GMAT-Geometry-1. Lines, Polygons, Triangles, Quadrilaterals and Circles

Transcription

1 GMAT-Geometry-1 Lines, Polygons, Triangles, Quadrilaterals and Circles Lines: 1. In the figure given, if a = b = c = d, which of the following must be true? I. x = w II. x = z III. y = w A. I only B. II only C. I and II D. I, II and III E. None Answer is E. Because these lines are not parallel. We know that x= y and z=w. But y is not necessarily be equal to z, because they are not parallel. So the answer is E. 2. In the figure given, line 1 is parallel to line 2. If the measure of x is less than 90 0, is the length of line segment PQ equal to line segment RS? I. x + y = II. x < y 1

2 Statement I is sufficient: x+y = 180 degrees. And x < 90 degrees. Y = 180-x is greater than 90 degrees. Suppose if x = 60, y becomes 120(angle SRQ is 60 degree, because y+ anglesrq = 180 degrees). Since they are parallel. PQ =SR. So it is sufficient: Statement II is insufficient: x<y. Suppose if x = 60, y becomes 120(angle SRQ is 60 degree, because y+ anglesrq = 180 degrees). Since they are parallel. PQ =SR. Suppose if x = 60, and y is equal 70(angle SRQ is 110 degree, because y+ anglesrq = 180 degrees). Not PQ is not equal to SR. So not sufficient. So the answer is A. 3. In the figure, lines l and k are parallel. If a is an acute angle, then which one of the following must be true? A. b> 10 B. b> 15 C. b< 20 D. b< 30 E. b> 45 2

3 Since l and k are parallel, if we extent a line from point O, the diagram looks like this, We know that (b+b a) = 180 degrees. 2b+30-a = 0. Because a < 90 degrees(a is an acute angle) 2b <0. So b < 30 So the answer is D. Polygons: 1. The measures of the interior angles in a polygon are consecutive integers. The smallest angle measures 136 degrees. How many sides does this polygon have? A. 8 B. 9 C. 10 D. 11 E. 13 Sum of interior angles for n sides: (n-2)*180 And each interior angle are increasing by 1. This can be written as : = 136+(136+1)+(136+2)+(136+3)+... 3

4 For n sides 136n+( n-1) 136n+(n-1)n/2 --- Sum of n-1 natural number Equate: 136n+(n-1)n/2 = (n-2)*180 n^2-89n+720=0 n^2-80n-9n+720=0 n(n-80)-9(n-80)=0 (n-9)(n-80)=0 n=9 and n=80 Since 9 is one of the options. So the answer is "B". Triangles: 1. If the three interior angles of a triangle are 2y, 15x, and 9x, what is the value of y? I. x = 3 II. x = (270 3y)/36 Sum of the interior angles of a triangle is 180 degrees. So 2y+15x+9x = 180 degrees. Then 2y+24x = 180 degrees. Statement I is sufficient: x=3, so clearly we can find the value of y. So sufficient. Statement II is insufficient: 36x = 270 3y. 4

5 Which is similar to the equation given in the question (2y+24x=180 degrees). So not sufficient. So the answer is A. 2. In the figure shown, the measure of angle ADC is how many degrees greater than the measure of angle ABC? I. The measure of angle BAD is 30 degrees. II. The sum of angles ABD and ADB is 150 degrees. Let's call ADC x. Let's call BAD z so ABC 'y'. Where x = y+z. In a triangle external angle is equal to sum of the internal opposite in internal angles. We are asked to find ADC - ABC = x - (y) = y+z-y = z. Statement 1 is sufficient: BAD is 30, therefore z=30. So it is Sufficient Statement 2 is sufficient: ABD + ADB = y x = 150. Therefore x-y = 30. Which is z = 30. So this is sufficient. So the Answer is D. 3. If AD is 6 3, and ADC is a right angle, what is the area of triangular region ABC? I. Angle ABD = 60 5

6 II. AC = 12 Given: AD=6 3. Then area of ABC=12 AD BC= (BD+DC)=3 3 (BD+DC)=? Statement I is insufficient: Angle ABD = 60 --> triangle ABD is triangle, so the sides are in ratio 1:3 :2 --> as AD=6 3 (larger leg opposite 60 degrees angle) then BD=6 (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient. Statement II is insufficient: AC=12, we can find DC. Using again DC length will be 6. But we still don't know BD. Not sufficient. (1)+(2) We know both BD and DC, hence we can find area. So it is Sufficient. So the Answer is C. 4. If x q = s y, what is the value of z? I. xq + sy + sx + yq = zr II. zq ry = rx zs x-q = s-y Thus x+y = q + s 6

7 Therefore z = r. Statement I is sufficient: x(q+s)+y(q+s) = zr. We get (x+y)(q+s)=z*r Putting the above 2 equations from the question in this we get (x+y)^2= z^2 x+y=z 180-z=z z=90 Thus it is sufficient. Statement II is insufficient: zq ry = rx zs Simplifying we get. z(q+s)=r(x+y). This is no additional information also. So not sufficient. So the answer is A. Quadrilaterals: 1. KLNM represents a lawn area divided into 3 separate lawn areas. What is ratio of side KM to MN? I. Perimeter of KLMN is 30 II. All three lawn areas have the same dimensions Let plot points in the rectangular field like below. 7

8 We need to find the ratio of KM/MN? Statement I is insufficient: Perimeter of the figure is 30. Not sure about the lengths of KM and MN. It can be anything. So not sufficient. Statement II is sufficient: All three lawn areas have the same dimension. Let KO = x, then OL=PR=RQ=PM=QN = x. KP=OR=LQ=MN=y. 8

9 y =2x(because MN =KL). KM/MN=( KP+PM)/MN =(y+x)/ y =(2x+x)/2x = 3/2. So sufficient. So the answer is B. 2. Given that ABCD is a rectangle, is the area of triangle ABE > 25? I. AB = 6 II. AE = 10 Area=12 AB BE Statement I is insufficient: AB = 6 --> clearly insufficient: BE can be 1 or 100. Statement II is sufficient: AE = 10 --> now, we should know one important property: for a given length of the hypotenuse a right triangle has the largest area when it's isosceles, so for our case area of ABE will be maximized when AB=BE. So, let's try what is the largest area of a right isosceles triangle with hypotenuse equal to 10. Finding legs: x^2+x^2=10^2 (where x=ab=be) --> x= 50--> area max=1/2( 50)^2=25. Since it's the maximum area of ABE then the actual area cannot be more than 25. Sufficient. 9

10 So the answer is B. Circles: 1. If angle ABC is 40 degrees, and area of circle is 81 π, how long is arc AXC? A. 2π B. 3π C. 4π D. 9π E. 18π Area of circle πr^2=81π r^2=81 r=9 The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Thus; Arc AXC will subtend an angle of θ=2 40=80 at the center of the circle. Length of arc AXC AXC (arc length) = θ/360 2πr= 80/360 2π 9=4π 2. In the figure given, if PR is a line segment, what is the sum of the lengths of the curved paths from P to Q and from Q to R? (1) XQ = QY = 5 centimeters. 10

11 (2) Every point on arc PQ is 5 centimeters from point X, and every point on arc QR is 5 centimeters from point Y. Given is a figure that looks like two semi circles of equal areas with center X and center Y respectively. However we cannot assume any of this as we are not told this in the question. Statement I is insufficient: XQ=QY=5. This would have been sufficient if we were given X and Y are the center of the circles. If they were centers then XQ and Qy were radius of the circle and we could have calculated the length of the arc. However as we do not know if they are centers they could be any arbitrary points. Hence not sufficient. Statement II is sufficient: The meaning of the statement is that X and Y are the centres of the 2 semi circles each with radius of 5. We can calculate length of arc by 2* Pie* Radius. Hence it is sufficient. So the Answer is B. 3. If line segments AB, BC and CD are equal in length in the figure shown and A is the center of the circle, what is the value of x? A. 15 B. 30 C. 45 D

12 E. 75 Given AB=BC=CD, since AC is also radius, then AB=BC=CD=AC. Now ABC is an equilateral triangle so each angle is 60 degrees. ABCD is a rhombus, so diagonals intersect at 90 degrees in rhombus., given below in the diagram. So x is 30. So the answer is B. 4. In the circle given, PQ is parallel to diameter OR, and OR has length 18. What is the length of the minor arc PQ? A. 2π B. 9π/4 C. 7π/2 D. 9π/2 E. 3π 12

13 Let x is the centre of the circle. To find the length minor arc PQ. We need to find the central angle y. Since PQ and OR are parallel. Angle P + angle Q =180 degrees. Triangle OPR is right angle triangle (because it is inscribed in the semicircle-by the property). So the angle POR is 55 degrees. Then angle OPQ is 125 degrees (they are parallel.).since XO and XP are radius of the circle angle OPX is 55 degrees. So XPQ is 70 degree. Since XQP is 70 degrees. So the central angle y is 40. Arc length = ( central angle / 360)* 2(pi)r. = (40/360)*2(Pi)9. = 2 pi. So the answer is A. 13

14 5. A machine has two flat circular plates of the same diameter. Both plates have circular holes of one-inch diameter that are equally spaced and are the same distance from the edges, as shown in the figure above. One plate is placed on top of the other so that their centers are aligned and two of the holes are perfectly aligned. If one plate remains stationary, what is the least number of degrees that the other plate must be rotated so that a different pair of holes is perfectly aligned? A. 6 0 B C D E The plate which has 4 holes has holes which are 90 degrees away from each other. The plate which has 5 holes has holes which are 360/5 = 72 degrees away from each other. Assume that the holes at the top are aligned. This is how the rest of the holes are placed on the two plates: 5 hole plate - 72 degrees, 144 degrees, 216 degrees, 288 degrees 4 hole plate - 90 degrees, 180 degrees, 270 degrees To align the first holes with each other, we can move the 5 hole plate clockwise 18 degrees. Or the second hole of 5 hole plate 36 degrees clockwise. The minimum movement required to align any two holes is 18 degrees (the difference between any two angles of the two plates) 14

15 So the Answer is C. Overlapping Figures 1. A rectangle is inscribed in a circle of radius r. If the rectangle is not a square, which of the following could be the perimeter of the rectangle? A. 2r 3 B. 2r ( 3+ 1) C. 4r 2 D. 4r 3 E. 4r( 3+ 1) A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Since a rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle. Now, since each option has 3 in it, then let's check a right triangle where the angles are 30, 60, and 90 (standard triangle for the GMAT), because its sides are always in the ratio 1: 3:2. Notice that the smallest side (1) is opposite the smallest angle (30 ), and the longest side (2) is opposite the largest angle (90 ). So, in this case we would have: 15

16 The perimeter of the rectangle is 2r 3+2r=2r( 3+1). So the answer is B. 2. A circle is inscribed in equilateral triangle ABC such that point D lies on the circle and on line segment AC, point E lies on the circle and on line segment AB and point F lies on the circle and on line segment BC. If line segment AB=6, what is the area of the figure created by line segments AD, AE and minor arc DE? A. 3 3-(9/4)π B. 3 3-π C. 6 3-π D π E π Above is the diagram for the question given. Always we need to draw the diagram. We can see that radius of the circle is 3 (because of ). 16

17 Then the area of AD, AE and minor DE = (Area of triangle (ABC) Area of Circle)/3 = (½*6*3 3 - (π 3) ^2)/3 = (9 3-3 π)/3 = 3 3- π. So the answer is B. 17