Answer:
37 9/20
Step-by-step explanation:
Sorry but you didnt add a picture nor description. If you can, please re-ask this question with the following:
Details to your question
A picture or drawing
Thanks - Madilyn.
Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by

The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
Length = 2W
P= 2L + 2W
60 = 2(2W) + 2W
60= 4W +2W
60 = 6W
60/6W = 6W/6W
w = 10
Length = 2(10)
Length = 20
P= 2L + 2W
60= 2(20) + 2(10)
60= 40 +20
60 = 60
The missing length in the right triangle as given in the task content is; 156.
<h3>What is the missing length indicated?</h3>
It follows from the complete question that the triangle given is a right triangle and the missing length (longest side) can be evaluated by means of the Pythagoras theorem as follows;
x² = 144² + 60²
x² = 20736 + 3600
x² = 24,336
x = √24336
x = 156.
Remarks: The complete question involves a right triangle and the missing length is the longest side.
Read more on Pythagoras theorem;
brainly.com/question/343682
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