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When two circles touch, their centres and their point of contact are collinear. 6. If the two circles touch ... â ABC and â ACD = 2 â AEC. 5. In a right angle âABC, ...
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ADVANCE MATHS - GEOMETRY DIGEST Dear readers, Geometry is a very important topic in numerical ability section of SSC Exams. You can expect 14-15 questions from Geometry in SSC Exams. So, keeping the upcoming SSC CHSL Exam 2015, SSC CGL Tier 2 and FCI Exam in mind, below is the Geometry Digest which includes all the Important concepts, formulas and shortcuts you should be aware of, for the upcoming SSC CGL Tier 2, SSC CHSL Exam 2015, FCI Exam and other SSC Competitive Exams Regards, Sandeep Baliyan(Community Manager – SSC)

Fundamental concepts of Geometry Point: It is an exact location. It is a fine dot which has neither length nor breadth nor thickness but has position i.e., it has no magnitude. Line segment: The straight path joining two points A and B is called a line segment AB . It has and points and a definite length. Ray: A line segment which can be extended in only one direction is called a ray. Intersecting lines: Two lines having a common point are called intersecting lines. The common point is known as the point of intersection. Concurrent lines: If two or more lines intersect at the same point, then they are known as concurrent lines. Parallel Lines If two lines in the same plane do not intersect, they are parallel to each other. Lines AB and CD are parallel and denoted by AB⏐⏐CD. Parallel lines and a transverse: When two parallel lines are cut by a transversal (i.e., a third line intersects the two parallel lines), a number of relationships exist between the resulting angles.

   

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Alternate Interior Angles Are Equal: K = L; O = J Alternate Exterior Angles Are Equal: H = M; N = I Corresponding Angles Are Equal: K = N; J = M; H = O; I = L Non-Alternate Interior Angles Are Supplementary: L + J = 180; K + O = 180

www.gradestack.com/ssc Type of Angles: Angles: When two straight lines meet at a point they form an angle. Right angle: An angle whose measure is 900 is called a right angle. Acute angle: An angle whose measure is less than one right angle (i.e., less than 90), is called an acute angle. Obtuse angle: An angle whose measure is more than one right angle and less than two right angles (i.e., less than 180 and more than 90) is called an obtuse angle Reflex angle: An angle whose measure is more than 180 and less than 360 is called a reflex angle Complementary angles: If the sum of the two angles is one right angle (i.e., 90), they are called complementary angles. Therefore, the complement of an angle θ is equal to 90° − θ. Triangles and Their Properties A. On the basis of sides, triangles are classified into three categories: a) Scalene: Having all sides unequal. b) Isosceles: Having any two sides of same length. c) Equilateral: Having all the three sides of equal length. On the basis of angles, triangles are divided into three categories: a) Obtuse angled triangle: Largest angle greater than 900 b) Acute angled triangle: All angles less than 900 c) Right Angled Triangle: Largest angle equal to 900 Properties of a Triangle: 1. Sum of the all the three angles is 1800. 2. An exterior angle is equal to the sum of the interior opposite angles. 3. The sum of any two sides is always greater than the length of the third side. 4. The difference between any two sides is always less than that of the third side. 5. The side opposite to the greatest angle is the greatest side and the side opposite to the smallest angle is the shortest side Points inside or outside a triangle with their properties: A. Centroid: The point of intersection of the medians of a triangle

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1. The centroid divides each median from the vertex in the ratio 2 : 1. 2. Apollonius theorem gives the length of the median. AB2 + AC2 = 2(AD2 + BD2) 3. If x, y, z are the lengths of the medians through A, B, C of a triangle ABC, x2 + y2 + z2 = (a2 + b2 + c2 ). 4. Median always divides a triangle into two equal portions. B. Circumcentre: The point of intersection of perpendicular bisectors of the sides of a triangle

1. The circumcentre is equidistant from the vertices. 2. If a, b, c, are the sides of the triangle, Δ is the area & R is the radius of the circum-circle, then abc = 4R. Δ 3. In a right angled, the median to the hypotenuse is equal to its circumradius and is equal to half the hypotenuse. Orthocentre: The point of intersection of the altitudes of a triangle.

1. B, Z, Y, C lie on a circle and form a cyclic quadrilateral. 2. C is the orthocentre of the right angled triangle ABC right angled at C. 3. Centroid divides the line joining the orthocentre and circumcentre in the ratio of 2 : 1 Incentre: The point of intersection of angle bisectors of the angles

1. It is equidistant from the sides of the triangle.

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www.gradestack.com/ssc 2. According to Angle bisector theorem BM/MC = AB/AC Congruency of triangles: Two triangles ABC and DEF are said to be congruent, if they are equal in all respects (equal in shape and size). The notation for congruency is ≅ or ≡ If ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

AB = DE, BC = EF, AC = DF Then ΔABC ≡ ΔDEF or ΔABC ≅ ΔDEF Mid-Point Theorem: A line joining the mid points of any two sides of a triangle is parallel and equal to half of the third side. If in Δ ABC, D & E are the mid points of AB & AC respectively, then we have

 

DE || BC DE =1/2 BC

Similar triangles: Two figures are said to be similar, if they have the same shape but not the same size.

NOTE: Congruent triangles are similar but similar triangles need not be congruent. Properties of similar triangles: If two triangles are similar, the following properties hold true. (a) The ratio of the medians is equal to the ratio of the corresponding sides. (b) The ratio of the altitudes is equal to the ratio of the corresponding sides.

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www.gradestack.com/ssc (c) The ratio of the internal bisectors is equal to the ratio of corresponding sides. (d) The ratio of inradii is equal to the ratio of the corresponding sides. (e) The ratio of the circumradii is equal to the ratio of the corresponding sides. (f) Ratio of area is equal to the ratio of squares of the corresponding sides. (g) Ratio of area is equal to the ratio of squares of the corresponding medians. (h) Ratio of area is equal to the ratio of the squares of the corresponding altitudes. (i) Ratio of area is equal to the ratio of the squares of the corresponding angle bisectors. Basic Proportionality Theorem: In a triangle, if a line drawn parallel to one side of a triangle divides the other two sides in the same ratio. So if DE is drawn parallel to BC, it would divide sides AB and AC proportionally i.e.

AD/BD = AE/AC Pythagoras Theorem: The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides. i.e. in a right angled triangle ABC, right angled at B, AC2 = AB2 + BC2 Angle Bisector Theorem: If in ΔABC, CD is the angle bisector of ∠BCA, the ratio of the lines BD & AD is equal to the ratio of sides containing the angle.

BD/AD = BC/ AC Circles If O is a fixed point in a given plane, the set of points in the plane which are at equal distances from O will form a circle. Properties of a Circle: 1. If two chords of a circle are equal, their corresponding a 2. rcs have equal measure. 2. Measurement of an arc is the angle subtended at the centre. Equal arcs subtend equal angles at the center. 3. A line from centre and perpendicular to a chord bisects the chord. 4. Equal chords of a circle are equidistant from the centre.

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www.gradestack.com/ssc 5. When two circles touch, their centres and their point of contact are collinear. 6. If the two circles touch externally, the distance between their centres is equal to sum of their radii. 7. If the two circles touch internally, the distance between the centres is equal to difference of their radii. 8. Angle at the centre made by an arc is equal to twice the angle made by the arc at any point on the remaining part of the circumference. Let O be the centre of the circle. ∠BOC = 2 ∠P, when ∠BAC = ∠P

9. If two chords are equal, the arc containing the chords will also be equal. 10. There can be one and only one circle that touches three non-collinear points. 11. The angle inscribed in a semicircle is 90o. 12. If two chords AB and CD intersect externally at P, PA × PB = PC × PD

13. If two chords AB and CD intersect internally at P, PA × PB = PC × PD

14. If PAB is a secant and PT is a tangent, PT2 = PA × PB 15. The length of the direct common tangent (PQ)

Cyclic Quadrilateral If a quadrilateral is inscribed in a circle i.e. all the vertex lies on the circumference of the circle, it is said to be a cyclic quadrilateral. 1. In a cyclic quadrilateral, opposite angles are supplementary.

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www.gradestack.com/ssc 2. In a cyclic quadrilateral, if any one side is extended, the exterior angle so formed is equal to the interior opposite angle. Alternate angle theorem Angles in the alternate segments are equal. In the given figure, PAT is tangent to the circle and makes angles ∠PAC & ∠BAT respectively with the chords AB & AC. Then, BAT = ∠ ACB & ∠ ABC = ∠ PAC

Polygons A closed plane figure made up of several line segments that are joined together is called a polygon. Types of Polygons  Equiangular (All angles equal)  Equilateral (All sides equal)  Regular (All sides & angles equal Properties of Polygon: 1. Sum of all the exterior angles of any regular polygon is equal to 3600 2. Each exterior angle of an n sided regular polygon is (3600/ N) degrees. ( ) 3. Each interior angle of an n sided equiangular polygon is 4. Also as each pair of interior angle & exterior angle is linear. 5. Each interior angle = 1800– exterior angle. 6. The sum of all the interior angles of n sided polygon is (n – 2)1800

Some Important Short Cuts 1.

In a  ABC, if the bisectors of B and C meet at O then

BOC = 900 + A 2. In a ABC, if sides AB and AC are produced to D and E respectively and the bisectors of DBC and ECB Intersect at O, then

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 BOC = 900 – ½ A 3.

In a ABC, If AD is the angle bisector of BAC and AE BC, then

BAE = 1/2 ( ABC - ACB) 4.

In a ABC, If BC is Produced to D and AE is the Angle bisector of A, then

ABC and ACD = 2 AEC 5.

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In a right angle ABC, B = 900 and AC is hypotenuse. The perpendicular BD is dropped on hypotenuse AC from right angle vertex B, then

www.gradestack.com/ssc (i) BD = (AB x BC)/ (AC) (ii) AD = AB2 /AC (iii) CD = BC2/AC (iv) 1/BD2 = (1/AB2) + (1/BC2)

Coordinate Geometry 1. Equation of line parallel to y-axis X=b For Example: A Student plotted four points on a graph. Find out which point represents the line parallel to yaxis. a) (3, 5) b) (0, 6) c) (8, 0) d) (-2, -4) Solution: Option (C) 2. Equation of line parallel to x-axis Y=a For Example: A Student plotted four points on a graph. Find out which point represents the line parallel to xaxis. a) (3, 5) b) (0, 6) c) (8, 0) d) (-2, -4) Solution: Option (B) 3. Equations of line a) Normal equation of line ax + by + c = 0 b) Slope – Intercept Form y = mx + c Where, m = slope of the line & c = intercept on y-axis For Example: What is the slope of the line formed by the equation 5y - 3x - 10 = 0? Solution: 5y - 3x - 10 = 0, 5y = 3x + 10 Y = 3/5 x + 2 Therefore, slope of the line is = 3/5 c) Intercept Form x/A + y/B = 1, Where, A & B are x-intercept & y-intercept respectively For Example: Find the area of the triangle formed the line 4x + 3 y – 12 = 0, x-axis and y-axis? Solution: Area of triangle is = ½ * x-intercept * y-intercept. Equation of line is 4x + 3 y – 12 = 0 4x + 3y = 12, 4x/12 + 3y/12 = 1 x/3 + y/4 = 1

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www.gradestack.com/ssc Therefore area of triangle = ½ * 3 * 4 = 6 d) Trigonometric form of equation of line, ax + by + c = 0 x cos θ + y sin θ = p, Where, cos θ = -a/ √(a2 + b2) , sin θ = -b/ √(a2 + b2) & p = c/√(a2 + b2) e) Equation of line passing through point (x1, y1) & has a slope m y - y1 = m (x-x1) 4.

Slope of line = y2 - y1/x2 - x1 = - coefficient of x/coefficient of y

5. Angle between two lines Tan θ = ± (m2 – m1)/(1+ m1m2) where, m1 , m2 = slope of the lines Note: If lines are parallel, then tan θ = 0 If lines are perpendicular, then cot θ = 0 For Example: If 7x - 4y = 0 and 3x - 11y + 5 = 0 are equation of two lines. Find the acute angle between the lines? Solution: First we need to find the slope of both the lines. 7x - 4y = 0 ⇒ y = 74x Therefore, the slope of the line 7x - 4y = 0 is 74 Similarly, 3x - 11y + 5 = 0 ⇒ y = 311x + 511 Therefore, the slope of the line 3x - 11y + 5 = 0 is = 311 Now, let the angle between the given lines 7x - 4y = 0 and 3x - 11y + 5 = 0 is θ Now, Tan θ = ± (m2 – m1)/(1+ m1m2) = ±[(7/4)−(3/11)]/[1+(7/4)*(3/11)] = ± 1 Since θ is acute, hence we take, tan θ = 1 = tan 45° Therefore, θ = 45° Therefore, the required acute angle between the given lines is 45°. 6. Equation of two lines parallel to each other ax + by + c1 = 0 ax + by + c2 = 0 Note: Here, coefficient of x & y are same. 7. Equation of two lines perpendicular to each other ax + by + c1 = 0 bx - ay + c2 = 0 Note: Here, coefficient of x & y are opposite & in one equation there is negative sign. 8. Distance between two points (x1, y1), (x2, y2) D = √ (x2 – x1)2 + (y2 – y1)2 For Example: Find the distance between (-1, 1) and (3, 4). Solution: D = √ (x2 – x1)2 + (y2 – y1)2 = √ (3 – (-1))2 + (4 – 1)2 = √(16 + 9) = √25 = 5 9.

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The midpoint of the line formed by (x1, y1), (x2, y2)

www.gradestack.com/ssc M = (x1 + x2)/2, (y1 + y2)/2 10. Area of triangle whose coordinates are (x1, y1), (x2, y2), (x3, y3) ½ [ x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)] For Example: Find area of triangle whose vertices are (1, 1), (2, 3) and (4, 5). Solution: We have (x1, y1) = (1, 1), (x2, y2) = (2, 3) and (x3, y3) = (4, 5) Area of Triangle = ½ [ x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) ] =1/2 [(1(3−5) +2(5−1) + 4(1−3)) ] =1

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