Answer:
Step-by-step explanation:
I can't make specific statements about the proof because the midpoint is missing.
Givens
There are two right angles created by where the perpendicular bisector meats MN. Both are 90 degrees.
MN is bisected by the point on MN where the perpendicular meets MN
The Perpendicular Bisector is is common to both triangles.
Therefore the two triangles are congruent by SAS
PM = PN Parts contained in Congruent triangles are congruent.
Answer:
A = 1/2(2πr)(r)
Explanation:
The circumference of the circle, 2πr, is the measure completely around the circle. When the pieces of the circle are rearranged, half of this circumference will be on the top of the parallelogram and half will be on the bottom. This means the base will be 1/2(2πr).
The approximate height of the parallelogram is the radius of the circle; this makes the area
A = 1/2(2πr)(r)
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From the question we can deduce the formula to be 16 + 3n = 76
3n - 76 - 16
3n = 60
n = 20
20 is the answer.
