Answer :
see the picture attached to better understand the problem
we know that
[tex]AB=4cm \\ AC=4 cm[/tex]
in the right triangle ADB
[tex]angle DBA=angle DAB \\ =45 degrees[/tex]
[tex]cos (DAB)= \frac{AD}{AB} [/tex]
where
AD is the adjacent side angle DAB (altitude)
AB is the hypotenuse triangle ADB
Angle DAB=[tex]45[/tex]°
so
solve for AD
[tex]AD=AB*cos (45)[/tex]
[tex]AD=4* \frac{ \sqrt{2}}{2} \\ AD=2 \sqrt{2} cm[/tex]
therefore
the answer is
the length of the altitude is [tex]2 \sqrt{2} cm[/tex]
we know that
[tex]AB=4cm \\ AC=4 cm[/tex]
in the right triangle ADB
[tex]angle DBA=angle DAB \\ =45 degrees[/tex]
[tex]cos (DAB)= \frac{AD}{AB} [/tex]
where
AD is the adjacent side angle DAB (altitude)
AB is the hypotenuse triangle ADB
Angle DAB=[tex]45[/tex]°
so
solve for AD
[tex]AD=AB*cos (45)[/tex]
[tex]AD=4* \frac{ \sqrt{2}}{2} \\ AD=2 \sqrt{2} cm[/tex]
therefore
the answer is
the length of the altitude is [tex]2 \sqrt{2} cm[/tex]

The length of the altitude drawn from the right angle to the hypotenuse is 2√2 cm
The new triangle
Since in the isosceles right triangle, the length of the altitude drawn from the right angle to the hypotenuse side creates a new right angle with hypotenuse side the length of isoceles right triangle.
Length of altitude
Let
- h = length of altitude and
- L = length of leg of isosceles right triangle = 4 cm
Since the altitude bisects the right angle, we have that
cos(90°/2) = h/L
cos45° = h/L
h = Lcos45°
h = 4 × 1/√2
h = 4 × 1/√2 × √2/√2
h = 4√2/2
h = 2√2 cm
So, the length of the altitude drawn from the right angle to the hypotenuse is 2√2 cm
Learn more about isosceles right triangle here:
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