Download Explanation of Similarity

Mathematics
Class-X
2015-2016
CHAPTER WITH EXPLAINATION
SIMILARITY
Two figures are said to be similar if they have the same shape but different size.
In the above diagram, ABC, A’B’C’, A’’B’’C’’ and A'’’B’'’C’'’ are similar.
Two triangles are said to be similar if the angles of one triangle are equal to the corresponding
angles of the other and their corresponding sides are proportional.
Two triangles are said to be equiangular if the angles of one are equal to the corresponding
angles of another.
If two angles of one triangle are equal to the two angles of another triangle, the third angle will
be automatically equal and the triangles will be equiangular.
In the above diagram,
1.
are similar if
2.
Note : - Congruent triangles are always similar but converse is not true.
Similarity Criterian
1. SAS – Criteria : - If two triangles have a pair of equal angles and the sides containing these
equal angles are proportional, then the triangles are similar.
In the abvoe diagram, if
1.
2.
then
are similar.
2. AAA or AA Criteria : - If two triangles have two pairs of corresponding angles equal the
triangles are similar.
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Mathematics
Class-X
2015-2016
1.
2.
then
are similar.
3. SSS – Criteria : - If sides of two triangles are proportional, the triangles are similar:
In the above diagram, if
Theorem – 1. A line drawn parallel to one side of a triangle divides the other two sides in the
same ratio.
Given : - PQ is drawn parallel to BC meeting AB at P and AC at Q.
To Prove : AP : PB = AQ : QC
Construction : -
BQ and CD are joined.
Statement
Reason
1.
1.
2.
2.
3.
3.
are on the same base PQ
and between the same parallels PQ and BC.
4. Statements (1), (2) and (3)
4.
Theorem – 2. If a straight line meets two sides of a triangle proportionally it is parallel to the
third side.
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Mathematics
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2015-2016
Given: - A line PQ meets AB at P and AC at Q such that AP : PB = AQ : QC
To Prove: - PQ is parallel to BC
Proof: - If PQ is not parallel to BC, let PR is parallel to BC meeting AC at R.
Statement
Reason
1. Given
1.
2. PR || BC
2. Assumption
3. Statement (2) by B.P.T.
3.
4. Statement (1) and (3)
4.
5. Statement (4)
5.
6. Statement (5)
6.
7. Statement (6)
7.
8. QC = RC
8. Statement (7)
9. Q and R are same line
9. Statement (8)
10. PQ || BC QED.
10. Statement (9)
THEOREM - 3. If a perpendicular drawn from the right angle to the hypotenuse of a right –
angled triangle, the two triangle so formed are similar to the original triangle and to one another.
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Mathematics
Class-X
2015-2016
Given : - ABC is a right angle triangle right angled at A. AD is drawn perpendicular to the
hypotenuse BC.
To Prove : Proof : -
are similar to one another.
Statement
Reason
1. (i) Common angles
(ii) Each being a right angle.
1.
1.
2.
2.
2. Statement (1.) by AAA.
3.
3. (i) Common angle
(ii) Each being a right angle.
1.
2.
4.
4. Statement (3) by AAA
5.
5. Statements (2) and (4)
6.
QED
6. Statements (2), (4) and (5)
THEOREM – 4. The areas of similar triangles are in the ratio of the squares on their
corresponding sides.
Given : To Prove : Construction : AM and DN are drawn perpendiculars to BC and EF respectively.
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Mathematics
Class-X
2015-2016
Proof : Statement
Reason
1. Given
1.
2. Statement (1)
2.
3.
3.
4. Statement (3)
4.
5. Statement (2) and (4)
5.
6. Statement (5)
6.
7. Statement (1)
7.
8. Statement (6) and (7)
8.
Example – 1. A map is drawn to scale 1 cm to 10 km
i. What is the actual distance between two places which on the map is 2cm.
ii. If the actual distance is 1000 km what will be this distance on the map?
iii. What is the actual area which on map occupies 3.6sq.cm
Solution : 1cm on the map = 10km on the land
i. 2cm on the map = 2 x 10km on the land
= 20km on the land
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Mathematics
Class-X
2015-2016
ii.
10km on the land = 1cm on the map
1km on the land = 1/10cm on the map
1000 km on the land = 1/10 x 1000cm on the map
= 100cm on the map
iii. 1 cm on the map = 10km on the land
(1cm2 on the map) = (10km2 on the land)
1cm2 on the map = 100km2 on the land
3.6cm2 on the map = 3.6 x 100km2 on the land
= 360km2 on the land
Example – 2 : On a map drawn to a scale of 1 : 1,25,000. One triangle plot of land has the
following measurements : PQ = 10cm, QR = 8cm and
i. The actual length of PQ in km.
ii. The area of the plot in square km.
Solution : By Pythagoras Theorem
i.
Calculate:
According to the scale
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Mathematics
Class-X
ii.
2015-2016
1cm on the map = 1.25km on the land
(1cm)2on the map = (1.25km)2 on the land
1 sq.cm on the map = 1.5625 sq km on the land
24 sq cm on the map = 24 X 1.5625 sq km on the land = 37.5 sq.km on the land
Example – 3 : In
, right angled at A, AD is drawn in such a way that
BD = 9cm and DC = 7cm, find AB.
If
Solution :
Example – 4 : In the adjoining diagram, ABC is a triangle right angled at A. AD is
perpendicular to BC.
Prove that
i.
ii.
iii.
Solution :
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Mathematics
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i.
ii.
iii.
From (i) and (ii)
Example - 5 : In the figure alongside, AD and CE are two medians of the triangle ABC.
DF is drawn parallel to CE meeting AB at F. Prove that
i. FE = FB
ii. AG : GD = 2 : 1
Solution :
i.
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Mathematics
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2015-2016
FE = FB
ii.
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Page 9