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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The total amount that Makayla needs to pay is equal to $530 + $760 which is equal to $1290. This value is lesser compared to the amount that Makayla is able to take home. Therefore, Makayla will not be in danger of credit overload and she will be able to provide for her other needs.
Answer:A fine of $0.75 is paid if the book returned by the due date.
No fine is paid if the book is returned 1 day after the due date.
A fine of $0.75 is paid if the book is returned 1 day after the due date.
Bit by bit clarification:
A fine of $0.75 is paid if the book returned by the due date.
No fine is paid if the book is returned 1 day after the due date.
A fine of $0.75 is paid if the book is returned 1 day after the due date.
I think its shear, because that is the only one in which the shape actually changes shape and size
Answer:G(-5)=21, G(-1)=5 G(0)=1 G(2)=-7 G(10)=-39
Step-by-step explanation:
