2022 States Theta Quad Test n Sol

lOMoARcPSD|23780900 Theta Quadrilaterals Test “NOTA” is an abbreviation for “None of the above” and should be chosen when choices A, B, C and D are incorrect. Diagrams may not be drawn to scale. If no units are given, assume linear measurements are cm and area is sq. cm. R R 5 4. RSTU is a kite. U 5 5 S Its diagonals X intersect at X. RX=UX=XS=5. ST=UT=13. Find 13 the area of the kite. 13 S U A. 85 B. 90 C. 92 D. 102 E. NOTA T 1. In parallelogram RSTU, m∠RSU = (5 x + 5)o , m∠UST = (3 x + 9)o and m∠RUT = (2 x + 56)o . Find the degree measure of ∠URS . A. 105 B. 110 C. 112 D. 120 E. NOTA R 2. 2022 FAMAT State Convention 12 R 5. 3 U P S P 4 S 12 U T T Convex quadrilateral RSTU above has RS=ST=12. m∠S = 90o . Give the least possible integer value of the perimeter of RSTU. A. 40 B. 41 C. 46 D. 48 E. NOTA 3. A rhombus has diagonal lengths 2 cm and 4 cm. What is the perimeter of the rhombus in cm? A. 20 B. 12 C. 4 5 D. 4 3 E. NOTA T Quadrilateral RSTU is inscribed in a circle with center P and radius 5. UR=RS and UT=TS. If chords UR and UT are a distance of 3 and 4 from P, respectively, then give the perimeter of RSTU. A. 20 B. 21 C. 28 D. 30 E. NOTA R 6. Rhombus RSUT and T U rectangle TUCK share side TU as shown. TU=8 and K C TK=5. If the two quadrilaterals have the same area, 2 find ( RU ) . A. 128 − 16 39 C. 133 − 20 39 E. NOTA B. 89 D. 8 S lOMoARcPSD|23780900 Theta Quadrilaterals Test R 7. 90o U 65o S T 2022 FAMAT State Convention 10. Given linear equations L1 : y = x + 20 , L2 : y = 30 − x , L3 : y = 12 and L4 : y = 0 , find the area of the quadrilateral bounded by the four graphs. (Note: increments shown on the grid below are 5.) Quadrilateral RSTU is inscribed in a circle above, which has radius 18 feet. m∠R = 90o and m∠U = 65o . Find the length of y 45 40 35 30 major arc TUR in feet. A. 11.5π B. 12π C. 13.5π D. 23π E. NOTA 25 20 15 10 5 x −25 8. −20 −15 −10 −5 5 10 15 20 25 −5 R A. 464 B. 456 C. 396 D. 336 E. NOTA U S T ∆RST is an equilateral triangle. SU=UT. m∠SUT = 140 . In quadrilateral RSUT, give m∠RSU . A. 20o B. 30o C. 40o D. 45o E. NOTA S T 9. RSTU is a X rectangle with diagonals intersecting at X. XS= 2n , R U RX= 4m − 4 and TX= 3m − 1 . Find the value of n . A. 2 B. 3 C. 4 D. 5 E. NOTA S 11. 12 T 10 60o R U Trapezoid RSTU has a right angle at T, and m∠R = 60o . RSTU is the floor diagram (aerial view) of a barn. Gary the goat is secured to point S by a 12foot leash outside of the barn. Ignoring the dimensions of Gary, find the area outside of the barn that Gary can roam, in square feet. 292 π 3 149 C. π 3 E. NOTA A. 292 π 5 148 D. π 3 B. lOMoARcPSD|23780900 Theta Quadrilaterals Test 12. RSTU and MNPQ are rectangles with correspondence RSTU~MNPQ. Each side of MNPQ is one-fourth the length of the corresponding side of RSTU. If UT = 3(UR ) then find sin ( ∠PMQ ) . A. 1 2 3 2 E. NOTA C. 4 5 3 10 D. 10 B. 2022 FAMAT State Convention 15. The bases of a trapezoid are each increased by 10%. The area of the trapezoid is to remain the same. By what percent will be altitude be decreased? 3 B. 9 % A. 10% 4 1 C. 9 % D. 9% 11 E. NOTA R 13. R U S 6 16. 2 P U M T RSTU is a square with M the midpoint of ST . RU=2. Find the area of quadrilateral RSMP. 5 4 B. A. 3 3 6 5 C. D. 5 6 E. NOTA P 5 T 5 S ∆PRS has a right angle at R. RS=6 and PS=10. T is the midpoint of PS , and UT ⊥ PS . Give the area of quadrilateral RSTU. 33 A. 17 B. 2 31 117 C. D. 2 8 E. NOTA R 5 12 14. A building has 20 all rectangular walls, floor and ceiling. The floor has dimensions 20 yards by 12 yards. The height of the building is 5 yards. All of the outside and inside walls and inside ceiling will be painted. The floor and the (outside) roof will not be painted. The underside of the building (below the floor) is inaccessible. How many square yards will be painted? A.640 B. 680 C.800 D. 880 E. NOTA 17. Square USQP is drawn with rhombus U RSTU, so that US is a diagonal of the rhombus and T lies on PQ . UP=6 then find the P T perimeter of RSTU. (Diagram is not to scale.) A. 54 B. 24 3 C. 12 5 D. 24 E. NOTA S Q lOMoARcPSD|23780900 Theta Quadrilaterals Test 18. A square RSTU is graphed on the xy-coordinate plane and R lies on (10, 0 ) . The area of RSTU is 100. Which of the following could NOT be coordinates of a vertex of the square? B. ( 20,10 ) A. 9, 199 ( C. ( 4,8 ) E. NOTA ) D. (5,10) P U 19. In ∆PQR , T three S V lines parallel Q R to QR intersect the other two sides of the triangle as shown at S, T and U, so that QS=ST=TU=UP. If the area of the trapezoid SVRQ (shaded) is 35, then tell the area of the original triangle, ∆PQR . A. 26.25 B. 64 C. 70 D. 80 E. NOTA 20. A rectangular prism aquarium has dimensions 3 feet by 4 feet by a height of 2 feet. It is 90% full (by volume) of water. No fish! Roland’s baby brother drops sugar cubes into the tank, and because the water is too cold, the sugar does not dissolve but sinks to the bottom of the tank. If each cube is 1 inch on each edge, then how many cubes will it take before the tank first overflows? That is, if n cubes is insufficient and (n+1) is too much, then the answer is (n+1). A. 4001 C. 4140 E. NOTA B. 4032 D. 4148 2022 FAMAT State Convention 21. R S P 4 U not drawn to scale T Square RSTU is drawn with each side of length 4, and two quartercircles are drawn as shown with centers U and T, each with radius 4. The arcs meet at point P. Give the distance from P to RS . A. 2 3 − 3 B. 4 − 2 3 C. 4 − 2 2 D. 4 3 3 E. NOTA 22. A rectangle has one side of length 12 and one diagonal of length 13. What is the perimeter of the rectangle? A. 34 B. 36 C. 50 D. 54 E. NOTA lOMoARcPSD|23780900 Theta Quadrilaterals Test R 23. Four squares share a 16 P Q common vertex, the 25 intersection 100 of chords RS and PQ S of the circle. The intersection of the chords is not the center of the circle. The square with vertex R shown has area 16. The square with vertex Q has area 25, and the square with vertex S has area 100. Each square has at least part of a side on the two chords. The a area of the circle is π for relatively b prime positive integers. Give the value of a + b . A. 196 B. 205 C. 209 D. 211 E. NOTA 24. 2022 FAMAT State Convention Q R S U 100 T 24 D P A R, S, T and P are on the sides of square QUAD as shown. The quadrilateral RSTP is a square and has area 100. The area of ∆RDP is 24. Find the length of one side of QUAD. A. 14 B. 3 31 C. 18 D. 19 E. NOTA R 25. 4 U ? P 2 1.5 S T A quadrilateral RSTU has diagonals which intersect at point P. Triangles RUP, TUP and STP have areas 4, 2 and 1.5 respectively. Give the area of triangle RPS. A. 4 B. 3.8 C. 3.5 D. 3 E. NOTA 26. Square RSTU R S has side RU 2 tangent to a circle. M N S and T are on the circle. The U T midpoint of ST is 2 units from the midpoint of arc ST , N. Find the area of RSTU. A. 96 + 4 5 B. 48 5 C. 32 5 D. 64 E. NOTA lOMoARcPSD|23780900 Theta Quadrilaterals Test R 27. 2022 FAMAT State Convention S S T P Q Two externally tangent circles with centers P and Q have radii 4 cm and 3 cm respectively. R is on circle P and S is on circle Q, as shown. RS is an external common tangent line to the circles. Find the perimeter of quadrilateral PRSQ in cm. R 30 U 29. The (interior) angles of a quadrilateral measure o m∠R = (10n + 10 ) , m∠T = (15n)o , o m∠S = (12n − 6 ) and o m∠U = ( 8n − 4 ) . If RU = 30 cm A. 14 + 5 2 C. 7 + 6 5 E. NOTA B. 14 + 4 3 D. 14 + 3 2 28. A quadrilateral has vertices at coordinates P ( −1,5) , Q (9, −3) , S (−5, 0) and R (15, −16) . Which describes convex quadrilateral PQRS (note that vertices are named clockwise from P around the quadrilateral)? A. rectangle, but not a square B. rhombus, but not a square C. trapezoid D. not a quadrilateral, since two vertices are collinear E. NOTA and ST = TU then give the area of RSTU is square cm. A. 25 3 + 175 B. 250 3 C. 75 3 + 425 D. 400 3 E. NOTA 30. A quadrilateral RSTU is similar to quadrilateral ABCD so that the correspondence is RSTU~ABCD. If RS=12 cm and AB=3 cm, and the area of RSTU is 36 sq. cm, then find the area of ABCD in sq. cm. A. 2.25 B. 4 C. 8 D. 64 E. NOTA lOMoARcPSD|23780900 Theta Quadrilaterals Test 1. B 2. B 3. C 4. A 5. C 6. A 7. D 8. C 9. C 10. B 11. A 12. D 13. A 14. D 15. C 16. D 17. C 18. D 19. D 20. D 21. B 22. A 23. C 24. A 2022 FAMAT State Convention 25. D 26. D 27. B 28. C 29. B 30. A Solutions: R 1. B. 5x+5 3x+9 S 2x+56 U T 5 x + 5 + 3 x + 9 = 2 x + 56 . 6 x = 42 . x=7. So m∠RUT = (2(7) + 56)o =70 deg. And ∠URS is the supplement of this, and has 110 degrees. 2. B. Connect R and T to make a right triangle RST with hypotenuse length 12 2 . By the Triangle Inequality Theorem, we know that RU + UT > 12 2 . The perimeter of RSTU is 12+12+RU+UT> 24+12 2 . 12 2 = 288 and the integer value greater than this is 289 = 17 . So the least integer value of the perimeter is 24+17= 41. 3. C. 1 2 Diagonals bisect each 2 1 other and are perpendicular so we look at four right triangles as shown, each with hypotenuse 5 . That makes the rhombus perimeter four times this. 4. A. Use the Pythagorean Theorem to get XT= 12. The top triangle RUS 1 has area (10)5 = 25 . Lower ∆ TUS 2 1 has area (12)10 = 60 .The kite area is 2 25+60=85. (note: diagonals of a kite are perpendicular) R 5. C. M Draw radii 5 PR and 3 PT. U S P That 4 gives 5 N ∆MRP and ∆PNT are 3-4-5 T triangles. The chord UR has length 8. UT has length 6. So the perimeter of quadrilateral RSTU is 8+6+8+6=28. 6. A. Area of TUCK R 8 S is 8(5)=40. Area of 8 5 RSUT = TU(h)=40. Since TU=8 T Y U because a 5 rhombus has congruent sides, K C h=5. So the triangle formed by the height and RT shown (RYT) above has leg TY of length 64 − 25 = 39 . YU= 8 − 39 . Now use the Pythagorean Theorem with ∆RYU : 52 + (8 − 39)2 = ( RU ) 2 . 25 + (64 − 16 39 + 39) = 128 − 16 39 . 7. D. Opposite angles R of a quadrilateral 90o inscribed in a circle are U 65o supplementary. S So m∠S = 115o . The major arc T TUR has degree measure 230 degrees. The length is a proportion of the circumference: 230 R ( 36π ) = 23π 360 60 8. C. Equilateral triangles have all angles of 60 degrees. U Isosceles 140 ∆SUT has 20 20 base angles S T at S and T of 20 degrees. So m∠RSU = 60-20=40 degrees. lOMoARcPSD|23780900 Theta Quadrilaterals Test 2022 FAMAT State Convention 9. C. In a rectangle, diagonals bisect each other and are also congruent so XS=XR=XT=XU. 4m − 4 = 3m − 1 so m=3.Each “half-diagonal” is then 8. XS= 2n = 8. n=4. 10. B. Since L3 : y = 12 and L4 : y = 0 are parallel, we have a trapezoid. Find the intersection of y=12 and the other lines, shown below. The distance between these points is 26, which is the length of the smaller base. The x-intercepts of L1 : y = x + 20 , L2 : y = 30 − x are -20 and 30 so the larger base is 50. The height of the trapezoid is 1 12. A = (12)(26 + 50) = 6(76) = 456 2 y 12. D. The sides can be drawn in a ratio as follows: S T x 4x N U M R 4x/3 Px Q x/3 If the smaller rectangle is ¼ of the larger, then let PQ=x and TU=4x. If UT = 3(UR ) then let RU= 4x/3 and MQ= x/3. ( x) 2 + x 3 2 = ( PM ) 2 x 10 . sin ( ∠PMQ ) = 3 x 3 3 10 = = 10 10 x 10 3 PM= 45 40 35 30 25 20 15 10 5 x −25 −20 −15 −10 −5 5 10 15 20 25 −5 11. A 10 120 120 12 60o The goat will be able to roam a 240 degree sector of radius 12, and then a 120 degree sector of radius 2. 240 120 (144π ) + (4π ) = 360 360 4 292 96π + π = π 3 3 13. A. UM = 5 using ∆MUT and the Pythagorean Theorem. The area of RSMP is the area of RSTU minus ∆MUT , minus ∆PUR . Let the height of ∆PUR be h. 1 2−h ∆PUR ~ ∆PMT . = . 2 h 4 4 − 2h = h . h = . Area of PUR is 3 1 4 4 R S (2) = . 2 3 3 2 1 Area of RSMP h P M is 1 1 4 5 4 − (2)(1) − = U 2 T 2 3 3 14. D. The walls inside and outside will be 2Ph = 2(64)(5)=640. The inside ceiling will be 20(12)=240. Total is 880. 1 15. C. The original area is (a + b)h . 2 1 The new area is (1.1a + 1.1b) H . 2 For area to remain the same, set them equal 1 1 (a + b)h = (1.1a + 1.1b) H . Divide by 2 2 lOMoARcPSD|23780900 Theta Quadrilaterals Test ½ and (a+b) to get h = 1.1H . That is 10 x h = H which is h − (h) = H as 11 100 10 x “percent decrease”. = 1− 11 100 x 1 1 = . 100/11 = 9 % 100 11 11 16. D. By AA~ Theorem, ∆PUT ~ ∆PSR . ∆PSR is a 6-8-10 Pythagorean Triple. Area of RSTU is area of PSR minus area of PUT. UT 5 15 = so UT= . Area of RSTU 6 8 4 1 1 15 is then (6)(8) − (5) = 2 2 4 75 117 24 − = . 8 8 17. C. Diagonal US will divide rhombus RSTU into congruent triangles RUS and TSU by SSS Postulate. By HL Theorem, ∆PUT ≅ ∆QST . So T is the midpoint of PQ . UT= 3 5 by the Pythagorean Th. in ∆PUT . Perimeter of RSTU is 12 5 . 18. D. If the area of the square is 100 then the side is 10. If we know one vertex location then the other vertices can be either 10 units away or 10 2 away if they are on the diagonal. So from (10,0) we get the distance to each choice given. Choice A gives distance 9, 199 to ( ) (10,0) to be 200 = 10 2 . Choice B is (20, 10) and distance from (10,0) is 10 2 . Choice C is (4, 8) and distance is 10. Choice D is (5,10) and distance to (10, 0) is 125 so this is not the location of a vertex. 19. D. The 4 triangles with vertex P are all similar by AA~ Th, due to the parallel lines. Consider ∆PSV and ∆PQR . The ratio of the sides PS to PQ is 3:4. The ratio of the areas is 9:16. The area of the trapezoid is area of ∆PQR less area of ∆PSV . 2022 FAMAT State Convention Let PQR have area 16k and PSV have area 9k. Then trapezoid SVRQ has area 7k. 7k = 35 gives k=5. So the area of ∆PQR is 16(5) = 80. 20. D . We need 10% of the tank to 3 1 x equal sugar. 0.10(24) = 12 12 x 12 4 = . x= . Since this is 5 12(12)(12) 5 a decimal, the answer is one more than the integer part, which gives 4148. 21. B. Let U be at the origin. The R S equation of the circle P 4 with center U shown is x 2 + y 2 = 16 . The T equation of U the circle with center T is ( x − 4) 2 + y 2 = 16 . Subtract. 2 2 x2 − ( x − 4) = 0 . x2 = ( x − 4) . ± x = x − 4 . 2 x = 4 , x=2. y = 12 = 2 3 . Distance is 4−2 3 . 22. A. Use the Pythagorean Theorem to get missing side length 5. Perimeter is 2(5+12)=34. 23. C. Let the vertex S R be at the origin. 16 That means P Q X R is at (0, 14). 25 Q is at (5, 10). 100 Using the chord intersection theorem, PX(XQ)= S RX(XS) . PX(5)=4(10) so PX=8. Now we need to find the circle equation through those coordinates. Consider x 2 + y 2 + Ax + By + C = 0 . Use S to get C=0. Use R(0, 14) to get 0 + 196 + 0 A + 14 B = 0 . This gives B= −14 . Use Q(5, 10) to get 25 + 100 + 5 A + 10 B = 0 which lOMoARcPSD|23780900 Theta Quadrilaterals Test 2022 FAMAT State Convention gives 125+5A-140=0. A=3. So the equation is x 2 + y 2 + 3 x − 14 y = 0 . Complete the square to get 205 205 ( x + 1.5)2 + ( y − 7) 2 = . Since 4 4 2 is radius , the area of the circle is 205 π . This gives a=205 and b=4 for 4 a final answer of 209. 24. A. Each triangle is congruent, so the area of the large square is 100+4(24)=196. Then one side is 14. (Let one acute angle of the triangle be x, and the other angle is 90-x deg. Then let the three angles at vertex P which form180 degrees be x+90+y=180. That third angle is also 90-x. So we have similar right triangles with an equal hypotenuse. Thus congruent triangles.) 25. D. R 26. D. S(a.a) Let U be at the origin and T at 1 a + 2, a 2 (a, 0) for a the length of the side of the U(0,0) T(a,0) square. That puts the point at mid-arc of minor arc ST to be (a+2, 0.5a). So the radius of a+2 the circle is and the equation of 2 the circle is 2 2 ( a + 2) . a+2 1 + ( y − a)2 = 2 2 4 Since (a,a) is on the circle, x− a+2 a− 2 a 4 h U P ? 2 H 1.5 T Drop the height from R to US . Call it h. The area of ∆RUP is 1 8 UP (h) = 4 . So h = . Similarly 2 UP 2(?) h= . Drop the height H from T PS 4 to US . We get H = and UP 3 4 3 H= . Substitute: = PS UP PS 8 2(?) and = . From these we UP PS UP 4 8 2(?) get = and so = . ? =3. PS 3 4 3 S 2 2 ( a + 2) 1 + (a − a) 2 = 2 4 2 2 2 a + 2) ( 1 2 ( a + 2) − a (a + 2) + + a = 4 4 4 1 a 2 − a 2 − 2a + a 2 = 0 . Since a is not 4 0, divide by a and solve to a=8 and square area is 64. 27. B. R S 3 1 Q 7 P Draw the radii to the points of tangency. There will be right angles at R and S. Draw from Q perpendicular to radius RP . PQ=7 as the sum of the radii. Use the Pythagorean Th. on the right triangle shown to get leg 7 2 − 12 = 4 3 and RS is the same. Perimeter of PRSQ = 4+ 4 3 +3+7 = 14 + 4 3 . lOMoARcPSD|23780900 Theta Quadrilaterals Test 2022 FAMAT State Convention 4 28. C. PQ has slope − . QR has 5 13 4 slope − . RS has slope − . 6 5 5 SP has slope . We have two 4 opposite sides parallel. Of the other two sides, one is perpendicular to the two parallel sides and one is not. So this is a trapezoid. 29. B. S 30 T 60 120 90 R 30 30 30 U (10n + 10 ) + (15n) + (12n − 6 ) + ( 8n − 4 ) =360 So 45n=360. n=8. Angles have measures 90, 120, 90 and 60 deg. for R, T, S and U respectively. Draw SU and we have isosceles triangle STU with base angles 30 degrees each. Use the angle measures found to get the diagram above RU=30 so we have a 30-60-90 triangle with long leg 30. RS=10 3 and SU= 20 3 . Drop an altitude form T to SU and we have two more 30-60-90 triangles and can fill in ST=TU=20. Area of RSTU is 1 1 h(b1 + b2 ) = (10 3)(20 + 30) = 2 2 250 3 RS 12 30. A. = so the ratio of the sides AB 3 is 4:1. The ratio of the areas is 16:1. 16 36 = where area = area of 1 area ABCD. This gives 9/4 = 2.25.