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Geometry (Triangle (POINTS OF CONGREUNCY (4) INCENTRE(I): Meeting point of…
Geometry
Triangle
Basic Rules:
In any triangle ABC,
Angles: A,B,C; Sides opposite to each angle: a,b,c
(i) a + b > c (and) b + c > a (and) c + a > b => Sum of 2 shorter sides is always greater than the longer side.
(ii) A + B + C = 180
(iii) If A>B <=> a>b {within the same triangle}
Classification
Based on Angles
Obtuse Angle: A or B or C > 90
Right Angle: A or B or C = 90
Acute Angle: A < 90, B < 90, C < 90
Based on sides
Isosceles [a=b;c]
Scalene [a!=b!=c]
Equilateral [a=b=c]
Pythagoras Theorem
In a right angled triangle, sum of the squares of 2 sides is equal to the square of the hypotenuse.
Obtuse Angle: a^2 + c^2 < b^2
Acute Angle: a^2 + c^2 > b^2
Right Angle: a^2 + c^2 = b^2
Pythagorean Triplets:
(i) (1,1,Sqrt 2)
(ii) (3,4,5)
(iii) (5,12,13)
(iv) (7,24,25)
(v) (8,15,17)
(vi) (9,40,41)
Theorems
Angle Bisector Theorem:
Internal Angle Bisector Theorem:
BD/DC = AB/AC
Here AD is the angular bisector to interior angle A
External Angle Bisector Theorem:
BE/CE = AB/AC
Here, AE is the angular bisector to exterior angle A
Basic Proportionality Theorem:
(i) CL/LA = CM/MB (ii) CL/CA = CM/CB (iii) LA/CA = MB/CB
Q: ABC is a triangle. BD is the median drawn to side AC. F is the mid-point of BD. If CE passes through F to meet AB in E, find BE if AB = 120
Sol:
When given mid-point and asked to find the length, draw a line parallel to the line passing through mid-point and apply basic proportionality theorem.
Apply this theorem to triangles ACE and BGD and find out the length of BE.
Area of a Triangle:
(i) Area of ABC = (1/2)(BC)(AE)
(ii) Area of ABD = (1/2)(BD)(AE)
(iii) Area of ADC = (1/2)(CD)(AE)
[&]
Height of the triangle can lie inside or outside the triangle.
Note: There can only be 1 perpendicular to a line from a particular point
Comparing Triangles
Congruence
Two triangles are congruent iff
(i) Same size and Same shape
=> In such triangles, all metrics are equal
How to identify Congruency?
There are
4 and only 4 ways
ASA: Angle Side Angle
If 2 angles and the included side between these 2 angles of one triangle are equal to another, then such triangles are congruent.
SAS: Side Angle Side
If 2 sides and an included angle between these 2 sides of one triangle is equal to another, then the 2 triangles are said to be congruent.
RHS: Right-angle Hypotenuse Side
In 2 right angle triangles, if hypotenuse and a side are equal, then such triangles are said to be congruent.
SSS: Side Side Side
If all the sides of 2 triangles are equal, then the 2 triangles are said to be congruent.
Similarity
Two triangles are similar iff
(i) All angles of 2 triangles are equal
(ii) Corresponding sides are in same proportion
=> In such triangles, all angle metrics are equal and all side metrics are in proportion
AAA: Angle Angle Angle
If all the angles of 2 triangles are equal, then the 2 triangles are said to be similar.
If ABCD is a rectangle such that BE is 20 units, what is the area of triangle FAD?
Sol:
(i) In order to find area of FAD, we need base AD and altitude. We already have base as 72, but altitude? (OR)
(ii) We can find area of FAD as FAD = ACD - DFC
We know ACD = 1/2(100x72) = 3600. But how to find DFC?
=> In both cases (i) and (ii), we need to find altitudes. But consider case (i), there is no regularity in shape CDEB. This becomes complex to deal with. Whereas in case(ii), we have 2 triangles AFE and DFE which can be compared. Hence, proceed with case (ii)
Note: Avoid dealing with irregular shapes and then converting them to regular. Instead look out for regular shapes.
(iii) Angle AFE = CFD {Vertically opposite angles},
FAE = FCD and CDF = FEA {Alternate interior angles}
Hence, triangles AFE and CFD are similar
(iv) Applying similarity eqn
=> AE/CD = perpendicular from (F to AE)/(F to CD)
=> (100 - 20)/100 = FM/FN (Say) => FM/FN = 4/5
=> Perpendicular from F to CD = FN = 5/9 (72) = 40
=> Area of FCD = 100x40/2 = 2000
Therefore, area of AFD = ACD - DFC = 3600 - 2000 = 1600
SAS: Side Angle Side
If the 2 corresponding sides of 2 triangles are in proportion and an included angle between these 2 sides is equal, then the 2 triangles are said to be similar.
SSS: Side Side Side
If the corresponding sides of the triangle are in equal proportion, then the triangles are said to be similar.
Similarity Equation:
AB/DE = BC/EF = AC/DF
In order to apply this equation,
orientation of 2 triangles must be same. This is the thumb rule!
After orienting them in same manner, then apply similarity equation.
POINTS OF CONGREUNCY
1)
ORTHOCENTRE(O)
: Meeting point of
Altitudes
=> Altitude:
Perpendicular
drawn from vertex to the opposite side
In a
right-angled triangle
ABC with right angle at A, if '
D
' is the foot of the perpendicular onto BC from A, then
(i) AD^2 = BD.DC
(ii) AB^2 = BC.BD
(iii) AC^2 = BC.CD
In triangle ABC, if O is the orthocentre, then
Angle BOC
= 180 - A
2)
CENTROID(G):
Meeting point of
Medians
Median: Line drawn from vertex to opp. side and
bisects
the side => Centroid (G) divides median in the ratio 2:1
Apollonius Theorem:
In Triangle ABC, if AD is the median drawn from A to BC, then
2(AD^2 + BD^2) = (AB^2 + AC^2)
Properties:
=> A median divides the triangle into 2 equal triangles (i.e., equal area)
=> Area of triangle ABC = 6(Area of triangle BGD) where D is the mid-point of side BC = 3(Area of triangle BGC)
1) In triangle PQR, A, B and C are mid points of sides PQ, QR and PR respectively. If QR = 17, QC = 22.5, RA = 24, find area of PQR.
Sol:
Since PB, QC and AR are medians, their POI is centroid and it divides the median in the ratio 2:1. We know QC and AR. Calculate QG and RG and then area of triangle QGR.
Area of triangle PQR = 3(Area of triangle QGR)
4)
INCENTRE(I)
: Meeting point of
internal angular bisectors
Internal Angular Bisector: Line that
bisects the internal angle
of a triangle
(i) Circle with centre I is called the
Incircle
(ii) Radius of incircle is
Inradius(r)
(iii)
r = ID = IE = IF
Note: ID is perpendicular to BC and not AD
Angle Bisector Theorem:
In triangle ABC, if AD is the angular bisector of angle A, then,
BD/DC = AB/AC
Q: In triangle ABC, AB = 20, BC = 24 and AC = 22. A semi-circle is drawn touching the sides AB and BC and its diameter lies on side AC. Find OC - OA
Sol:
AB and BC are tangents to the semi-circle => OM = ON
Therefore, O lies on the angular bisector of B
=> O divides AC in the ratio of AB and BC
=> AB/BC = AO/OC => AO/OC = 5/6 => OC - OA = 2
Incircle can be imagined as a circle to which, 3 tangents are drawn from 3 different points => CD = CE; AE = AF; BD = BF
In triangle ABC, if I is the incentre then,
Angle BIC
= 90 + (A/2)
3)
CIRCUMCENTRE(S):
Meeting point of
perpendicular bisectors
=> Perpendicular Bisector: Line drawn to a side that is
perpendicular to the side and bisects it
. It need not pass through the opp. vertex
(i) Circle with centre S is
Circumcircle
(ii) Radius of the circle is
Circumradius(R)
R = SA = SA = SC
Important Triangles
Equilateral Triangle
1) Area = [Sqrt(3)/4]a^2
2) Altitude = [Sqrt(3)/2]a
3) Perimeter = 3a
In an equilateral triangle, all points of congruency will coincide i.e., O, G, S and I will coincide. Therefore,
4) Circumradius(R) = (2/3)([Sqrt(3)/2]a) = a/(sqrt(3))
5) Inradius(r) = (1/3)([Sqrt(3)/2]a) = a/(2xsqrt(3))
=>
R : r = 2 : 1
Right Angled Triangle
1) O is the right angle vertex
2) S is the mid-point of hypotenuse
In a right angle triangle ABC with B at 90 and AC as hypotenuse, if BD is the median to AC then,
=> BD = R (Circumradius)
=> BD = Median = (1/2)(hypotenuse)
=> BD = Line joining O and S
1) PQR is a right-angled triangle right angled at P. QAR, the arc subtends max angle possible at P and a circle drawn with QR as diameter. If QR = 8(Sqrt 2), calculate the area of teh sector common to the arc and the circle?
Sol:
The figure looks as follows. We should calculate area of shaded region.
Since QR = 8(Sqrt 2) = Diameter of circle, radius = 4(Sqrt 2)
=> Area of semi - circle PQR = 1/2 (Pi) (4 Sqrt 2)^2 = 16Pi
=> Since QR = 8(Sqrt 2), PQ and PR can be 8
=> Area of triangle PQR = 32
=> Area of shaded region = 16Pi - 32
Isosceles Triangle
1) In an isosceles triangle, S, O, G and I are collinear.
In an isosceles triangle ABC with AB = AC, if AD is the altitude from A to BC, then it is also the median to BC => AD will also be the perpendicular bisector to BC
2) Area = (b/4)(Sqrt(4a^2 - b^2)) where 'a' is the common side
// Not so important as it is easy to derive this formula than to remember
Area of Triangle
1) (1/2)(Base)(Height)
2) Sqrt[(s)(s-a)(s-b)(s-c)]
3) rxs
4) abc/4R
5) (1/2)(bc)(Sint), where 't' is the angle between sides b and c
Angles
Supplementary:
A + B = 180
Complementary:
A + B = 90
Reflex:
180 - A
Line Segment
When 2 line segments are cut by a transversal,
Alternate Interior angles: d = e
Eg:
x + 32 = 71
Alternate Exterior angles: a = h
Corresponding Angles: a = f
Vertically opposite angles: a = c
This is very important idea
Whenever 2 straight lines intersect, pair of opposite angles are equal.
Angles on straight line: a + b = 180
Consecutive interior angles: d + f = 180
If 3 or more parallel lines are cut by 2 transversals, these 2 transversals make equal intercepts between these 3 lines.
Bisector:
A line that divides an angle or a line segment into 2 equal parts
Perpendicular:
A line drawn to a line segment at an angle of 90
Question 1
There are 8 points lying on a plane such that no two of the lines joining them are parallel and no three of them are collinear. Calculate the number of points of intersections of these lines excluding the original 8 points (Note: All possible lines are drawn through every pair of these points)
Sol: No. of lines drawn with these 8 points = 8C2 = 28
=> No. of points of intersection with these 8 lines = 28C2
Assume these 8 points to be the vertices of an octagon
=> 7 lines pass through each of these 8 vertex points
=> No. of points of intersections at these 8 vertices = 8(7C2)
=> No. of distinct points of intersection excluding vertices = 28C2 - 8(7C2) = 210
Question 2:
ABCD and CDEF are trapeziums. The lines AB, CD and EF are parallel and measure 1.2m, 2.7m and 7.2m respectively. If ED = 3m and the points B, C, F are collinear and A, D, E are collinear. Calculate AD?
Sol:
From the figure, triangles BPC and BQF are silimar (AAA rule)
=> BP/BQ = PC/QF => BP/(3 + BP) = (1.5)/6 => BP = 1m => AD = 1m
CIRCLES
(i) Area of the circle = (Pi)(r^2)
(ii) Area of a sector = lr/2 = (1/2)(r^2)(t); where l is the length of the arc and r is the radius of the circle, t is the angle subtended by the arc at the centre
Basic Theory
1) Line drawn from center to a chord is always the perpendicular bisector to the chord.
2) 'n' chords will divide the circle into a maximum of [n(n+1)/2] + 1 parts
Touching Circles
(i) Direct Common Tangent(DCT) = Sqrt[(d^2) - (R - r)^2]
// Touches 2 circles are 2 different points
(ii) Transverse Common Tangent(TCT) = Sqrt[(d^2) - (R + r)^2]
// Touches 2 circles at the same point
where d: Dist. between centres, R and r: radii of 2 circles
1) Circles touching externally
Dist btw centres = R + r
(i) No. of DCT = 2 and TCT = 1
2) Circles touching internally
Dist btw centres = R - r
(i) No. of DCT = 0 and TCT = 1
3) Intersecting circles
Here, O1O2 is the perpendicular bisector of AB
(i) No. of DCT = 2 and TCT = 0
Note: |R - r| < d < |R + r|
For non-intersecting and non touching circles,
(i) No. of DCT = 2 and TCT = 2
Circle Angle Theory
1) Angles in the same segment are equal
2) Angle subtended by an arc at the centre is twice the angle subtended by the arc at any point on the circle in the same segment
Eg: Angle in a semi - circle is 90
Circle, Chord & Tangent Properties
1) If 2 chords intersect either internally or externally, product of their lengths from the point of intersection will be equal
AE x EB = CE x ED
Tangent Secant Theorem:
PT^2 = PAxPB
Alternate Segment Theorem
Angle PAB = Angle ACB
Angle between tangent and a chord is equal to the angle subtended by the chord on the alternate segment
Note: Chords of same length are equidistant from the center.
Questions
1) If AB, CD and EF are diameters of a circle with centre O and H is any point on the circle, calculate the number of triangles formed with these 8 points.
Sol: Number of triangles formed with 8 points = 8C3.
But, AOB, COD and EOF are collinear => we should remove these 3 combinations
Therefore, no. of triangles = 8C3 - 3 = 53
POLYGONS
Any geometric figure formed with >= 3 lines is a polygon
Quadrilateral
(i) Sum of interior angles = 360
(ii) A quadrilateral inscribed in a circle is called a cyclic quadrilateral
(iii) Sum of opp. angles in a cyclic quadrilateral is 180
Note: When the perimeter of a quadrilateral is fixed, the greatest possible area occurs when the quadrilateral is a square. This concept can be used while eliminating options.
1)
Parallelogram
(i) Opposite sides are parallel and equal
(ii) Diagonals bisect each other
(iii) Area = bh
2)
Rhombus
(i) All sides are equal
(ii) Diagonals are perpendicular and bisect each other
(iii) Area = (1/2)(product of length of diagonals)
3)
Rectangle
(i) Opposite sides are equal
(ii) All angles are 90
(iii) Diagonals are of equal length and bisect each other
(iv) Area = bh
4)
Square
(i) All sides are equal and all angles are equal to 90
(ii) Diagonals are of equal length and are perpendicular bisectors to each other
(iii) Area = a^2
5)
Trapezium
(i) Only 1 pair of opposite sides are parallel
(ii) An isosceles trapezium is one in which 2 non-parallel sides are equal
(iii) Every trapezium inscribed in a circle is an isosceles trapezium and a cyclic quadrilateral
(iv) Area = (1/2)h(a+b); h = height, a,b: lengths of parallel sides
1) An isosceles trapezium is circumscribed about a circle. Find the area of the trapezium, if perimeter = 8 cm
'n' sided polygon
(i) Sum of all the exterior angles = 360
(ii) Sum of all the interior angles = (n-2)180
(iii) Number of diagonals = n(n-3)/2
Regular Polygon
(i) In a regular polygon, all interior angles are equal; all exterior angles are equal; all sides are equal
(ii) Each exterior angle = 360/n
(iii) Each interior angle = 180 - (360/n)
Regular Hexagon
(i) Each interior angle = 120
(ii) Each exterior angle = 60
(iii) If all the vertices of a regular hexagon are joined at the centre, then 6 equilateral triangles are formed at the centre with base as one of the sides of the regular hexagon
=> Area of a regular hexagon = 6(Area of equilateral triangle)
Moot Points:
(1) The number of equilateral triangles formed by joining the vertices of an n-sided regular polygon = (n/3)
Typical Questions
1) ABC is a triangle in which quadrilateral WXZY is drawn. W is a point on AB such that AW = WB = 5 units, X is a point on AC such that AX = 4 units and XC = 6 units. Y and Z are points on BC such that BY = 3, YZ = 3 and ZC = 4 units. Find the area of quadrilateral WXZY?
Sol: As we add up the given numbers, we find that ABC is an equilateral triangle with side 10 units.
=> Each angle is 60 degrees
=> In order to find the area of an irregular shape, we need to find the area of regular shapes and subtract these values from the total area.
Here, find areas of triangles AWX, CZX and BWY. Subtract these areas from area of ABC and we get area of WXZY
2) A polygon has 65 diagonals. Find the number of points of intersections of these diagonals inside the polygon?
Sol: 65 diagonals are there in a polygon => the polygon has 13 sides
=> Max number of points of intersection inside the polygon = 13C4 = 715
Note: If the number of sides of a polygon = even => No. of points of intersection = nC4 - (n/2)C1 + 1
If the number of sides of a polygon = odd => No. of points of intersection = nC4
Why is this difference? Because, if n = even, multiple diagonals intersect at centre and hence same points is counted more than once. In case of n = odd, diagonals don't intersect at the centre.
Note that these formulas are for points of intersection inside the polygon.
Counting the points of intersection of diagonals is as good as counting the number of quadrilaterals formed from n points!
