Question

If the area of an isosceles right triangle is 8, what is the perimeter of the triangle?

Answer 1

ok that was wrong. since
\[\frac{1}{2}bh =8\]
\[bh=16\]
\[b=h=4\]

Answer 2

so both legs of the right triangle are 4
and the hypotenuse is
\[\sqrt{4^2+4^2}=\sqrt{32}=4\sqrt{2}\]

Answer 3

\[a. 8+\sqrt{2}\]

Answer 4

add up to get the perimeter, no it is
\[8+4\sqrt{2}\]

Answer 5

b. \[b. 8+4\sqrt{2}\]

Answer 6

yeah, that one

Answer 7

1/2 x^2 = 8

x^2 = 16

x = 4
Perimeter = 4 + 4 + 4√2 = 8 + 4√2

Answer 8

A = (1/2) a^2 = 8 ( isocelses)
-> a^2 = 16
=> a = 4
=> hyp = sqrt( 16 + 16) = 4 sqrt (2)
Thus P = 4 + 4 + 4V2 = 8 + 4V2 = 13.66

Answer 9

Note, this is isosceles *right triangle*, hence we can conclusively say that \( bh=16 \implies b=h=4\)