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In a certain isosceles right triangle, the altitude to the hypotenuse is 4sqrt{2}. What is the area of the triangle?
I suck at geometry... please help
 Jan 29, 2014
 #1
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Soup:

In a certain isosceles right triangle, the altitude to the hypotenuse is 4sqrt{2}. What is the area of the triangle?
I suck at geometry... please help



Here's the hint I'll give you

triangle.png

we know the left up corner and the right down corner are 45 degrees because the two sides adjecent to the right angle are of equal length.

We know the altitude always makes a 90 degree angle with the hypotunes,

From there we know the other two 45 degrees,

now with the use of properties of isosceles right angled triangles and pythagoras theorem you should be able to solve this.

If you're still in trouble or dont know pythagoras theorem, let me know
 Jan 29, 2014
 #2
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The legs are both 4. Add 4 + 4 + 4 (root 2). 8 + 4 (root 2) is your answer.
 Jan 29, 2014
 #3
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Halper:

The legs are both 4. Add 4 + 4 + 4 (root 2). 8 + 4 (root 2) is your answer.



I am sorry halper but I believe that that is not entirely the correct answer.

The two triangles that were formed by the altitude line are also isosceles triangles because the angles are 90, 45 and 45 degrees.

Therefore, we can use pythagoras to calculate

a 2 = b 2 + c 2

a = sqrt(b 2 + c 2)

We use a for the bottom, b = 4sqrt(2) for the altitude line and c = 4sqrt(2) for half of the hypotenuse.

Then the bottom is given by

[input]sqrt((4sqrt(2))^2 + (4sqrt(2))^2))[/input]

We know that the bottom is equal to the left side since it is an isosceles triangle,

The area of a triangle is given by base*height*1/2 which then amounts to 8*8*1/2 = 32

Hence the area of the triangle is 32
 Jan 29, 2014

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