In the previous section we saw the third method for obtaining scaled versions. In this section, we will see the relation between perimeters of scaled versions. Later we will see the relation of areas also. We will also see angle bisectors, medians and circumradii.
• Let
the length of sides a given ΔABC be a, b and c cm. So
perimeter will be equal to (a+b+c).
• Let the scale factor be ‘k’.
Then the length of sides of the scaled version will be ka, kb and kc
cm. So the perimeter of the scaled version will be equal to
(ka+kb+kc). This is same as k(a+b+c).
We can write:
■ The perimeter of
the scaled version is equal to ‘k’ times the perimeter of
original triangle.
Next we will see the relation
between areas.
1. Fig.19.25(a) below shows a ΔABC. Length of sides are a, b and c cm.
 |
| Fig.19.25 |
2. Fig.b shows the scaled version ΔA’B’C’. The scale
factor is ‘k’. So the length of it’s sides are ka, kb and kc.
3. Drop perpendiculars CD and C’D’ for both triangles.
4. Now consider
the two smaller triangles on the left sides: ΔADC and ΔA’D’C’. Both have
xo and 90o the same. So the third angle (90-x)o will also be the same.
That means: ΔADC and ΔA’D’C’ are 'similar' or 'scaled versions'.
5. In the smaller triangles ΔADC and ΔA’D’C, the side with length b is transformed to a side with
length kb. In 'scaled versions' the scale factor will be the same for all sides. So the height h will be
transformed to kh. That means, the altitude of ΔA’D’C
is kh.
6. Now we can calculate the areas of ΔADC and ΔA'D'C. We have: Area = 1⁄2 × base × altitude.
• So area of ΔADC = 1⁄2 × c × h = (ch)⁄2
• Area of ΔA’D’C = 1⁄2 × kc × kh = (k2ch)⁄2 .
7. We can write:
■ Area of the scaled version is equal to ‘ k2 ’ times the Area of original triangle.
Now we will see angle bisectors. In fig.19.26(a), ΔABC is the given triangle. It's scaled version ΔPQR is given in fig.b.
1. Consider ∠ACB. It's value
is xo. CD is the angle bisector of this angle. So we get: ∠ACD = ∠BCD =
(x/2)o
7. We can write:
■ Area of the scaled version is equal to ‘ k2 ’ times the Area of original triangle.
Now we will see angle bisectors. In fig.19.26(a), ΔABC is the given triangle. It's scaled version ΔPQR is given in fig.b.
 |
| Fig.19.26 |
2. Since ΔABC and ΔPQR are similar (scaled versions), ∠ACB = ∠PRQ
3. This ∠PRQ is bisected by RS. So
we get ∠PRS = ∠QRS = (x/2)o
4. Now consider the two triangles ΔACD and ΔPRS. Two angles yo and (x/2)o are present in both the
triangles. So the third angle must be the same. That is., ∠ADC = ∠PSR
5. So all the three angles are
same. They are similar triangles.
6. We have information about one
pair of sides. They are AC and PR. We know that PR = k × AC. So the other two sides must
also be scaled by the same factor.
7. Thus we get: RS = k × CD and PS
= k × AD
8. We want the first result in (7).
It shows that, when ΔABC is scaled by a factor 'k', the angle bisector
at C is also scaled by the same factor. The same result can be
obtained at the other vertices A and B also. In general we can write:
■ When a triangle is scaled by a
factor 'k', it's angle bisectors are also scaled by the same factor
Now we will see medians
In fig.19.27(a), ΔABC is the given
triangle. It's scaled version ΔPQR is given in fig.b.
 |
| Fig.19.27 |
1. CD is the median from vertex C
and RS is the median from vertex R. That means D is the midpoint
of AB, and S is the midpoint of PQ
2. Consider the two triangles ACD
and PRS:
PR is the scaled version of
AC. (since ΔABC and ΔPQR are scaled versions) That is., PR = k × AC
3. What about sides AD and PS?
(i) We know that PQ = k × AB.
(ii) So (1/2 × PQ) = [1/2 × (k × AB)] ⇒ (1/2 × PQ) = [k × (1/2 × AB)]
(iii) But 1/2 × PQ = PS and 1/2 × AB =
AD
(iv) Substitute these values in (ii).The result is: PS = k × AD
4. Consider the results in 3(iv) and (2)
• We have two pairs: ♦ AD is scaled to PS ♦ AC is scaled to PR
• The angle between them remains
unchanged. So ΔPRS is a scaled version of ΔACD. The scale factor is 'k'
• So we get RS = k × CD
• The same result can be
obtained for medians drawn from other vertices also. Thus we can
write:
■ When a triangle is scaled by a
factor 'k', it's medians are also scaled by the same factor
So in this section, we saw perimeter, area, angle bisector and median.
There is one more: Circumradius. But we already saw it in solved example 19.11 in the previous section. We can write:
■ When a triangle is scaled by a factor 'k', it's circumradius is also scaled by the same factor
Now we will see some solved examples related to the topics that we saw in this chapter in general.
Solved example 19.12
In the fig.19.28(a) below, ∠ABC = ∠CEF
 |
| Fig.19.28 |
Prove
that EC × AC = FC × BC
Solution:
1. The fig.a is separated into
two different triangles. The inner ΔCEF (fig.b) and the outer ΔABC (fig.c)
2. We find that the two new
triangles have xo in common. The angle at vertex C, whose value is
shown as yo, is also common to both the triangles. This is because,
vertex C is common.
3. Thus we have two angles same
in the two triangles. So the third angle, whose value is shown as zo,
will also be same for both.
4. That means, all the three
angles are same. The triangles are scaled versions of each other. We
can apply theorem 19.2:
• CF⁄AC = EF ⁄AB = CE⁄BC
Take the first and the third ratios. We get: FC⁄AC = EC⁄BC ⇒EC × AC = FC × BC
• CF⁄AC = EF ⁄AB = CE⁄BC
Take the first and the third ratios. We get: FC⁄AC = EC⁄BC ⇒EC × AC = FC × BC
Solved example 19.13
The length of the shadow of a
tree of height 10 m is 4 m. At the same time, the length of the
shadow of a tower is 14 m.
(i) Draw a rough sketch based on
the given details
(ii) Calculate the height of the
tower
Solution:
1. In fig.a, RQ is the tree. RP
is the ray of the sun. It passes through the top end R of the tree,
and then falls on the ground at P. So PQ is the shadow of the tree.
 |
| Fig.19.29 |
2. In fig.b, BC is the tower. CA
is the ray of the sun. It passes through the top end C of the tower,
and then falls on the ground at A. So AB is the shadow of the tower.
3. The ground on which shadow is
formed is horizontal
4. The tree and tower are 'upright'. That is., the tree and tower makes 90o with the ground
5. The shadows are measured at
the same time. That means, the sun’s rays that causes the shadows
will be making the same angle with the horizontal. This is marked as
xo
6. Thus we can complete the rough
sketch. The height of the tower is denoted as ‘h’
7. xo and 90o is present in both
the triangles. So the third angle (90-x)o will also be the same in
both the triangles.
8. We have two triangles. ⊿ABC and ⊿PQR. The angles are same. So we can apply theorem 19.2.
• Let x = x, y = (90-x) and z = 90. Then we can write:
• h⁄10 = 14⁄4 = AC⁄PR = k ⇒ h⁄3 = 2⁄h. This is solution of part (ii)
Take the first and the second. We get: h⁄10 = 14⁄4 ⇒h = 140⁄4 = 35 m
8. We have two triangles. ⊿ABC and ⊿PQR. The angles are same. So we can apply theorem 19.2.
• h⁄10 = 14⁄4 = AC⁄PR = k ⇒ h⁄3 = 2⁄h. This is solution of part (ii)
Take the first and the second. We get: h⁄10 = 14⁄4 ⇒h = 140⁄4 = 35 m
Solved example 19.14
In fig.19.30(a), ∠ADB = ∠BCD = xo.
Prove
that AB × AC = AD2
 |
| Fig.19.30 |
Solution:
1. The fig.a is separated into two different triangles. The inner ΔABD (fig.b) and the outer ΔACD (fig.c)
2. We find that the two new triangles have xo in common. The angle at vertex A, whose value is shown as yo, is also common to both the triangles. This is because, vertex A is common.
3. Thus we have two angles same in the two triangles. So the third angle, whose value is shown as zo, will also be same for both.
4. That means, all the three angles are same. The triangles are scaled versions of each other. We can apply theorem 19.2:
• AB⁄AD = BD⁄CD = AD⁄AC
Take the first and the third ratios. We get: AB⁄AD = AD⁄AC ⇒AB × AC = AD2
• AB⁄AD = BD⁄CD = AD⁄AC
Take the first and the third ratios. We get: AB⁄AD = AD⁄AC ⇒AB × AC = AD2
Solved example 19.15
In ΔABC (fig.19.31), D is the midpoint of
AB. E, F G are the midpoints of AD, BD and CD respectively.
Prove
that ΔABC and ΔEFG are similar
 |
| Fig.19.31 |
Solution:
1. Consider ΔADC and it's inner ΔEDG on the left.
2. E is the midpoint of AD. So ED is half of AD. That means:
■ ED is scaled by a factor '2' to get AD
3. Similarly, since G is the midpoint of CD, We can write:
■ GD is scaled by a factor '2' to get DC
4. The angle at D remains the same. So, applying theorem 19.4, ΔADC is a scaled version of ΔEDG. The scale factor is '2'
5. Thus we get the result:
■ EG is scaled by a factor '2' to get AC
6. Considering ΔBDC, and it's inner ΔFDG, we will get the result:
■ FG is scaled by a factor '2' to get BC
7. Now consider the base AB
• We can write: AD = 2ED and BD = 2FD
• But AB = AD + BD = 2ED + 2FD ⇒ AB = 2(ED+FD) ⇒ AB = 2 EF
That means:
■ EF is scaled by a factor '2' to get AB
8. From (5), (6) and (7), we get the result:
Each side of ΔEFG is scaled by the factor '2' to get corresponding sides of ΔABC. So they are similar.
1. Consider ΔADC and it's inner ΔEDG on the left.
2. E is the midpoint of AD. So ED is half of AD. That means:
■ ED is scaled by a factor '2' to get AD
3. Similarly, since G is the midpoint of CD, We can write:
■ GD is scaled by a factor '2' to get DC
4. The angle at D remains the same. So, applying theorem 19.4, ΔADC is a scaled version of ΔEDG. The scale factor is '2'
5. Thus we get the result:
■ EG is scaled by a factor '2' to get AC
6. Considering ΔBDC, and it's inner ΔFDG, we will get the result:
■ FG is scaled by a factor '2' to get BC
7. Now consider the base AB
• We can write: AD = 2ED and BD = 2FD
• But AB = AD + BD = 2ED + 2FD ⇒ AB = 2(ED+FD) ⇒ AB = 2 EF
That means:
■ EF is scaled by a factor '2' to get AB
8. From (5), (6) and (7), we get the result:
Each side of ΔEFG is scaled by the factor '2' to get corresponding sides of ΔABC. So they are similar.
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