Solid Geometry
Let's walk through the solution step by step:
First, copy the figure into your scratchpad. Include the given lengths, any other given data, and mark the segment in question - in this case CM. By investing those 20-25 seconds (yes, we measured) you achieve 2 crucial goals: It makes you notice every detail in the figure and in the text. From now on all the data is in one place and you don't need to wander with your eyes from the text to the figure, a wandering that both takes time and exposes you to risk of missing important data. Next, pick an anchoring length. Wait a minute, I'm still not convinced that copying the figure isn't wasting time. What is an anchoring length? A....... On the contrary. The time you think you're "saving" by not copying the figure is saved because you run through the question and rush to solution. That means "saving" time by compromising your level of understanding of the data presented to you. GMAT questions use efficient minimal phrasing, so that it's easy to miss a word or a number that are crucial for solving the question. Time is saved by solving efficiently, not by compromising the understanding of the question. As any other skill, the skill of copying the figure is an acquired one, and the more you practice, the better and faster you perform it. First, copy the figure into your scratchpad. Include the given lengths, any other given data, and mark the segment in question - in this case CM. By investing those 20-25 seconds (yes, we measured) you achieve 2 crucial goals: It makes you notice every detail in the figure and in the text. From now on all the data is in one place and you don't need to wander with your eyes from the text to the figure, a wandering that both takes time and exposes you to risk of missing important data. Next, pick an anchoring length. What is an anchoring length? BC=3 AM=1.6 B........ Indeed so. AM looks close in size to CM. The closer in length your anchoring line is to the line you want, the better your estimation gets. In fact, AM is so close to CM that it's hard to decide whether they are identical or slightly different. It's necessary to decide between the two options, as the second answer choice CM=1.6 would have been correct if they were identical, while the third answer CM=1.8 is true if CM is slightly longer than AM. Can I measure with a ruler? Well, now I turn to traditional geometry to decide between the two. I'll guess between the two remaining options. I'll find a creative way to measure sizes without a ruler. C......... If left with no other choice you may do that and have a 50% chance of getting it right. But this is only a last resort and you're not there yet. B........ Don't. You'll waste all the time you saved. D......... Ok, we like creativity. In the GMAT, figures are drawn to scale unless stated otherwise using the sentence: note: figure not drawn to scale. The exception is DS questions, where the figure is not necessarily drawn to scale, and is usually supplied to make you think that you know something about the figure which is not stated in the statements. The data is right in front of you. it would be mad not to use it just because you don't have a ruler. When not sure, take your scratchpad, place it against the screen, mark on it the ends of CM and now move it to AM so you can compare the sizes. Don't be shy, go ahead and do that. As the title of a Woody Allen feature says: "Whatever Works". Now you see for yourself that CM is slightly longer than AM, and therefore the correct answer, by elimination is CM=1.8 The more you practice this approach, the more efficient you get.
To sum up:
Many of the solid Geometry problems are really about one-dimensional and two-dimensional concepts, such as lines, perimeters, and areas. A face diagonal is found on the face of a solid and is the same as a "regular" diagonal.
The volume of any solid is how much three-dimensional space it occupies. Many solid Geometry problems require you to calculate the volume of a solid.
The volume formula of a prism is given in the formula: Volume= [Base Area]×Height. The volume of a pyramid is a third of the volume of a prism with similar dimensions. The Volume of a pyramid is given in the formula: Pyramid Volume=1/3 [Base Area]×Height.
A pyramid created within a cylinder is termed a cone. It has similar dimensions as the cylinder that defines it, namely height and radius.
To calculate the volume of a cone, follow the volume formula of a pyramid: Pyramid Volume=1/3 [Base Area]×Height, hence Cone Volume=1/3·πr2·H.
Q26...ABEF and BCDE are congruent squares. G is a midpoint of AF. Which of the following best approximates the angle ACG? 14° 24° 34° 44°
A Correct. Ballpark or guesstimate like so: Cut right angle C in half. Then guesstimate for angle ACG. Now you can see that angle ACG is about a third of 45°, i.e., about 15°.
Q10...In the figure above, a right cone of height x is placed on top of a cylinder of height y, so that they share the same base. If the cone and the cylinder are of equal volume, which of the following must be true? y=3x x=y+3 x=3 x=3y y=x+3
D Correct. Remember that the volume of a cone is one third of the volume of a cylinder of the same height. Here, the volume of the cylinder, πr2·y, is equal to the volume of the cone, πr2·x/3: --> πr2·y = πr2·x/3 reduce both sides by πr2 : --> y = x/3 --> x = 3y
Q9...Right triangle ABC is the base of the prism in the figure above. If AB=AC=√2 and the height of the prism is 3, what is the volume of the prism? 1 3/2 2 3 6
D Correct. The volume of a prism is given in the formula: Volume= [Base Area]×Height. The base area is the area of right triangle ABC. AB and AC are perpendicular, and therefore can be considered the base and height for the purposes of finding the area of the right triangle. Since AB=AC=√2, the area =(√2·√2)/2 = 2/2 = 1. The height of the prism is 3, so the volume of the prism = 1·3 = 3
Q32... If a circle is circumscribed around a regular pentagon, as shown above, what is the value of x? 72 108 135 140 144
E Correct. COPY the figure, INDICATE the measure of every arc next to it: The pentagon's vertices divide the circle's circumference (360°) into 5 equal arcs of 360°/5=72° each. The angle x is an inscribed angle that defines an arc that is made of four 72° arcs. Hence, x defines a 4×72° arc. Recall that the arc is twice the inscribed angle that defines it, therefore the angle x is half the 4×72° arc, --> x=2×72° --> x=144°
The volume of a right cone is 6. If the area of the base of the cone is 6, what is the height of the cone? 1/π 1/3 3/π 1 3
E Correct. Cone Volume=1/3·[base area]·H. 6 = 1/3*6*H 6=2H H=3
A prism is a solid with a bottom and top identical, parallel bases. The bases are connected by faces, also termed planes. These joining faces are parallelograms.
The side of a solid is also termed edge. The GMAT tests only right prisms, in which adjacent edges and faces are perpendicular. Very rarely, you may need to deal with a pyramid. It has one base, but its faces end in a point or apex.
Q4... Note: Figure not drawn to scale. The bases of a certain prism are equilateral triangles, and the sides of that prism are rectangles. If the area of each of the bases of the prism is 9√3 and the volume of the prism 18√3, then what is the area of one side of the prism? 12 12√3 36 36√3 72
A Correct. The volume of a prism is given in the formula: Volume= [Base Area]×Height. The area of a rectangular face is length⋅width. Use the Volume formula and the base area to find the height, which is also the length. Use the area of the equilateral triangle base to find the side of the triangle, which is also the width of the rectangular face. The base area = 9√3, and the volume is 18√3, thus: --> 18√3 = 9√3·height --> height = (18√3)/(9√3) = 2 The height of the prism is also the length of the rectangular faces. The width of the faces is also a side of the equilateral base, whose area is given by: --> --> a2 = 36 --> a = 6 Thus, the area of one of the prism's sides (a rectangle) = length × width = 6·2 = 12
Q18...Arc FE is inscribed within isosceles right triangle ABC (∠A=90°), as shown above. The arc is centered at A. If AC=2, what is EB? √2-1 √2(√2-1) 2(√2-1) 1 √2
A Incorrect. Avoid uncomfortable calculations. Trust the diagram and guesstimate for EB. Halve AB, making a mark on the diagram you inscribed to your draft board. Now each half equals 1. Is EB slightly larger or smaller than half of 1? When checking the answer remember that √2≈1.4. POE (eliminate) any answer choice that is not in the right ballpark. B Correct. EB seems to be a little over a half of half of AB, i.e. a little over 0.5. √2(√2-1)≈1.4×0.4 Which is between 0.5-0.6, and thus fits our ballpark. Still you need to eliminate all other answers to be certain.
A cylinder of radius R and height H is reshaped. Which of the following changes results in the cylinder of greatest volume? a 20% decrease in H and a 50% increase in R. a 500% increase in H and a 50% decrease in R. a 300% increase in H and a 30% decrease in R. a 70% decrease in H and a 100% increase in R. a 100% increase in H and a 10% decrease in R.
A Incorrect. Wouldn't it be nice to work with real numbers, rather than with abstract radius and height? Treat this question as an invisible plug in. Since the problem asks about percents, plug in a nice number that allows percent calculations. Normally we'd use 100, but since the volume of a = πr2·h, which can get up to really big numbers, we'll use H = 10 and R = 10. Thus, the volume = π·102·10 = 1000π For each answer, apply the required changes, and look for the greatest modified volume. A 20% decrease in H will make it 8, and a 50% increase in R will make it 15. The new volume = π·152·8 = 225·8·π = 1800·π --> not the greatest of all answers --> eliminated C Correct. A 300% increase in H will make it 40, and a 30% decrease in R will make it 7. The new volume = π·72·40 = 49·40·π = 1960π. This is the greatest volume among all answer choices, so C is the right answer.
Q33...Two tangent arcs of equal radii are centered at opposite corners of a square, as shown above. If the side of the square is 6, what is the area of the shaded region? 9(4-∏/2) 9(4-∏) 9(2-∏/2) 9(2-∏) 9(4-2∏)
B Correct. Avoid tedious calculation that will probably get you to the wrong answer anyway. Start with a snapshot ballpark. Your given value is the fact that the side of the square = 6. From that you easily calculate the area of the square to be 6²=36. The target value is the shaded region. To better compare between the shaded region and the square you'd better subdivide the square into four quarters. In order to help guesstimating the shaded area, divide the square into four square quadrants: Each quadrant is ¼ of the area of the square, i.e., ≈9. The shaded region in each quadrant is about half of a fourth of the square. So eventually the shaded areas sum up to about one quarter of the square, i.e. ≈9. POE any answer that is not in the right ballpark.
Q24...ABCD is a square. E and F are midpoints of AD and BC respectively. Which of the following best approximates the angle FEB? 16° 26° 36° 46°
B Correct. Ballpark like so: Cut right angle AEF in half: Now you can see that angle FEB is about half of 45°.
Q30...ABCD is a square. E and F are the midpoints of sides CD and AD, respectively. What is the best approximation of angle x? 10° 20° 30° 40° 45°
B Correct. Draw line EG, so that angle DEG is a right angle. DEF is a 45 degree angle in right isosceles triangle FED. Thus, the sum of angles AEG and x is the remaining 45°. Focus on the difference between angle AEG and x - use this visualization to estimate angle x. Angle AEG looks bigger than x so x must be smaller than half of the total 45º. 1/2 of 45 is 22.5, so you can immediately eliminate answer choices C, D and E. Now, is x 10 degrees or 20 degrees? If x is 10 degrees, than AEG must be the remaining 35°, which would make it more than 3 times x. The difference between the two angles is not that big, so your best bet is answer choice B of 20°.
What is the greatest possible straight line distance between any two points on a cube of width 3? 3√2 3√3 6 6√2 9
B Correct. The greatest distance between any two corners of a box is termed space diagonal. It can be calculated by using The super Pythagorean theorem: a²+b²+c²=d², where a, b, c are the dimensions of the box, and d is the length of the space diagonal. --> a²+b²+c²=d² --> 32+32+32 = 3·32 = d2 --> d = 3√3 Note: As a general rule, the main diagonal of a cube is always √3 times larger than the cube's width.
What is the surface area of a cylinder of diameter 1 and height 4? 5π 4.5π 4.25π 4π π
B Correct. The surface area of a cylinder is given in the formula Surface Area=2·πr2+2πr·H. The diameter is 1, so the radius (r) is 1/2. The height (H) is 4. Plug in these values into the surface area formula. --> 2·πr2+2πr·H = 2·π·(1/2)2+2π·(1/2)·4 = --> 2·π·(1/4) + π·4 = --> π/2 + 4π = --> 4.5π
If all the faces of pyramid P (including the base) are equilateral triangles of side 2, what is the surface area of pyramid P? 4+(3√3)/4 4√3 4+3√3 6√3 12
B Correct. The surface area of a pyramid is the sum of the areas of all four faces. Since the faces of the pyramid referred to in the question are equilateral triangles, calculate the area of one triangle using the formula . Multiply the area of one face by 4 to get the surface area of the pyramid. Area of one equilateral triangle with side 2 = (22·√3) / 4 = √3. Thus, area of four equal equilateral triangles is 4√3.
Q27...ABCD is a square. E is the midpoint of AD. Which of the following best approximates angle BEC? 30° 43° 53° 75°
B Incorrect. Ballpark or guesstimate like so: Draw a right angle at E. Cut it in half if necessary. Then guesstimate for angle BEC. C Correct. The lines added reveal that angle BEC is slightly bigger than 45°.
The base of pyramid P is an isosceles right triangle whose leg is 3. If the height of P is 4, what is the volume of P? 36 18 12 8 6
B Incorrect. Remember, the volume of a pyramid with a triangular base is (1/3) x Area of base x Height of pyramid. E Correct. First calculate the area of the base i.e. (1/2) x base x height = (1/2) x 3 x 3 = 9/2. Now use the formula for the volume of a pyramid with a triangular base i.e. (1/3) x Base area x Pyramid height = (1/3) x (9/2) x 4 = 36/6 = 6. Hence, this is the correct answer.
Q31...In the triangle ABC shown above, AB=AC. If DA=DB and CD=CB, what is the closest approximation of angle x? 51° 65° 77° 103° 115°
B You underestimated the time this question took you. You actually solved it in 4 minutes and 6 seconds. Incorrect. Avoid the cumbersome equations - Ballparking quickly solves this problem. Use the figure to guesstimate for x. A Correct. In order to better ballpark, draw a right angle at D: It seems that x is a little more than 45°. Eliminate (POE) any answer choice that is not in the right ballpark. Since all other answer choices are much greater than 45°, this is the right answer choice.
The radius of cylinder C is 5 inches, and the height of cylinder C is 5 inches. What is the greatest possible straight line distance, in inches, between any two points on a cylinder C? 5√2 5√3 5√5 10 15
C Correct. The greatest possible straight line distance, in inches, between any two points on a cylinder C is line segment AB. Drawing the cylinder shows that this problem is actually a 2 dimensional problem in a 3D disguise - finding the hypotenuse of a right triangle: AB is the hypotenuse of right triangle ABC, whose legs are AC=5 - the height of the cylinder, and BC=5+5=10 - the diameter of the cylinder. Use the Pythagorean Theorem to find AB. --> (AB)2 = 52 + 102 --> (AB)2 = 125 --> AB = √125 = 5√5
Q5...Two identical cubes of surface area 12 each are combined to form a rectangular box as portrayed in the figure above. What is the surface area of the newly created rectangular box? 10 18 20 22 24
C Correct. The surface area of a cube is the joint area of the six square faces that create the cube. Thus, each of this cube's faces has an area of 12/6=2. To find the surface area of the rectangular box, count the number of cube faces, each of area 2. Note that after gluing the cubes side by side, the two glued sides are no longer counted in the surface area. The surface area now contains only 10 squares - two squares for each of the four larger faces of the solid and one square in each of the smaller faces. Each of the squares has an area of 2, so the surface area is 20.
Q14...Circle P is the base of a right circular cone whose apex is D. Points A and B lie on circle P such that line segment AB is a diameter of P. If ∠ADB is a right angle, and AD=3√2, then what is the volume of the cone? 3π 6π 9π 18π 27π
C Correct. This is actually a 2D question in a 3D disguise. In order to find the volume of the cone, you need its radius and its height. Look at triangle ADB. What kind of a triangle is it? The cone is a right cone, so AD=DB. Since ∠ADB is a right angle, triangle ADB is a right isosceles triangle. The ratio of the sides in a right isosceles triangle, a 45-45-90 triangle, is x:x:x√2, so if the legs are of length 3√2, the hypotenuse AB, which is also the diameter of the base of the cone, is 3√2*√2=3*2=6. To calculate the height of the cone, connect points D and O. If radius OA is half of the diameter, or 3, and DA=3√2, then triangle DAO is also a right isosceles triangle. Thus, DO=OA=3. Now you can find the volume of the cone: πr²·h/3=9π*3/3=9π [fig]
Q17...The circle above is inscribed inside right triangle ABC. What is the radius of the circle, if BC=8 ? √8 / 4 √2 / √2+1 8 - 4√2 4√2 4√2 / (√2-1)
C Correct. Trade in complicated calculations for quick Ballparking. Don't hassle with uncomfortable numbers. Be efficient - guesstimate! How many times does the radius of the circle fit in the leg of the triangle? Use the circle's horizontal diameter (equal to twice the radius) to compare lengths. The radius fits in approximately 3 times in side BC, so it should be somewhere around 8/3 - 2.5 . Remember, √2≈1.4. Ballpark 4√2 as 4∙1.4, which is slightly less than 6. The result of subtracting this value from 8 should be slightly greater than 2, which is pretty close to our goal value of 2.5. This answer is a valid option - move on and try to eliminate the other options. Since all of the other answer choices can be eliminated by dint of being too small or too big, this is the correct answer.
Q8...A cube and a pyramid of square base share the same dimensions length, width height. If the volume of the cube is equal to the surface area of the cube, then what is the volume of the pyramid? 9 64/3 27 72 216
D Correct. If the side of the cube is s, then: Volume of a cube = s3 Surface area a cube = 6 s2 Volume of a pyramid = 1/3 area of the base x height = 1/3 × s2 × s. Since the volume of the cube is equal to the surface area of the cube, you get: s3 = 6 x s2 /:s2 ---> s = 6 Plug in the value s = 6 in the equation: Volume of the pyramid = 1/3 x s3 = 1/3 x 63 = 72 Hence, the answer is correct.
Rita went on vacation thrice in 2005, four times in 2004, twice in each of the years 2002, 2003 and once in 2006. What is the sum of the median, mode and range of her annual number of vacation trips from 2002 to 2006 inclusive? 2 3 5 7 9
D Correct. Remember the range is a rough estimate for the span of a set of elements. In order to find the range of a list of numbers subtract the smallest number from the largest number. List down Rita's trips as a list of numbers: {1, 2, 2, 3, 4} 2 appears most often, so it is the mode. The middle value is also 2, so the median is 2. The range is 4 - 1 = 3. Hence, the sum of all three is 2 + 2 + 3 = 7
Q7...The base of pyramid P is a rectangle of width 27 and length 7. If the height of P is 4, what is the volume of P? 28 189 224 252 756
D Correct. Remember, the volume of a pyramid = 1/3 x base area x height. Plug in the values of width, length and height in the equation of the volume of a pyramid. Volume = 1/3 x width x length x height --> 1/3 x 27 x 7 x 4 = 252. Hence, this is the correct answer.
Q16...In the figure above, ABC is an isosceles right triangle. Line segment AC is the diameter of semicircle S1. If the length of line segment AC is 1, then what is the perimeter of the shaded region? (2/√2) + ∏ 2 + (√2/2) ∏ 1/√2 + ∏ √2 + ∏/2 1/√2 + ∏/2
D Correct. The perimeter of the figure is composed of the perimeter of the semicircle plus the two legs of the right angle AB and BC. Concentrate on the circular part. It is a semicircle with a diameter of 1, thus the length of arc AC is 1/2·πd = (1/2)·π·1 = π/2. POE A, B and C, because they do not have π/2. Then POE E, as 1/√2 ≈ 1/1.4 is smaller than 1, whereas the two legs of the triangle, added together must be longer than the hypotenuse 1.
The surface area of a certain cylinder is 20π+10. If the area of the base of that cylinder is 10π, what is the volume of the cylinder? (√10)/(2π) 5 10 5√10 50
D Correct. The volume of a cylinder is given in the formula Volume=πr2·H. Use the area of the base to find r, then use r and the given surface area to find H. The area of the circular base is: --> πr2 = 10π --> r2 = 10 --> r = √10 The surface area of a cylinder is given in the formula Surface Area=2·πr2+2πr·H. The area of the base of that cylinder is 10π and the surface area is 20π+10 : --> Surface Area = 2·10π + 2πr·H = 20π+10 --> 2πr·H = 10 --> πr·H = 5 Now at this point, you could plug in r=√10 into this equation to find H, but you don't really need to: The volume of a cylinder is given in the formula Volume=πr2·H. Rewrite this expression so it resembles the expression above: --> πr2·H = {πrH}·r Since πr·H = 5, {πrH}·r = 5·√10
Q6...A four inch by four inch square sheet of paper was folded to form a cylinder, as shown above. What is the volume of the cylinder? 16/π 16 64/π 16π 64π
D Incorrect. The volume of a cylinder is given in the formula Volume=πr2·H. The base circumference (which is one of the sides of the square) is equal to 4. 2πr = 4, and thus the base radius is: r = 4/(2π) = 2/π The height of the cylinder is also 4. A Correct. The volume of the cylinder is: --> π·(2/π)2·4 = 16π/π2 = 16/π
Q1...The figure above was cut out of paper and folded to form a pyramid with a square base and four equilateral triangles. If the side of the pyramid's base is 1, what is the height of the pyramid? 1/2 √2/2 (√3)/2 (√5)/2 √3
D You grossly underestimated the time this question took you. You actually solved it in 4 minutes and 5 seconds. Incorrect. First, find the height of an equilateral triangle. Then, use the calculated height as the hypotenuse of a triangle connecting the tip of the pyramid to the center of the base. The height of this triangle is the height of the pyramid. E Incorrect B Split one of the equilateral triangle at the center to mark the height, as shown in the figure. The split results in two 30, 60, 90 right triangles with sides of ratio 1, √3, 2. Given that the base of the square is 1, short leg of each of the 30:60:90 triangles is 0.5, and the height of the equilateral triangles is 0.5√3. Now calculate the height of the pyramid through the triangle formed by connecting the center of the base, point x and the tip of the pyramid. The base and hypotenuse of this triangle are 0.5 and 0.5√3. Use the Pythagorean theorem to calculate the height: (0.5√3)2 = 0.52 + h2 --> 0.52(√3)2 = 0.52 + h2 1/22 is 1/4, and (√3)2 is simply 3 (as the square root and power cancel each other out), so --> 1/4⋅3 = 1/4 + h2 --> h2 = (3/4) - (1/4) --> h2 = 1/2 --> h = 1/√2 To make the final transition to √2/2, multiple the top and bottom by √2: (1/√2) × (√2/√2) = √2/(√2·√2) = √2/2
A cylinder is placed inside a cube so that it stands upright when the cube rests on one of its faces. If the volume of the cube is 16, what is the maximum possible volume of the cylinder that fits inside the cube as described? 16/π 2π 8 4π 8π
D You slightly underestimated the time this question took you. You actually solved it in 1 minutes and 35 seconds. Correct. Volume of a cylinder = πr2·H. In order to find the maximum possible volume of the cylinder, you want to find the maximum radius and height of the cylinder, subject to the limitations the cube imposes on these two parameter. Volume of a cube = s3 = 16 Thus, the side of the cube = 3√16 = 3√8·2 = 2·3√2. The maximum height of the cylinder can thus be 2·3√2. Since the base diameter of the cylinder is limited by the dimensions of the base of the cube, 2·3√2 is also the maximum base diameter for the cylinder (the maximum diameter is equal to the side), and thus the base radius is 3√2, and the base area is πr2 = π·(3√2)2 Now that you have the maximum height and radius, plug in 3√2 for r and 2∙3√2 for H: πr2·H= ---> π·(3√2)2 · (2·3√2) = Rewrite the roots as fractional exponents: ---> π·22/3∙2∙21/3= Simplify according to the rule for multiplying powers with the same base: ---> 2π∙22/3+1/3 = ---> 2π∙2 = 4π
If a rectangular box has two faces, each with an area of 30, two faces with areas of 60 each, and two faces with areas of 72 each, what is its volume? 60 90 162 300 360
E Correct. Remember, the volume of a rectangular solid is Length x Width x Height. Form the equations of the three different faces of the box, and isolate the 3 required dimensions. Based on the question assume that: Length x Width = 30 --> Length = 30/Width Width x Height = 72 --> Height = 72/Width Length x Height = 60 --> (30/Width) x (72/Width) = 60 Based on this (30 x 72)/60 = Width² --> 36 = Width² --> 6 = Width. Plug in width=6 into the above equations. The volume is Length x Width x Height or (30/6) x 6 x 12 = 5 x 6 x 12 = 360. Hence, this is the correct answer.
Q19...Both triangles ABC and ADE are equilateral. The shaded area enclosed in BCED is equal to the area of ΔADE. If AE=5, what is the length of BD? 5(√2+1) 2√5 3 5(2-√2) 5/(√2+1)
E Correct. Shaded region problems are best dealt with Ballparking. Leave the confusing calculations behind and guesstimate the length of BD by comparing it to AD. Then eliminate the irrelevant answer choices. Look closely at the figure, you can approximate how many times BD fits into AD. Your target is a little less than half of AD, i.e., ~2. Remember √2≈1.4. POE any answer choice that is not in the right ballpark, and you are left with the correct answer, E.
Q3...If the perimeter of the square above is 1, what is the circumference of the inscribed circle? π 3π/4 π/2 π/3 π/4
E Correct. When two or more Geometric shapes overlap - find the missing link. It may be a radius a diameter a diagonal any other line that is functional in both shapes What's the connection between the side of the square and the diameter of the circle? [fig] The radius of the circle is half the square's side. The side is 1/4, therefore the radius is 1/8. The circumference is 2πr=2*π*(1/8)=π/4.
Q2...What is the value of a+b+c+d+e in the figure above? 72 90 144 180 360
E Incorrect. Each of the five inscribed angles lies on an arc, a part of the circle. D Correct. [fig] Each of the five inscribed angles lies on an arc, a part of the circle: Angles a, b, c, d, and e lie on the arcs CD, DE, EA, AB, and BC respectively. The sum of those five arcs constitutes the circumference of the entire circle. Since an inscribed angle equals half of the arc length defined by its rays, the sum of the five inscribed angles is half of the sum of the angles of an entire circle = 360/2=180°
Now let's summarize:
In the GMAT - drawings in problem solving questions are always drawn to scale, unless explicitly stated otherwise. Start by copying the drawing from the screen to your draft board. Use that draft to add markings, sizes so that your drawing includes all the textual data in the question. Mark the requested value on your draft. Ballpark for the estimated value of the requested length. Quickly eliminate answer choices that are not within the range of your ballpark. If left with more than one answer choice, solve geometrically to decide between the answers left. Next choose an anchoring length - that is a reference segment, to which you compare the requested size. If few known lengths are available, select the closest in size. When comparing two similar sizes - use your scratchpad as an improvised ruler - do that by marking the ends of the anchoring length as shown on screen on the scratchpad. Now you can compare any other segment to your anchoring length.
The greatest distance between any two corners of a box is termed space diagonal. But GMAC has a subtle way of saying things, so be on the lookout for it. Whenever the question asks for the space diagonal use the super pythagorean theorem: a²+b²+c²=d², where a,b,c are the dimensions of the box, and d is the length of the space diagonal.
Let's check it out with the above problem. The dimensions of the box are 5, 12, 13, therefore √(5²+12²+13²)=d=√(25+144+169)=√338=13√2. To sum up: The greatest distance between any two corners of a box is termed space diagonal. The super Pythagorean theorem is a²+b²+c²=d², where a,b,c are the dimensions of the box, and d is the length of the space diagonal.
Only 1-2 problems per test are 3-D Geometry problems. Most of these problems are really about the rules of Geometry that have nothing to do with 3-dimensional figures. Only the lesser part of these problems are about solid Geometry.
Solid Geometry is about 3-dimensional shapes, i.e., a glass of water, a shoebox,etc. In other word, objects or bodies that contain volume. The GMAT tests only selected solids which will be explained in detail, later on. In general, the solids in the GMAT may be a prism or a pyramid.
Surface area is the measure of how much exposed area a solid has. It is expressed in square units (i.e., square meters, m2, square feet, etc.). If you cut open a prism along it sides and flatten it, the surface area is the area of the resulting shape.
The surface area of a prism is the sum of the areas of its faces.
A pyramid with a square base is created within a cube, such that the two objects have similar dimensions, length, width, and height.
To calculate the volume of a square based pyramid, follow the volume formula of a pyramid Pyramid Volume=1/3 [Base Area]×Height, hence Pyramid Volume=1/3·s3.
In the GMAT, triangular based pyramids may be created within a rectangular box or a cube, such that the pyramid and the prism share some of the dimensions, length, width, and height, but not necessarily all of them.
To calculate the volume of a triangular based pyramid, follow the volume formula of a pyramid Pyramid Volume=1/3 [Base Area]×Height. The height is the same as the prism that defines the pyramid. The base area may change according to type and size of the triangle.
Q13...Consider the following example: In the rectangular solid depicted above AB=4, AC=12, and CD=5. What is the area of rectangle ABDE?
To find the area of a rectangle, you need its dimensions, namely the width and the length. The width DE=4, but what about the length? I can find the length using the Pythagorean theorem in triangle ACD. I'd better look for a recycled right triangle in ACD. B... It is always wiser to find a corresponding recycled triangle before using the Pythagorean theorem. Look closely to see that it is the 5:12:13 triangle. The length of AD is 13. Therefore, the area of ABDE is length * width, or 13×4=52 There was nothing three-dimensional about the area of rectangle ABDE, was there?
If the surface area of cube A is 4 times the surface area of cube B, then the volume of cube A is how many times the volume of cube B? 8 16 24 64 512
A Correct. Remember, the surface area of a cube is 6(Side²). Also, the volume of a cube is Side³. There is an invisible variable in the question, side of cube B. Plug in your own number for that to find out the volume of cube B is how many times the volume of cube A. Plug in 1 for the side of cube B. Based on this the surface area of cube B = 6(1²) = 6(1) = 6. Hence, the surface area of cube A is 24 (4 x 6). Since the surface area of a cube is given in the formula 6*surface of one square face, the area of one face of cube A is 24/6=4, and any side of cube A = √4 = 2. The volume of cube B = 1³ = 1 and the volume of cube A = 2³ = 8. Hence, the volume of cube A is 8 times the volume of cube B.
Q25...What is the approximate measure of angle x? 15° 25° 30° 35° 40°
A Correct. This non-GMAT problem intentionally does not include the needed information to calculate x. Instead, Create the right angle as shown in the figure above. x lies within the difference of 30 degrees, between the given 60º angle and the right angle. Estimate the measure of angle x. Now you can see that x is about half of the remaining 30°.
If the dimensions of a rectangular box are a by b by c, the surface area of the box is how many times the volume of the box? (2/a)+(2/b)+(2/c) (1/a)+(1/b)+(1/c) 1/(a·b·c) 2(a+b+c) 2/(a+b+c)
A You grossly underestimated the time this question took you. You actually solved it in 4 minutes and 47 seconds. Correct. The volume of a rectangular box is given in the formula: Volume=w·l·h=a×b×c The surface area of a rectangular box is given in the formula: Surface Area=2wl+2hw+2lh=2bc+2ac+2ab Note that the answer choices contain algebraic expressions - this is a sign for you to plug in to make this question easier! Plug in for example a=1, b=2, c=4. The volume of the box is thus 1*2*4=8 The surface area of the box is 2(1*2)+2(1*4)+2(2*4)=28 The surface area is therefore 28/8=3.5 times the volume of the box. Plug in the same numbers in the answer choices and check which one yields 3.5. Remember - when plugging in you have to check all answer choices. (2/a)+(2/b)+(2/c)=2/1+2/2+2/4=2+1+0.5=3.5 This is the correct answer choice because the other answer choices do not yield anything close to 3.5 when plugging in 1,2 and 4.
Q28...In triangle ABC shown above, AB=AC. If BC=BD, what is the value of x? 18° 36° 60° 72° 108°
B Correct. Ballparking quickly solves this problem. Use the figure to guesstimate for x. In order to better ballpark, draw a right angle at A: It seems that x is a little less than 45°. Now go to the answer choices and eliminate (POE) any answer choice that is not in the right ballpark.
The edges of a cube are made out of wire. If the surface area of that cube is 24, what is the total length of the wire? 12 24 36 48 96
B Correct. Remember, the surface area of a cube is 6 x Side². Use the surface area of the cube to find the side of the cube. The length of wire is the sum of all twelve sides of a cube. Given that the surface area of the cube is 24, 6(side²) = 24 or side² = 4 or side = 2. The total length of wire is the sum of all twelve sides of the cube i.e. 12 x side = 12 x 2 = 24. Hence, this is the correct answer.
Q20...Two circular regions of equal radii are centered at adjacent corners C and D of square ABCD, as shown above. If shaded regions, X and Y, have equal areas, and if DE=CF=√18 , what is the length of AB? 3 / √∏ (3/√2) ×√∏ 3√∏ 3√2√∏ 3∏
B You grossly underestimated the time this question took you. You actually solved it in 5 minutes and 11 seconds. Incorrect. Trade in complicated calculations for quick Ballparking. Don't hassle with uncomfortable numbers. Be efficient - guesstimate! AB is a side of the square. As such it is slightly larger than DE. DE=√18 Orientate yourself using known square numbers as upper and lower borders of your ballpark: √16<√18<√25 √16<DE<√25 4<DE<5 So DE is slightly over 4, and AB is slightly over DE. Now check the answers to see what is slightly larger than 4? Remember π≈3+. Now if we're looking for an answer greater than 4, do we have it here? Let's tear the masks over the numbers here: (3/√2) ×√∏ Which is approximately 3*1.7/1.4 - and that's just a little over 3. C Since √18 is slightly more than 4. If DE is ≈4, then AB is about one quarter longer, i.e., AB≈5. Our answer is approximately 3*1.7, which is more than 4, close to 5. All other answers but this one were eliminated based on that criterion.
If the space diagonal of cube C is 5 inches long, what is the length, in inches, of the diagonal of the base of cube C? 5/(√6) 5·√(2/3) 5·√(3/2) 5·√3 5·√6
B You underestimated the time this question took you. You actually solved it in 1 minutes and 55 seconds. Correct. The space diagonal of a rectangular solid is calculated using the super Pythagorean Theorem: a²+b²+c²=d² , where a, b, and c are the three dimensions of the box, and d is the space diagonal. Since in a cube all three dimensions are equal, the square of the space diagonal, d2, actually equals 3a2, where a is any one of the equal dimensions. --> 3a2 = 52 --> a2 = 52/3 --> a = 5/√3 --> this is the side of the cube The base diagonal in a cube is actually a diagonal of a square, or the hypotenuse of a 45:45:90 triangle composing half of the square. Using the 45:45:90 recycled triangle ratio, the diagonal is is equal to √2·(side) = (5/√3)·√2. Enter both bases under the same root to get 5·√(2/3)
Q21...The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? 3 4 4.5 5 5.5
C Correct. Trade in complicated calculations for quick Ballparking. Divide the side of the square to estimate what is the fraction of DI in the whole side. The answer choices are not far apart. Better ballpark by dividing the side of the square: It is clear now that each of the four rectangles has width 1.5. DI is 3 times the width of the rectangle, i.e. 3×1.5=4.5. The correct answer is C.
Q23...ABEF and BCDE are squares. G is a midpoint of AF. Which of the following best approximates the angle GHE? 85° 75° 65° 55°
C Incorrect. Ballpark or guesstimate like so: Draw a right angle at H. Cut it in half if necessary. Then guesstimate angle GHE. B Correct. Now, you can see that angle GHE is a little less than 90°. How much less? Using the 45° reference it is possible to approximate that: the difference of angle GHE from 90° is about a third of 45°, therefore your target is 75°.
Q29...In the figure above, AB=AC and CB=CD. What is x? 50 80 100 130 160
C Incorrect. Ballparking quickly solves this problem. Use the figure to guesstimate for x. D Correct. To better guesstimate draw a right angle at D: It seems that angle x is 90° plus 45°, i.e., approximately 135°. Eliminate (POE) any answer choice that is not in the right ballpark.
Q22...O is the midpoint of AC in the regular octagon shown above. If OB bisects a side of the octagon, what is angle x? 105° 120° 135° 157.5° 210°
C Incorrect. Ballparking quickly solves this problem. Use the figure to guesstimate for x. D Correct. To better guesstimate, extend DO to the perimeter. It seems angle x is 90° +( 90° - 90°/4) (actually, that is exactly the case.) That is approximately 90°+90°-20°=160°. Eliminate (POE) any answer choice that is not in the right ballpark.
What is the surface area of a rectangular box whose dimensions (length, width, height) are (a−1, a, a+1)? 3a2 3a2−1 6a2−1 6a2−2 6a2
D Correct. The surface area of a rectangular box is given in the formula: Surface Area=2wl+2hw+2lh. In this case, the dimensions (length, width, height) are (a-1, a, a+1). Instead of simplifying a monstrous expression, note that the answer choices all include variables, a surefire sign to plug in and eliminate. Plug in a=2, so that the length equals a-1=1 and the height equals a+1=3. Now find the surface area of the rectangular box of dimensions (1, 2, 3): 2wl+2hw+2lh = 2·1·2 + 2·2·3 + 2·1·3 = 4 + 12 + 6 = 22. That's you Goal. Plug a=2 into the answer choices, and eliminate those which do not match your Goal. --> 6a2-2 = 6·22-2 = 22. Thus, this answer choice cannot be eliminated for a=2. all the other answer choices do not match your goal of 2 for a=2, so this is the right answer choice.
If the two bases of a certain prism are equilateral triangles, and the sides of that prism are squares, then how many edges does that prism have? 18 12 10 9 8
D Correct. Draw the prism and count the edges: Both triangular bases have 3 edges each. The prism also has 3 squared 'standing edges', for a total of 9 edges.
In the figure above, what is the slope of the line that passes through points P and Q? −7 −4/3 −3/4 3/4 4/3
D Correct. Remember, the slope of a downward sloping line is always negative. The slope is given in the formula Use the graph to read the coordinates of points P and Q. Based on the graph P and Q have x, y coordinates (-3, 0) and (0, -4) respectively. calculate the slope of line PQ using the above coordinates i.e. -4 - 0 / 0 - (-3) = -4/3. Hence, the slope of PQ is -4/3.
Q11...A right cone is to be placed within a rectangular box so that the cone stands upright when the box is placed on one of its sides. If the dimensions of the box are 2 inches by 2 inches by 4 inches, then what is the greatest possible volume of such a cone? (2/3)·π (4/3)·π 2π (8/3)·π 4π
E You grossly underestimated the time this question took you. You actually solved it in 4 minutes and 44 seconds. Incorrect. You need to insert a cone into a box with given dimensions. In how many ways can you perform this task? Since the problem doesn't supply a figure, draw one yourself. D Incorrect. Look at the figure: Just because one dimension of the box is 4, doesn't mean you have a 4*4 side. C Incorrect. You need to insert a cone into a box with given dimensions. In how many ways can you perform this task? Since the problem doesn't supply a figure, draw one yourself. B Correct. Since the radius is limited in any case by at least one side of 2, the greatest volume will be achieved by maximizing the height. Since that is the case, the base of the box is 2x2, so the cone has a base radius of 1 and a base area of πr²=π⋅1²=π. The height is 4, thus the Maximum Cone Volume=1/3·πr2·H=(4/3)π
Once in a few problems in solid Geometry, GMAC really tests you on three-dimensional concepts. Consider the following problem What is the longest, rigid, pole that can be put between any two corners, inside a rectangular box, with dimensions of 5 inches, by 12 inches, by 13 inches? (The diameter of the pole is negligible) (A) 18 (B) 13√2 (C) √361 (D) 20 (E) 25
First, draw your own box since the problem doesn't provide one. Put all the labels and information the problem provides. The longest pole that can be put inside the box extends from corner to corner, such that it traverses the space within the box. Focus on the question, what is the length of the pole? C Incorrect. A Incorrect. 5+13=18 but that has nothing to do with the length of the pole. B Correct. You may calculate the length of the pole, using the face diagonal and recycled right triangle 5:12:13. Form an isosceles right triangle, inside the box. Its legs are 13, each, therefore the hypotenuse or pole is 13√2.
Ballparking is a powerful technique, especially in geometry problems. Since the figures in Geometry problems are drawn to scale (unless otherwise stated), it is advisable to ballpark and save precious time. Let's try out what you've learned regarding Ballparking.
Let's review Ballparking Angles: Remember: the figure in geometry problem solving is accurate, unless stated otherwise. In order to Ballpark for Angles easily, FIX THE REQUESTED ANGLE IN A BOX: Continue one of the rays to form 180 degrees - OR- Draw a line at right angle for one of the rays to form 90 degrees. For improved accuracy half the 90 degrees box by a bisector.
Many of the solid Geometry problems are really about one-dimensional and two-dimensional concepts, such as lines, perimeters, and areas. Let's review some examples:
Q12...What is the length of HC in the rectangular box shown above, if GE=12 and FC=9? HC is a face diagonal, which is exactly the same as a "regular" diagonal. It is on the face of the box. How can you find the length of HC? Use the Pythagorean theorem Look for a right recycled triangle. It is always wiser to find a corresponding recycled triangle before using the Pythagorean theorem. Look closely to see that it is the 3:4:5 triangle, multiplied by 3, 3×3:4×3:5×3. The length of HC is 15. There was nothing three-dimensional about HC, was there?
Ballparking is a very powerful tool to override unnecessary calculations. Remember- figures in problem solving questions are drawn to scale unless stated otherwise. Use that fact to your benefit! By using Ballparking you demonstrate that you are a target-focused test taker who understands that the name of the game is circling the correct answer in minimum time, and not producing the most detailed mathematical solution.
Q15... Let's see how you deal with the following question. Remember, you only have 2 minutes: If in the figure above AM=1.6, DM=1.2 and CB=3. If AC and DB are perpendicular, what is the length of CM? 0.8 1.6 1.8 2.4 2.8 Very good! How did you solve it? Geometrically, using similiar triangles. I ballparked for the estimated length of AM based on the drawing. B......... That's the way to do it!
The surface area of a rectangular box is composed of six rectangles, two opposite faces of each type. Therefore, Surface Area=2wl+2hw+2lh.
Rectangular box B is of width 2, length 3 and height 5. If the width of rectangular box C is 3 times the width of B, the length of C is 5 times the length of B and the height of C is 2 times the height of B, then the volume of C is how many times the volume of B? 900 300 90 30 10 D Correct. The volume of a rectangular box is given in the formula: Volume=w·l·h. The volume of B = 2·3·5 = 30 Therefore, The volume of C = (3·2)·(5·3)·(2·5) = 900 Thus, the volume of C is 900/30 = 30 times the volume of B.
Cube is not only a name of a film, but also an object made up of six perpendicular squares.
The three dimensions of a cube, width, length, and height, are equal. Volume of a Cube = S³ Surface Area = 6 S² To calculate the volume of a cube with side s, follow the volume formula of a prism Volume= [Base Area]×Height, hence Volume=s3. (reads: "s cubed") The surface area of a cube is composed of six identical squares, hence, Surface Area=6·s2.
A rectangular box or rectangular prism is made of 6 rectangles, with all right angles. It is sometimes referred to as a crate.
The three dimensions of a rectangular box are width, length, and height. To calculate the volume of a rectangular box, follow the volume formula of a prism Volume= [Base Area]×Height, hence Volume=w·l·h.
A cylinder is an object, a prism, whose bases are circles. Imagine a can of coke. A cylinder has a height dimension, but instead of width and length, the cylinder has a radius (like any circle)
To calculate the volume of a cylinder with radius r, follow the volume formula of a prism: Volume= [Base Area]×Height, hence Volume=πr2·H. The surface area of a cylinder is composed of the area of its surfaces: bottom, top, and lateral surface. In order to understand what the surface area of a cylinder is, let's roll open a cylinder. It turns out to be a rectangle. Its vertical dimension is the height of the cylinder, and the other dimension is the circumference of the cylinder's base, calculated as 2πr. Don't forget to add the bottom and top bases, namely 2·πr2. And so the surface area of a cylinder is given in the formula Surface Area=2[base area] + area of lateral rectangle = 2·πr2+2πr·H
A pyramid with a rectangular base is created within a rectangular box, such that the two objects have similar dimensions, length, width, and height.
To calculate the volume of a rectangular base pyramid, follow the volume formula of a pyramid Pyramid Volume=1/3 [Base Area]×Height, hence Pyramid Volume=1/3·w·l·h.
