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Alosies George
IIM Calcutta
director@georgeprep.com
+91- 9985-372-371
www.georgeprep.com
1
Geometry Concept Session 2
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2
Triangles – Congruency and
Similarity
Congruency
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3
- Corresponding sides and corresponding angles are equal
- Congruency conditions
- SSS
- SAS
- AAS
- ASA
- RHS
- Properties of two congruent triangles
Similarity
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4
- Corresponding sides proportional;
corresponding angles are equal
- Any congruency condition is also a similarity
condition(the vice versa is not true)
- Properties of two similar triangles
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5
ABC is a right angled triangle and E is a point chosen on AC such that the rectangle formed inside ABC
by dropping the perpendiculars ED and EF from E has the maximum area possible. What is the area of
the triangle EFC in that case, if AB = 12cm and AC = 13cm?
1. 12
2. 12.5
3. 7.5
4. 10
5. None of these
Problems
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6
ABCD is a trapezium with the two parallel sides measuring 15cm and 9cm respectively. A line
segment PQ, parallel to the parallel sides, is drawn such that P is on AB and Q is on CD and
PQ divides the trapezium into two figures of equal area. The length of the line segment PQ is
1. 168
2. 153
3. 8
4. None of these
Problems
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7
Special Triangles and their
Properties
Special Triangles – Right Angled Triangle
Pythagoras theorem
Pythagoras triplets
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8
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9Problems
Special Triangles – Right Angled
Triangle
 Circumradius
 Inradius
 Area
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11
R =
ℎ
2
A =
𝑎 ∗ 𝑏
2
r =
𝑎+𝑏−ℎ
2
=
𝑎𝑏
𝑎+𝑏−𝑐
h a
b
Special Triangles – Right Angled Triangle –(Altitude-on-
Hypotenuse Theorem)
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12
In a right triangle, the altitude that’s
perpendicular to the hypotenuse creates two
smaller right triangles that are both similar to
the original right triangle.
Problems
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13
- CAT 2006
Special Triangles – Right Angled Triangle
 Special Right Angled triangles
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14
90-60-30 triangle 90-45-45 triangle
Side Ratios 2: √3:1 √2 : 1: 1
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15
- CAT 2007
Answer : Option 1
Special Right Triangles ( 90-45-45)
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16
Answer : Option 1
Problems
Special Triangles – Equilateral Triangle
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18
In terms of
side, a
In terms of
altitude, h
Side - 2h/ √3
Altitude √3a/2 -
Circumradius a/√3 2/3 h
Inradius a/2√3 1/3 h
Area √3a2 /4 h2/√3
Special Triangles – Equilateral Triangle
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19
AD = median, perpendicular bisector, Angle
Bisector and altitude
E = centroid, incenter, orthocenter and
circumcenter
For a given perimeter, Equilateral Triangle has the
largest Area
For a given area, the equilateral triangle has the least
perimeter.
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20Problems
Answer : Option B
Special Triangles – Isosceles Triangle
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21
• Exactly two sides are equal
• Exactly two angles are equal
• AD = median, angle bisector, altitude and
perpendicular bisector
• Area = [b√(4a2 –b2)]/4
Area of Triangles
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22
 ½ breadth x height
 Heron’s formula
 ab sin(included angle)/2
 abc/4R (where R is circumradius)
 rs ( where r is inradius and s is semiperimeter)
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23Problems
Answer : Option B
PQ and PR are two sides of a triangle that make an angle of 120 degrees between them. If the angle
bisector of angle QPR is of length 4cm, what is the harmonic mean of the lengths of PQ and PR?
1. 4
2. 8
3. 4/ 3
4. None of these

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