Please show us what you have tried and exactly where you are stuck.
In a right triangle the length of the altitude, \({\bf a}\), to the hypotenuse forms a mean proportional the parts.
Please show us what you have tried and exactly where you are stuck.
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Please share your work/thoughts about this assignment.
Hint: You know: AD = 3√2 & AB = 2√5 \(\displaystyle \ \ \to \ \ \) BD = √2?
Area = ?
But ABC is not a right triangle.In a right triangle the length of the altitude, \({\bf a}\), to the hypotenuse forms a mean proportional the parts.
Thus \(\dfrac{3\sqrt2}{{\bf a}}=\dfrac{{\bf a}}{\sqrt2}\)
right angleHere is another hint.
What kind of angle is angle BDA?
What can you conclude therefore about triangle BDA?
Base is 3√2 so b x h = √2 x 3√2 = 4√2You 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?
Missed that! thanksBut ABC is not a right triangle.
NO! Of course it is solvable. \(\Delta ABD\) is a right triangle.Does it make it unsolvable?
EXACTLY.A right
right angle
Close. We want triangle ABC, whose base is not 3√2, but ... what?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
You do realize that 3sqrt(2) = 3*sqrt(2)√2 x 3√2 = 4√2
That is exactly correct: \(AB=2\sqrt6\) makes the mean proportional work.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.
ABC is the triangle whose area you need to FIND.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
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.That is exactly correct: \(AB=2\sqrt6\) makes the mean proportional work.
Then \(BD=\sqrt6\) does not seem too far off.
When I said you were close, I missed the fact that you added where you said you were multiplying ...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