Pythagoras with surds

Here is another hint.

What kind of angle is angle BDA?

What can you conclude therefore about triangle BDA?
 
You said: BD = √2？

Correct! That is the height of the triangle.

What is the base of this triangle?

Now use the known formula for area of a triangle.

What do you get?
 
Subhotosh Khan said:
You said: BD = √2？

Correct! That is the height of the triangle.

What is the base of this triangle?

Now use the known formula for area of a triangle.

What do you get?
Click to expand...
Base is 3√2 so b x h = √2 x 3√2 = 4√2
Then you have to divide by 2 which will be 2√2
 
The picture is very misleading, perhaps intentionally.

But on thinking about it more, it's also possible that they intended it to be a right triangle, but wrote [MATH]2\sqrt{5}[/MATH] where they meant [MATH]2\sqrt{6}[/MATH]. Of course, the proportions would still be off.
 
Student31 said:
Does it make it unsolvable?
Click to expand...
NO! Of course it is solvable. \(\Delta ABD\) is a right triangle.
Therefore \(BD^2=(2\sqrt5)^2-(3\sqrt2)^2=20-18=2\). So \(BD=\sqrt2\)
Area \(\Delta ABC=\frac{1}{2}(AC)(BD)\)
 
Student31 said:
A right

right angle
Click to expand...
EXACTLY.

So you know the length of two sides of a right triangle. How can you compute the length of the third side?

And that side represents the what of the overall triangle?

EDIT: For some reason, students, when first introduced to surds, forget that they are just numbers that cannot be expressed exactly in decimal form (like most fractions). That does not mean that you cannot work with them like any other number. And a square root of a rational number when squared is just a rational number, which you have been working with since third grade.
 
Student31 said:
√2 x 3√2 = 4√2
Click to expand...
You do realize that 3sqrt(2) = 3*sqrt(2)
Suppose 4sqrt(2) =sqrt(2)x3sqrt(2) = (sqrt(2)x3)xsqrt(2) by associative law. But only 4*sqrt(2) = 4sqrt(2). So then sqrt(2)x3 = 4 but that is not true. Please try again and as pointed out already use the correct base. Please post back.
 
Dr.Peterson said:
But on thinking about it more, it's also possible that they intended it to be a right triangle, but wrote [MATH]2\sqrt{5}[/MATH] where they meant [MATH]2\sqrt{6}[/MATH]. Of course, the proportions would still be off.
Click to expand...
That is exactly correct: \(AB=2\sqrt6\) makes the mean proportional work.
Then \(BD=\sqrt6\) does not seem too far off.
 
Student31 said:
Base is 3√2 so b x h = √2 x 3√2 = 4√2.........................................Incorrect
Then you have to divide by 2 which will be 2√2
Click to expand...
ABC is the triangle whose area you need to FIND.
Base of ABC = AC = ?
 
pka said:
That is exactly correct: \(AB=2\sqrt6\) makes the mean proportional work.
Then \(BD=\sqrt6\) does not seem too far off.
Click to expand...
Of course, even then, you couldn't use the mean proportional to solve the problem without first showing that it is a right triangle, which I did by using ABD to find the altitude and check it against the mean proportional, so it becomes a little circular.
 
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