Pythagoras with surds

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Does anyone know how to do this?

....................cm^2
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this assignment.

Hint: You know: AD = 3√2 & AB = 2√5     \displaystyle \ \ \to \ \ BD = ?

Area = ?
 
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?
 
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?
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.
 
Does it make it unsolvable?
NO! Of course it is solvable. ΔABD\Delta ABD is a right triangle.
Therefore BD2=(25)2(32)2=2018=2BD^2=(2\sqrt5)^2-(3\sqrt2)^2=20-18=2. So BD=2BD=\sqrt2
Area ΔABC=12(AC)(BD)\Delta ABC=\frac{1}{2}(AC)(BD)
 
A right

right angle
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.
 
√2 x 3√2 = 4√2
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.
 
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.
That is exactly correct: AB=26AB=2\sqrt6 makes the mean proportional work.
Then BD=6BD=\sqrt6 does not seem too far off.
 
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
ABC is the triangle whose area you need to FIND.
Base of ABC = AC = ?
 
That is exactly correct: AB=26AB=2\sqrt6 makes the mean proportional work.
Then BD=6BD=\sqrt6 does not seem too far off.
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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