We can find the diameter by dividing the circumference by 3.14
d = 22
Thus, our radius is 11.
r = 11
Now, let's use the area formula (
) to solve for the area.
a =
Answer:
A) mPQ = 71º
B) mSR = 161º
C) mQRT = 199ª
D) mPSR = 270º
E) mPS = 109º
Step-by-step explanation:
We know that mST is 19º and QR is also 19º because they're opposite angles. We also know that angle PUR is 90º.
If we subtract 19º from 90º we get 71º for mPQ.
We also know that TR is a 180º angle. Using this if we take 19º from 180º we're left with 161º for mSR.
Using the 180º from angle TR, as well as the 19º from mQR we know that mQRT has to be 199º.
A circle is 360º and a right angle is 90º. That means that mPSR is 270º.
Knowing that TR is 180º and that PR is 90º, PTº must be supplementary making it also 90º. Adding the 19º from ST to the 90º from TP we know that PS is 109º
Answer:
Areas of the colored parts are equal
Step-by-step explanation:
Given Mauricio divided two identical rectangles into equal parts.
Now given Mauricio colored one part of each rectangle. we have to tell about the true statement of colored parts.
Two identical rectangles means the rectangle having equal areas. He divides these two into equal parts and shaded one of each part
Let area of rectangle is x
∴ Colored part area = → (1)
Similarly, other rectangle area identical to above rectangle is x
∴ Colored part area = → (2)
From above eq (1) and (2), we get
Areas of the colored parts are equal which is equals to