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GEOMETRY
This chapter comes under Paper - II. This topic plays very important role in getting 100%
marks for students in the public exam. From this chapter 4 marks problem-1 (1×4=4M), 2
Marks Problem -1 (1×2 = 2M), 1 Mark Problem-1 (1×1=1M), 5 Marks Problem -1 (1×5=5M),
and 6 objective bits (6×1/2 = 3M) altogether 15 Marks can be scored. Students need to be
very very cautious in the preparation of Geometry and also lot of hard work is required to
get good score in this chapter. The following information will help for the S.S.C. exam
appearing students.
GEOMETRY
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"THALES" introduced the study of Geometry in Greece.
Triangles having same shape are called similar triangles
Traingles having the same shape and same size are called congruent triangles.
Congruent triangles are similar but converse need not be true.
Q. Write the properties of ''similar triangles"
- Two triangles are said to be similar if and only if their
i) Corresponding angles are equal
ii) Corresponding sides are in proportion
- Two squares, Two circles, two equilateral triangles, two right angled isosceles triangles
are always similar.
Q. State and Prove ''Basic proportionality theorem'' or Thales Theorem.
- Statement: In a triangle, a line drawn parallal to one side, will divide the other two sides
in the same ratio.
Given: In ∆ABC,
DE || BC and DE intersects AB in D and AC in E.
AD AE
=
Required to Prove:
DB EC
Construction: Join BE and CD and also draw EF⊥ AB and DG ⊥ AC
Proof: Consider
1
× AD × EF
area (∆ADE ) 2
=
area (∆BDE ) 1
× BD × EF
2
AD
.................( 1)
DB
also consider
1
× AE × DG
area (∆ADE ) 2
=
area (∆CDE ) 1
× EC × DG
2
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=
AE
.................( 2)
EC
∆BDE and ∆CDE are on the same base and between the same parallel lines
area (∆BDE ) = area (∆CDE ) ..............(3)
From (1) (2) & (3)
area (∆ADE ) area (∆ADE )
=
area (∆BDE ) area (∆CDE )
⇒
AD AE
=
DB EC
Hence the theorem is proved
Q. State vertical angle bisector theorem.
1M
- The vertical angle bisector of a triangle divides the base in the ratio of the other two
sides.
A.A. Similarity: If two angles of one triangle are equal to the corresponding angles of
antoher traingle, then the two triangle are similar.
Q. State and prove Converse of Pythagorean theorem.
4M
Statement: In a trianlge, if the square of one side is equal to the sum of the squares of
the other two sides, then the angle opposite to the first side is a right angle.
Given: In ∆ABC, AC2 = AB2 + BC2
RTP: ∠ABC = 90°
Construction: Construct ∆PQR,
Such that PQ = AB, QR = AB and ∠PQR = 90°
Proof: In ∆PQR,
PR2 = PQ2 + QR2 [ ∵ By pythagorean theorem]
PR2 = AB2 + BC2 [ ∵ Given]
PR2 = AC2 [ ∵ Given]
PR = AC ..................(1)
Consider
∆ ABC
∆ PQR
AB=
PQ (S) [∵Construction ]
BC=
QR (S) [∵Construction ]
AC=
PR (S) [∵from(3)]
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By S.S.S. axiom
∆ABC ≅ ∆PQR
by corresponding parts of congruent triangles
∠ABC = ∠PQR
∠ABC = 90° [ ∵ construction]
Hence theorem proved
Q. In ∆ABC, ∠B is an obtuse angle and AD ⊥ CB. Prove that AC2 = AB2 + BC2 + 2BC.BD
Given: In ∆ABC ∠B > 90°, AD ⊥ CB
RTP: AC2 = AB2 + BC2 + 2BC.BD
Proof: In ∆ADC, ∠D = 90°
by pythagorean theorem
⇒ AC2 = AD2 + DC2
⇒ AC2 = AD2 + (DB + BC)2
⇒ AC2 = AD2 + DB2 + BC2 + 2DB.BC
⇒ AC2 = AB2 + BC2 + 2DB.BC [∴ AB2 = AD2 + DB2]
⇒ AC2 = AB2 + BC2 + 2.BC.BD
Hence Proved
Note: Phythagorean theorem was originally proposed by "Baudhayana"
-
Rigorous proof for pythagorean theorem was given by "Bhaskarudu-II"
Note: The ratio of the areas of two similar triangles is equal to the ratio of the squares of
any two corresponding (i) sides (ii) medians (iii) altitudes (iv) angle bisector segments of
triangle.
Q. In ∆ABC b2 = a2+c2 then right angle at B
Q. ∆ABC~ ∆PQR ∠A=50° then ∠Q + ∠R = 130°
Tangents to a circle
-
A straight line touches the circle at only one point is called tangent to the circle.
A straight line intersects the circle in two distinct point is called secant to the the circle
The point that is common to the tangent and the circle is called points of contact.
-
The tangent is always perpendicular to the radius through the point of Contact.
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Q. Prove that the lengths of two tangents drawn from an external point to a circle are
equal.
Given: PA and PB are two tangents of a circle with the Centre 'O'.
RTP: PA = PB
A
Con: Join, OA,OB and also OP
Proof: Consider
P
∆ OAP
OA=
o
∆ BOP
OB (S)= [∵ same radii]
B
∠OAP=90° ∠OBP =90°(R)= [∵ radiusisperpendicularto tan gent ]
OP=
OP (A)= [∵ Common side ]
by R.H.S axiom
∆OAP ≅ ∆OBP
By corresponding parts of congruent triangles, PA = PB
Hence Proved
Q. If PAB is a secant of a circle and PT is a tangent then show that PA.PB = PT2
Proof:
Given: PAB is a Secant of a circle. PT is a tangent
RTP: PA.PB = PT2
Con: Draw on ⊥ AB and join OA, OP, OB and also OT.
Proof: Consider PA.PB = (PM – AM) (PM + BM)
= (PM – AM) (PM + AM) [∵ AM = MB ]
=
=
=
=
PM2 – AM2
PM2 – [OA2 – OM2]
PM2+OM2-OA2
OP2 – OA2
[PM2+OM2= OP2]
= OP2 – OT2 [∵ OA = OT ]
= PT2
Hence proved
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Alternate segment theorem: If a chord is drawn through an end point of a chord of a
circle, then the angles which this chord makes with the given tangent an equal
respectively to the angles formed in the corresponding alternate segments.
Here ∠BAQ = ∠ACB and ∠BAP = ∠ADB
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If two chords of a circle intersect inside or out side of the circle when produced, the
rectangle formed by the two segments of one chord is equal in the area to the rectangle
formed by the two segments of the other.
e.g. In the adjacent figure x = 2 2
x×x=1×8
x2 = 8
x= 8
x = 2 2 units
e.g. In the adjacent figure x = 10
PA.PB = PC. PD
3 (3+5) = 2 (2+x)
9+15 = 4+2x
24 – 4 = 2x
20/2 = x
x = 10 units
Common tangent: The tangent that is common to two or more circles is called common
tangent.
Direct Common tangent: If two circles are lying same side of common tangent with two
points of contacts then the common tangent is called as direct common tanget (DCT)
d2 − (r1 − r2 )
2
The lenght of DCT =
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Here PQ & RS are DCT's
d - distance between centres of circles
r1 & r2 - radii of two circles
Transverse Common tangent: If two circles are lying either side of the common tangent
with two points of contact then the common
tangent is called Transverse common tangent.
(TCT)
d2 − (r1 + r2 )
2
The length of TCT =
PQ − Lengthof TCT
d - distance between centres
r1 & r2 - radii of circles
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'd' is the distance between the centre of circle to an external point, r is the radius of the
circle, l is the length of the tangent then the relation between
l, d and r is i) d =
ii) l =
(d
2
− r2
(l
2
+ r2
)
)
iii)r = d2 − l2
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The tangent at the ends of a diameter of a circle are parallel
The angle between the tangent to a circle and the radius drawn at the point of contact is
90°
Two circles of radii 3 cm and 1 cm. The distance between their centres is 5 cm then
i) length of TCT - 3 cm
ii) length of DCT -
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21cm
The number of tangents that can be drawn to a circle are infinite.
The number of tangents that can be drawn to a circle from an external point are 2
Angle in a semicircle is 90°
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If a trapezium is cyclic then it is isosceles trapezium
Cyclic rhombus is square
Cyclic parallelogram is rectangle
S.No
Figure
1.
Condition
others
Two circles are touching
each other externally
CT =3
DCT = 2
TCT = 0
Two circles are intersecting
in two points
CT =2
DCT = 2
TCT = 0
Two circles are not at all
touching and one is not in
the other
CT =4
DCT = 2
TCT = 2
touching
4.
Two circle
internally
CT = 1
DCT = 0
TCT = 0
5.
Two are concentric same
centred
CT = 0
DCT = 0
TCT = 0
6.
One circle is lying in the
other and without touching
and also not concentric
CT = 0
DCT = 0
TCT = 0
2.
3.
are
d < r2 – r1
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Circumcentre of a circle is equidistant from vertices of the vertices
Incentre of a circle is equidistant form sides of the vertices.
Assignment:
1. State and prove converse of basic proportionality theorem.
2. State and prove phythagorean theorem.
4m
4m
3. In an equilateral triangle with side 'a' units then prove that i) its height (h) =
ii) its area (A) =
3
.a units.
2
3 2
.a sq. units.
4
2m
=
+
– 2.BC.BD
4. ∠B of ∆ABC is an acute angle and AD ⊥ BC then prove that
5. A vertical stick 12 cm long casts a shadow 8 cm long on the ground. At the same time a
tower casts the shadow 40 m long on the ground. Determine the height of the tower.
2m
6. Construct a ∆ ABC in Which BC = 4cm, A = 50° and altitude through A = 3 cm.
5m
AC2
AB2
BC2
Construction - No. 1
1. Construct a traingle in which BC = 5 cm, ∠A = 70° and median AD through A =3.5 cm.
Given: BC = 5 cm, ∠A = 70°, median AD = 3.5 cm.
RTC: ∆ ABC
Rough Sketch:
Construction:
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Steps of Construction:
1. Draw a line segment BC = 5cm and draw BX such that ∠CBX = 70°
2. Draw BY such that ∠XBY = 90° and draw the perpendicular bisector of BC which
intersects BY at O and BC at D.
3. Taking 'O' as centre and OB as radius draw a circle which passes through B and C
4. Taking 'D' as centre and with the radius of 3.5 cm. Cut the circle at A and A'.
5. Join AB, AC and A 'B, A 'C , Here ∆ ABC, ∆ A'BC are required triangles.
Proof:
By alternate segment theorem ∠BAC = ∠CBX = 70°
Construction - No. 2
2. Construct a a triangle ABC in which BC = 7 cm, A = 70° and foot of the
perpendicular 'D' on BC from A is 4.5 cm away from B.
Given: BC = 7cm, ∠A = 70°, BD = 4.5 cm
Required to construct: ∆ ABC
Rough sketch:
Construction:
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Steps of Construction:
-
Draw a linesegment BC = 7 cm and draw BX such that ∠CBX = 70° and also draw BY
such that ∠XBY = 90°
Draw the perpendicular bisector of BC such that it intersects BY at 'O' and taking 'O' as
centre, OB = OC as radius draw a circle.
Taking 'B' as centre and cut BC at D with the radius of 4.5 cm and also by using
-
protractor draw a perpendicular through D such that it intersects the circle at A.
Join AB and AC, here ∆ ABC is required triangle.
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Proof: By alternate segment theorem,
∠BAC = ∠CBX = 70°
Construction - No. 3
3. Construct the cyclic quadrilateral ABCD in which AC = 4 cm, ∠ABC = 57°, AB = 1.5 cm,
AD = 2 cm.
RTC: Cyclic quadrilateral ABCD.
Rough Sketch:
Construction:
Steps of Construction:
-
Draw a line segment AC = 4 cm and draw AX such that ∠CAX = 57° and also draw AY
such that ∠XAY = 90°
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Draw the P.b of AC which intersects AY at O and AC at P.
Taking "O" as centre and Oa as radius draw a circle that passes "A", and "C"
Taking A as centre cut the circle upwards at B with the radius of 1.5 cm and downwards
at 'D' with the radius of 2 cm.
Join AB, BC, CD and AD. Here ABCD is the required cyclic quadrilateral
Proof: By Alternate segment theorem
∠ABC = ∠CAX = 57°
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