2. Geometry
Introduction
In Geometry, we have studied triangles and many of their properties in earlier classes. We also have studied
congruence and similarity of two triangles. Recall that the two figures are said to be congruent, if they
have the same shape and same size whereas the two figures are said to be similar, if they have the same
shape (and not necessarily the same size). All circles with the same radii are congruent, all squares with
the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent.
Note that all congruent figures are similar but the similar figures need not be congruent.
We can state the similarity of two triangles as:
Two triangles are similar, if
(i) their corresponding angles are equal.
(ii) their corresponding sides are in the same ratio (or proportion).
Note that if corresponding angles of two triangles are equal, then they are known as equiangular
triangles. We can say that the ratio of any two corresponding sides in two equiangular triangles is always
the same. Now, here we shall start with:
Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle to intersect the other
two sides in distinct points, the other two sides are divided in the same ratio.
It is also called Thales Theorem as it is given by the Thales, a mathematician.
The Centroid: The centroid of a triangle is the point
where the three medians of the triangle meet and is
often described as the triangle’s centre of gravity.
In the triangle ABC, AD, BE and CF are medians.
Point O is the centroid of DABC.
Pythagoras Theorem: In a right-angled triangle, the
square of the hypotenuse is equal to the sum of the
squares of the other two sides.
DABC shown alongside is a right-angled triangle,
in which ∠B = 90°.
According to Pythagoras Theorem,
(Hypotenuse)2=(Base)2 + (Perpendicular)­2
i.e.,AC2=BC2 + AB2.
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Activity 2.1 Triangles: Basic Proportionality Theorem
If a line DE is drawn parallel to side BC to intersect other two sides AB and AC at the points D and E,
AD AE
=
respectively of a DABC, then
DB EC
Fig. 1
This is called ‘Basic Proportionality Theorem’ or ‘Thales Theorem’.
Objective
Materials Required
To verify the ‘Basic Proportionality Theorem’ using parallel line
board and triangle cut-outs.
Pre-requisite Knowledge





Glazed papers of different colours
White chart paper
Geometry box, sketch pens, fevicol
A pair of scissors
Parallel-line board (ruled paper
sheet)
(i ) The statement of basic proportionality theorem: If a line is drawn
parallel to one side of a triangle to intersect the other two sides
in distinct points, the other two sides are divided in the same ratio.
(ii ) Drawing a line parallel to a given line which passes through a given point.
Procedure
(i ) Take three rectangular sheets of ruled paper.
(ii ) Paste the ruled sheets on the white chart paper.
(iii ) Draw an acute-angled triangle, a right-angled triangle and an obtuse-angled triangle on glazed papers
and cut these using scissors.
Fig. 2
(iv ) Paste the acute-angled ∆ABC on a ruled sheet such that the base of triangle coincides with a ruled
sheet. Draw a line l1 parallel to base such that it meets the other two sides at P and Q.
Fig. 3
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(v ) Measure the line segments AP, AQ, BP and CQ with the help of a scale.
(vi ) Repeat the above activity for the right-angled triangle as well as for the obtuse-angled triangle.
Fig. 4
Fig. 5
(vii ) Record the readings in the following table:
Length of the line segment
Triangle ABC
AP
PB
AQ
QC
AP
AQ
PB
QC
Acute-angled
......
......
......
......
......
......
Right-angled
......
......
......
......
......
......
Obtuse-angled
......
......
......
......
......
......
Observation
From the table for each triangle, we find that
AP AQ
.
=
PB QC
Conclusion
For each triangle, Basic Proportionality Theorem is verified by using parallel-line board.
Learning Outcomes
The students will learn that in all the three triangles, the Basic Proportionality Theorem is verified.
Remark
Basic Proportionality Theorem can be applied to acute/obtuse/right-angled triangle.
Suggested Activity
1. Draw an acute-angled DABC, obtuse-angled DDEF and right DPQR. Take two points on the base (or
base produced) of each triangle and draw equal arcs cutting the other two sides. Draw lines l, m,
and n touching the arcs at B1, C1, and D1, F1, and Q1, R1, respectively (as shown in the figures given
on the next page).
Here, l || BC, m || EF and n || QR.
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Measure the line segments AB1, B1B, AC1, C1C, in figure (i ), DD1, D1E, DF1, F1F in figure (ii ), PQ1,
Q1Q, PR1, R1R in figure (iii ) and verify the Basic Proportionality Theorem.
Fig. (i )
Fig. (ii )Fig. (iii )
Viva Voce
Q1.What is the other name of the “Basic Proportionality Theorem”?
Ans. Thales Theorem.
Q2. Are two quadrilaterals having their corresponding angles equal, similar?
Ans. Yes.
Q3. If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct
points, then what is the relation between the ratio in which the two sides of the triangle are
divided?
Ans. Their ratios are same.
Q4.If a line divides any two sides of a triangle in the same ratio, then what is the relation between the
third side of the triangle and the given line?
Ans. They are parallel to each other.
Q5. Can a rhombus and a square be similar?
Ans. No, because they are not equiangular.
Q6. Can a rectangle and a square be similar?
Ans. No, because their sides are not proportional.
Q7. Are the two similar triangles always congruent?
Ans. No.
Q8. Are the two congruent triangles always similar?
Ans. Yes.
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Activity 2.2 Triangles: The Centroid
(a)The centroid is always inside the triangle. It is a point where
the three medians of the triangle intersect.
(b)Each median divides the triangle into two smaller triangles of
equal area.
(c)The centroid is exactly two-third the way along each median.
Objective
To illustrate that the medians of a triangle occur at a point
called the centroid, which always lies inside the triangle.
Fig. 1
Pre-requisite Knowledge
(i )Median of a triangle: The line segment joining a vertex of the triangle
to the mid-point of the opposite side.
(ii )Concurrent point: The point where three or more lines intersect.
(iii )Interior and exterior of a triangle.
(iv )Finding the mid-point of a line segment using paper folding method.
Materials Required
 Coloured
Scale
Pencil
 A pair of
Fevicol
paper
scissors
Procedure
(i)Draw three types of triangle:
(a ) acute-angled triangle
(b )right-angled triangle and
(c ) obtuse-angled triangle.
(ii)Fold the side AB of the acuteangled triangle such that vertex
A falls on B.
(iii)Press it to obtain a crease.
Fig. 2
Fig. 3
(iv)Unfold and mark the fold on AB as K. We obtain a crease as shown by dotted line KC and K as the
mid-point of AB.
(v)Similarly, obtain mid-points of BC and AC as L and M, respectively.
Fig. 4
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(vi)Draw line segments KC, LA and MB which are medians with respect to sides AB, BC and AC,
respectively.
Fig. 5
(vii)Repeat the same activity with the
right-angled triangle DEF and the
obtuse-angled triangle PQR as shown
alongside.
Observations
(i )In each type of triangle, the three
medians are concurrent at a point G as:
(a)in acute-angled ∆ABC, AL, BM and
CK intersect at G.
(b)in right-angled ∆DEF, DL, EM and
Fig. 6
FK intersect at G.
(c)in obtuse-angled ∆PQR, PL, QM and RK intersect at G.
(ii )The point of concurrence of the medians lies in the interior of the triangle.
Conclusion
Medians of all types of triangles are concurrent at a point which lies in the interior of the triangle.
Learning Outcomes
The students will learn that medians of all types of triangles are concurrent at a point which lies in the
interior of the triangle is called centroid.
Remark
Centroid of a triangle always lies inside the triangle.
Suggested Activity
1. Using paper folding method, show that medians of a triangle are concurrent at a point (inside the ∆),
which is the point of trisection of each median.
Viva Voce
Q1.What do we call the point of concurrence of the medians of triangle?
Ans.Centroid.
Q2.Can centroid of a triangle divides the medians into 3 : 1?
Ans.No.
Q3.Can centroid always lies inside of triangle?
Ans.Yes.
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Activity 2.3 Triangles: Pythagoras Theorem
Always consider a right-angled triangle. By using Pythagoras Theorem, always follow the rule:
(Hypotenuse)2 = (Base)2 + (Perpendicular)2
Objective
To verify the Pythagoras Theorem by the method of paper folding,
cutting and pasting.
Pre-requisite Knowledge
Materials Required

(i ) Pythagoras theorem: In a right-angled triangle, the square of
the hypotenuse is equal to the sum of the squares of the other
two sides.
(ii ) Area of a square = Side × Side
(iii ) Algebraic identity: (a + b)2 = a2 + b2 + 2ab.
Chart paper
A pair of scissors
 Fevicol
 Geometry box
 Sketch pens

Procedure
(i ) Draw a right-angled triangle on a chart paper such that its
hypotenuse is ‘c’ and other two sides are ‘a’ and ‘b’.
(ii ) Paint it with green colour.
(iii ) Make 8 copies of this right triangle on green glazed paper.
Fig. 1
(iv ) Construct squares of side
a, b and c as shown in the
adjoining figure. Paint them
as blue, yellow, and red,
respectively.
Fig. 2
(v ) Take 4 copies of the given right-angled triangle along with yellow
and blue squares. Paste them on a drawing sheet as shown in
the given figure.
Fig. 3
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(vi ) Take the remaining 4 copies of the right-angled triangle along with the red square. Paste all these
shapes on a drawing sheet as shown in the following figure.
Fig. 4
Observations
(i ) In square PQRS, each side is (a + b) units.
(ii ) Also, in square ABCD each side is (a + b) units.
⇒ Area of square PQRS = Area of square ABCD.
“  Area of the   Area of the  ’’
(iii ) On removing four triangles from PQRS, we are left with blue square  +  yellow square  .

 

(iv ) On removing the four triangles from ABCD, we are left with “Area of red square”.
(v ) From steps (iii ) and (iv ), we get
 Area of the 
 Area of the   Area of the 
red square  = blue square  +  yellow square 
⇒
 Area of square 
 Area of square   Area of square 
 with side ' c '  =  with side ' a '  +  with side ' b ' 
2
⇒c = a 2 + b 2
a 2 + b 2 = c 2
⇒
⇒
Sum of the squares 
=
of the two sides

of a right ∆


Square of the 
hypotenuse of 

a right ∆


which is the Pythagoras theorem.
Conclusion
Pythagoras Theorem is verified.
Learning Outcomes
The students will learn that in a right-angled triangle, the sum of the squares of two sides is equal to the
square of its hypotenuse.
Remark
Pythagoras Theorem is applicable only on right-angled triangle.
Suggested Activities
1. To verify that the area of an equilateral triangle described on the hypotenuse of a right-angled
triangle is equal to the sum of the areas of equilateral triangles described on the other two sides by
performing an activity.
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Activity Plus in Mathematics-10
2. To verify that the area of a semi-circle described on the hypotenuse of a right-angled triangle is equal
to the sum of the areas of semi-circles described on the other two sides of right-angled triangle.
Viva Voce
Q1. What is called the side opposite to 90° in right-angled triangle?
Ans. Hypotenuse.
Q2. Can a right-angled triangle have two equal sides?
Ans. Yes.
Q3. What is the another name of Pythagoras Theorem which was given by an ancient Indian mathematician?
Ans. Baudhayan Theorem.
Q4. Is Pythagoras Theorem valid for equilateral triangles?
Ans. Yes.
A. Multiple Choice Questions (MCQs)
1. Sides of two similar triangles are in the ratio 3 : 2. Areas of these triangles are in the ratio:
(i )3 : 2
(ii )27 : 8
(iii )9 : 4
(iv )
3: 2.
2. If P, Q and R are the mid-points of sides of BC, CA and AB, respectively of ∆ABC, then the ratio of
the areas of ∆PQR and ∆ABC is:
(i )4 : 5
(ii )2 : 3
(iii )1 : 2
(iv )1 : 4.
3. If in two triangles ABC and PQR,
AB BC CA
, then
=
=
QR PR PQ
(i ) ∆PQR ~ ∆ABC(ii )∆CBA ~ ∆PQR
(iii ) ∆BCA ~ ∆PQR
4. In the given figure, if DE || BC, AC = 4.8 cm and AD = 3 , then measure of
DB 5
AE is:
(i )0.9 cm
(ii ) 1.5 cm
(iii )2.5 cm
(iv ) 1.8 cm.
5. In the given figure, if AT = AQ = 6 cm, AS = 3 cm, TS = 4 cm, then:
(i ) x = 4 cm, y = 5 cm
(iii ) x = 1 cm, y = 2 cm
(iv ) ∆PQR ~ ∆CAB.
(ii ) x = 2 cm, y = 3 cm
(iv ) x = 3 cm, y = 4 cm.
6.In ∆ABC, DE || BC. If BC = 8 cm, DE = 6 cm and area of ∆ADE = 45 cm2, then
the area of ∆ABC is:
(i ) 45 cm2
(iii ) 125 cm2
Activity Plus in Mathematics-10
(ii ) 80 cm2
(iv ) 35 cm2.
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7. If the diagonals of a rhombus are 12 cm and 16 cm, then the length of the side of the rhombus is:
(i ) 20 cm
(ii ) 10 cm
(iii ) 9 cm
(iv ) 8 cm.
8.In ∆ABC and ∆DEF,
(i ) ∠B = ∠D AB BC
. They will be similar if
=
DE FD
(ii ) ∠B = ∠E(iii ) ∠A = ∠F(iv ) ∠A = ∠D.
9. If the ratio of the corresponding sides of two similar triangles is 3 : 2, then the ratio of their
corresponding altitudes is:
(i )3 : 2
(ii )2 : 3
(iii )81 : 16
(iv )9 : 4.
10. In the given figures, ∆ABC ~ ∆PQR.
Then x + y is:
(i ) 4 + 3 3 (ii ) 3 + 4 3 (iii ) 2 + 3 (iv ) 4 + 3 .
11. In the given figure, two line segments AC and BD intersect each other at the point P such that PA
= 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to:
(i ) 100°
(ii ) 60°
12. In the given figure, PQ || BC. If
(i )PQ = BC
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(iii ) 50°
(iv ) 30°.
AP AQ 1
=
= , then
PB QC 2
(ii )PQ2 = BC2(iii ) PQ =
BC
BC
(iv ) PQ =
.
3
2
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13. In the given figure, if ∠BAC = 90° and AD ⊥ BC, then:
2
2
2
2
(i ) BC = BD.CD(ii ) BC = AB.AC(iii ) AD = BD.CD(iv ) AD = AB.AC.
14. If S is a point on side PQ of a ∆PQR such that PS = QS = RS, then
2
2
2
2
2
2
2
2
2
2
(i ) RS = PR.QR
(ii )QR = QS + RS (iii )PQ = PR + QR (iv ) PR = PS + RS .
ar ( ∆ABC ) 9
= , AB = 18 cm and BC = 15 cm, then PR is equal to:
ar ( ∆PQR ) 4
20
(iv ) 8 cm.
(i ) 12 cm
(ii ) 10 cm
(iii )
3
ar ( ∆PQR )
BC 1
is equal to:
16.If ∆ABC ∼ ∆PQR, with
= , then
QR 3
ar ( ∆ABC )
(i ) 1 (ii ) 1 (iii ) 3
(iv ) 9.
9
3
15.If ∆ABC ∼ ∆QRP,
17.In ∆ABC and ∆DEF, ∠B = ∠E, ∠F = ∠C and AB = 3DE. Then, the two triangles are:
(i )neither congruent nor similar
(iii )congruent but not similar
(ii )congruent as well as similar
(iv )similar but not congruent.
B. State ‘true’ or ‘false’ for each of the following statements:
1. Two similar triangles are always congruent.
2. Two congruent triangles are always similar.
3. If in two triangles one pair of corresponding sides are proportional and the included angles are equal,
then the two triangles are similar.
4. The ratio of the areas of similar triangles is equal to the ratio of their corresponding sides.
5. In a right triangle, the square of hypotenuse is equal to the sum of the other two sides.
6. In a triangle, if the sum of the squares of two sides is equal to the square of the third side, then
the angle opposite to the third side is 90°.
7. The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two
corresponding medians.
8. Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.
9. All circles are not similar.
10. All equilateral triangles are similar.
C. Match the following:
Column A
(i ) If in a triangle, a line is drawn parallel to one of its sides
then the other two sides are divided in the same ratio.
(ii ) In a right-angled triangle, the square of the hypotenuse is
equal to the sum of the squares of the other two sides.
(iii ) The sides of two similar triangles are in the ratio of 4 : 9.
The ratio of areas of these triangles:
Activity Plus in Mathematics-10
Column B
(a ) SSS similarity criterion
(b ) 2 : 3
(c ) Pythagoras Theorem
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(iv ) In ∆ABC and ∆PQR, ∠A = ∠P, ∠B = ∠Q
⇒ ∆ABC ~ ∆PQR
(v ) In ∆ABC and ∆PQR, AB = AC = BE
PQ PR QR
⇒ ∆ABC ~ ∆PQR
(d ) Thales Theorem
(e ) AAA similarity criterion
D. Complete the following crossword puzzle, using the given hints.
ACROSS →
1.All congruent triangles are
.......... .
2.The corresponding sides of
two similar triangles are in the
same .......... .
3.The longest side of a rt ∆.
4.The ratio of ......... of two
similar ∆ is equal to the ratio
of squares of corresponding
sides.
Down ↓
5.All sides of an ........... triangle
are equal.
6.Two circles of equal radii are
.......... .
7.According to ............ theorem,
(Hypotenuse)2 = [Sum of
squares of the remaining two
sides].
Answers
A.Multiple Choice Questions (MCQs)
1.(iii )
2.(iv )
3.(iv )
4.(iv ) 5.(iv )
11.(i )
12.(iii ) 13.(iii )
14.(iii ) 15.(ii )
6.(ii )
16.(iv )
7.(ii )
17.(iv )
8.(i )
B.Write ‘true’ or ‘false’ for each of the following statements:
1.False
2.True
3.True
4.False
6.True
7.True
8.True
9.False
9.(i )
10.(i )
5.False
10.True
C.Match the following:
(i ) → (d )(ii ) → (c )(iii ) → (b )(iv ) → (e )(v ) → (a )
D.Crossword Puzzle
1.Similar
6.congruent
2.ratio
7.PYTHAGORAS
3.hypotenuse
4.areas
5.equilateral
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