It should've been:
(6.2-1.6)-4.4+7.8
6,2-1.6-4.4+7.8=4.6-4.4+7.8=0.2+7.8=8
If you were working with a flat rectangle and you were given the diagonal,
you'd want to use the Pythagorean theorem to choose two sides whose
squares would add up to the square of the diagonal.
It works exactly the same with a 3-D box. We need three dimensions for
the box, whose squares add up to the square of the diagonal between
opposite corners. That's (4)² = 16.
So (L)² + (W²) + (H²) = 16 . From there, you're completely free to pick any
numbers you want, just as long as their squares add up to 16. There are
an infinite number of possibilities. Here are a few:
1 x 1 x √14
1 x 2 x √11
2 x 2 x √8
2 x 3 x √3
1 x 3 x √6
Answer:
Two possible lengths for the legs A and B are:
B = 1cm
A = 14.97cm
Or:
B = 9cm
A = 12cm
Step-by-step explanation:
For a triangle rectangle, Pythagorean's theorem says that the sum of the squares of the cathetus is equal to the hypotenuse squared.
Then if the two legs of the triangle are A and B, and the hypotenuse is H, we have:
A^2 + B^2 = H^2
If we know that H = 15cm, then:
A^2 + B^2 = (15cm)^2
Now, let's isolate one of the legs:
A = √( (15cm)^2 - B^2)
Now we can just input different values of B there, and then solve the value for the other leg.
Then if we have:
B = 1cm
A = √( (15cm)^2 - (1cm)^2) = 14.97
Then we could have:
B = 1cm
A = 14.97cm
Now let's try with another value of B:
if B = 9cm, then:
A = √( (15cm)^2 - (9cm)^2) = 12 cm
Then we could have:
B = 9cm
A = 12cm
So we just found two possible lengths for the two legs of the triangle.
I’m confused?? Do you just need 1/3 of 12? That’s 4 because 12 x 1 =12 and 12/3 is 4 lol