What is the surface area of the cylinder with height 8 in and radius 2 in? Round your answer to the nearest thousandth.
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Using the midpoint formula
So, your answer would be (2,6.5).
So, your answer would be (2,6.5).
Let's begin noting that a triangle is isosceles if and only if two of its angles are congruent. We can thus find the angle <ABP, recalling that the sum of the interior angles of a triangle is equal to 180°.
Finally, let point K be the intersection between segments BC and PQ, and let's note that the triangle PQB is a right isosceles triangle, since all the angles in a square are equal to 90°, and the two triangles APB and BQC are congruent.
Therefore, the angle BKQ is equal to 180-50-45=85°.
Of course angle BKP=180-85=95°.
Hope this helps :)