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TRIANGLES
1.TRIANGLES
1 Introduction
You are familiar with triangles and many of their properties from your earlier classes.
In Class IX, you have studied congruence of triangles in detail. Recall that two figures
are said to be congruent, if they have the same shape and the same size. In this
chapter, we shall study about those figures which have the same shape but not necessarily
the same size. Two figures having the same shape (and not necessarily the same size)
are called similar figures. In particular, we shall discuss the similarity of triangles and
apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.
Can you guess how heights of mountains (say Mount Everest) or distances of
some long distant objects (say moon) have been found out? Do you think these have
been measured directly with the help of a measuring tape? In fact, all these heights
and distances have been found out using the idea of indirect measurements, which is
based on the principle of similarity of figures (see Example 7, Q.15 of Exercise 6.3
and also Chapters 8 and 9 of this book).
2 Similar Figures
In Class IX, you have seen that 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.
Now consider any two (or more)
circles [see Fig. 6.1 (i)]. Are they
congruent? Since all of them do not
have the same radius, they are not
congruent to each other. Note that
some are congruent and some are not,
but all of them have the same shape.
So they all are, what we call, similar.
Two similar figures have the same
shape but not necessarily the same
size. Therefore, all circles are similar.
What about two (or more) squares or
two (or more) equilateral triangles
Fig 6.1
[see Fig. 6.1 (ii) and (iii)]? As observed
in the case of circles, here also all
squares are similar and all equilateral
triangles are similar.
From the above, we can say
that all congruent figures are
similar but the similar figures need
not be congruent.
Can a circle and a square be
similar? Can a triangle and a square
be similar? These questions can be
answered by just looking at the
figures (see Fig. 6.1). Evidently
these figures are not similar. (Why?)
Fig 6.2
What can you say about the two quadrilaterals ABCD and PQRS
(see Fig 6.2)?Are they similar? These figures appear to be similar but we cannot be
certain about it.Therefore, we must have some definition of similarity of figures and
based on this definition some rules to decide whether the two given figures are similar
or not. For this, let us look at the photographs given in Fig. 6.3:
Fig 6.3
You will at once say that they are the photographs of the same monument
(Taj Mahal) but are in different sizes. Would you say that the three photographs are
similar? Yes,they are.
What can you say about the two photographs of the same size of the same
person one at the age of 10 years and the other at the age of 40 years? Are these
photographs similar? These photographs are of the same size but certainly they are
not of the same shape. So, they are not similar.
What does the photographer do when she prints photographs of different sizes
from the same negative? You must have heard about the stamp size, passport size and
postcard size photographs. She generally takes a photograph on a small size film, say
of 35mm size and then enlarges it into a bigger size, say 45mm (or 55mm). Thus, if we
consider any line segment in the smaller photograph (figure), its corresponding line
segment in the bigger photograph (figure) will be (45/35) or (55/35) of that of the line segment.
This really means that every line segment of the smaller photograph is enlarged
(increased) in the ratio 35:45 (or 35:55). It can also be said that every line segment
of the bigger photograph is reduced (decreased) in the ratio 45:35 (or 55:35). Further,
if you consider inclinations (or angles) between any pair of corresponding line segments
in the two photographs of different sizes, you shall see that these inclinations(or angles)
are always equal. This is the essence of the similarity of two figures and in particular
of two polygons. We say that:
Two polygons of the same number of sides are similar, if (i) their
corresponding angles are equal and (ii) their corresponding sides are in the
same ratio (or proportion).
Note that the same ratio of the corresponding sides is referred to as the scale
factor (or the Representative Fraction) for the polygons. You must have heard that
world maps (i.e., global maps) and blue prints for the construction of a building are
prepared using a suitable scale factor and observing certain conventions.
In order to understand similarity of figures more clearly, let us perform the following
activity:
3 Similarity of Triangles
What can you say about the similarity of two triangles?
You may recall that triangle is also a polygon. So, we can state the same conditions
for the similarity of two triangles. That is:
Two triangles are similiar, if
their corresponding angles are equal and
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. A famous Greek
mathematician Thales gave an important truth relating
to two equiangular triangles which is as follows:
The ratio of any two corresponding sides in
two equiangular triangles is always the same.
It is believed that he had used a result called
the Basic Proportionality Theorem (now known as
the Thales Theorem) for the same.
To understand the Basic Proportionality
Theorem, let us perform the following activity:
EXAMPLE 1 :
If a line intersects sides AB and AC of a Δ ABC at D and E respectively
and is parallel to BC, prove that (AD / DB) = (AE / EC) (see Fig. 6.13).
Fig. 6.12
Solution :
Fig. 6.13
DE || BC (Given)
So, (AD / DB) = (AE / EC) (Theorem 6.1)
or, (DB / AD) = (EC / AE)
or, [(DB / AD) + 1] = [(EC / AE) + 1]
or, (AB / AD) = (AC / AE)
So, (AD / AB) = (AE / AC)
EXAMPLE 2 :
ABCD is a trapezium with AB || DC.
E and F are points on non-parallel sides AD and BC
respectively such that EF is parallel to AB
(see Fig. 6.14). Show that (AE / ED) = (BF / FC).
Fig. 6.14
Solution :
Let us join AC to intersect EF at G
(see Fig. 6.15).
Fig. 6.15
Let us join AC to intersect EF at G
(see Fig. 6.15).
AB || DC and EF || AB (Given)
So,
EF || DC (Lines parallel to the same line are
parallel to each other)
Now, in Δ ADC,
EG || DC (As EF || DC)
So, (AE / ED) = (AG / GC) (Theorem 6.1)
(1)
Similarly, from Δ CAB,
(CG / AG) = (CF / BF)
i.e., (AG / GC) = (BF / FC) (2)
Therefore, from (1) and (2),
(AE / ED) = (BF / FC)
EXAMPLE 3 :
In Fig. 6.16 (PS / SQ) = (PT / TR) and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle.
Therefore, PQ = PR
(Sides opposite the equal angles)
i.e.,
PQR is an isosceles triangle.
6.4 Criteria for Similarity of Triangles
In the previous section, we stated that two triangles are similar, if (i) their corresponding
angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).
That is, in Δ ABC and Δ DEF, if
∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F and
(AB / DE) = (BC / EF) = (CA / FD) then the two triangles are similar (see Fig. 6.22).
Fig. 6.22
Here, you can see that A corresponds to D, B corresponds to E and C
corresponds to F. Symbolically, we write the similarity of these two triangles as
‘Δ ABC ~ Δ DEF’ and read it as ‘triangle ABC is similar to triangle DEF’. The
symbol ‘~’ stands for ‘is similar to’. Recall that you have used the symbol ‘≅’ for
‘is congruent to’ in Class IX.
Now a natural question arises : For checking the similarity of two triangles, say
ABC and DEF, should we always look for all the equality relations of their corresponding
angles (∠ A = ∠ D, ∠ B = ∠ E, ∠ C = ∠ F) and all the equality relations of the ratios
of their corresponding sides [(AB / DE) = (BC / EF) = (CA / FD)]? Let us examine. You may recall that
in Class IX, you have obtained some criteria for congruency of two triangles involving
only three pairs of corresponding parts (or elements) of the two triangles. Here also,
let us make an attempt to arrive at certain criteria for similarity of two triangles involving
relationship between less number of pairs of corresponding parts of the two triangles,
instead of all the six pairs of corresponding parts. For this, let us perform the following
activity:
Remark :
If two angles of a triangle are respectively equal to two angles of another
triangle, then by the angle sum property of a triangle their third angles will also be
equal. Therefore, AAA similarity criterion can also be stated as follows:
If two angles of one triangle are respectively equal to two angles of another
triangle, then the two triangles are similar.
This may be referred to as the AA similarity criterion for two triangles.
You have seen above that if the three angles of one triangle are respectively
equal to the three angles of another triangle, then their corresponding sides are
proportional (i.e., in the same ratio). What about the converse of this statement? Is the
converse true? In other words, if the sides of a triangle are respectively proportional to
the sides of another triangle, is it true that their corresponding angles are equal? Let us
examine it through an activity :
Remark :
You may recall that either of the two conditions namely, (i) corresponding
angles are equal and (ii) corresponding sides are in the same ratio is not sufficient for
two polygons to be similar. However, on the basis of Theorems 6.3 and 6.4, you can
now say that in case of similarity of the two triangles, it is not necessary to check both
the conditions as one condition implies the other.
Let us now recall the various criteria for congruency of two triangles learnt in
Class IX. You may observe that SSS similarity criterion can be compared with the SSS
congruency criterion.This suggests us to look for a similarity criterion comparable to
SAS congruency criterion of triangles. For this, let us perform an activity.
We now take some examples to illustrate the use of these criteria.
Example 4 :
In Fig. 6.29, if PQ || RS, prove that Δ POQ ~ Δ SOR.
Fig. 6.29
Solution :
PQ || RS (Given)
So,
∠P= ∠S (Alternate angles)
and ∠Q= ∠R
Also, ∠ POQ = ∠ SOR (Vertically opposite angles)
Therefore, Δ POQ ~ Δ SOR (AAA similarity criterion)
Example 5 :
Observe Fig. 6.30 and then find ∠ P.
Fig. 6.30
Solution
In Δ ABC and Δ PQR,
(AB / RQ) = 3.8/7.6 = 1/2, (BC / QP) = 6/12 = 1/2 and (CA / PR) = (3√3) / (6√3) = 1/2
That is, (AB / RQ) = (BC / QP) = (CA / PR)
Δ ABC ~ Δ RQP
So,
(SSS similarity)
Therefore, ∠C= ∠P (Corresponding angles of similar triangles)
But ∠ C = 180° – ∠ A – ∠ B (Angle sum property)
= 180° – 80° – 60° = 40°
So, ∠ P = 40°
Example 6 :
In Fig. 6.31,
OA . OB = OC . OD.
Fig. 6.31
Solution :
OA . OB = OC . OD (Given)
So, (OA / OB) = (OC / OD) (1)
Also, we have ∠ AOD = ∠ COB (Vertically opposite angles) (2)
Therefore, from (1) and (2), Δ AOD ~ Δ COB (SAS similarity criterion)
So, ∠ A = ∠ C and ∠ D = ∠ B (Corresponding angles of similar triangles)
Example 7 :
A girl of height 90 cm is
walking away from the base of a
lamp-post at a speed of 1.2 m/s. If the lamp
is 3.6 m above the ground, find the length
of her shadow after 4 seconds.
Solution :
Let AB denote the lamp-post
and CD the girl after walking for 4 seconds
away from the lamp-post (see Fig. 6.32).
From the figure, you can see that DE is the
shadow of the girl. Let DE be x metres.
Fig. 6.32
Now, BD = 1.2 m × 4 = 4.8 m.
Note that in Δ ABE and Δ CDE,
∠B= ∠D (Each is of 90° because lamp-post
as well as the girl are standing
vertical to the ground)
and ∠E= ∠E (Same angle)
So, Δ ABE ~ Δ CDE
(AA similarity criterion)
Therefore, (BE / DE) = (AB / CD)
i.e., (4.8 + x) / x = (3.6/0.9) (90 cm = 90/100 m = 0.9 m)
i.e., 4.8 + x = 4x
i.e., 3x = 4.8
i.e., x = 1.6
Example 8 :
In Fig. 6.33, CM and RN are
respectively the medians of Δ ABC and
Δ PQR. If Δ ABC ~ Δ PQR, prove that :
Fig. 6.33
Δ AMC ~ Δ PNR
(CM / BN) = (AB / PQ)
Δ CMB ~ Δ RNQ
Solution :
Δ ABC ~ Δ PQR (Given)
So, (AB / PQ) = (BC / QR) = (CA / RP) (1)
and ∠ A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R (2)
But AB = 2 AM and PQ = 2 PN (As CM and RN are medians)
So, from (1), (2AM / 2PN) = (CA / RP)
i.e., (AM / PN) = (CA / RP)
Also, ∠ MAC = ∠ NPR [From (2)] (4)
So, from (3) and (4), Δ AMC ~ Δ PNR (SAS similarity) (5)
From (5), (CM / RN) = (CA / RP) (6)
But (CA / RP) = (AB / PQ) [From (1)] (7)
Therefore, (CM / RN) = (AB / PQ)[From (6) and (7)] (8)
[Note : You can also prove part (iii) by following the same method as used for proving
part (i).]
6.5 Areas of Similar Triangles
You have learnt that in two similar triangles, the ratio of their corresponding sides is
the same. Do you think there is any relationship between the ratio of their areas and
the ratio of the corresponding sides? You know that area is measured in square units.
So, you may expect that this ratio is the square of the ratio of their corresponding
sides. This is indeed true and we shall prove it in the next theorem.
EXAMPLE 9
n Fig. 6.43, the line segment
XY is parallel to side AC of Δ ABC and it
divides the triangle into two parts of equal
areas. Find the ratio (AX / AB).
Fig. 6.43
Solution :
We have XY || AC (Given)
So, ∠ BXY = ∠ A and ∠ BYX = ∠ C
(Corresponding angles)
So, [ar (ABC) / ar (XBY)] = (AB / XB)^{2} (Theorem 6.6)
(1)
Also, ar (ABC) = 2 ar (XBY) (Given)
So, [ar (ABC) / ar (XBY)] = 2/1
Therefore, from (1) and (2),
(AB / XB)^{2} = 2/1 i.e., (AB / XB) = (√2 / 1)
or, (XB / AB) = (1 / √2)
or, 1–(XB / AB) = 1–(1 / √2)
or, (AB – XB) / AB = (√2 – 1) / √2, i.e (AB / XB) = (√2 – 1) / √2 = (2 – √2) / 2
6 Pythagoras Theorem
You are already familiar with the Pythagoras Theorem from your earlier classes. You
had verified this theorem through some activities and made use of it in solving certain
problems. You have also seen a proof of this theorem in Class IX. Now, we shall prove
this theorem using the concept of similarity of
triangles. In proving this, we shall make use of
a result related to similarity of two triangles
formed by the perpendicular to the hypotenuse
from the opposite vertex of the right triangle.
Now, let us take a right triangle ABC, right
angled at B. Let BD be the perpendicular to the
hypotenuse AC (see Fig. 6.45).
Fig. 6.45
You may note that in Δ ADB and Δ ABC
∠A= ∠A
and ∠ ADB = ∠ ABC (Why?)
So, Δ ADB ~ Δ ABC (How?) (1)
Similarly, Δ BDC ~ Δ ABC (How?) (2)
So, from (1) and (2), triangles on both sides of the perpendicular BD are similar
to the whole triangle ABC.
Also, since Δ ADB ~ Δ ABC
and Δ BDC ~ Δ ABC
So, Δ ADB ~ Δ BDC
(From Remark in Section 6.2)
The above discussion leads to the following theorem :
Let us now apply this theorem in proving the
Pythagoras Theorem:
Note : Also see Appendix 1 for another proof of this theorem.
Let us now take some examples to illustrate the use of these theorems.
Solution
Fig. 6.48
EXAMPLE 10
In Fig. 6.48, ∠ ACB = 90° and CD ⊥ AB. Prove that (BC^{2} / AC^{2}) = (BD / AD)
Δ ACD ~ Δ ABC (Theorem 6.7)
So, (AC / AB) = (AD / AC)
or, AC^{2} = AB . AD (1)
Similarly, Δ BCD ~ Δ BAC (Theorem 6.7)
So, (BC / BA) = (BD / BC)
or, BC^{2} = BA . BD (2)
Therefore, from (1) and (2), (BC^{2} / AC^{2}) = (BA ⋅ BD) / (AB ⋅ AD
)= (BD / AD)
EXAMPLE 11
A ladder is placed against a wall such that its foot is at a distance
of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the
length of the ladder.
Fig. 6.49
Solution
Let AB be the ladder and CA be the wall
with the window at A (see Fig. 6.49).
Also, BC = 2.5 m and CA = 6 m
From Pythagoras Theorem, we have:
AB^{2} = BC^{2} + CA^{2}
= (2.5)^{2} + (6)^{2}
= 42.25
So, AB = 6.5
EXAMPLE 12
Fig. 6.50
In Fig. 6.50, if AD ⊥ BC, prove that
AB^{2} + CD^{2} = BD^{2} + AC^{2}
Solution
From Δ ADC, we have AC^{2} = AD^{2} + CD^{2} (Pythagoras Theorem) (1)
From Δ ADB, we have
AB^{2} = AD^{2} + BD^{2}
(Pythagoras Theorem) (2)
Subtracting (1) from (2), we have
AB^{2} - AC^{2} = AD^{2} + BD^{2}
or,AB^{2} + CD^{2} = BD^{2} + AC^{2}
EXAMPLE 13
BL and CM are medians of a
triangle ABC right angled at A. Prove that
4 (BL^{2} + CM^{2}) = 5 BC^{2} .
Fig. 6.51
BL and CM are medians of the
Δ ABC in which ∠ A = 90° (see Fig. 6.51).
From Δ ABC,
BC^{2} = AB^{2} + AC^{2} (Pythagoras Theorem) (1)
From Δ ABL,
BL^{2} = AL^{2} + AB^{2}
or, BL^{2} = (AL / 2)^{2} + AB^{2} (L is the mid-point of AC)
or, BL^{2} = AL^{2} / 4 + AB^{2}
or, 4 BL^{2} = AL^{2} + AB^{2} (2)
From Δ CMA, CM^{2} = AC^{2} + AM^{2}
or, CM^{2} = AC^{2} + (AB / 2)^{2} (M is the mid-point of AB)
or, CM^{2} = AC^{2} + AB^{2} / 4
or, 4 CM^{2} = AC^{2} + AB^{2} (3)
Adding (2) and (3), we have
4 (BL^{2} + CM^{2}) = 5 (AC^{2} + AB^{2})
i.e., 4 (BL^{2} + CM^{2}) = 5 BC^{2} [From (1)]
EXAMPLE 14
O is any point inside a
rectangle ABCD (see Fig. 6.52). Prove that OB^{2} + OD^{2} = OA^{2} + OC^{2}.
Fig. 6.52
Solution
Through O, draw PQ || BC so that P lies on
AB and Q lies on DC.
Now, PQ || BC
Therefore, PQ ⊥ AB and PQ ⊥ DC (∠ B = 90° and ∠ C = 90°)
In this chapter you have studied the following points :
Two figures having the same shape but not necessarily the same size are called similar
figures.
All the congruent figures are similar but the converse is not true.
Two polygons of the same number of sides are similar, if (i) their corresponding angles
are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion).
If a line is drawn parallel to one side of a triangle to intersect the other two sides in
distinct points, then the other two sides are divided in the same ratio.
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the
third side.
If in two triangles, corresponding angles are equal, then their corresponding sides are in
the same ratio and hence the two triangles are similar (AAA similarity criterion).
If in two triangles, two angles of one triangle are respectively equal to the two angles of
the other triangle, then the two triangles are similar (AA similarity criterion).
If in two triangles, corresponding sides are in the same ratio, then their corresponding
angles are equal and hence the triangles are similar (SSS similarity criterion).
If one angle of a triangle is equal to one angle of another triangle and the sides including
these angles are in the same ratio (proportional), then the triangles are similar
(SAS similarity criterion).
The ratio of the areas of two similar triangles is equal to the square of the ratio of their
corresponding sides.
If a perpendicular is drawn from the vertex of the right angle of a right triangle to the
hypotenuse, then the triangles on both sides of the perpendicular are similar to the
whole triangle and also to each other.
In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the
other two sides (Pythagoras Theorem).
If in a triangle, square of one side is equal to the sum of the squares of the other two
sides, then the angle opposite the first side is a right angle.
If in two right triangles, hypotenuse and one side of one triangle are
proportional to the hypotenuse and one side of the other triangle,
then the two triangles are similar. This may be referred to as the
RHS Similarity Criterion.
If you use this criterion in Example 2, Chapter 8, the proof will become
simpler.