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# geometry

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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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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 $$\frac{1}{2} \times 2\sqrt2 \times 2\sqrt2$$

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

Jan 5, 2021