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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I do this in 3 ways:
Divide it into 10 equal pieces: So it will be .1 , .2, .3, .... .9 or 1/10 2/10
Divide it into 4 equal pieces: .25 ; .50 ; .75;
Divide it into 2 equal pieces: .5 or 1/2
Answer:
- domain: x ≥ 7
- range: f(x) ≥ 9
Step-by-step explanation:
The domain is the set of values of x for which the function is defined. Those are the values of x that make the square root argument non-negative:
x -7 ≥ 0
x ≥ 7
__
Since the square root cannot be negative, the sum of it and 9 cannot be less than 9. The range is ...
f(x) ≥ 9
Answer:
36.7
Step-by-step explanation:
let angle be x
tan x = 0.8
x = tan⁻¹ 0.8 (use calculator)
x = 36.6598
x = 36.7 (nearest tenth)