Question

the width of a rectangle is 5 feet and the diagonal is 9 feet. what is the area of the rectangle?

Answer 1

\[2\sqrt{14}\]\[10\sqrt{14}\]2

14

none of above

Answer 2

@mathstudent55

Answer 3

@mathmate

Answer 4

This is what the problem states.
The width is 5 and the diagonal is 9.
Do you know the Pythagorean theorem?
You can use it to find the length.
Then the area = length * width

Answer 5

ok

Answer 6

would it be\[\sqrt{81}\]

Answer 7

81 is part of the equation but it's not the answer.

\(5^2 + x^2 = 9^2\)

\(25 + x^2 = 81\)

\(x^2 = 56\)

\(x = \sqrt{56} = 2\sqrt{14} \)

Answer 8

Now that you know the length and width, just multiply them together.

Answer 9

would you divide 56 and 2?

Answer 10

\(A = LW\)

\(A = 2\sqrt{14} \times 5\)

\(A = 10\sqrt{14} \)

Answer 11

ty

Answer 12

Use the Pythagorean theorem again.

\(3^2 + 4^2 = x^2\)

Square the 3 and the 4 and add the squares together.
Then take the square root to find x.