4sqrt(3)$ [since AC=hypotenuse]

$P = A C + A C + 4 \sqrt{3}$ $P=AC+AC+4sqrt(3)$ [since it is a isosceles triangle]

$p = 2 A C + 4 \sqrt{3}$ $p=2AC+4sqrt(3)$ $(1)$ $" "color(blue) ((1))$

![http://images.tutorvista.com/cms/images/67/area-of-right-triangle]( useruploads.socratic.org )

$sin θ$ $sintheta$ =opposite side/hypotenuse

${sin 45}^{\circ} = \frac{A C}{C B}$ $sin45^@=(AC)/(CB)$

$A C = (B C) \times {sin 45}^{\circ}$ $AC=(BC)xxsin45^@$

$A C = (B C) \times \frac{1}{\sqrt{2}}$ $AC=(BC)xx1/(sqrt(2))$ [from angle table]

$A C = (4 \sqrt{3} \times \frac{1}{\sqrt{2}})$ $AC=(4sqrt(3)xx1/sqrt2)$

$A C = \frac{4 \sqrt{3}}{\sqrt{2}}$ $AC=(4sqrt3)/(sqrt2)$

substitute the above value in $(1)$ $" "color(blue) ((1))$

$P = (2 \times \frac{4 \sqrt{3}}{\sqrt{2}}) + (4 \sqrt{3})$ $P=(2xx(4sqrt3)/sqrt2)+(4sqrt3)$

$P = (\frac{8 \sqrt{3}}{\sqrt{2}}) + 4 \sqrt{3}$ $P=((8sqrt3)/sqrt2)+4sqrt3$

$P =$ $P=$ $16.72616$ $color(blue)16.72616$ units.