In this chapter, we will be discussing specifically the use of similar triangles which is of utmost importance where if it is beyond our reach to physically measure the distance and height with simple measuring instruments.

**Similarity**

Two figures are said to be similar if every aspect of one figure
is proportional to other figure. For example:

The above houses
look the same but different in size. Both the mobile
phones are the same but they vary in their sizes. Therefore,
mathematically we say that two objects are similar if they are of same shape
but not necessarily they need to have the same size. The ratio of the
corresponding measurements of two similar objects must be
proportional.

Here is a box of geometrical shapes. Collect the similar objects
and list out.

In this chapter, we will be discussing specifically the use of
similar triangles which is of utmost importance where if it is beyond our reach
to physically measure the distance and height with simple measuring instruments. The concept of similarity is widely used in the fields of engineering, architecture
and construction.

Here are few applications of similarity

(i) By analyzing the shadows that make triangles, we can
determine the actual height of the objects.

(ii) Used in aerial photography to determine the distance from sky
to a particular location on the ground.

(iii) Used in Architecture to aid in design of their work.

In class IX, we have studied congruent triangles. We can say that
two geometrical figures are congruent, if they have same size and shape. But,
here we shall study about geometrical figures which have same shape but
proportional sizes. These figures are called “similar”.

Congruency is a particular case of similarity. In both the cases,
three angles of one triangle are equal to the three corresponding angles of the
other triangle. _{DABC} _{DPQR} But in congruent
triangles, the corresponding sides are equal. While in similar triangles, the
corresponding sides are proportional.

**Note**

The traingles *ABC* and *PQR* are similar can be
written as ΔABC ~ ΔPQR

The following criteria are sufficient to prove that two triangles
are similar.

If two angles of one triangle are respectively equal to two angles
of another triangle, then the two triangles are similar, because the third
angle in both triangles must be equal. Therefore, AA similarity criterion is same as the AAA similarity
criterion.

So if ∠*A* = ∠*P* = 1 and ∠*B* = ∠*Q* = 2 then Δ*ABC ~ *Δ*PQR.*

If one angle of a triangle is equal to one angle of another
triangle and if the sides including them are proportional then the two
triangles are similar.

Thus if ∠*A* = ∠*P* = 1 and

*AB*/*PQ* = *AC*/*PR* then Δ*ABC ~ *Δ*PQR.*

If three sides of a triangle are proportional to the three
corresponding sides of another triangle, then the two triangles are similar.

So if, *AB /PQ* = *AC /PR* = *BC /QR* then Δ*ABC* *~*
Δ*PQR*

**Thinking Corner:** Are any two right angled triangles
similar? If so why?

1. A perpendicular line drawn from the
vertex of a right angled triangle divides the triangle into two triangles
similar to each other and also to original triangle.

ΔADB *~ *ΔBDC, ΔABC *~ *ΔADB,
ΔABC *~ *ΔBDC

2. If two triangles are similar, then
the ratio of the corresponding sides are equal to the ratio of their
corresponding altitudes.

i.e. if ΔABC *~ *ΔPQR then

AB/PQ = BC/QR = CA/RP = AD/PS = BE/QT =
CF/RU

3. If two triangles are similar, then
the ratio of C the corresponding sides
are equal to the ratio of the
corresponding perimeters.

ΔABC *~ *ΔDEF then

AB/DE = BC/EF = CA/FD = [AB + BC +CA] /
[DE + EF + FD]

4. The ratio of the area of two similar
triangles are equal to the ratio of the squares of their corresponding sides.

5. If two triangles have common vertex and their bases are on the same straight line, the ratio between their areas is equal to the ratio between the length of their bases.

Here, area( ABD) / area( BDC) = AD/DC

Two triangles are said to be similar if their corresponding
sides are proportional.

The triangles are equiangular if the corresponding angles are
equal.

Two triangles,** **Δ*XYZ*** **and** **Δ*LMN*** **are similar because the corresponding angles are equal.

*(i) *∠*X *=* *∠*L*,* *∠*Y *=* *∠*M*,∠*Z *=* *∠*N *(by angles)

*(ii) XY/LM* = *YZ/MN* = *XZ/LN* (by sides)

Here the vertices *X, Y, Z* correspond to the vertices *L,
M, N* respectively. Thus in symbol Δ*XYZ* ~ Δ*LMN*

**Note**

(i) A pair of equiangular triangles are similar.

(ii) If two triangles are similar, then they are equiangular.

Show that Δ*PST* ~ Δ*PQR *

(i) In Δ*PST* and Δ*PQR* ,

Thus, and ∠*P* is common

Therefore, by *SAS* similarity, Δ*PST* ~ Δ*PQR*

(ii) In Δ*PST* and Δ*PQR* ,

Thus, and ∠*P* is common

Therefore, by SAS similarity,

Δ*PST* ~ Δ*PQR*

**Example 4.2**

Is Δ*ABC* ~ Δ*PQR* ?

*Solution*

In** **Δ

The corresponding sides are not proportional.

Therefore Δ*ABC* is not similar to Δ*PQR*

If we change exactly one of the four given lengths, then we can
make these triangles similar.

Observe Fig.4.18 and find ∠*P* .

In Δ*BAC* and Δ*PRQ* , *AB/RQ* = 3/6 = 1/2;

Therefore, *AB/RQ* = *BC/QP* = *PR/CA*

By *SSS* similarity, we have Δ *BAC ~ *Δ*QRP *

∠*P *= *C *(since the
corresponding parts of similar triangle)

∠*P *= *C *=* *180°−*
*(∠*A *+ ∠*B*)* *=* *180° −*
*(90° +* *60°)

∠*P* = 180° − 150° =
30°

**Example 4.4**

A boy of height** **90cm is walking away from the base of a lamp post at a** **speed of 1.2m/sec. If
the lamppost is 3.6m *A * above the ground, find the length
of his shadow cast after 4 seconds.

*Solution*

Given, speed = 1.2 m/s,

time = 4 seconds

distance = speed × time = 1.2×4 = 4.8 m

Let *x* be the length of the shadow after 4 seconds

Since, Δ *ABE* ~ Δ*CDE* , = 4
(since 90 cm = 0.9 m)

4.8 + *x* =4*x* gives 3*x* =4.8 so, *x* =1.6 m

The length of his shadow *DE* = 1.6 m

In Fig.4.20 ∠*A* = ∠*CED* prove that Δ*CAB ~ *Δ*CED. *Also find the value
of *x*.

In Δ*CAB* and Δ*CED* , ∠*C* is common, ∠*A* = *CED*

Therefore, Δ*CAB* ~ Δ*CED* (By *AA* similarity)

**Example 4.6**

In Fig.4.21,** ***QA*** **and** ***PB*** **are perpendiculars to** ***AB*. If** ***AO*** **=** **10** **cm,** ***BO*=* *6* *cm and* PB*=* *9* *cm.
Find* AQ*.

*Solution*

In** **Δ

∠*AOQ *= *BOP *(Vertically
opposite angles)

Therefore, by *AA* Criterion of similarity, Δ*AOQ* ~
Δ*BOP*

The perimeters of two similar triangles** ***ABC*** **respectively 36 cm and
24 cm. If *PQ* = 10 cm, find *AB*.

The ratio of the corresponding sides of similar triangles is same
as the ratio of their perimeters.

Since Δ*ABC* ~ Δ*PQR* ,

**Example 4.8**

If Δ*ABC* is similar to Δ*DEF* such that *BC*= 3
cm, *EF*= 4 cm and area of Δ*ABC* = 54 cm^{2}. Find the area
of Δ*DEF* .

*Solution*

Since the ratio of area of two similar triangles is equal to the
ratio of the squares of any two corresponding sides, we have

Two poles of height ‘*a*’ metres and ‘*b*’ metres are ‘*p*’
metres apart. Prove that the height of the point of intersection of the lines
joining the top of each pole to the foot of the opposite pole is given by *ab* / *a+b*
metres.

Let* **AB*** **and

Let *CL* = *x* and *LA* = *y*.

Then, *x* + *y* = *p*

In Δ*ABC* and Δ*LOC* , we have

∠*CAB *=* *∠*CLO *[each equal to* *90°* *]

∠*C *=* *∠*C *[*C *is common]

Δ*CAB* ~ Δ*CLO* [By *AA* similarity]

In Δ*ALO* and Δ*ACD* , we have

∠*ALO *= *ACD *[each equal
to* *90°* *]* *

∠*A *=* *∠*A *[A is common]

Δ *ALO* ~ Δ*ACD* [by AA similarity]

Therefore, *h* = *ab*
/ *a*+*b*

Hence, the height of the intersection of the lines joining the top
of each pole to the foot of the opposite pole is *ab* / *a*+*b* metres.

So far we have discussed the theoretical approach of similar
triangles and their properties. Now we shall discuss the geometrical
construction of a triangle similar to a given triangle whose sides are in a
given ratio with the corresponding sides of the given triangle.

This construction includes two different cases. In one, the
triangle to be constructed is smaller and in the other it is larger than the
given triangle. So, we use the following term called “scale factor” which
measures the ratio of the sides of the triangle to be constructed with the
corresponding sides of the given triangle. Let us take the following examples
involving the two cases:

Construct a triangle similar to a given triangle PQR with its
sides equal to 3/5 of the corresponding sides of the triangle PQR (scale factor
3/5 < 1)

Given a triangle* **PQR*** **we are required to construct another triangle
whose sides are 3/5 of the corresponding sides of the triangle

1. Construct a Δ*PQR* with any measurement.

2. Draw a ray *QX* making an acute angle with *QR* on
the side opposite to vertex *P.*

3. Locate 5 (the greater of 3 and 5 in 3/5 ) points.

*Q*_{1},* Q*_{2}* *,*Q *_{3}* *,*Q*_{4}*
*and* Q*_{5}* *on* QX *so that* QQ*_{1}*
*=* Q*_{1}*Q*_{2}* *=* Q*_{2}*Q*_{3
}= *Q* _{3}*Q*_{4} = *Q* _{4}*Q*_{5}

4. Join *Q*_{5} *R* and draw a line through *Q*_{3}
(the third point, 3 being smaller of 3 and 5 in 3/5 ) parallel to *Q*_{5}
*R* to intersect QR at *R*¢ .

5. Draw line through *R*’ parallel to the line *RP* to
intersect QP at *P*’.

Then, Δ*P’QR *is the required triangle each of whose sides is three-fifths
of the corresponding sides of Δ*PQR* .

**Example 4.11**

Construct a triangle similar to a given triangle *PQR *with
its sides equal to 7/4 of the corresponding sides of the*P *triangle *PQR*
(scale factor 7/4 > 1)

*Solution*

Given a triangle* **PQR*, we are required construct another triangle
whose sides are 7/4 of the corresponding sides of the triangle *PQR*.

**Steps of construction**

1. Construct a Δ*PQR* with any measurement.

2. Draw a ray *QX* making an acute angle with *QR* on
the side opposite to vertex *P*.

3. Locate 7 points (the greater of 7 and 4 in 7/4)

*Q*_{1},* Q*_{2}* *,*Q *_{3}* *,*Q*_{4},*Q*_{5},*Q*_{6}*
*and* Q*_{7}* *on* QX *so that

*QQ*_{1}* *=* Q*_{1}*Q*_{2}* *=* Q*_{2}*Q*_{3}*
*=* Q *_{3}*Q*_{4}* *=* Q *_{4}*Q*_{5}*
*=* Q*_{5}*Q*_{6}* *=* Q*_{6}*Q*_{7}

4. Join *Q*_{4} (the 4th point, 4 being smaller of 4
and 7 in 7/4 ) to *R* and draw a line through

*Q*_{7} parallel to *Q* _{4} *R*, intersecting the
extended line segment *QR* at *R*’.

5. Draw a line through *R*’ parallel to *RP*
intersecting the extended line segment *QP* at *P*’.

Then Δ*P*′*QR*′ is the required triangle each of whose
sides is seven-fourths of the corresponding sides of Δ*PQR* .

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