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The perimeter of right isosceles ΔABC with midsegment DE is 16 + 8√2.
If right isosceles ΔABC has hypotenuse length h, then the two other sides are congruent.
side a = side b
Using Pythagorean theorem, c^2 = a^2 + b^2
h^2 = a^2 + b^2 a = b
h^2 = 2a^2
a = h/√2
If DE is a midsegment not parallel to the hypotenuse, then it is a segment that connects the midpoints of one side of a triangle and the hypotenuse. See photo for reference.
ΔABC and ΔADE are similar triangles.
a : b : h = a/2 : 4 : h/2
If a/2 = a/2, then b/2 = 4.
b/2 = 4
b = 8
If a = b, then a = 8.
If a = h/√2, then
8 = h/√2
h = 8√2
Solving for the perimeter,
P = a + b + h
P = 8 + 8 + 8√2
P = 16 + 8√2
P = 27.3137085
To learn more about midsegment: brainly.com/question/7423948
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Answer:
m∠ABC = 45°
Step-by-step explanation:
See the attached figure.
As shown in the figure
Line C B extends through point D to form the exterior angle that is 135 degrees.
So, m∠ABD = 135°
But m∠ABC + m∠ABD = 180°
∴m∠ABC = 180° - m∠ABD = 180° - 135°= 45°
Also, we should know that
m∠ABD = m∠BAC + m∠ACB
So, m∠ACB = 135° - 75° = 60°
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<u>Very important note:</u>
If we replaced the location of B and C
SO, m∠ABC = 60° and m∠ACB = 45°
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