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Sagot :
To determine the length of one leg of an isosceles right triangle when given the altitude to the hypotenuse, let's break down the problem step by step.
An isosceles right triangle has two equal legs, and the hypotenuse is the side opposite the right angle. If we draw an altitude from the right angle to the hypotenuse, this altitude splits the hypotenuse into two equal segments and forms two smaller right triangles, which are also isosceles right triangles.
Let's denote each leg of the large right triangle as [tex]\( L \)[/tex].
In an isosceles right triangle, where the legs have length [tex]\( L \)[/tex], the hypotenuse [tex]\( H \)[/tex] can be calculated using the Pythagorean theorem:
[tex]\[ H = L\sqrt{2} \][/tex]
Since the altitude [tex]\( x \)[/tex] is drawn to the hypotenuse and it bisects the hypotenuse into two equal segments, each segment of the hypotenuse will have a length of:
[tex]\[ \frac{L\sqrt{2}}{2} = \frac{L}{\sqrt{2}} \][/tex]
Considering one of the smaller right triangles formed by the altitude, the leg of the smaller triangle perpendicular to the hypotenuse is of length [tex]\( x \)[/tex], and the other leg is [tex]\( \frac{L}{\sqrt{2}} \)[/tex]. Because this is a 45-45-90 triangle (isosceles right triangle), the altitude itself (being one leg of the small right triangle) is equal to the other leg of the small triangle, i.e., [tex]\( x = \frac{L}{\sqrt{2}} \)[/tex].
Solving for [tex]\( L \)[/tex]:
[tex]\[ x = \frac{L}{\sqrt{2}} \][/tex]
[tex]\[ L = x\sqrt{2} \][/tex]
Thus, the length of one leg of the large right triangle in terms of [tex]\( x \)[/tex] is:
[tex]\[ \boxed{x\sqrt{2}} \][/tex]
An isosceles right triangle has two equal legs, and the hypotenuse is the side opposite the right angle. If we draw an altitude from the right angle to the hypotenuse, this altitude splits the hypotenuse into two equal segments and forms two smaller right triangles, which are also isosceles right triangles.
Let's denote each leg of the large right triangle as [tex]\( L \)[/tex].
In an isosceles right triangle, where the legs have length [tex]\( L \)[/tex], the hypotenuse [tex]\( H \)[/tex] can be calculated using the Pythagorean theorem:
[tex]\[ H = L\sqrt{2} \][/tex]
Since the altitude [tex]\( x \)[/tex] is drawn to the hypotenuse and it bisects the hypotenuse into two equal segments, each segment of the hypotenuse will have a length of:
[tex]\[ \frac{L\sqrt{2}}{2} = \frac{L}{\sqrt{2}} \][/tex]
Considering one of the smaller right triangles formed by the altitude, the leg of the smaller triangle perpendicular to the hypotenuse is of length [tex]\( x \)[/tex], and the other leg is [tex]\( \frac{L}{\sqrt{2}} \)[/tex]. Because this is a 45-45-90 triangle (isosceles right triangle), the altitude itself (being one leg of the small right triangle) is equal to the other leg of the small triangle, i.e., [tex]\( x = \frac{L}{\sqrt{2}} \)[/tex].
Solving for [tex]\( L \)[/tex]:
[tex]\[ x = \frac{L}{\sqrt{2}} \][/tex]
[tex]\[ L = x\sqrt{2} \][/tex]
Thus, the length of one leg of the large right triangle in terms of [tex]\( x \)[/tex] is:
[tex]\[ \boxed{x\sqrt{2}} \][/tex]
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