Question

Look at the rectangle and the square

Answer 1

Answer:

No Ada is not correct

Step-by-Step Explanation:

The length of diagonal SQ is not two times the length of diagonal OM. When you split the rectangle using the diagonal you can make a right triangle and use the Pythagorean Theorem to the rest of it. 

This would be the Rectangle:

a^2+b^2=c^2

8^2+16^2=c^2

64+256+c^2

320=c^2

√320

17.8

This would be the Square: 

a^2+b^2=c^2

8^2+8^2=c^2

64+64=c^2

128=c^2

√128

11.3

Now when I divide these answers that I came up with I get 1.7 which is not equal to 2 therefore SQ is not two times bigger than OM. 

I hope this helped :3

Answer 2

Answer:

NO. Ada is not correct.

Step-by-step explanation:

Using Pythagorean Theorem, find the length of the diagonal of the rectangle and the square, respectively.

✔️Diagonal of the Rectangle:

$a^2 + b^2 = c^2$

Where,

a = 8 in.

b = 16 in.

c = hypotenuse (longest side of a right ∆)

Plug in the values into the equation

$8^2 + 16^2 = c^2$

$64 + 256 = c^2$

$320 = c^2$

Take the square root of both sides

$\sqrt{320} = \sqrt{c^2}$

$17.9 = c^2$ (nearest tenth)

Length of diagonal SQ = 17.9 in

✔️Diagonal of the Rectangle:

$a^2 + b^2 = c^2$

Where,

a = 8 in.

b = 8 in.

c = hypotenuse (longest side of a right ∆)

Plug in the values into the equation

$8^2 + 8^2 = c^2$

$64 + 64 = c^2$

$128 = c^2$

Take the square root of both sides

$\sqrt{128} = \sqrt{c^2}$

$11.3 = c^2$ (nearest tenth)

Length of diagonal OM = 11.3 in.

SQ is not two times the length of OM.

Therefore, Ada is not correct.