Given the perimeter a and the area b^2 of a right triangle, find its hypotenuse.?

Question

Please give a little detailed description, I get confused easily!

Answer 1

l+l´+h = a ( l ,l´are the legs)
l*l´= 2b^2
and l^2+l´^2=h^2
l^2+l´^2 = (l+l´)^2 -2l*l´= (a-h)^2-4b^2
so
h^2=(a-h)^2-4b^2 where the only unknown is h
h^2=a^2-2ah+h^2-4b^2
so
h=(4b^2-a^2)/2a

Answer 2

a2+b2=c2 is for excellent triangles if 2 facets are unknown 0.33 side could be got here across... Perimeter upload the dimensions of the three facets i assume you recommend section not quantity on the grounds that triangles are 2-d not 3-D and section is got here across a million/2( base length) (top) and ultimately sin cos tan in basic terms undergo in innovations soh cah toa soh- sin=(opposite side)/Hypotnuse cah- cos= (adjoining side)/Hypotnuse toa- tan= (opposite side)/(adjoining side) basically reliable for excellent triangles wish this helped