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:
D. 17.
Step-by-step explanation:
The nth term of an arithmetic sequence (an) is a1 + d(n - 1) where a1 = first term and d = the common difference.
In this sequence d = a2 - a1 = 5 - 2 = 3.
a6 = 2 + 3(6 - 1)
= 2 + 3*5
= 17.
There are 2 ways to do this....
(1) 20(1.15) = 23
(2) 20 + 0.15(20) = 20 + 3 = 23