super hard area question

[hot pulsar]

what do you have so far

[spring arrow]

But instead it’s supposed to be 1/3^2 + 1/5^2 = 1/c^2
ab = sqrt 2
1/a^2+1/b^2 = 1/c^2
c^2/a^2b^2=1/c^2
c^4=a^2b^2
c^4=2
c=forth root of 2
a^2+b^2=sqr2
a^2b^2=2
Solving we get
(a+b)^2-2ab=sqr(2)
(a+b)^2=2ab+aqr(2)
(a+b)^2=2sqr(2)+sqr(2)=-sqr(2)
so the 2nd triangle will have area 1/2 (1/c )(1/b)
and (1/c)^2+(1/b)^2 = (1/a)^2

this makes me wanna die
To know our working is right Hyp should be longest side and a and b together should be equal to c squared then reverse and root the c2 value to find C, then a+b+c is in the proximity of 4

[hot pulsar]

why have you set ab = 1/ab

this assumption is wrong

[spring arrow]

Yeah I changed. It in. The working here

To solid solution in text

But I didn’t get. The answer

Can you help here?

[hot pulsar]

im ngl i dont get the soln either

[spring arrow]

I got no real solutions?

It’s a negative

I’m stuckkkkkkkk

Can you try this one?

I worked through it too.

[spring arrow]

(a+b)^2=a^2+b^2+2ab=c^2+2ab=3ab=3sqrt(3)

@hot pulsar What do i do from ehre

[indigo cargo]

Why did u write out the whole q can’t u just screenshot it

[gleaming mauve]

1/c is not the hypotenuse

once you choose the side lengths one of triangle the other can’t be chosen arbitrarily

r u still stuck on this 1 lol

[spring arrow]

yh