Geometry Geometry is one of the most important topics of Quantitative Aptitude section. Lines and Angles Sum of the angles in a straight line is 180 Vertically opposite angles are always equal. If any point is equidistant from the endpoints of a segment, then it must lie on the perpendicular bisector. When two parallel lines are intersected by a transversal, corresponding angles are equal, alternate angles are equal and co-interior angles are supplementary. (All acute angles formed are equal to each other and all obtuse angles are equal to each other) Fact The ratio of intercepts formed by a transversal intersecting three parallel lines is equal to the ratio of corresponding intercepts formed by any other transversal. = = Triangles Sum of interior angles of a triangle is 180 and sum of exterior angles is 360. Exterior Angle = Sum of remote interior angles.
Sum of two sides is always greater than the third side and the difference of two sides is always lesser than the third side. Side opposite to the biggest angle is longest and the side opposite to the smallest angle is the shortest. = ½ x Base x Height = ½ x Product of sides x Sine of included angle = ( )( )( ) here s is the semi perimeter [s = (a+b+c)/ ] = r x s [r is radius of incircle] = 4 [R is radius of circumcircle] Median A Median of a triangle is a line segment joining a vertex to the midpoint of the opposing side. The three medians intersect in a single point, called the Centroid of the triangle. Centroid divides the median in the ratio of :1 Altitude An Altitude of a triangle is a straight line through a vertex and perpendicular to the opposite side or an extension of the opposite side. The three altitudes intersect in a single point, alled the Orthocenter of the triangle. Perpendicular Bisector A Perpendicular Bisector is a line that forms a right angle with one of the triangle's sides and intersects that side at its midpoint. The three perpendicular bisectors intersect in a single point, called the Circumcenter of the triangle. It is the center of the circumcircle which passes through all the vertices of the triangle. Angle Bisector An Angle Bisector is a line that divides the angle at one of the vertices in two equal parts. The three angle bisectors intersect in a single point, called the Incenter of the triangle. It is the center of the incircle which touches all sides of a triangle.
Theorems Mid Point Theorem: The line joining the midpoint of any two sides is parallel to the third side and is half the length of the third side. Apollonius Theorem: + = ( + ) Basic Proportionality Theorem: If DE BC, then AD/DB = AE/EC Interior Angle Bisector Theorem: AE/ED = BA/BD Special Triangles Right Angled Triangle: ABC ADB BDC = AD x DC and AB x BC = BD X DC
Equilateral Triangle: All angles are equal to 60. All sides are equal also. Isosceles Triangle: Angles equal to opposite sides are equal. Area = 4 4 30-60 -90 Triangle Area = 3 45-45 -90 Triangle Area =
30-30 -10 Triangle Area = 3 4 Similarity of Triangles Two triangles are similar if their corresponding angles are congruent and corresponding sides are in proportion. Tests of similarity: (AA / SSS / SAS) For similar triangles, if the sides are in the ratio of a:b Corresponding heights are in the ratio of a:b Corresponding medians are in the ratio of a:b Circumradii are in the ratio of a:b Inradii are in the ratio of a:b Perimeters are in the ratio of a:b Areas are in the ratio a : b Congruency of Triangles Two triangles are congruent if their corresponding sides and angles are congruent. Tests of congruence: (SSS / SAS / AAS / ASA) All ratios mentioned in similar triangle are now 1:1 Polygons Sum of interior angles = (n - ) x 180 = (n - 4) x 90 Sum of exterior angles = 360 Number of diagonals = n ( 3) C n = Number of triangles which can be formed by the vertices = n C 3
Regular Polygon : If all sides and all angles are equal, it is a regular polygon. All regular polygons can be inscribed in or circumscribed about a circle. Area = ½ Perimeter Inradius {Inradius is the perpendicular from centre to any side} Each Interior Angle = ( )180 ; Exterior = 360 Quadrilaterals : Sum of the interior angles = Sum of the exterior angles = 360 Area for a quadrilateral is given by 1 d 1 d Sin Cyclic Quadrilateral If all vertices of a quadrilateral lie on the circumference of a circle, it is known as a cyclic quadrilateral. Opposite angles are supplementary Area = ( )( )( )( ) where s is the semi perimeter s = + + + Parallelogram
Opposite sides are parallel and congruent. Opposite angles are congruent and consecutive angles are supplementary. Facts Diagonals of a parallelogram bisect each other. Perimeter = (Sum of adjacent sides); Area = Base x Height = AD x BE Each diagonal divides a parallelogram in two triangles of equal area. Sum of squares of diagonals = Sum of squares of four sides o + = + + + A Rectangle is formed by intersection of the four angle bisectors of a parallelogram. Rhombus A parallelogram with all sides equal is a Rhombus. Its diagonals bisect at 90. Perimeter = 4a; Area = 1 d 1 d
Area = d x ( ) Rectangle A parallelogram with all angles equal (90 ) is a Rectangle. Its diagonals are congruent. Perimeter = (l+b) Area = lb Square A parallelogram with sides equal and all angles equal is a square. Its diagonals are congruent and bisect at 90. Perimeter = 4a Area = Diagonals = a Fact: From all quadrilaterals with a given area, the square has the least perimeter. For all quadrilaterals with a given perimeter, the square has the greatest area. Kite Two pairs of adjacent sides are congruent. The longer diagonal bisects the shorter diagonal at 90. Area = Trapezium / Trapezoid
A quadrilateral with exactly one pair of sides parallel is known as a Trapezoid. The parallel sides are known as bases and the non-parallel sides are known as lateral sides. Area = 1 (Sum of parallel sides) Height Median, the line joining the midpoints of lateral sides, is half the sum of parallel sides. Fact Sum of the squares of the length of the diagonals = Sum of squares of lateral sides + Product of bases. + = + + AB CD Isosceles Trapezium The non-parallel sides (lateral sides) are equal in length. Angles made by each parallel side with the lateral sides are equal. Facts: If a trapezium is inscribed in a circle, it has to be an isosceles trapezium. If a circle can be inscribed in a trapezium, Sum of parallel sides = Sum of lateral sides. Hexagon (Regular) Perimeter = 6a; Area = 3 3
Sum of Interior angles = 70. Each Interior Angle = 10. Exterior = 60 Number of diagonals = 9 {3 big and 6 small} Length of big diagonals (3) = a Length of small diagonals (6) = 3 a Area of a Pentagon = 1.7 Area of an Octagon = ( + 1) Facts: A regular hexagon can be considered as a combination of six equilateral triangles. All regular polygons can be considered as a combination of n isosceles triangles. Circles Diameter = r; Circumference = πr; Area = π Chords equidistant from the centre of a circle are equal. A line from the centre, perpendicular to a chord, bisects the chord. Equal chords subtend equal angles at the centre. The diameter is the longest chord of a circle. A chord /arc subtends equal angle at any point on the circumference and double of that at the centre. Chords / Arcs of equal lengths subtend equal angles.
Chord AB divides the circle into two parts: Minor Arc AXB and Major Arc AYB Measure of arc AXB = AOB = Length (arc AXB) = 360 πr Area (sector OAXB) = 360 π Area of Minor Segment = Shaded Area in above figure Area of Sector OAXB - Area of OAB [ 360 - ] Properties of Tangents, Secants and Chords The radius and tangent are perpendicular to each other. There can only be two tangents from an external point, which are equal in length PA = PB
PA PB = PC PD = 1 [m(arc AC) m(arc BD)] PA PB = PC PD = 1 [m(arc AC) + m(arc BD)] Properties PA PB = = 1 [m(arc AC) - m(arc BC)] Alternate Segment Theorem
The angle made by the chord AB with the tangent at A (PQ) is equal to the angle that it subtends on the opposite side of the circumference. BAQ = ACB Common Tangents Two Circles No. of Common Tangents Distance Between Centers (d) One is completely inside 0 < r1 - r other Touch internally 1 = r1 - r Intersect r1 - r < d < r1 + r Touch externally 3 = r1 + r One is completely outside other 4 > r1 + r Length of the Direct Common Tangent (DCT) AD = BC = ( 1 ) Length of the Transverse Common Tangent (TCT) RT = SU = ( 1 + ) Example with Solution
Example 1: In following figure, CE is perpendicular to AB, ACE = 0 and ABD = 50. Find BDA: Solution: To find: angle BDA For this what we need, BAD Because, Sum of all angles = 180 Consider, ECA, CEA + EAC + ACE = 180 i.e. 90 + 0 + EAC = 180 Therefore, EAC = 70 Now, come to ABD, ABD + BDA + BAD = 180 70 + 50 + BAD = 180 Therefore, BAD = 60 Example : In given figure. BC is produced to D and BAC = 40 and ABC =70.Find ACD: Solution: In above figure, ACD is an exterior angle, and according to property, Exterior angle = Sum of interior angles Therefore, ACD = 70 + 40 =110 This is not the end of this chapter. These are just the basics. In next session, I will discuss some important results, properties (congruency, similarity) and much more. Always remember, Geometry needs practice and time. Exercise
1) If (5, 1), (x, 7) and (3, -1) are 3 consecutive verticles of a square then x is equal to : a) - 3 b) - 4 c) 5 c) 6 e) None of these ) What is the area of an obtuse angled triangle whose two sides are 8 and 1 and the angle included between two sides is 150? a) 4 sq units b) 48 sq units c) 4 3 d) 48 3 e) Such a triangle does not exist 3) What is the measure of the radius of the circle that circumscribes a triangles whose sides measure 9, 40 and 41? a) 6 b) 4 c) 4.5 d) 0.5 e) 1.5 4) Verticles of a quadrilateral ABCD are A (0, 0), B (4, 5), C (9, 9) and D (5, 4). What is the shape of the quadrilateral? a) Square b) Rectangle but not a square c) Rhombus d) Parallelogram but not a rhombus e) None of these 5) If the sum of the interior angles of a regular polygon measures upto 1440 degrees, how many sides does the polygon have? a) 10 sides b) 8 sides c) 1 sides d) 9 sides e) None of these 6) What is the radius of the in circle of the triangle whose sides measure 5, 1 and 13 units? a) units b) 1 units c) 6.5 units d) 6 units e) 7.5 units 7) How many diagonals does a 63 sided convex polygon have? a) 3780 b) 1890 c) 3843 d) 3906 e) 1953 8) If 10, 1 and x are sides of an acute angled triangle, how many integer values of x are possible? a) 7 b) 1 c) 9 d) 13 e) 11 9) Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 6 inches and 8 inches. a) 10 inches b) 11 inches c) 18 inches d) 0 inches e) None of these
10) Find the length of one side of a right triangle if the length of the hypotenuse is 15 inches and the length of the other side is 1 inches. a) 8 inches b) 7 inches c) 9 inches d) 13 inches e) None of these 11) Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both 3 inches. a) 5 inches b) 3 4 inches c) 6 inches d) 3 inches e) None of these 1) Find the lengths of the other two sides of a right triangle if the length of the hypotenuse is 4 inches and one of the angles is 45. a) 4 inches b) 9 inches c) 8 inches d) 7 inches e) None of these 13) Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 4 inches and 4 3 inches. a) 8 inches b) 9 inches c) 10 inches d) 11 inches e) None of these 14) Find the lengths of the other two sides of a right triangle if the length of the hypotenuse is 8 inches and one of the angles is 30. a) 4, 4 3 inches b) 5, 6 inches c), 4 d) 3, 4 inches e) None of these 15) What is the area of the following square, if the length of BD is? a) 1 b) c) 3 d) 4 e) 5
16) In the figure below, what is the value of y? a) 40 b) 50 c) 60 d) 100 e) 10 17) Two circles both of radii 6 have exactly one point in common. If A is a point on one circle and B is a point on the other circle, what is the maximum possible length for the line segment AB? a) 1 b) 15 c) 18 d) 0 e) 4 18) A right circular cylinder has a radius of 3 and a height of 5. Which of the following dimensions of a rectangular solid will have a volume closest to the cylinder? a) 4, 5, 5 b) 4, 5, 6 c) 5, 5, 5 d) 5, 5, 6 e) 5, 6, 6 19) Note: Figures not drawn to scale In the figures above, x = 60, How much more is the perimeter of triangle ABC compared with the triangle DEF. a) 0 b) c) 4 d) 6 e) 8 0) A right triangle has one other angle that is 35. What is the size of the third angle? a) 55 b) 65 c) 90 e) 180 e) None of these
1) An equilateral triangle has one side that measures 5 in. What is the size of the angle opposite that side? a) 55 b) 70 c) 110 d) 60 e) None of these ) An isosceles triangle has one angle of 96. What are the sizes of the other two angles? a) 4 b) 34 c) 4 d) 96 e) None of these 3) 4) 5) Find the circumference of the circle with a diameter of 8 inches? a) 5 inches b) 5.163 inches c) 9.45 inches d) 35.6 inches e) None of these Find the area of the circle with a diameter of 10 inches? a) 55.78 sq. inches b) 99.75 sq. inches c) 9 inches d) 78.55 sq. inches e) None of these Find the area of the circle with a radius of 10 inches? a) 314. sq. inches b) 115 inches c) 30.9 sq. inches d) 56.1 sq. inches e) None of these 6) The parallelogram shown in the figure below has a perimeter of 44 cm and an area of 64. Find angle T in degrees? a) 43.4 b) 44. c) 34.8 d) 48.1 e) None of these 7) Find the area of the quadrilateral shown in the figure. (note: Figure not drawn to scale) a) 169 b) 185 c) 199 d) 144 e) None of these
8) In the figure below triangle OAB has an area of 7 and triangle ODC has an area of 88. Find x and y. a) x = 0, y = 14 b) x = 14, y = 0 c) x = 41, y = d) x = 4, y = 1 e) None of these 9) Find the circumference of a circular disk whose area is 100π square centimeters? a) 40 π b) 10 π c) 0π d) 30π e) None of these 30) The semicircle of area 150 π centimeters is inscribed inside a rectangle. The diameter of the semicircle coincides with the length of the rectangle. Find the area of the rectangle? a) 4000 b) 5000 c) 3000 d) 000 e) None of these 31) If in a triangle ABC cos = cos = cos, then what can be said about the triangle? a) Right angled triangle b) Isosceles triangle c) Equilateral triangle d) Obtuse triangle e) None of these Solutions: 1. Option A For the verticles to form a square, we know that the length of each side of the square should be equal. Therefore, ( 5) + (7 1) = ( 3) + (7 + 1) [ + 5 - (x) (5) ] + [36] = [ + 3 - (x) (3) ] + [64] [5 + 36] - [9 + 64] = (10-6) x x = - 1 = - 3 4 This gives the side of the square x = - 3.. Option A
If two sides of a triangle and the included angle y is known, then the area of the triangle can be found using the formula 1 * (product of sides) * sin y Substituting the values in the formula, we get 1 * 8 * 1 * sin 150 = 4 sq units 3. Option D From the measure of the length of the sides of the triangle 9, 40 and 41 we can infer that the triangle is a right angled triangle. 9, 40, 41 is a Pythagorean triplet. In a right angled triangle, the radius of the circle that circumscribes the triangle is half the hypotenuse. In the given triangle, the hypotenuse = 41 Therefore, the radius of the circle that circumscribes the triangle = 41 = 0.5 units 4. Option C The lengths of the four sides AB, BC, CD and DA are all equal to 41 Hence, the given quadrilateral is either a Rhombus or a Square. Now let us compute the lengths of the two diagonals AC and BD. The length of AC is 16 and the length of BD is As the diagonals are not equal and the sides are equal, the given quadrilateral is a Rhombus. 5. Option A We know that the sum of an exterior angle and an interior angle of a polygon = 180 We also know that sum of all the exterior angles of a polygon = 360 The question states that the sum of all interior angles of the given polygon = 1440 Therefore, sum of all the interior and exterior angles of the polygon = 1440 + 360 = 1800 If there are n sides to this polygon, then the sum of all the exterior and interior angles = 180 n = 10 6. Option A The triangle given is a right angled triangle as its sides are 5, 1 and 13 which is one of the Pythagorean triplets. Note: In a right angled triangle, the radius of the incircle is given by the following relation h As the given triangle is a right angled triangle, radius of its incircle = 5 + 1 13 = 1 units
7. Option B The number of diagonals of an n-sided convex is This polygon has 63 sides. Hence, n = 63 Therefore, number of diagonals = 63 60 = 1890 ( 3) 8. Option C For any triangle sum of any two sides must be greater than the third side. The sides are 10, 1 and x. From Rule, x can take the following values : 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15, 16, 17, 18, 19, 0, 1 A total of 19 values. When x = 3 or x = 4 or x = 5 or x = 6, the triangle is an OBTUSE angled triangle. The smallest value of x that satisfies both conditions is 7. (10 + 7 > 1 ) The highest value of x that satisfies both conditions is 15. (10 + 1 + 15 ) When x = 16 or x = 17 or x = 18 or x = 19 or x = 0 or x = 1, the triangle is an OBTUSE angled triangle. Hence, the values of x that satisfy both the rules are x = 7, 8, 9, 10, 11, 1, 13, 14, 15. A total of 9 values. 9. Option A Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio. 6 : 8 :?= 3 () : 4 () :? Yes, it is a 3-4-5 triangle for n = Calculate the third side 5n = 5 = 10 The length of the hypotenuse is 10 inches. 10. Option C Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.? : 1 : 15 =? : 4 (3) : 5 (3) Yes, it is a 3-4-5 triangle for n = 3 Calculate the third side 3n = 3 3 = 9 The length of the side is 9 inches. 11. Option D This is a right triangle with two equal sides so it must be a 45-45 - 90 triangle. You are given that the both the sides are 3. If the first and second value of the ratio n : n : n is 3 then the length of the third side is 3 The length of the hypotenuse is 3 inches. 1. Option A
This is a right triangle with a 45 so it must be a 45-45 - 90 triangle. You are given that the hypotenuse is 4. If the third value of the ratio n : n : n is 4 then the lengths of the other two sides must 4. The lengths of the two sides are both 4 inches. 13. Option A Test the ratio of the lengths to see if it fits the n : n 3 : n ratio. 4 : 4 3 :? n : n 3 : n Yes, it is a 30-60 - 90 triangle for n = 4 Calculate the third side n = 4 = 8 The length of the hypotenuse is 8 inches. 14. Option A 15. Option D This is a right triangle with a 30 angle so it must be a 30-60 - 90 triangle. You are given that the hypotenuse is 8. Substituting 8 into the third value of the ratio n: n 3 : n, we get that n = 8 n = 4 Substituting n = 4 into the first and second value of the ratio we get that the other two sides are 4 and 4 3 The lengths of the two sides are 4 inches and 4 3 inches. 16. Option C We need to find the length of the side of the square in order to get the area. The diagonal BD makes two 45-45 - 90 triangles with the sides of the square. Using the 45-45 - 90 special triangle ratio n: n : n. If the hypotenuse is then the legs must be. So, the length of the side of the square is. Area of square = 5 = = 4
Vertical angles being equal allows us to fill in two angles in the triangle that y belongs to. Sum of angles in a triangle = 180 So, y + 40 + 80 = 180 y + 10 = 180 y = 60 17. Option E Sketch the two circles touching at one point. The furthest that A and B can be would be at the two ends as shown in the above diagram. If the radius is 6 then the diameter is 6 = 1 and the distance from A to B would be 1 = 4 18. Option E Write down formula for volume of cylinder V = πrh Plug in the values V = π 3 5 = 45 π V = 45 3.14 = 141.39 We now have to test the volume of each of the rectangular solids to find out which is the closest to 141.39 (A) 4 5 5 = 100 (B) 4 5 6 = 10 (C) 5 5 5 = 15 (D) 5 5 6 = 150 (E) 5 6 6 = 180 19. Option A Note: Figures not drawn to scale Since x = 60, triangle ABC is an equilateral triangle with sides all equal. The sides are all equal to 8. Perimeter of triangle ABC = 8 + 8 + 8 = 4 Triangle DEF has two angles equal, so it must be an isosceles triangle. The two equal sides will be opposite the equal angles.
So, the length of DF = length of DE = 10 Perimeter of triangle DEF = 10 + 10 + 4 = 4 Subtract the two perimeters. 4-4 = 0 0. Option A A right triangle has one angle = 90. Sum of known angles is 90 + 35 = 15 The sum of all the angles in any triangle is 180. Subtract sum of known angles from 180. 180 15 = 55 The size of the third angle is 55 1. Option D Since it is an equilateral triangle all its angles would be 60. The size of the angle does not depend on the length of the side. The size of the angle is 60.. Option C Since it is an isosceles triangle it will have two equal angles. The given 96 angle cannot be one of the equal pair because a triangle cannot have two obtuse angles. Let x be one of the two equal angles. The sum of all the angles in any triangle is 180. x + x + 96 = 180 x = 84 x = 4 The sizes of the other two angles are 4 each. 3. Option B Formula C = πd C = 8π The circumference of the circle is 8π = 5.163 inches 4. Option D Formula A = Change diameter to radius r = 1 d = 1 10 = 5 Plug in the value: A = 5 = 5 The area of the circle is 5 78.55 sq. inches 5. Option A Formula A =
6. Option C Plug in the value A = 10 = 100 The area of the circle is 100 314. sq. inches 44 = (3x + ) + ( (5x + 4), solve for x x = height = = 64 = 3 cm 14 7 Sin(T = (3/7) / 8 = 3 = 4 7 56 7 T = arcsin [ 4 ] = 34.8 7. Option D ABD is a right triangle: hence = 15 + 15 = 450 Also + = 1 + 3 = 450 The above means that triangle BCD is also a right triangle and the total area of the quadrilateral is the sum of the areas of the two right triangles. Area of quadrilateral = [ 1 ] 15 15 + [ 1 ] 1 3 = 144 8. Option A Area of OAB = 7 = [ 1 ] sin (AOB) * OA * OB Solve the above for sin (AOB) to find sin (AOB) = [ 1 ]
Area of ODC = 88 = [ 1 ] sin (DOC) * OD * OD Note that sin(doc) = sin(aob) = [ 1 ], OD = 18 + y and OC = 16 + x and substitute in the above to obtain the first equation in x and y 115 = (18 + y) (16 + x) We now use the theorem of the intersecting lines outside a circle to write a second equation in x and y 16 (16 + x) = 14 (14 + y) Solve the two equations simultaneously to obtain x = 0 and y = 14 9. Option C Let r be the radius of the disk. Area is known and equal to 100π ; hence 100π = π r.r = 10 Circumference = π = 0π 30. Option B Let the radius of the semicircle. Area of the semicircle is known; hence 150π = 1 π r.r = 50 Length of rectangle = r = 100 Width of rectangle = r = 50 Area = 100 50 = 5000 31. Option C The sine rule of triangle says that = = = k sin sin sin Therefore, a = k(sin A), b = k(sin B) and c = k(sin C) Hence, we can rewrite cos = cos = cos as cos sin = cos sin = cos sin
Or Cot A = Cot B = Cot C Or A = B = C Or the triangle is an equilateral triangle