( ) ( ) Geometry Team Solutions FAMAT Regional February = 5. = 24p.

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Matthew Price
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1 . A 6 6 The semi perimeter is so the perimeter is 6. The third side of the triangle is 7. Using Heron s formula to find the area ( )( )( ) 4 6 = B Draw the altitude from Q to RP. This forms a triangle with the altitude opposite the 45 degree angle. Since the hypotenuse is, the altitude would be. The area is 5 0 =.. C 5 Find the length of each side. AB = 0, BC = 0, AC = 5. Now to find the length of the altitude wanted, find the area and then set that equal to ½ altitude side. The area is so area = = 5, 5. h h=. D 48 Let the base equal. That makes the legs each equal to 6. Using one right triangle formed by drawing the altitude, use Pythagorean Theorem to find the value of. ( ) 6 = + 64, = 6, to. The area will be = 5. making the base equal ( )( 8 ) = 48.. A BC Since the triangle is isosceles, set two sides equal to each other to see which ones satisfy the requirements of the perimeter being less than 45, and that the sides of the triangle are positive. AC = AB, 8= 0, = 9. This makes the sides of the triangle 0, 0, and 6. The perimeter is 6. This fits the requirements. AC = BC, 8= + 7, = 5. which makes the sides of the triangle 0,,. This makes the perimeter 54. Does NOT meet requirements. AB = BC, + 7= 0, =. sides of the triangle are 0, 0,. NOT a triangle reject. So the base is. B There are triangles that meet the conditions. The side lengths are 4,45, 45.. C The third side must meet the conditions < <. Find the average of the integers through inclusive.. D 8. A 48 The apothem is. Draw a radius of the heagon. Now there is a triangle with the side opposite the 60 as. To find the side opposite the 0, divide by which gives 4. Double this to find the length of a side of the heagon and them multiply by 6 to get a perimeter of 48.. B 88 The side of the heagon was found in A as BC. ( ) ( ) 80 = , = To find the area s (8 ) 6 = 6 = C 9p The length of a side is the radius of ( ) the circumscribed circle. 8 p = 9p.. D 4p The apothem of the heagon is the radius of the inscribed circle so the circumference is p = 4p. 4. A 7 The angle formed by the tangents and the minor arc they intercept are supplementary so mcd! = 88, so the mcbd! = 7.

2 4. B p The 40 arc is of the circle. The 9 radius of the circle is 9. The length of the arc is 8 p = p C 0 Draw the diagram. From the center of the smaller circle, draw a segment perpendicular to the radius of the larger circle. This forms a rectangle. The longer sides are both 6. The shorter sides are both 5 so the longer radius is divided into two sections of 5 and. The right triangle formed has legs of length 6 and making the hypotenuse, which is the distance between the centers, D 44p pr = 6 pr, r = making the area 44p. 5. A 5 The sum of ÐAand ÐBis 90. Drawing the angle bisectors, it can be seen that the sum of ÐDABand ÐBDAis equal to the sum of ÐCAD and ÐCBD. This makes each sum 45. The mð ADB = = B 6 The hypotenuse is and one leg is 6 making the triangle The other leg AC has length 6. Given that AD = 4 makes CD =. Use Pythagorean Theorem to find the length of ( ) BD, + 6 = 4. BD + CD = C The third side of the triangle is 0, use Pythagorean triples. Since the length of CDis wanted, let BD =. That makes AD = 0. Use the angle bisector theorem to set up the proportion =, =. Now use Pythagorean theorem 0 to find the length of CD. æ5 ö ç = =. è ø D Draw the altitude from A to BC. It is opposite an angle of 45 so its length is 0 5. The area is = 5 5 =. 6. A 50 Diagonals are perpendicular and bisect each other. So one right triangle formed has 5 the legs with lengths and 0. Use Pythagorean Theorem to find the hypotenuse, which is a side of the rhombus = =. The 4 4 perimeter is B 7 The other angle of the rhombus would be 0. Draw the altitude and since it is opposite the angle of 0, the altitude is 6. The area would be base times height which is 6 = C Finding the length of the other 9 diagonal would be 48 = 4 d., d = Use Pythagorean Theorem to find the length of a side. æ9 ö 47 + ç =. è ø 6. D p The area of the rhombus is The radius of the circle would be = ½ of the altitude of the rhombus. The side of the rhombus would be 5, use Pythagorean triples. Using the area formula backwards, 600 = 5 hh, = 4, r=, A= 44 p. 7. A 5 The triangles are similar by AA. AE 5 =, AE = 8., BC = 0 BC = BD = 0 = 8. AE + BC + BD = 5.

3 7. B 40 The triangles are similar by AA. 6 4 AC 9, AC 9 6. AC = 6, = = = 8 4, 7. AB = 6 AB = EC + AB = C 4 The triangles are similar by AA. DE =, DE = 9., AC = 0 AC = DE + AC = D The triangles are similar by AA. 4 AD =, AD = 8, BD = 8 = DE =, DE =. BD + DE = 5 + = A 5 The formula is ½ apothem times perimeter. The perimeter is 50 which makes each side 0. Drawing the radius and the apothem gives a right triangle with leg 5 and hypotenuse Pythagorean Theorem, the apothem is is =. Using. The area 8. B 95 The interior angle is 56 making the eterior angle 4. The sum of all eterior angles is 60 so 60 divided by 4 gives 5 angles/sides. The perimeter is 5 = C 40 The radius of the circle is 0 which is also the apothem of the circumscribed heagon. Draw the radius and apothem which gives a triangle with 0 opposite the angle of 60 so the 0 side opposite the angle of 0 is. The side of the heagon would be =. The perimeter is 8. D 648 The apothem is 8. A side is 8. The area is = ( ) 6 = A 5 The altitude of the trapezoid is opposite the angle of 0 making the altitude 5. The area is 5 ( 8+ 5 ) = B 8 The other side of the rectangle is (Pythagorean triples) so the area of the rectangle is 6 = 9. To find the side of the square take the square root of 9 which simplifies to C gives area of D 5k Let the diagonals be represented by and 4. The area using ½ the product of the diagonals is 4 = 4. Set that equal to k k and solve for gives k = 4, =. Now use Pythagorean theorem to find the length of a side in

4 æ k ö æ k ö terms of k. s =. ç + ç è ø è ø 0. A 96 Draw the altitude of the trapezoid through the center of the circle. This length needs to be found in two parts. First, draw the radius from a verte of the lower base to the center of the circle. This forms a right triangle with hypotenuse 0 and leg of 8. This makes the lower part of the altitude of the trapezoid 6. Now draw the radius from a verte of the upper base to the center of the circle. This gives a right triangle with hypotenuse 0 and leg of 6. This makes the upper part of the altitude 8. The altitude is 4. To find the area 4 ( 6 ) = 0. B 6 The angle formed by the tangent and secant is ½ of the intercepted arc. The arc is /5 of the 60 degrees in a circle which would be 7 making the angle measure C Let the apothem be. Since that is the side opposite the angle of 60, to find the side opposite the 0 divide by then multiply by to find the hypotenuse which is the radius. That makes the radius k 5k 5k s = + k, s =, s=. 4 4 = =.. The ratio wanted is 0. D 4 When a quadrilateral circumscribes a circle, the sum of the opposite sides are equal. This can be proved by using the tangents from the same eternal point theorem. AB + CD = 7so AD + BC = 7making the perimeter 4.. A Radius drawn to the tangent is perpendicular. Using Pythagorean theorem, RO =4. The triangles are similar by AA so 4 50 proportional sides can be used.,. SP = 75 SP =. B 756 ad STP = SP ST. Find SPas described in.a. Again, the triangles are similar by AA so to find ST, 48 50, 7. Now the ST = 75 ST = ad STP = 7 = C 84 prspo = SR + RO + PO + SP. Using the solutions above, we can find SR = 4, making the perimeter 84.. D 40 RSPOis a trapezoid so the area is SR ( RO + SP) = 4 ( 4 + ) = 40.. A 6 =, = B 7 The triangles are similar by AA. The ratio of the sides of the triangles is. The ratio of 4 4 the area is. =, = C 50p After drawing the chords and radii, there are two right triangles. One has legs of and 4, the other had legs 0 and + 8. Using Pythagorean to find the radius, ( ) 4 + = Solving gives = 7. Using Pythag again, find the radius to be 5 which makes the circumference 50 p.. D 60p Let equal the number of revolutions. p + 0 = p, = 0. And p 0 = 60 p. ( )

5 . A 40 A is an inscribed angle so it is ½ the intercepted arc which would be 40.. B 50 An angle formed by a tangent and secant is ½ the intercepted angle so the arc must be 50.. C 8 The measure of an angle formed by two secant segments is the average of the intercepted arcs. Working backwards, 8 = (80 + ), = 8.. D 7 Let be the distance from D to the intersection of segments AB and CD. Using the theorem about two secant segments intersecting in a circle gives 6 = 8, = 9,9 + 8 = A 45 The length of BPis needed so let that be represented by and AB = 75. Set up the 75 proportion =, = B 8 Let one leg be represented by and the other would be 5. Using the Pythagorean Theorem gives + 5 =. Solving gives the values of 7 and 8. The area will be. 4. C Drawing the diagonals gives triangles. Let the side opposite the 0 be which makes the side opposite the 60 ratio simplifies to ( ) =. The 5. A Find the midpoint of BC æ6 4 46ö ç, = (5,) è ø Find the distance from this midpoint to 6, using the distance formula B Find the length of the diameter using the distance formula: 5 making the radius. The area of the circle would be ( ) ( ) ( ) =. 4 p ( ) ( ) æ 5 ö p 45 ç = p. è ø 4 5. C The midpoint wanted is æ ö ç, = (, ). To find the intercept, è ø set y equal to 0 and solve for which makes the æ7 ö intercept ç,0. The distance between these two è ø æ7 ö 8 points is ç + ( 0 ) = + 9 =. è ø 4 5. D 0p Write the equation of the circle by completing the square to find the radius y + 8y+ 6 = The right hand side of the equation gives 5 which is the radius squared making the radius 5. The circumference of the circle is 0p = 5 ( ) ( ) ( ) ( ) 4. D 64 AC = BC makes Ð A@ÐB.Solving for : 5 70= 80, = 50, mð A= 64.

Indicate whether the statement is true or false.

PRACTICE EXAM IV Sections 6.1, 6.2, 8.1 8.4 Indicate whether the statement is true or false. 1. For a circle, the constant ratio of the circumference C to length of diameter d is represented by the number.