Quadrilaterals Solutions FAMAT State Convention 2002 1. C With the given information, 5. A AB+BN+NM+AM=36 mADE∠=50, so mBAC∠=130. CD+CN+NM+MD=36, adding m∠+∠ C m BAC +∠= m B 180 these two equations and using the the 32++∠= 70mB 180, m ∠= 78, facts AD=AM+MD,BC=BN+NC, mEDCmEBC∠+∠=+=130 78 208 AB+CD+BC+2NM+AD=72 AB+CD+BC+AD =52; 2. C Let x=length of a side of the square. subtracting these two equations gives Since triangles are similar by AA, set up 2NM=20 which makes NM=10 15 − xx the proportion = , x = 6, 6. C Let the shorter base = x. Then − xx10 drawing the altitudes from the upper area is 36. base vertices, let the base of the 8 − x 15-x rectangle =x. AE =FD= . BD=7. 2 Use a system using the Pythagorean 2 x 8 − x Theorem. 322+=h  and 2 10-x x 8 − x 2 hx22++ =7.Solving each for 3. D CD=x,AB= 46x + legs = 21x + , 2 9xx+= 8 62. = 6, bases are 6 and 30. h2 and solving the system, x = 5 . Use Pythagorean theorem to find height. B x C Using the triangle with hypotenuse 13, base of triangle is 12 so height is 5. Area equals median (18) times height(5)=90. h x A D EFx 2x+1 2x+1 7. B mB∠=∠ mD , 4 x + 15 = 6 x − 27, 4x+6 x = 21 making mB∠=99 . ∠A is the ∠ ∠= !""# supplement of B making mA 81 4. B Draw altitude from B to CD , since 7798. A Longest altitude is drawn to shortest mC∠=60, BE = 3, CE = , ED = . 222side 4 3 .That altitude is opposite a 60 Use Pythag to find BD2 = 57. Draw 1 !""# degree angle, so is 362•• . altitude from A to CD . Since altitudes 2 77 9. A Quadrilateral formed is a rectangle. are congruent, AF==3, DF .Use Using the midline formula, shorter side 22 1 1 Pythag in triangle ACF to find is •=63. Longer side is •=10 5, 2 =−= 2 2 AC 169.169 57 112 . Area =15.

Quadrilaterals Solutions FAMAT State Convention 2002 10. E 44 Triangle AXB is isosceles so 17. B Triangle EBC is right isosceles. m∠=∠= XAB m XBA 22. ∠BXC is an Since hypotenuse =5 2 , legs = 5. area exterior angle of that triangle so is the 25 sum of the two remote interior angles. of triangle CBE= , area of trapezoid 2

95 11. D Draw the square. Using one of the is . So ratio is 5:19. trapezoids formed, draw the altitude of 2 the trap from the vertex of the octagon. Let this length be x. The triangle formed 18. C Distance to midpoint of each side by drawing the altitude is isosceles right containing the vertex is 2, since there are so the other leg of the triangle is x. The two of these the distance is 4. Drawing 5 the segment from the vertex to the hypotenuse is 5, x = 2 . Side of the midpoint of the other two sides forms 2 2 congruent right triangles with legs 4 and square is 10+ 5 2 , area = 150+ 100 2 2 making the hypotenuse 2 5 ,there are

12. D If a quadrilateral is inscribed in a 2 of these so that sum is 4 5 . 4+ 45 circle, opposite angles are supplementary. m∠=100. d 2 19. D Area of square is = 72. 2 13. A Only II is true for CD|| AB . 20. A is the only true statement

14. B DEAF is a parallelogram (opp 21. B Opposite angles in an inscribed sides parallel). ∠≅∠BC by Isos quadrilateral are supplementary making Triangle Thm. ∠≅∠A DFC ≅∠ BED - mABC∠=112 . ∠EBC is the corresponding angles of parallel lines, supplement of ∠ABC . m∠= A m ∠ BED = m ∠ EFG = m ∠ DFC =x. mB∠=∠= mC y, 22. C Since the perimeter is 28 and the mEDBmDCz∠=∠=, bases are 8 and 12, the legs of the xyz++=180, 2 zx += 180, solving this trapezoid have to be 4. Drawing the altitudes to form triangles, the leg that is system, yz= , so BE=DE=AF. Since not the altitude must be 2 so the triangles AB=7, AE+ED=7 and DF+AF=7. p=14 are 30-60-90 making the altitude 23.

!""# The area is the median 10 times the 15. D Extend ED so it intersects AB . height 2 3= 20 3 . altitude of triangle AEB= 2 3 , so

EF=BC −=−23 4 23. 23. A Find the equations of the diagonals, then solve the system. !""# 16. E 10+ 10 3 Draw both diagonals Equation of AC=+52 x y = 5, equation !""# forming 30-60-90 triangles since of BD== x411 y = . Point of diagonals in a rhombus are ⊥ and bisect 21 25 4 angles. Legs of triangles formed are 5 intersection is , − . Sum is − . and 53. 11 11 11

Quadrilaterals Solutions FAMAT State Convention 2002

24. C EF=4 making FB=8. Use Pythagorean Theorem to find AF= 413

25. E Using Pythag, AC=15 making the area of ABC=60, AD=12 making the area of triangle ACE=84. Sum is 144.

26. B Diagonals are perpendicular and bisect each other forming 4 congruent triangles. Hypotenuse = 25, let = 7 so other leg is 24. Area is product of 48• 14 diagonals divided by 2. = 336 2 27. C Triangle AEF is similar to triangle CEF by AA (sides are ||). So CE:EA=BC:AF. Since DF=4x, FA=3x, BC=7x (opposite sides congruent) making the ration 7:3.

28. A Altitudes are congruent, so the ratio of the areas is the ratio of the bases 15:18=5:6.

29. B Let the longer side equal x. The shorter side is 77 − x since perimeter is 154, the sum of 2 consecutive sides is 77. Set the areas equal (77−=xx )(12) 10 . Therefore x = 42 . Since 42 is the longer side, its altitude is 10 so area is 420.

30. C Set xx+=42 − 1 and x = 5 . Making AB=AD=8 and BC=CD=9. Perimeter is 34.