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>
Answer:
-10, -3, -2.001, 1, -7, -2.1, -5, -2.01
Step-by-step explanation:
Answer:
120 sq. units
plus them all and add 100
Answer:
x= 7/5
Step-by-step explanation:
The marked angles are opposite angles, and as such they have the same measure. So, we have
Subtract 2a and 11 from both sides to get
Divide both sides by 4 to get
Now that we know the value of a, we can compute the measure of the angles. We can also verify that the solution we found is correct by verifying that both expressions actually give the same result: