![\"gdgq_imgq1\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq1.png\")
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1) In quadrilateral ABCD, is angle D \u2264 100 degrees?<\/p>\n
Statement #1<\/u>: AB = BC<\/p>\n
Statement #2<\/u>: angle A = angle B = angle C<\/p>\n
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![\"gdgq_imgq2png\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq2png.png\")
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2) Point P is a point inside triangle ABC.\u00a0 Is triangle ABC equilateral?<\/p>\n
Statement #1: Point P is equidistant from the three vertices A, B, and C.<\/p>\n
Statement #2<\/u>: Triangle ABC has two different lines of symmetry that pass through point P.<\/p>\n
<\/p>\n
3) ABC is an equilateral triangle, and point D is the midpoint of side BC.\u00a0 A is also a point on circle with radius r = 3.\u00a0 What is the area of the triangle?<\/p>\n
Statement #1<\/u>: The line that passes through A and D also passes through the center of the circle.<\/p>\n
Statement #2<\/u>: Including point A, the triangle intersects the circle at exactly four points.<\/p>\n
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![\"gdgq_imgq3\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq3.png\")
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4) ABCD is a square, and EFGH is a square, each vertex of which is on a side of ABCD.\u00a0 What is the ratio of the area of square EFGH to the area of square ABCD?<\/p>\n
Statement #1<\/u>: AE:AB = 4:7<\/p>\n
Statement #2<\/u>: The ratio of the area of triangle AHE to the area of square EFGH is 0.24<\/p>\n
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![\"gdgq_imgq4\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq4.png\")
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5) In the diagram above, the four triangles ABE, CBE, ADE, and CDE are all equal, and CD = 5.\u00a0 What is the area between the two circles?<\/p>\n
Statement #1<\/u>: AE = 3<\/p>\n
Statement #2<\/u>: angle BEC = 90 degrees<\/p>\n
<\/p>\n
![\"gdgq_imgq5\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq5.png\")
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6) In trapezoid JKLM, KL\/\/JM, and JK = LM = 5.\u00a0 What is the area of this trapezoid?<\/p>\n
Statement #1<\/u>: KL = 10 and JM = 15<\/p>\n
Statement #2<\/u>: angle J = 60 degrees<\/p>\n
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![\"gdgq_imgq6\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq6.png\")
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7) In the diagram above, ADF is a right triangle.\u00a0 BCED is a square with an area of 12.\u00a0 What is the area of triangle ADF?<\/p>\n
Statement #1<\/u>: angle DCF = 75 degrees<\/p>\n
Statement #2<\/u>: AB:EF = 3<\/p>\n
<\/p>\n
8) FGHJ is a rectangle, such that FJ = 40 and FJ > FG.\u00a0 Point M is the midpoint of FJ, and a Circle C is constructed such that M is the center and FJ is the diameter.\u00a0\u00a0 Circle C intersects the top side of the rectangle, GH, at two separate points.\u00a0 Point P is located on side GH.\u00a0 What is the area of triangle FJP?<\/p>\n
Statement #1<\/u>: One of the two intersections of Circle C with side GH is point P, one vertex of the triangle FJP.<\/p>\n
Statement #2<\/u>: One of the two intersections of Circle C with side GH is point R, such that RH = 7<\/p>\n
<\/p>\n
9) Points A, B, and C are points on a circle with a radius of 6.\u00a0 Point D is the midpoint of side AC.\u00a0 What is the area of triangle ABC?<\/p>\n
Statement #1<\/u>: Segment BD passes through the center of the circle.<\/p>\n
Statement #2<\/u>: Arc AB has a length of 4(pi)<\/p>\n
<\/p>\n
![\"gdgq_imgq7\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq7.png\")
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10) In the figure, ABCD is a trapezoid with BC\/\/AD, AB = CD, BE\/\/CG, and angle AEB = 90\u00b0.\u00a0 Point M is the midpoint of side BC.\u00a0 Point F, not shown, is a vertex on triangle EFG such that EF = FG.\u00a0 Is point F inside the trapezoid?<\/p>\n
Statement #1<\/u>: BE = EG<\/p>\n
Statement #2<\/u>: FG\/\/CD<\/p>\n
Full solutions will come at the end of this article.<\/p>\n
<\/p>\n
Here are two previous blogs on GMAT DS questions about Geometry.<\/p>\n
GMAT Data Sufficiency: Congruence Rules<\/a><\/p>\n GMAT Geometry: Is It a Square? <\/a><\/p>\n One big difference between Geometry on the PS questions and Geometry on the DS questions is that for all the PS questions, unless otherwise noted, you know that all diagrams are drawn as accurately as possible<\/a>.\u00a0\u00a0 That is the written guarantee of the test writers.\u00a0 By contrast, no guarantee at all accompanies the diagrams on the DS questions.\u00a0\u00a0 Consider the following diagram.<\/p>\n This triangle appears equilateral.\u00a0 There is no guarantee that it is exactly equilateral, with three exactly equal sides and angles exactly equal to 60 degrees.\u00a0 If this were diagram given on a PS questions, we would know that the triangle is at least close<\/em> to equilateral: all the side lengths are close to one another, and the angles are close to 60 degrees.\u00a0 We would know that much on a PS question.\u00a0 If this diagram were given on a DS question, then triangle ABC could be absolutely any triangle on the face of the Earth<\/em>.\u00a0\u00a0 It could be a right triangle, or a triangle with a big obtuse angle, or a tall & thin triangle, or a short & wide triangle, or etc.\u00a0 It could be any triangle at all.\u00a0 Aside from the bare fact that ABC is some kind of triangle, we can deduce nothing from the diagram on a DS question.\u00a0 Other than the bare facts of what’s connected to what, you can deduce nothing about lengths, angles, and shapes of figures given on DS questions.\u00a0 They may be 100% accurate or they may look nothing like the the shape described by the two statements.<\/p>\n Because of this, some DS questions are a real test of your capacity for spatial reasoning and geometric imagination.\u00a0 Many DS Geometry questions, including ones here, test your capacity to imagine how different the spatial scenario might be.<\/p>\n If this is not a natural gift for you, I strong recommend drawing out shapes on paper.\u00a0 Even get a ruler, compass, and protractor, and practice constructing specific shapes.\u00a0 Use straws or some other straight items to construct triangles in which you can adjust the sides and the angles.\u00a0 Strive to visualize and picture physically every rule of geometry you learn.\u00a0 By working with shapes you can see, and working with your hands, you will be engaging multiple parts of your brain that will give you a much deeper understanding of geometry.<\/p>\n <\/p>\n If the above discussion gave you some insights, you may want to look back at those practice problems before jumping into the explanations below.\u00a0 If you don’t understand something said in an explanation here, draw it yourself, and explore the different possibilities within the constraints.\u00a0\u00a0 The point of geometry is to see.<\/p>\n <\/p>\n 1) The figure is drawn as a square, but on GMAT DS, there’s no reason to assume the figure is drawn anywhere to scale.<\/p>\n If both statements are true, then the figure could be a square, in which the answer to the prompt question would be “yes,” or it could be this figure:<\/p>\n For this figure, all the conditions are met, and angle D is considerably larger than 100\u00b0; thus, the answer to the prompt question is “no.”<\/p>\n We could get either a “yes” or a “no” to the prompt consistent with these conditions, even with both statements put together.<\/p>\n Answer = (E)<\/strong><\/p>\n <\/p>\n 2) Statement #1<\/u>: As it turns out, for any triangle of any shape, there is some<\/em> point that is equidistant from all three vertices: this is center of the circle that passes through all three vertices.<\/p>\n If all three angles of the triangle are acute, then the point is inside the triangle.\u00a0 If the triangle is a right triangle, then this center is always the midpoint of the hypotenuse.\u00a0 If the triangle has an obtuse angle, then the center is outside the triangle. \u00a0All Statement #1 tells us is that triangle ABD has three acute angles.\u00a0 Beyond that, we know nothing.\u00a0 Statement #1, alone and by itself, is not sufficient<\/strong>.<\/p>\n Statement #2<\/u>: A triangle that has a line of symmetry is isosceles.\u00a0 Let’s say that one line of symmetry goes through vertex A and point P.\u00a0 This would mean that AB = AC and that angle B = angle C.\u00a0 Now, let’s say that another line of symmetry goes through vertex B and point B.\u00a0 This would mean that AB = BC and angle A = angle C.\u00a0 Putting those together, we get three equal angles and three equal sides: an equilateral triangle.\u00a0 If a triangle has two separate lines of symmetry, it must be an equilateral triangle.\u00a0 We can give a definitive “yes” to the prompt question on the basis of this statement.\u00a0\u00a0 Statement #2, alone and by itself, is sufficient<\/strong>.<\/p>\n Answer = (B)<\/strong><\/p>\n <\/p>\n 3)\u00a0 Statement #1 tells us that the center of the circle is on the line of symmetry of the triangle through point A, but the triangle could be any size.\u00a0 In the diagram below, this line of symmetry is blue, and triangles of four different sizes are shown.<\/p>\n There actually would be an infinite number of possible triangle sizes on the basis of this statement alone.\u00a0 This statement is wildly insufficient<\/strong>.<\/p>\n Forget about Statement #1.\u00a0 With Statement #2 alone, a variety of off-center triangles with four intersection points are possible:<\/p>\n Notice AB is a chord of the circle as well as a side of the triangle.\u00a0 This chord could be a medium length chord or anything up to the full diameter, and different sides of the triangle would result in different areas.\u00a0 We still cannot give a definitive answer to the prompt question.\u00a0 This statement, alone and by itself, is insufficient<\/strong>.<\/p>\n Combined statements.\u00a0 If the center of the circle is on the line of symmetry of the triangle, then this places significant constraints on the number of intersections.\u00a0 For tiny triangles, they would simply intersect at point A and not reach the circle on the other side: one point of intersection, so this doesn’t work.<\/p>\n Larger triangles would touch the circle in three places, at the three vertices: this also doesn\u2019t work.<\/p>\n Slightly larger, and those two vertices at B and C would “poke out” of the triangle, producing five points of intersection: Point A plus four other points.<\/p>\n The only way we will get exactly four points is when the sides get long enough and the side BC drops low enough that it is tangent to the circle.<\/p>\n The altitude of this triangle, AD is exactly equal to the diameter.\u00a0 We could use the ratios of the 30-60-90 triangle<\/a> to figure out the sides, and thus figure out the area.\u00a0 If the sides get any longer, then side BC would break contact with the circle, and there would be only three points of intersection.\u00a0 This triangle, with the point of tangency at D, is the only triangle on this line of symmetry that has exactly four intersection points, and we can compute its area.<\/p>\n The combined statements allow us to give a numerical answer to the prompt question, so together, the statements are sufficient.<\/p>\n Answer = (C)<\/strong><\/p>\n <\/p>\n 4) Statement #1: since we care only about ratios, we can set any lengths that are convenient.\u00a0 Let AE = 4 and AB = 7: then BE = 3.\u00a0 The figure is symmetrical on all four sides, so, for example, AH = 3.\u00a0 This means AEH is a right triangle with legs of 3 and 4\u2014that is, a 3-4-5 triangle!\u00a0 The hypotenuse HE = 5.\u00a0 That’s the side of the smaller square, and 7 is the side of the larger square.\u00a0 The ratio of areas is 25\/49.\u00a0\u00a0 This statement leads directly to a numerical answer to the prompt question.\u00a0\u00a0 This statement, alone and by itself, is sufficient<\/strong>.<\/p>\n Now, forget all about statement #1.<\/p>\n Statement #2: triangle to small square = 0.24 = 24\/100 = 6\/25.\u00a0 Let’s say that the central square has an area of 25 and one triangle has an area of 6.\u00a0 This means that four triangles together would have an area of 24.\u00a0 The big square equals the central square plus four triangles: 24 + 25 = 49.\u00a0 The ratio of the two squares = 25\/49.\u00a0 This statement also leads directly to a numerical answer to the prompt question.\u00a0\u00a0 This statement, alone and by itself, is sufficient<\/strong>.<\/p>\n Each statement sufficient on its own.\u00a0 Answer = (D)<\/strong><\/p>\n <\/p>\n 5) Statement #1<\/u>: If AE = 3, then it must be true that EC = 3, because the triangles are all equal.\u00a0 Also, AB = BC = CD = AD = 5.\u00a0 Because the four angles meeting at point E are all equal, it must be true that each one equals 90 degrees.\u00a0 Thus, we have four right triangles, and each one has a leg of 3 and an hypotenuse of 5.\u00a0 Thus, each must be a 3-4-5 triangle.\u00a0 This allows us to see that the radius of the smaller circle is EC = 3 and the radius of the larger circle is BE = 4.\u00a0 From these, we could figure out the areas and then subtract these areas to find the area between them.\u00a0\u00a0 This statement allows us to arrive at a numerical answer to the prompt question.\u00a0 Statement #1, alone and by itself, is sufficient<\/strong>.<\/p>\n Statement #2<\/u>: This statement tells us something we already could figure out from the prompt information.\u00a0 Technically, this statement is tautological.\u00a0 A tautological statement is one that contains no new information, nothing new that we couldn’t figure out on our own; examples of tautologies are “My favorite flavor of ice cream is the flavor I like most” or “Today is the day after yesterday.”\u00a0 Like those statements, Statement #2 adds nothing to our understanding.\u00a0\u00a0 Statement #2, alone and by itself, is not sufficient<\/strong>.<\/p>\n Answer = (A)<\/strong><\/p>\n <\/p>\n 6) This is question that demands visual insight.<\/p>\n Statement #1<\/u>:\u00a0 Think about these lengths.\u00a0 The top, KL is twice the length of the slanted sides, and the bottom, JM, is three times the length.\u00a0 This means that we could build this trapezoid from five equilateral triangles.<\/p>\n With other combinations of four lengths, we would be able to get different quadrilaterals resulting (e.g. changing the tilt of a rhombus).\u00a0 With these lengths (5, 10, 5, 15), there is no other quadrilateral possible.\u00a0 (Try this with physical items with lengths in the ratio 1:2:1:3 to see for yourself.)\u00a0 Thus, we know all the angles.\u00a0 We know that each equilateral has side of 5, so we could figure out the area of each equilateral<\/a>, then multiply by five.\u00a0\u00a0 Thus, we can find the area on the bases of this statement alone.\u00a0 Statement #1, alone and by itself, is sufficient<\/strong>.<\/p>\n Statement #2<\/u>:\u00a0 If we know just this, then the shape could have any width.\u00a0 It could be relative narrow or a mile-wide.\u00a0 We cannot determine a unique area on the basis of this statement alone.\u00a0 Statement #2, alone and by itself, is not sufficient<\/strong>.<\/p>\n Answer = (A)<\/strong><\/p>\n <\/p>\n 7) We know the area of the square, so the side of the square is<\/p>\n Thus, we know the length of the vertical leg, CE, in right triangle CEF, and we know the horizontal leg, BC, in right triangle ABC.\u00a0 Furthermore, these two triangles must be similar to teach other and similar to the larger triangle, ADF, because all the angles are the same.<\/p>\n Statement #1<\/u>: We know triangle CDE is a half a square, so it’s a 45-45-90 triangle<\/a>.\u00a0 Angle DCE = 45 degrees.\u00a0 Well,<\/p>\n (Angle ECF) = (Angle DCF) \u2013 (Angle DCE) = 75 \u2013 45 = 30 degrees<\/p>\n This means that CEF is a 30-60-90 triangles<\/a>, and so is triangle ABC because they are similar.\u00a0\u00a0 In each, we know the length of one side, so we could find the other sides and solve for the areas.\u00a0 Thus, we could find the area of the entire triangle ADF.\u00a0 This statement leads directly to a numerical answer to the prompt question.\u00a0 Statement #1, alone and by itself, is sufficient<\/strong>.<\/p>\n Statement #2<\/u>: This is interesting.\u00a0 We know that triangles ABC and CEF are similar, so they are proportional.\u00a0 Let AB:BC = r.\u00a0 Then CE:EF = r as well.<\/p>\n Now, notice that both BC and CE are sides of the square.\u00a0 Let BC = CE = s.<\/p>\n Now, multiply those two fractions together, and the s terms will cancel.<\/p>\n![\"gdgq_imgq8\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq8.png\")
<\/p>\nSummary<\/h2>\n
![\"gdgq_imgq9\"](\"https:\/\/magoosh.com\/gmat\/files\/2016\/05\/gdgq_imgq9.png\")
<\/p>\nText Explanations<\/h2>\n
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