- Similar Figures
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- Similar Polygons
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- Basic Proportionally Theorem (or Thales Theorem)
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- Criteria for Similarity of Triangles
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- Different Criterion for similarity of the triangles
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- Areas of Similar Triangles
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- Pythagoras Theorem
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- Converse of Pythagoras Theorem

Two figures having the same shape but not necessarily the same size are called similar figures.

Examples:

All circles are similar

All square are similar

All equilateral triangles are similar

All the congruent figures are similar but the converse is not trueAll circles are similar

All square are similar

All equilateral triangles are similar

Two polygons with same number of sides are said to be similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in proportion (or are in the same ratio).

If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

In the ΔABC , if DE || BC,

Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB

Consider ΔADE and ΔBDE.

ar(ΔADE) = ½ x AD x EN

ar(ΔBDE) = ½ x DB x EN

ar( ? ADE) /ar( ? BDE) = ( 1/ 2 x AD x DM )/(1/ 2 x DB x DM) = AD/ DB ...........(1)

Consider ΔADE and ΔDEC

ar(ΔADE) = ½ x AE x DM

ar(ΔDEC) = ½ x EC x DM

ar( ? ADE) /ar( ? DEC) = (1/ 2 x AE x DM)/ (1/ 2 x EC x DM) = AE/ EC ..........(2)

Consider ΔBDE and ΔCED

ar(ΔBDE) = ar(ΔCED) .........(3)

[Since ΔBDE and ΔCED are on the same base, DE, and between the same parallels, BC and DE.]

Therefore

AD/DB = AE/EC [From (1), (2) and (3)]

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Now triangle is also type of polygon and we already know the similarity criteria for that. So Two triangles are said to be similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in proportyion (or are in the same ratio).

Corresponding angles are equal

∠A=∠D,∠B=∠E,∠C=∠F

Corresponding sides are in proportyion (or are in the same ratio).

AB/DE = BC/EF = AC/DF

Symbolically, we write the similarity of these two triangles as

Δ ABC ~ Δ DEF

The symbol ‘~’ stands for ‘is similar to’. Recall that you have used the symbol ‘≅’ for ‘is congruent to’ in Class IX

We should keep the letters in correct order on both sides

We may recall that the below two conditions

(i) corresponding angles are equal and

(ii) corresponding sides are in the same ratio

are required for two polygons to be similar

However, on the basis of last two SSS and AAA criterio 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.

(i) corresponding angles are equal and

(ii) corresponding sides are in the same ratio

are required for two polygons to be similar

However, on the basis of last two SSS and AAA criterio 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.

1) if PQ || RS, prove that Δ POQ ~ Δ SOR

So, ∠ P = ∠ S (As per Alternate angles)

and ∠ Q = ∠ R

Also, ∠ POQ = ∠ SOR (As they are Vertically opposite angles)

Therefore, Δ POQ ~ Δ SOR (AAA similarity criterion)

2) The side lengths of ΔPQR are 16, 8, and 18, and the side lengths of ΔXYZ are 9, 8, and 4. Find out if the triangle is similar

For these questions , it is recommended to compare the ratio of largest side and shortest side and then remaining side

So 18/9=2 ( Longest side)

8/4=2 ( Shortest side)

16/8=2 ( Remaining side)

Since all ratio's are equal , triangles are similar

3) In ΔABC, ∠A = 22° and ∠B = 68°. In ΔDEF, ∠D = 22° and ∠F = 90°.

For ΔABC,three angles are

∠A = 22° and ∠B = 68°

∠C= 180 - ( ∠A+ ∠B) = 90

For ΔDEF,three angles are

∠D = 22

∠E= 180 - ( ∠D+ ∠F) = 68

So from AAA similarity criterion, triangles are similar

ΔABC ~ ΔDEF

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

.

Given Δ ABC ~ Δ DEF ar (ABC)/ar (DEF) =(AB/DE)

Also

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes.

The ratio of the areas of two similar triangles is equal to the sum of the squares of their corresponding angle bisectors.

1) Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm

We know that

ar( ABC)/ar( DEF)= (BC/EF)

64/121=(BC/15.4)

BC= 11.2 cm

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

AC

In a triangle, if 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 a perpendicular is drawn from the vertex of the right angle of a right trianglr to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and similar to each other.

i. e., if in triangle ABC,

<B = 90 and BD - AC, then

(i) ADB ~ ABC

(ii) BDC ~ ABC

(iii) ADB ~ BDC

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 by RHS similarity criterion.

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