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In a certain iscoceles right triangle, the altitude to the hypotenuse has length 4. What is the area of the triangle?

 Jan 4, 2021
 #1
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Fin side length  

s^2 + s^2 = 4^2
2s^2 = 16

s = sqrt 8

 

area = 1/ 2   Base * height 

        = 1/2   Sqrt8   * sqrt 8  =  4 units2

 Jan 4, 2021
 #2
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*** error ****   Misread Q...

Guest Jan 4, 2021
 #3
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...if see if reflec answers correctly....

Guest Jan 4, 2021
 #5
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Since altitude to hypotenuse is 4 

tan45 = 4/(1/2b)

1 = 4/(1/2b)

b = 8

 

area = 1/2 b * h = 1/2 (8)(4)= 16 units 2

Guest Jan 4, 2021
 #4
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So the first thing to notice is that it is a 45 - 45 - 90 right triangle.

This means that the ratio of the sides are 1 - 1 - \(\sqrt2\)

This means that one leg has the length of\(4\over\sqrt2\) which simplifies to \(2\sqrt2\)

Now because it is a right triangle, you get \(\frac{1}{2} \times 2\sqrt2 \times 2\sqrt2\)

So the answer is 4.

 Jan 4, 2021
 #6
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In a certain isosceles right triangle, the altitude to the hypotenuse has length 4. What is the area of the triangle?

 

Area = 42 laugh

 Jan 5, 2021

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