What is the sum of the measures, in degrees, of the interior angles of a 16- sided polygon?
1 answer:
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Notice that every pair of point (x, y) in the original picture, has become (-y, -x) in the transformed figure.
Let ABC be first transformed onto A"B"C" by a 90° clockwise rotation.
Notice that B(4, 1) is mapped onto B''(1, -4). So the rule mapping ABC to A"B"C" is (x, y)→(y, -x)
so we are very close to (-y, -x).
The transformation that maps (y, -x) to (-y, -x) is a reflection with respect to the y-axis. Notice that the 2. coordinate is same, but the first coordinates are opposite.
ANSWER:
"<span>a 90 clockwise rotation about the origin and a reflection over the y-axis</span>"
Let ABC be first transformed onto A"B"C" by a 90° clockwise rotation.
Notice that B(4, 1) is mapped onto B''(1, -4). So the rule mapping ABC to A"B"C" is (x, y)→(y, -x)
so we are very close to (-y, -x).
The transformation that maps (y, -x) to (-y, -x) is a reflection with respect to the y-axis. Notice that the 2. coordinate is same, but the first coordinates are opposite.
ANSWER:
"<span>a 90 clockwise rotation about the origin and a reflection over the y-axis</span>"
Area inside the semi-circle and outside the triangle is (91.125π - 120) in²
Solution:
Base of the triangle = 10 in
Height of the triangle = 24 in
Area of the triangle =
Area of the triangle = 120 in²
Using Pythagoras theorem,
Taking square root on both sides, we get
Hypotenuse = 23 inch = diameter
Radius = 23 ÷ 2 = 11.5 in
Area of the semi-circle =
Area of the semi-circle = 91.125π in²
Area of the shaded portion = (91.125π - 120) in²
Area inside the semi-circle and outside the triangle is (91.125π - 120) in².