Triangles

1.Having identified the special recycled triangle 45°:45°:90° in a problem, smile. You're half way through. Use the fixed ratio of the sides to get you going.

1.A diagonal in a square defines a 45°:45°:90° triangle, or an isosceles right triangle. This is a special recycled right triangle. 1Sometimes, GMAC will require that you solve for one of the legs, given the hypotenuse. Have no fear the 45°:45°:90° is here. Use the fixed ratio of the sides to get you going, with a slight variation. If b is the hypotenuse of the triangle, the sides opposite the angles are ...........45°................45°.................90° ..........b√2/2........b√2/2.................b

Common forms in the GMAT: Similar triangles can appear in many forms, but three specific recycled figures are more popular. If you encounter the following figures in a GMAT question, think "similarity" - check if the concept of constant ratio between respective sides of the triangles can help you quickly solve the problem: 1) "Triangle within a triangle", parallel bases: If DE||AC, small triangle BDE and large triangle ABC are similar, as all their angles are equal to each other: ∠B - common angle for both triangles. ∠D = ∠A - small angles formed by two parallel lines and a line AB intersecting them. ∠E = ∠C - small angles formed by two parallel lines and a line BC intersecting them.

2) "Hourglass" form: If AB||CD, small triangle ABE and large triangle DCE are similar, as all their angles are equal to each other: ∠AEB = ∠CED - opposite angles are equal to each other. ∠D = ∠A - small angles formed by two parallel lines and a line AD intersecting them. ∠B = ∠C - angles formed by two parallel lines and a line BC intersecting them. 3) Right triangle with height drawn to the hypotenuse This figure actually presents three similar right triangles: small triangle BDC, medium triangle ABD, and large triangle ABC. Proving that this is so will take too long here and is not that important, but go ahead and try it on your own (Hint: Show that the small and medium triangles have the same angles as the large triangle).

Q11...Right triangle ABC is divided into five identical right triangles as shown above. What is the ratio, by length, of AC to BC? √5 : 2 2√5 : 5 √5 : 1 √3 : 1 3 : 2√3

A Correct. At first glance, this question looks unsolvable, as there are no numbers. But then again, since the question only asks for a ratio, the actual numbers are unimportant. Plug in your own value for a side of triangle ABC, then calculate the other sides. From these values, you can find the ratio of AC to BC. There's no indication that the right triangle ABC is a special recycled triangle, but you can always use the Pythagorean theorem. Plug in AB=1. AB is the hypotenuse of small triangle ABE in the figure below. All other instances of lines equal to AB are marked in blue. If AB=1, then BC=2 (BF and FC are both equal to AB) From AB and BC, it is possible to use the Pythagorean Theorem to find AC: AB2 + BC2 = AC2 --> 12 + 22 = AC2 --> AC2 = 5 --> AC = √5 Thus, the ratio of AC to BC is √5:2.

If a and b are the roots of x2−3x−10=0, then (a,b) may be which of the following? (−2,5) (2,5) (2,−5) (−3,10) (10,3)

A Correct. Factor x2−3x−10=0: 1) Write in factorized form (x+e)(x+f)=0. 2) Find e and f: a pair of integers whose product is c=-10 and whose sum is b=-3. Begin with e·f=c=-10 - the possible pairs of integers whose product is -10 are {-2·5}, {2·-5}, {-1·10}, {1·(-10)}. Out of these four pairs, only e=2 and f=-5 satisfy the second condition of e+f=b=-3 . 3) Write e and f in the appropriate places in the factors: (x+2)(x-5)=0 4) Set each factor to zero and solve: (x+2) = 0 --> x1=-2 (x-5) = 0 --> x2=5

369/722 is approximately: 50% 58% 60% 65% 66%

A Correct. Percent problems are always a wonderful opportunity to ballpark. Avoid tedious calculations by replacing uncomfortable "ugly" numbers with rough estimates. 369/722 is approximately 360/720, which is 1/2 or 50%. The best answer choice is the one closest to 50%.

Which of the following is an obtuse triangle? a triangle whose angles are 2x, 2x and 5x a triangle whose angles are 3x, 2x and x a triangle whose angles are 10x, 10x and x a triangle whose angles are 9x, 4x and 5x a triangle whose angles are 3x, 3x and 3x

A Correct. Remember that an obtuse triangle is a triangle in which one of the angles is obtuse. Given that the sum of all angles in a triangle is 180, 2x + 2x + 5x = 180 or 9x = 180 i.e. x = 20. Hence, 5x (5 X 20 = 100) is an obtuse angle and the resulting triangle is an obtuse triangle.

Q9...In the figure below, what is the value of x? 60° 80° 100° 140° 160°

A Correct. Remember that the sum of the angles inside any triangle is 180°. Also, angles on a straight line sum up to 180°. Acute angle BCD = 180° - 140° = 40°. Based on this, obtuse angle DBC = 180° - 40° - 20° = 120°. Since x is the supplementary angle of angle DBC, x = 180° - 120° = 60°.

Q17...In the figure below, what is the value of x? 30° 45° 60° 90° 120°

A Correct. Remember that the sum of the angles inside any triangle is 180°. Also, angles on a straight line sum up to 180°. Remember an equilateral triangle has all sides and angles equal. Given that all three sides of triangle ABD are equal, all three angles are also equal i.e. 60° (180°/3). Also, BDC is the supplementary angle of BDA so BDC = 180° - 60° = 120°. Since BDC is an isosceles triangle and side DB = DC, x = DBC = [(180° - 120°) / 2] = 60° / 2 = 30°.

Q14...What is the area of triangle ABC shown above? 60 65 78 120 130

A Correct. The area of a triangle is AREA= ½×BASE×HEIGHT. If a triangle is not a right triangle, draw a line yourself to identify and calculate the height of the triangle. Look at the triangle closely to identify that it can be split into two 5:12:13 right triangles. Remember to use recycled right triangle ratios to solve triangle questions. Draw a line joining point B and the midpoint of AC, creating a right triangle with a leg of 5 and a hypotenuse of 13. This line is the height of triangle ABC. Use the recycled triangle 5:12:13 ratio to calculate that the height of triangle ABC is 12. Now find the area of the triangle using AREA= ½×BASE×HEIGHT = (1/2) x 10 x 12 = 60. Hence, the area of triangle ABC is 60.

If a+b=−8, and a=20/b, what is the value of a2+b2? 24 32 34 40 104

A Correct. The question asks for a2+b2, which is close to Recycled quadratic I: (a+b)2 = a2+2ab+b2. The only component that's missing is 2ab, but that too can be found in the question. (a+b)2 = a2+2ab+b2 Isolate a2+b2: --> a2+b2 = (a+b)2-2ab From the first equation: --> a+b = -8 --> (a+b)2 = 64 From the second equation: --> a=20/b --> ab=20 --> 2ab=40 Therefore, --> a2+b2 = (a+b)2-2ab = 64-40 = 24

In the equation x2+6x+m=0, x is a variable, and m is a constant. If x=−2 is one solution of the equation x2+6x+m=0, what is the other solution of the equation? −4 −3 3 4 8

A Correct. Translate the information presented in the question: If x=-2 is a solution of the equation, then x+2=0 must be a Factor of the equation. Factor the quadratic in the question stem: x2+6x+m = (x+e)(x+f) = (x+2)(x+f) In an equation of the form of ax2+bx+c=0, the sum of e and f must equal b. In our case, b=6, so e+f = 2+f = 6 --> f = 4 So (x+f) = (x+4) is the other factor of the equation. Set the factor to zero to find the other solution: x+4 = 0 --> x = -4

Having identified the special recycled triangle 45°:45°:90° in a problem, smile. You're half way through. Use the fixed ratio of the sides to get you going.

A diagonal in a square defines a 45°:45°:90° triangle, or an isosceles right triangle. This is a special recycled right triangle. Use the fixed ratio of the sides to get you going. If a is the leg of the triangle, the sides opposite the angles are ...........45°...........45°.................90° ...........a...............a..................a√2 Memorize this ratio box so you can deal with the 45°:45°:90° triangle in real life GMAT situations. Get used to identifying recycled triangles and using their respective fixed ratios in order to save time and prevent careless calculation errors caused by using the Pythagorean theorem. One more thing - this fixed ratio means that only one side is necessary to calculate the other two sides of a 45°:45°:90° triangle. In the above example, the fact that one of the legs is 5 is enough to calculate the other two sides

Having identified the special recycled triangle 30°:60°:90° triangle in a problem, smile. You're half way through. Use the fixed ratio of the sides to get you going.

A height in an equilateral triangle defines a 30°:60°:90° triangle. This is a special recycled right triangle. Use the fixed ratio of the sides to get you going: If a is the smallest leg of the triangle, the sides opposite the angles are .............30°......60°...........90° ..............a........ a√3..........2a Memorize this ratio box so you can deal with the 30°:60°:90° triangle in real life GMAT situations. Get used to identifying recycled triangles and using their respective fixed ratios in order to save time and prevent careless calculation errors caused by using the Pythagorean theorem. One more thing - this fixed ratio means that only one side is necessary to calculate the other two sides of a 30°:60°:90°. In the above example, the fact that the hypotenuse is 8 is enough to calculate the other two sides.

Amongst triangles, which are themselves fairly common in GMAT Geometry problems, right triangles are the most common.

A right triangle has a right angle, with two legs perpendicular to each other. The side opposite the right angle is called the hypotenuse. It is the largest side in a right triangle.

Equilateral means equal-sided.

An equilateral triangle has three equal sides, and consequently three equal angles. Hence, the angles in an equilateral are 60° each. In an equilateral triangle any height is also the median and the bisector.

The area formula for any triangle is AREA= ½×BASE×HEIGHT.

Any side in the triangle may be considered as the base, as long as you use the height that is perpendicular to it.

If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true? 5y−7x+4 < 0 5y+7x−4 > 0 7x−5y−4 > 0 4+5y+7x < 0 7x−5y+4 > 0

B Correct! This is an inequalities question, so the first thing that should pop in your head is that inequalities are like equations in all aspects, except for cases when you divide\multiply both sides by a negative number; in these cases you must flip the inequality sign. Knowing that x is negative (<0) and therefore that (x-1) is also negative, you can multiply both sides by (x-1) and flip the sign, getting: (3+5y)/(x−1) < −7 /·(x-1) 3+5y > -7(x-1) --> 3+5y > -7x+7 --> 5y+7x-4 > 0

Q24...The height of an equilateral triangle is the side of a smaller equilateral triangle, as shown above. If the side of the large equilateral triangle is 1, what is AB? 1-√3/2 0.25 2-√3 1/3 1-√3/4

B Correct. Draw your own diagram and label every piece of information that you know. Your drawing should look like this: To find AB, subtract the height of the smaller triangle from the length of one side of the larger triangle. Drawing a height down the large triangle forms two 30°:60°:90° recycled right triangles of sides 1, √3/2 and 1/2. Drawing a height down the smaller triangle forms two 30°:60°:90° special recycled right triangles. Use the fact that the height of the large triangle (which is also the hypotenuse of the smaller recycled triangle)is √3/2, combined with the recycled a:a√3:2a recycled ratio to find the sides of the smaller 30:60:90 triangles: √3/2, 3/4 and √3/4. Subtract the height of the smaller triangle, 3/4, from the side of the larger triangle, 1, to get the answer .25. Alternative explanation: examine the diagram more closely. The little triangle, with side AB, is also a 30°:60°:90° special recycled right triangle. Since you know that the side opposite the 90° angle in that triangle is 1/2 (as the height of the large triangle is also the median), you also know that side AB is equal to half of 1/2, or .25. drawing the figure and noting everything you know can help find shortcuts to save precious time on the GMAT.

John is 4 years older than Paul is. If the product of their ages (in years) is 96, how old is John? 8 years 12 years 16 years 18 years 19 years

B Correct. Numbers in the answer choices and a specific question ("...how old...?") call for Plugging In The Answers. You may feel like jotting down an equation or two. This is just your algebraic urge, which is another stop sign for Reverse PI problems. Assume the amount in the answer choice is the age of John, and then follow the story in the problem. If everything fits - stop. Pick it. Otherwise - POE and move on, until you find an answer that works. Start with answer choice C. Assume John is 16 years old. Then, Paul must be 12 year old. The product of their ages (in years) is too big, so POE C. In order to reduce the product of their ages, the answer must be smaller than 16, so POE D and E. Now, plug in A or B to check which is correct. Plugging in B works: If John is 12, Paul is 8 and their age's product is 96.

Q2...In isosceles triangle CEB shown below, what is the value of x? 40° 50° 60° 80° 100°

B Correct. Remember that the sum of the angles inside any triangle is 180°. Also, angles on a straight line sum up to 180°. Remember that in a triangle angles opposite equal sides are also equal. Since angles CEB and AEB are supplementary angles, angle CEB = 180° - 100° = 80°. Based on this, x = (180° - 80°) / 2 = 50°.

(√8−3)(3+√8)(√5−2)(√5+2)=? −3 −1 1 5 15

B Correct. The issue of the question is recycled quadratics. Use the formula for Recycled quadratic III: (a+b)(a-b) = a2-b2 to simplify the product. Note that (√8)2=8, and (√5)2=5. (√8−3)(3+√8)·(√5−2)(√5+2) = (√8−3)(√8+3)·(√5−2)(√5+2) = [8−9]·[5−4] = −1·1 = −1

Q23...If the area of the equilateral triangle above is 0.75, what is the area of the adjacent square? (√3)/2 √3 √6 3 2√3

B Correct. This problem is quickly solved by using the formula for area of an equilateral triangle: According to the question, where a is the side of the triangle. Note that a2 is already the area of the square. Solve for a2: or a2=3/√3 3 is equal to (√3)2, so reduce top and bottom by √3: 3/√3 = √3/1 = √3.

(2−√5)^2=? 7−4√5 9−4√5 1 7−2√5 9−2√5

B Correct. Use recycled quad II. --> (2−√5)2 = --> 22-2×2×√5+√52 = --> 4-4√5+5 = --> 9-4√5

Q34...The diagonal of a square is the height in an equilateral triangle. If the side of the square is x, what is the side of the equilateral triangle in terms of x? √2/3 × x 2/√3 × x √3x √8/3 × x 2√3 × x

B Incorrect. A square can be divided into two 45 45 90 triangles and an equilateral triangle can be divided into two 30 60 90 triangles as follows: Remember the ratio of the sides of a 45, 45, 90 triangle is 1, 1, √2. Also, the ratio of the sides of a 30, 60, 90 triangle is 1,√3, 2. D Correct. If the side of the square is x, the diagonal is x√2. Also, The height of an equilateral triangle with side 2s is s√3. Based on the figure given in the question, the height of the equilateral triangle is also the diagonal of the square, so s√3 = x√2 --> s = x√2 / √3. Given that the side of the equilateral triangle is 2s 2s = 2 (x√2 / √3) = --> x2√2 / √3 = Rewrite 2 as √4 to get --> x√4√2 / √3 = Combine all the roots (in the top and bottom) under the same root sign: --> x√(8 / 3) Hence, this is the correct answer.

Q30...What is x in the figure above? √2 2 2√2 √5 √6

B Remember the ratio of the sides of a 45-45-90 degree triangle is 1, 1, √2. Correct. Since the top triangle is a right isosceles triangle, and the legs of the right angle are equal 1, the hypotenuse is √2. Since the hypotenuse of the first triangle is the side of the second right triangle, the side of the second triangle is √2 and the hypotenuse is √[12+(√2)2]=√3. Likewise, the sides and hypotenuse of the third triangle are 1, √3 and 2. Hence x = 2.

Triangles: Impossible Triangles and The Third Side Rule (Diag on Excel) Consider the following triangle: What is the perimeter of the triangle? 15 It cannot be determined

B This triangle has a major flaw. This flaw is so fundamental that the triangle is, in fact, impossible. To illustrate the point, collapse the sides that measure 3 and 4, so that they are almost one with the base. This is no triangle at all! Since sides 3 and 4 cannot meet, they cannot form a valid triangle. The lengths of the two sides added together must be longer than the length of the third side if they are to meet and form a triangle.

What is so special about the special recycled right triangles? Well for starters, you don't need the measure of the sides to identify them, only their angles.

By drawing a diagonal inside a square you get an isosceles right triangle, namely the 45°:45°:90° triangle. By drawing a height inside an equilateral triangle you get the 30°:60°:90° triangle. Memorize the special recycled triangles as GMAC uses them frequently in Geometry problems.

Q25...If the perimeter of ΔACD is 3+3√3, what is the perimeter of equilateral triangle ΔABC? 6 9 6√3 9+3√3 9√3

C Correct! ΔACD is half of an equilateral triangle, so use the 30°:60°:90° special triangle to solve this problem. Remember the ratio box: The perimeter of ΔACD is a + a√3 + 2a = 3+3√3. Solve for a: 3a+a√3=3+3√3 --> a(3+√3)=3+3√3 /:(3+√3) --> a = (3+3√3) / (3+√3) = 3(1+√3) / √3(√3+1) --> a = 3 / √3 = --> 31 / 31/2 = --> 31-1/2 = 31/2 = √3 The hypotenuse of ΔACD, which is also the side of ΔABC, is then 2a = 2√3 , according to the recycled ratio. Hence, the perimeter is 3×2√3 = 6√3.

Q6...If ABC is an equilateral triangle and AB||CD, what fraction of the area of quadrilateral ABDC is shaded? 1/4 1/(2√3) 1/3 1/2 2/3

C Correct. A triangle's area is 1/2·height·side. Triangle ABC is an equilateral triangle, so it base AB = AC = 3. Therefore, the base of the unshaded triangle is twice the base of the shaded triangle (6 and 3 respectively.) Since the shaded triangle and the unshaded triangle share the same height between parallel lines, but the shaded triangle has half the base of the unshaded triangle, the shaded triangle has half the area of the unshaded triangle. Plug in 1 for the area of the shaded triangle - the unshaded area is twice 1, or 2, so the And the total area of ABCD is 1+2=3. Therefore, the shaded area 1/3 of ABCD.

Q32...What is the length of line segment AB in the figure above? (√2)/2 (√3)/2 √2 3/2 √3

C Correct. Draw a height AD. Since the sum of the angles in a triangle equal 180 degrees, ∠CAD = 45º and ∠BAD= 60º. ∆ACD is therefore a recycled 45:45:90 triangle. Use the recycled ratio for a given hypotenuse: AC=1, so the legs AD and CD=1/√2. ∆ADB is a 30:60:90 recycled triangle, with a corresponding ratio of 1:√3:2. AD, the little leg is 1/√2. so AB, the hypotenuse is 2·(1/√2)=2 / √2 = √2.

Q8...What is the total number of vertices formed when nine squares fill a larger square, as shown above? 36 27 16 12 9

C Correct. Every point in the figure where two or more lines intersect is a vertex. To determine the total number of vertices, treat the figure as a matrix: multiply the number of vertices in the top row by the number of vertices on one of the sides. 4·4=16.

Q12...What is BD in rectangle ABCD above? 1.4 2.5 2.6 6.5 13

C Correct. For every right triangle in the GMAT, check for recycled right triangles first. If there aren't any use the Pythagorean Theorem. Check for recycled right triangles first. The 5:12:13 will do just fine. Nasty GMAC divided each of the numbers by 5. And so, the third side must b 13:5=2.6. If you can't see the division by 5, multiply everything by 10 to avoid the decimals and get a clearer picture: If AD=10 and AB=24, we get a ratio of 10:24:BD, which fits the 5:12:13 ratio times two. It follows that BD must equal 13*2=26, for a ratio of 10:24:26. Now divide everything back by 10 to get the proper ratio of 1 : 2.4 : 2.6

What is the closest approximation for 34% of 5,950? 1500 1750 2000 2250 2500

C Correct. Percent problems are always a wonderful opportunity to ballpark. Avoid tedious calculations by replacing uncomfortable "ugly" numbers with rough estimates. 34% of 5950 is roughly 33% of 6000, which is about a third, hence, 2000.

It is impossible to draw a triangle whose sides are 1, 9 and 9 5, 7 and 6 1, 2 and 3 1/2, 1/3 and 1/4 4, 4, and 4

C Correct. Remember that any side of a triangle must be more than the difference and less than the sum of the other two sides. If the numbers in the answer choice satisfy this condition, then the triangle is possible - eliminate this answer choice. 1 = 3 -2 and 3 = 1 + 2. Hence, this triangle is NOT possible.

Q22...In the figure below, what is the value of angle EAD? 20° 40° 80° 140° 160°

C Correct. Remember that the sum of the angles inside any triangle is 180°. Also, angles on a straight line sum up to 180°. Remember that in a triangle angles opposite equal sides are also equal. Since angle AED and angle DEC are supplementary angles, angle AED = 180° - 160° = 20°. Given that EAD is an isosceles triangle, angle EAD = (180° - 20°) / 2 = 80°.

x²+4x−12=? (x−4)(x−12) (x−4)(x+8) (x−2)(x+6) (x−6)(x+2) (x−4)(x+3)

C Correct. Remember the method for factoring quadratics: Factoring a quadratic of the form ax2+bx+c=0: 1) Write in factorized form (x+e)(x+f)=0. Each of the pairs of parentheses is called a factor of the equation. 2) Find e and f: a pair of integers whose product is c and whose sum is b. 3) Write e and f in the appropriate places in the factors. e.g. (x+5)(x+3)=0 4) Set each factor to zero and solve. The quadratic x2+4x−12 is already of the form of ax2+bx+c, where a=1, b=4 and c=-12. You're looking for e and f: a pair of integers whose product is c=-12 and whose sum is b=4. Start with the product -possible pairs of numbers include (1, -12), (3, -4), (2, -6) and of course their opposite sign counterparts, e.g (-1, 12). Out of these pairs, only the sum of the pair (-2, 6) is equal to b=4, so e and f are -2 and 6. Write them in the factored form of the equation: (x+e)(x+f) = (x-2)(x+6). Since this expression appears in answer choice C, this is the right answer.

What is the perimeter of a regular polygon with sides of length 12 cm and internal angles measuring 144° each? 96 cm 108 cm 120 cm 132 cm 144 cm

C Correct. Remember the perimeter of a figure is the sum of all sides enclosing the figure. The sum of all interior or internal angles of a polygon with n sides is 180(n - 2). Begin with assuming that the polygon has n sides. Given that each internal angle is 144 and the sum of all internal angles is 180(n - 2) i.e.180(n - 2) = 144n. Solving further 180n - 360 = 144n or 180n - 144n = 360 or 36n = 360 or n = 10. Since the number of sides is 10 and the length of one side is 12 cm, the perimeter of the polygon is 12 x 10 = 120 cm.

Q5...In the figure above EB||DC, the height of point A in ΔABF is 2, and the height of point A in ΔACD is 10. What is the height of point D in ΔEDF? 12 10 8 4 2

C Correct. The height of a triangle is the distance between any vertex and the opposing side. The height is by definition perpendicular to the opposing side. Since EB||DC, the height of ΔEDF is equal to the difference between the heights of ΔACD and ΔABF. Given that the height of ΔABF is 2, and the height of ΔACD is 10, the height of ΔEDF = 10 - 2 = 8.

Q10...If ∠C=90°, what is DB in the figure above? √3−√2 1 √8−√3 √2 √5

C Correct. When a geometry problem introduces a right triangle, check for recycled right triangles first. If there aren't any use the Pythagorean Theorem. Since there is no recycled triangle available, use the Pythagorean Theorem: In ΔABC, BC = √(AB2-AC2) = √(32-12) = √(9-1) = √8 In ΔADC, CD = √(AD2-AC2) = √(22-12) = √(4-1) = √3 DB = BC - CD = √8 - √3

Q28...What is x in the figure above? 24 25 30 32 48

C For any right triangle in the GMAT check for a recycled right triangle first. If there isn't any, use the Pythagorean theorem. Correct. Always check for recycled right triangles first. Divide the sides by 6 to get 72:6=12, 78:6=13. This triangle must be the recycled 5:12:13, Hence, the third side must be 5×6=30. Two parts of a recycled triangle always reveal the third.

Q26...If the perimeter of ΔACD is 9+3√3, what is the perimeter of equilateral triangle ΔABC? 9 18−3√3 18 18+3√3 27

C Good work! Use the 30°:60°:90° special triangle to solve this problem. Split equilateral triangle ΔABC into two 30°:60°:90° triangles. Focus on ΔACD: the sides of ΔACD match the 30°:60°:90° ratio of a : a√3 : 2a. Therefore: a+a√3+2a = 9+3√3. --> 3a+a√3 = 9+3√3 Extract a common factor on both sides: ---> a(3+√3) = 3(3+√3) And divide both sides by (3+√3): --> a=3 Therefore, ΔACD triangles has sides 3, 6 and 3√3. Since the hypotenuse (6) of the ΔACD is the side of the bigger equilateral triangle, each side of the equilateral triangle is 6. Hence, the perimeter is 3·6=18.

The difference between the roots of x2−x−1=0 is which of the following? 0 1 √2 √5 5

C Incorrect. The issue of this question is the difference between the roots of a quadratic equation. Factoring this equation is difficult, so use the quadratic equation roots formula to find the roots of x2−x−1=0. First, define the coefficients a, b and c of the quadratic of the form ax2+bx+c=0: a=1; b=-1; c=-1. Plug these into the solutions formula: -b±√b²-4ac / 2a Now, calculate the difference between x1 and x2 carefully. D Correct.

Q20...If CD is a bisector in triangle ABC, what is ∠ABC? 25° 30° 35° 40° 45°

C Remember a bisector divides an angle into two equal parts. Correct. Given that CD is a bisector, angle ACD = angle BCD. Since angle CDA is the supplementary angle of angle CDB, CDA = 180 - 95 = 85. Based on the fact that sum of all angles in a triangle is 180, angle ACD = 180 - 85 - 45 = 180 - 130 = 50. As mentioned in the begining angle ACD = angle BCD = 50. Using the 180 degree triangle rule, angle ABC = 180 - 50 - 50 - 45 = 35. Hence, angle ABC = 35.

Q16...If AB=20 and BC=25, what is the length of AD in the figure above? 6 9.6 12 20 24

C You grossly underestimated the time this question took you. You actually solved it in 4 minutes and 10 seconds. Nice job! The two sides of ABC are AB=20=4·5, BC=25=5·5. Use the recycled ratio 3:4:5 expanded times 5 to find the third side AC=3·5=15. Notice that AD is one of the heights of the triangle, which is part of area of a triangle formula {AREA= ½×BASE×HEIGHT}. To find AD, first find the area with the formula AREA= ½×BASE×HEIGHT. Use AC as the base and AB as the height. Area=½×15×20=150. To find AD, use the same formula with BC as the base and AD as the height: 150=½×25×AD. --> 300=25×AD --> AD=12

Q31...If the height of an equilateral triangle ABC is equal in length to the hypotenuse of an isosceles right triangle DEF, a side of ΔABC is how many times as long as a leg of the right triangle DEF? (√2)/(√3) (√3)/(√2) √2 (2√2)/(√3) 2/(√3)

D Correct. Look at the figure. If the side AC is 2, then AG, the height of ΔABC = √3 (according to the 30:60:90 recycled ratio of 1:√3:2, if the hypotenuse is 2, the longer leg must be √3). According to the question, AG is equal to the hypotenuse in DEF, so DF = √3. DEF is a right isosceles triangle with a ratio of 1:1: √2. Thus, in order to transition from the hypotenuse to one of the legs, we need to divide the hypotenuse by √2, and the legs of ΔDEF are therefore √3/√2. The ratio between AC and DE is therefore 2/(√3/√2)=(2√2)/(√3)

Five congruent equilateral triangles are combined to form the shape shown above. If the perimeter of this shape is 210, then what is the perimeter of one such triangle? 30 35 42 90 105

D Correct. Remember that congruent triangles have equal angles and equal sides. The perimeter is the sum of all sides. Given that the perimeter of the shape shown is 210, the length of one side of any of the equilateral triangles is 30 (perimeter / no. of sides = 210 / 7). Hence the perimeter of one triangle is 90 (30 x 3 = 90)

Which of the following equations has two different roots? x2−x+4=0 x2−4x+4=0 x2−3x+9=0 x2−5x+5=0 x2+3x+10=0

D Correct. For a quadratic equation to have two solutions, its discriminant (b2-4ac) has to be positive. Find the discriminant and see for which equation it is positive. --> b2-4ac=(-5)2-4×1×5=25-20=5>0

(2x)³·(3/2)³=? 3³x 3³·x³/2³ 3³·x³·2³ 27x³ 8x

D Correct. Remember that if two or more bases have the same exponent, they can be combined under the same exponent. Combine (2x)3·(3/2)3 under exponent 3. (2x)3·(3/2)3 = --> (2x·3/2)3 = --> (3·x)3 = --> 33·x3 = --> 27x3 Hence, this is the correct answer.

Q18...In the figure below, what is the value of x? 20° 40° 60° 70° 80°

D Correct. Remember that the sum of the angles inside any triangle is 180°. Also, angles on a straight line sum up to 180°. Remember an equilateral triangle has all sides and angles equal. Since AB = AC = BC, angle ABC = angle BCA = angle CAB = (180° / 3) = 60°. Based on this, angle ACD = 180° - 60° - 40° = 80°. Given that angle CAB = 60°, angle CAD = 90° - 60° = 30°. Hence, x = 180° - 80° - 30° = 70°.

Q3...In the figure below, what is the value of x? 10° 20° 80° 140° 160°

D Correct. Remember that the sum of the angles inside any triangle is 180°. Also, angles on a straight line sum up to 180°. Remember that in a triangle angles opposite equal sides are also equal. Since angles DCB and DCE are supplementary angles, angle DCB = 180° - 170° = 10°. Based on this, angle BDC = 180° - 2(10°) = 160°. Given that angles ADB and CDB are also supplementary angles, angle ADB = 180° - 160° = 20°. Hence x = 180° - 2(20°) = 140°.

Q29...The figure above is a square divided into four identical squares. If the perimeter of the larger square is 1, then what is the perimeter of one such smaller square? 1/16 1/6 1/4 1/2 3/4

D Correct. Remember the perimeter of a figure is the sum of all sides enclosing the figure. If the perimeter of the larger square is 1, the length of each large side is 1/4 (perimeter / # of sides). Based on the figure, each side of the small squares is equal to half the length of a side of the larger square. Hence, the side of the smaller square is 1/8 [(1/4) / 2], and the perimeter of the smaller square is {1/2 [(1/8)+(1/8)+(1/8)+(1/8)] = 4(1/8) = 1/2}. Alternatively, you can "open up" on of the smaller squares, to see that its sides are in fact one-half of the larger square, hence 0.5.

Q21...In the figure above ED||AC. If the height of ΔAEC from AC is 2, what is the height of ΔCDB from AC? 0.25 0.5 1 2 4

D Correct. The height of a triangle is the distance between any vertex and the opposing side. The height is by definition perpendicular to the opposing side. Given that ED||AC and the height of ΔAEC from AC is 2, the height of ΔCDB from AC is also 2.

If x > 0 > y, and (x−10)/y > 0, then which of the following must be true? x/y > 10 x−10 > y x−10 < y x < 10 x > 10

D Correct. The issue of this question is manipulating an an inequality to make it match one of the answer choices. Start with the inequality in the question and simplify it according to the rules of inequalities. Try to get rid of the denominator, i.e., to multiply both sides of the inequality by y. Do not forget that y<0 - multiplying an inequality by a negative quantity requires flipping the direction of the sign. x-10/y >0 Multiply both sides by y. Since y is negative, flip the inequality sign - --> x-10<0 --> x<10

Q27...If the height of an equilateral triangle is the side of a right triangle, as shown above, what is the side of the equilateral triangle? 2/√3 √3 2 6/√3 3√6/2

D Correct. The triangle on the right is a special 3-4-5 right triangle. Therefore the height of the equilateral triangle, which is also the height of the right triangle, is 3. The height of the equilateral triangle splits it into two equal 30°:60°:90° triangles. The height of the equilateral triangle is also the medium side of the recycled right triangle. Use the a:a√3:2a recycled ratio to work out the sides of the 30:60:90 triangle: 3/√3, and 6/√3. The side of the equilateral triangle is opposite the 90° angle and therefore measures 6/√3.

In triangle ABC, the lengths of line segments AB, BC and AC are x, y, and z, respectively. If y=3.5, is x=7? (1) z ≥ 10.5 (2) z = 10.5

D Correct. This is a Yes/No DS question. Answering a definite "Yes" or a definite "No" means Sufficient. If the answer is sometimes "Yes" and sometimes "No", it means Maybe which means Insufficient. Since there's no indication that triangle ABC is a right triangle, it is impossible to use the pythagorean theorem to find the third side from the lengths of the two sides. Then again, since this is a Yes/No DS question, you may not have to. Remember the third side rule: Any side in a triangle must be smaller than the sum of the other sides, and greater than their difference. Stat. (1): Take the extreme case of z=10.5. According to the third side rule, x is smaller than 10.5+3.5 = 14 and greater than 10.5 - 3.5 = 7. If x>7, it follows that it cannot equal 7. Any z greater than 10.5 will just intensify the problem: For example, if z=11.5 then x must be greater than 11.5-3.5 = 8, and will definitely not equal 7. That's a definite "No", which is "Sufficient". Stat.(1)->S->AD. Stat. (2): According to the third side rule, x is smaller than 10.5+3.5 = 14 and greater than 10.5 - 3.5 = 7. If x>7, it follows that it cannot equal 7. That's a definite "No", so Stat.(2)->S->D.

Q33...In the figure above, D is the center of the circle, and DE||AB. If the radius of the circle is 4, what is the length of line segment AB? 4√2 6 4+3√2 8 6√2

D Correct. Triangles ABC and CDE are similiar triangles of the "triangle within a triangle" form : two triangles whose bases are parallel. Therefore, their sides maintain the same linear ratio. Use this fact to determine the ratio between the given side DE and the required length AB. Consider AC and DC, respective sides of the two similiar triangles: since AC, the circle's diameter, is twice DC, the radius, the linear ratio between the triangles is 1:2, respectively. Therefore, AB is twice DE (another radius)=2*4=8.

(√7-4)(√7+4)= 33 3 −3 −9 −81

D Correct. Use recycled quadratic III: (a+b)(a-b) = a2-b2 Where a = √7 and b=4. (√7-4)(√7+4)=(√7)²-4²=7-16=-9

Q7...What is the perimeter of the shape formed by attaching square ABDE and equilateral triangle BCD, as shown above, if the side length of the equilateral triangle is 3? 5 6 12 15 18

D Excellent! An equilateral triangle has sides of equal length. A square, by definition, also has sides of equal length. Since triangle BCD has 1 side in common with square ABDE, all sides in this figure are equal. Since perimeter is the sum of all external sides of a given figure, 3+3+3+3+3=15.

Q15...What is x in the figure above? 4 8 10 12 13

D For any right triangle in the GMAT check for a recycled right triangle first. If there isn't any, use the Pythagorean theorem. Correct. Always check for recycled right triangles first. Divide the sides by 3 to get 15:3=5, 9:3=3. This triangle must be the recycled 3:4:5, Hence, the third side must be 4×3=12. Two parts of a recycled triangle always reveal the third.

Q13...What is x in the figure above? 17 51 64 68 72

D For any right triangle in the GMAT check for a recycled right triangle first. If there isn't any, use the Pythagorean theorem. Correct. Always check for recycled right triangles first. Divide the sides by 4 to get 32:4=8, 60:4=15. This triangle must be the recycled 8:15:17, Hence, the third side must be 17×4=68. Two parts of a recycled triangle always reveal the third.

Q6... If AB=AC, BD=BC and BD is the bisector of ∠B, then what is x? 27 30 32 36 42

D Remember that a bisector divides an angles into two equal parts. Correct. If BD is the bisector of angle B, then B is divided into two equal angles of x, and angle B is a total of 2x. Since AB=AC, the two bottom angles are equal: If angle B is 2x, then angle C is also 2x. Since BD and BC are also equal, the two base angles (BCD and BDC) in this small isosceles triangle are also equal. If angle C is 2x, then angle BDC is also 2x. Thus, if we sum the angles in the small triangle (DBC + C + BDC) we get x+2x+2x=5x. Since the sum of the angles in any triangle is 180, we get 5x=180. Hence, x = 36.

Q1...If in the figure above, l1 and l2 are parallel, what is x? 14° 28° 38° 66° 76°

E Correct. Extend l2 backwards and draw a line to connect the line making the 48° angle with l2. The resulting triangle is shown in the figure. Given that l1 and l2 are parallel lines, the alternate angle of 48° is also 48°. Since the sum of all angles in a triangle is 180, the supplementary angle of x is 104 (180 - 48 - 28). Hence, x=180-104=76°.

ABCD is a square of perimeter 8. If E is the midpoint of AD, what is the altitude of E from CD? 8 6 4 2 1

E Correct. Remember that perimeter of a shape is the sum of all sides. Given that ABCD is a square with perimeter 8, each side of ABCD is 2 (8/4 = 2). Since E is the midpoint of AD, the perpendicular distance (altitude) of E from CD is 1 ( 2/2 = 1).

Q4...In the figure below, what is the value of x? 45° 50° 65° 70° 85°

E Correct. Remember that the sum of the angles inside any triangle is 180°. Also, angles on a straight line sum up to 180°. Remember that in a triangle angles opposite equal sides are also equal. Since some of all angles in a triangle is 180°, acute angle DCE = 180° - 90° - 40° = 50°. Also, angle ACB is 45° [(180 - 90)/2]. Based on this, 45° + x + 50° = 180° i.e. x = 85°.

If the area of an equilateral triangle, in square inches, is equal to its perimeter, in inches, then, which of the following is the triangle's side, in inches? (√3)/4 √3 2√3 6 4√3

E Correct. Save time by using the unique area formula for an equilateral triangle of side a: The perimeter of an equilateral triangle is 3 times its side. Therefore: (a2√3) / 4 = 3a (Divide both sides by a.) ---> (a√3) / 4 = 3 (Multiply both sides by 4 to get rid of the fraction.) ---> (a√3) = 12 (Divide both sides by √3.) ---> a = 12 / √3 (Rationalize the denominator: multiply by √3 / √3.) ---> a = 4√3

Practical Implications of similarity: Similar triangles are basically copies of the same triangle on a different scale. If two triangles are similar, their respective sides maintain a constant ratio.

For example, if the base of one triangle in a similar duo is three times the base of the triangle similar to it, then all other sides of the bigger triangle will also be three times the size of the respective counterpart in the smaller triangle. The large triangle is effectively the same triangle as the small triangle, only enlarged 3 times.

The area of an equilateral triangle is given in the triangle area formula AREA= ½×BASE×HEIGHT.

However, some GMAT problems require you to solve for the area of an equilateral triangle with as little amount of data as length of its side. You can certainly use the 30:60:90 special triangle ratio box (If this doesn't mean anything to you now - don't worry, it will) to calculate the height of the triangle. However, for the specific case of an equilateral triangle we have a quicker way. The area of an equilateral triangle is given in the formula Area= a²√3 / 4 Memorize this formula to quickly get rid of a GMAT problem involving the the area of an equilateral triangle. To sum up: Save time by using the unique area formula for an equilateral triangle of side a: .Area= a²√3 / 4 You can also use AREA= ½×BASE×HEIGHT, which works for any triangle.

Similar triangles are considered an advanced subject for GMAT geometry. Many questions involving similar triangles can also be solved without knowledge of the subject, but a working knowledge of the principles involved can sometimes quickly resolve a question.

Identifying similar triangles: Triangles are similar if all their angles are equal to each other. The figure below presents two similar triangles, as their angles satisfy the above condition.

This concept can be expressed in the following equation: d/a=e/b=f/c Thus, if b=2 and e=6, then the similar triangles in the figure below maintain a ratio of 6/2=3 to 1 between their respective sides. Therefore, if a=4, then d must equal 4·3=12. d/d=d/4=6/2=3

Important note: Only compare sides opposite the same angle! In the figure above, sides a and d can be compared because they are both opposite the 50° angle in their triangle. The ratio between a and d will equal the ratio between b and its respective counterpart e (b must be compared to e, as both of them are opposite the 30° in the respective triangles). The similar triangle data can be organized in a slightly modified Ratio Box, with the two triangles taking up the first and third columns, and the constant ratio as the multiplier: .....................Small ∆...Multiplier....Large ∆ Side opp 30°.....2............. ×3.............6 Side opp 50°.....4............. ×3.............12 Side opp 100°...................................... Total....................................................

An obtuse triangle has one obtuse angle. Since the angles in a triangle always sum up to 180°, there can be only one obtuse angle in a triangle.

In an obtuse triangle the height may be found outside the triangle, so that it is perpendicular to the extension of the side.

In any right triangle the relationship of the sides is given in the Pythagorean Theorem: if the legs of a right triangle measure a and b, and the hypotenuse measures c, then a2+b2=c2.

The Pythagorean Theorem applies only to right triangles. Use it to find the third side whenever you are given any two sides of a right triangle.

The height or altitude, in a triangle is the distance between any vertex and the opposing side.

The height or altitude is by definition perpendicular to the opposing side. To create a height or altitude, draw a straight line from a vertex to the opposing side. Make sure the line and the opposing side are perpendicular. In any triangle there are three heights.

GMAC recycles. Not for the sake of global environmentalism, but rather as a means to save time and money. Recycled questions abound in the GMAT, and you can benefit from it.

The subject that is most recycled by GMAC is right triangles. Memorize these simple recycled right triangles. GMAC uses these triangles or their multiples (i.e., GMAC may use the 3:4:5 or 6:8:10 or 9:12:15 etc.) To spot a simple recycled right triangle in a question you need two sides of the triangle to find the third. However, pay attention to the placing of the numbers, as they stand for the ratios of the sides. The largest number represents the hypotenuse. (In the recycled right triangle 6:8:10 the number 10 represents the hypotenuse) Whenever you see a right triangle check for a recycled ratio first. If the triangle does not fit one of the recycled patterns, use the Pythagorean Theorem.

A bisector is a line that cuts one of the angles in half. To create a bisector, draw a line from a vertex to the opposing side, such that the line halves the angle.

There are three bisectors in a triangle, bisecting each of the three angles.

The word median is derived from middle. It points to the midpoint of a side. To create a median in a triangle, draw a line from a vertex to the midpoint of the opposing side.

There are three medians in a triangle, connecting the mid-point of each side to the opposite vertex.

Let's look at the third side rule: in a triangle, any side has to be smaller than the sum of the other sides. In this case, the rule fails: 8>3+4, therefore this triangle cannot exist. The third side must be smaller than 7.

To put this in algebraic form: in a triangle with sides a, b, and c, the following inequalities must hold: a < b+c b < a+c c < a+b Yes, we know, algebra *shiver*. But notice an interesting thing here: By subtracting c from both sides of the second inequality, we are left with b-c < a Yeah? So? This effectively leaves us with a definite range for side a of a triangle: b-c < a < b+c This bit of inequality manipulation can be applied to all three inequalities, leaving us with an additional important aspect of the third side rule: Any side in a triangle must be smaller than the sum of the other sides, and greater than the difference of the other sides. ++++++++++Important note+++++++++++: the third side rule is the obscure rule that students tend to forget. When you work with any triangle, whether it's in a figure provided by the GMAT problem or one you draw yourself, always check if the triangle is "possible".

An isosceles triangle has two equal sides, and consequently two equal angles opposite the equal sides.

You need only one angle of an isosceles triangle to figure out all of its angles. For example, if the top angle is 40°, then the other two equal angles can be determined: each of them will equal half of (180-40)=140, or 70°. In an isosceles triangle the height is also the median and the bisector.