Question

@johnweldon1993

Answer 1

I'm gonna do it the way I did before okay? Won't be like your study guide :)

Answer 2

yeah thats fine hun

Answer 3

\[\large \frac{x}{12} = \frac{y}{30}\]
\[\large y = \frac{30x}{12}\]

Area is 15 so we have

\[\large \frac{1}{2}xy = 15\]
\[\large y = \frac{30}{x}\]

Set them equal to each other

\[\large \frac{30x}{12} = \frac{30}{x}\]
\[\large 30x^2 = 360\]
\[\large x^2 = 12\]
\[\large x = \sqrt{12}\]

Answer 4

i thought x would have a number infront of it like would it be 6square root of 2

Answer 5

Now if x = √12

Then

\[\large \frac{1}{2}xy = 15\]

becomes

\[\large \sqrt{12}y = 30\]
\[\large y = \frac{30}{\sqrt{12}}\]
Rationalize the denominator
\[\large y = \frac{30}{\sqrt{12}} \times \frac{\sqrt{12}}{\sqrt{12}} = \frac{30\sqrt{12}}{12}\]

Reduce the fraction

\[\large y = \frac{5\sqrt{12}}{2}\]

Answer 6

but i cant put y in like that

Answer 7

Lol grr, alright..let me use the method they did that last time..see if that's any better

But...mine checks out :)

Answer 8

okay

Answer 9

Alright so based on your last one..

Area of bigger = \(\large \frac{1}{2} 30 \times 12 = 180 \)

\[\large \frac{180}{30} = 12 \rightarrow \sqrt{12} = 2\sqrt{3}\]

If that is true \(\huge \frac{30}{2\sqrt{3}} \times \frac{2\sqrt{3}}{2\sqrt{3}}= \frac{60\sqrt{3}}{4 \times 3} = \frac{60\sqrt{3}}{12} = 5\sqrt{3}\)

so

\[\large x = 2\sqrt{3}\]
\[\large y = 5\sqrt{3}\]

Answer 10

thats better lol thanks

Answer 11

Lets check that too just to make sure

\[\large \frac{1}{2}xy = 15\]
\[\large xy = 30\]
\[\large 2\sqrt{3} \times 5\sqrt{3} = 30\]
\[\large 10 \times 3 = 30\]
\[\large 30 = 30\]

So yeah that works too...lol just different methods :)