Question

What is the perimeter of a triangle which adjacent sides of the square angle are 9" each and the opposite side of that angle has 12'9"?...And what is the area of the previous triangle?

Answer 1

12'9" is `153` inches, and there's no real triangle with sides 9, 9 and 153. In a real triangle the sum of any two sides must exceed the third.

Even if there were such a triangle, it wouldn't be a right triangle.

Explanation:

Answer 2

Making the assumption described below, perimeter 30.75', area `40.5 '^2` .

Explanation:

I will assume you confused the ' and " and therefore meant the question to be "What is the perimeter of a triangle which adjacent sides of the square angle are 9' each and the opposite side of that angle has 12'9"?...And what is the area of the previous triangle?"

The perimeter is 9'+9'+12.75' = 30.75'

The formula for the area of a triangle is

`"area" = (b*h)/2`

Since the two 9' sides are perpendicular, b = h = 9'. Therefore

`"area" = (b*h)/2 = (9'*9')/2 = (81'^2)/2 = 40.5 '^2`

Note: the `'^2` above is `"feet"^2`.

I hope this helps,
Steve