In a rectangle, opposite sides are equal, and each angle measures 90 degrees.Triangle ABD is a 5-12-13 right triangle, so BD = 13. Furthermore,BC = 12, and CD = 5.To find the perimeter of any figure, add the lengths of the sides. In this case, 5 + 12 + 13 = 30, so the answer is choice (A).

The interior angles of a triangle add up to 180°, therefore, x = 110.

A,B, and C

To solve this question, draw and label the figure. When drawn and labeled, the figure should look like this:

Since points E and F are midpoints of the sides of a square and ∠C is a right angle, triangle ECF is a 45:45:90 triangle; thus choice (A) must be correct. Since ∠CEF = 45°, if ∠FEA also equaled 45, then ∠CEA would be a right angle and would be perpendicular to ; instead, since ∠FEA + 45 > 90, then ∠FEA > 90 and choice (B) must be correct. If ∠EFA = 90, then ∠FEA = 90, and thus ∠EAF = 0; since ∠EAF > 0, ∠EFA < 90, so choice (C) must be correct. If ∠FAD = 30, then triangle FAD would be a 30:60:90 right triangle with legs of x and and a hypotenuse of 2x; since that is not the case, ∠FAD ≠ 30; eliminate choice (D). If ∠AEB = 60, then triangle EAB would be a 30:60:90 right triangle with legs of x and and a hypotenuse of 2x; since that is not the case, ∠AEB ≠ 60; eliminate choice (E). If ∠AFD = 45, then ∠FAD = ∠EAB = 45, which would mean that ∠EAF = 0; since ∠EAF > 0, ∠AFD cannot = 45; eliminate choice (F).

To answer this question, Plug In the Answers.Because \(\frac{\overline{BC}}{\overline{CE}}=\frac{\sqrt{3}}{1}\), triangle BCE is a 30:60:90 triangle and BCD is an equilateral triangle with all angle measures equal to 60. Start with choice (C). If ∠CAD = 30, then since triangle ACD is isosceles, ∠CAB =30 as well. Since both triangles ABC and ACD share \(\overline{AC}\), and triangle ABC is isosceles, then ∠CAB and ∠CBA are both equal to 30 as well. The angles within triangle ABD should add up to 180. However, ∠CAD + ∠CAB + ∠CDA + ∠CDB + ∠CBD + ∠CBA =30 + 30 + 30 + 60 + 60 + 30 = 240, which is too big; eliminate choices (C) and (D), which is bigger. Now try choice (B). If ∠CAD = 15, then ∠CDA = 15 as well, and ∠CAB and ∠CBA are both equal to 15 as well. Verify that the angles will add up to 180 for triangle ABD: ∠CAD + ∠CAB + ∠CDA + ∠CDB + ∠CBD + ∠CBA =15 + 15 + 15 + 60 + 60 + 15 = 180.Choice (B) is the correct answer.

To solve this question, plug in some easy values for the variables. For example, if the polygon were a square, then x = 4. Since each angle would equal 90, q = 90, your target answer.Check all the answers by plugging in x = 4. Only \(\frac{180(4-2)}{4}=90\) , so choice (D) is correct.

As soon as you see variables in the answer choices, draw your figure and set up your scratch paper to Plug In.The sum of d, e, and f has to be 180, so try d = 50, e = 60, and f = 70. Similarly, the sum of a, b, and c has to be 360, so try a = 110, b = 120, and c = 130. Since the problem asked for a + b + f, your target answer is 110 + 120 + 70 = 300. Now plug your values for c, d, and e into the answer choices; only choice (D) hits your target.

For this problem it helps to draw the figure. When you do, exaggerate the angles as shown in the diagram below. From the diagram, ∠ACE and ∠BCD are both smaller than 90°.The sum of any two angles less than 90° will be less than 180°; this sum is x from the question, making choice (D) correct.You can also plug in your own numbers to the diagram to test the choices. If ∠ACD is 120°, then ∠ACD and ∠BCD are both 60° and add up to 120°, which is now the value of x.You can eliminate choices (A) and (E).To eliminate further, try a really big number such as 170° for ∠ACD Angles ∠ACE and ∠BCD are now both 10° and add up to 20°, the new value of x.You can eliminate choices (B) and (C), leaving choice (D) as the correct answer.

Fill in the diagram little by little. Angle LON must be equal to 75 degrees, because ∠LON + 115° = 180° to make a straight line. The shape is a parallelogram, so it must also be true that ∠LON = ∠LMN.There are 360 degrees in the whole parallelogram, and ∠LON + ∠LMN = 150, so there are 360 – 150 = 110 degrees remaining in the parallelogram.Thus, (x + 10) + (y + 8) = 110. You can now simplify the expression to get x + y = 92.

The question is asking for a specific amount and there are no variables in the answer choices, so PITA. Starting with choice (C), b = 60.By vertical angles, b = 2a, so a = 30. If a = 30, then ∠SWU = 90.This won’t work because all four angles of a square equal 90° and ∠SWU must be smaller than 90.Eliminate choices (C), (D), and (E).Try a smaller value, such as in choice (B). Now b = 40 which means a = 20, ∠SWU = 60 and ∠UWV is 30°. A right triangle in which the hypotenuse is twice one of the sides is a 30:60:90 triangle.That means that triangle UWV is a 30:60:90 triangle in which ∠VUW is 60° and ∠UWV is 30°. Per our calculations, that’s what ∠UWV is supposed to be, so the correct answer is choice (B).

It’s a geometry problem with variables in the answer choices, so draw the figure and set up your scratch paper to Plug In.Try x = 60 and y = 70; the missing angle in the small triangle on top is now 50°, as is the missing angle in the small triangle in the middle. Since z combines with the 2 angles you just found to form a line, 2(50) + z = 180, and z = 80.The problem asked for the sum of x and y, so plug 80 in for z to all the answers and look for your target answer of 130. Only choice (B) works.