Squaring the Square Root of a Number?

[geekily]

I have the problem "find the area of a rhombus." It shows the figure cut into 4 right triangles, so I assume that the first step is to find the area of the triangle so I can multiply it by four and get the area of the rhombus. The base of 1 triangle is the square root of 8, and the hypotenuse is 4. I plugged it in to the Pythagorean Theorem as square root of 8 + b squared = 4 squared. This became b squared = 16 - square root of 8. To get b by itself, I take the square root of the other side, so 16 becomes 4. However, what does the square root of 8 become? I'm stuck!

Thanks for your help!

Oh, I forgot to add that my teacher wants the answers as exact as possible, so she doesn't want us to actually find the square root (so I can't just change square root of 8 to 2.828427125.)

 

[Loren]

The area of a rhombus can be found with the formula \(\displaystyle A=d_1\cdot d_2\) where the d_1 and d_2 are the diagonals. Does this take care of your problem?

 

[Loren]

I might add...

You say "b squared = 16 - square root of 8" which becomes "4 - whatever the square root of 8 is".

If I read that correctly, that is not right.

\(\displaystyle b^2 = 16 - \sqrt{8}\)

Take square root of each side getting...

\(\displaystyle b = \sqrt{16 - \sqrt{8}} = \sqrt{16 - 2\sqrt{2}}\) and that's about as simple as you can get it. you might take it to \(\displaystyle b = \sqrt{2(8 - \sqrt{2})}\) but most would leave it as is.

 

[geekily]

Sorry, but I'm still confused. This problem is from part B of my book (no answers), so I went back to part A to find a similar problem that would have answers. The first one I found is the exact same problem, except that 2 is the base of one of the triangles inside the rhombus and 4 is the hypotenuse. The book says the answer is 4 square root 3. In the second one, the base of one of the triangles in the rhombus is 1 and the hypotenuse is 4. The book says the answer is 2 square root 15. In both of these cases, multiplying the diagonals doesn't seem to give me the same answer.

Thanks for your help!

 

[Denis]

the base of one of the triangles in the rhombus is 1 and the hypotenuse is 4. The book says the answer is 2 square root 15. In both of these cases, multiplying the diagonals doesn't seem to give me the same answer.

Area of rhombus = product of diagonals / 2 ; I think Loren made 1st mistake this year

With hypotenuse = 4 and a leg = 1, then other leg = sqrt(16 - 1) = sqrt(15).
So area of 1 triangle = 1 * sqrt(15) / 2 = sqrt(15) / 2;
so area 4 triangles = 4sqrt(15) / 2 = 2sqrt(15) : OK :?:

The diagonals are 1+1 = 2 and sqrt(15) + sqrt(15) = 2sqrt(15);
so area is 2 * 2sqrt(15) / 2 = 2sqrt(15) : same as above

Are you ok now :?:

 

[Mrspi]

I have the problem "find the area of a rhombus." It shows the figure cut into 4 right triangles, so I assume that the first step is to find the area of the triangle so I can multiply it by four and get the area of the rhombus. The base of 1 triangle is the square root of 8, and the hypotenuse is 4. I plugged it in to the Pythagorean Theorem as square root of 8 + b squared = 4 squared. This became b squared = 16 - square root of 8. To get b by itself, I take the square root of the other side, so 16 becomes 4. However, what does the square root of 8 become? I'm stuck!

Thanks for your help!

Oh, I forgot to add that my teacher wants the answers as exact as possible, so she doesn't want us to actually find the square root (so I can't just change square root of 8 to 2.828427125.)

If one leg of the right triangle is b, one leg is sqrt(8), and the hypotenuse is 4, you can use the Pythagorean Theorem:

leg<SUP>2</SUP> + leg<SUP>2</SUP> = hypotenuse<SUP>2</SUP>

Substitute the values you're given:

b<SUP>2</SUP> + (sqrt(8))<SUP>2</SUP> = 4<SUP>2</SUP>

Here's the part I think you're missing....(sqrt(8)]<SUP>2</SUP> = sqrt(8*8) or just 8.....

b<SUP>2</SUP> + 8 = 16

b<SUP>2</SUP> = 8

b = sqrt(8)

ok...now you know that in one of the little right triangles, each leg is sqrt(8).

The area of one of those triangles can be found using the formula

A = (1/2)*leg*leg
A = (1/2)*sqrt(8)*sqrt(8)
A = (1/2)*sqrt(64)
A = (1/2)*8
A = 4

and you have four triangles like that....

You were also given in one of the previous responses one of my FAVORITE formulas...for any rhombus, Area = (1/2)(d<SUB>1</SUB>)(d<SUB>2</SUB>) where d<SUB>1</SUB> and d<SUB>2</SUB> are the lengths of the two diagonals.

This works nicely here.

 

[geekily]

Oh, Mrspi, thank you SO much - you are a lifesaver. That makes so, so much sense now!

Just out of curiosity: If I were to use your Area = (1/2)(d1)(d2) formula, since sqrt(8) is only the length of half of the diagonal, what would the whole diagonal be? I thought it'd be sqrt(16), but then, that's 4, and (1/2)(4)(4) is only 8, versus the 16 that I got with the other method.

Thank you so much - you don't know what a relief this is!