I tried to solve this one by calculating the length of one of the diagonals: √2 *x, then since half of the diagonal makes up the height of a triangle and the other half diagonal becomes the base, I get Area of ABC: ( (√2 * x)/2 ) ^2 = 2 then I get x=2, where x stands for a side of the square. To find the perimeter I multiply 2 by 4 and get 8. Isn’t this a valid method for this problem?
Sorry, let me retry that comment. I didn’t read yours carefully enough the first time!
I think you’re not dividing by 2 in your area calculation! The area of a triangle is (1/2)bh, and I think you’re just doing bh. Right? b = (x√2)/2 and h = (x√2)/2, so (1/2)bh should be (1/2)((x√2)/2)^2.
That said, I think you’re also starting from a harder place than necessary. Start with what you know and work from there, rather than trying to start from x. You know the area of the small triangle, and you know it’s an isosceles right triangle, so calculate the legs. Then multiply that by √2—the side of the square is the hypotenuse of the small triangle.
Comments (3)
I tried to solve this one by calculating the length of one of the diagonals: √2 *x, then since half of the diagonal makes up the height of a triangle and the other half diagonal becomes the base, I get Area of ABC: ( (√2 * x)/2 ) ^2 = 2 then I get x=2, where x stands for a side of the square. To find the perimeter I multiply 2 by 4 and get 8. Isn’t this a valid method for this problem?
Sorry, let me retry that comment. I didn’t read yours carefully enough the first time!
I think you’re not dividing by 2 in your area calculation! The area of a triangle is (1/2)bh, and I think you’re just doing bh. Right? b = (x√2)/2 and h = (x√2)/2, so (1/2)bh should be (1/2)((x√2)/2)^2.
That said, I think you’re also starting from a harder place than necessary. Start with what you know and work from there, rather than trying to start from x. You know the area of the small triangle, and you know it’s an isosceles right triangle, so calculate the legs. Then multiply that by √2—the side of the square is the hypotenuse of the small triangle.
thanks