Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Euler angles wikipedia , lookup
Rational trigonometry wikipedia , lookup
History of trigonometry wikipedia , lookup
Pythagorean theorem wikipedia , lookup
Integer triangle wikipedia , lookup
Perceived visual angle wikipedia , lookup
Area of a circle wikipedia , lookup
Category 2 Geometry Meet #4 February 2007 – Practice #3 D C 1) In the circle to the right angle ADC is 66 degrees and angle BAD is 110 degrees. How many degrees are in the measure of angle CBA? A B F 2) Circle O at the left has diameter EG. The measure of angle FEG is 17 degrees and the measure of angle HGE is 56 degrees. What is the sum of the measures of the angles FGE and GEH? O G E M H 3) In circle B at the right triangle AMC is an isosceles triangle with AM = MC. How many degrees are in the minor Arc AM? A C B 4) The two concentric circles to the left are designed such that the area of the shaded region is equal to half the area of the unshaded region. If radius AB is 4 cm, what is the length of the radius(AC) of the larger circle? Give your answer as a decimal to the nearest thousandth. Category 2 - Geometry - Meet #4 February 2007 – Practice #3 D 1) * Since AD = AC as they are both radii of the same circle making ADC an isosceles triangle. If ADC = 66 then ACD = 66 degrees as well. Since the angles in triangle ADC must add up to 180 degrees and two of the angles already add up to 66 + 66 = 132, the third angle DAC must equal 180 – 132 = 48 degrees. If angle DAC = 48 degrees then minor arc DC also equals 48 degrees since DAC is a central angle. Angle CBD is called an inscribed angle and is therefore equal to half the measure of the arc it intersects. Since minor arc DC is 48 degrees, inscribed angle CBD = half of 48 degrees or 24 degrees. F 2) O G E H A B *Any triangle inscribed in a circle that uses the circles diameter as one of its sides is a right triangle. In this case both triangles would be right triangles with angles EFG and angle EHG both right angles. Since the sum of the angle is every triangle must be 180 degrees, mFEG mEFG mFGE 180 and mGEH mHGE mGHE 180 . Since we know four of the angles already we can substitute what we know. 17 90 mFGE 180 and mHEG 56 90 180 . So M mFGE 73 and mGEH 34 . 3) * Since AC is the diameter of the circle and one side of triangle AMC, Angle AMC must be a right angle. Since triangle AMC is isosceles, angles MAC and MCA must be equal and both equal to 45 degrees. A Since angle MAC is an inscribed angle it is equal to half of the arc in intersects, in this case minor arc AM. So minor arc AM must be double angle MCA or 2(45) = 90 degrees 4) C C B *Since AB = 4cm, the inner circle has an area of 16 cm2. That makes the shaded region equal to half that or 8 cm2. The area of the larger circle then is 16 + 8 = 24 . If the area is 24 which is equal to r2, so r2 = 24 and r = 24 4.8989 which rounds to 4.899