Question

How would I do area and perimeter using radicals of square root? Please help!
A square has sides of length (sqrt(5)+sqrt(6)) inches each.
A) Find a simplified expression for the area of the square using square roots.
B) Find a simplified expression for the perimeter of the square using square roots.

Answer 1

area is:\[A=(\sqrt{5}+\sqrt{6})^2\]expand and simplify this.

Answer 2

So I would foil it to get the area then? @eseidl

Answer 3

yeah, then collect like terms.

Answer 4

So the area is 13sqrt30 @eseidl How would I find the perimeter?

Answer 5

A=25+36+2sqrt(30)
A=61+2sqrt(30)

Answer 6

Perimeter is just the sum of the lengths of all the sides. Since all four sides are equal, you have:\[P=4(\sqrt5 + \sqrt 6)\]Expand and collect like terms.

Answer 7

How did you get A=61 + 2sqrt(30) ? I dont understand... When I foiled it I got 11+2sqrt30 @eseidl

Answer 8

ok. \[(\sqrt5+\sqrt6)^2=\sqrt5\sqrt5+\sqrt5\sqrt6+\sqrt5\sqrt6+\sqrt6\sqrt6\]

Answer 9

\[25+2\sqrt30+36=...\]

Answer 10

25 and 36 are perfect squares don't you square them after? To get 5+6+2?

Answer 11

@eseidl

Answer 12

ah crap...yeah. sorry!

Answer 13

11+2sqrt(30) is correct.

Answer 14

Lol it's ok I thought I was losing my mind for a second... Question so is the answer 11+2sqrt30 or is 13sqrt30?

Answer 15

So I don't add the 2?

Answer 16

11+2sqrt(30). Not 13sqrt(30). You can't add the 11 and 2, they are "apples and oranges"

Answer 17

So then the perimeter is 4sqrt5 +4sqrt6 .... Right because the 5 and 6 are just numbers alone without the radical correct?
@eseidl

Answer 18

To see this, set them equal.
11+2sqrt30=13sqrt30
Now subtract 2sqrt30 from both sides
11=11sqrt(30)
obviously this is false.

Answer 19

correct.

Answer 20

Thank you for showing me that I like that..

Answer 21

YAY!!!! I did it!!! WOOOOOO! lol Thanks so much for helping me understand it better

Answer 22

no prob :)

Answer 23

:D