- 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 $ \Delta ABC \; , \;if \; DE \; || \; BC $

i) $\frac {AD}{DB}= \frac {AE}{EC}$

ii) $\frac {AD}{AB}= \frac {AE}{AC}$

iii) $\frac {DB}{AB}= \frac {EC}{AC}$

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

Consider $\Delta ADE \; and \; \Delta BDE$

$ar(\Delta ADE) =\frac {1}{2} \times AD \times EN$

$ar(\Delta BDE) =\frac {1}{2} \times DB \times EN$

$\frac {ar( \Delta ADE) }{ar( \Delta BDE)} =\frac {(1/ 2 \times AD \times DM )}{(1/ 2 \times DB \times DM)} =\frac {AD}{ DB}$ ...........(1)

Consider $\Delta ADE \; and \; \Delta DEC$

$ar(\Delta ADE) =\frac {1}{2} \times AE \times DM$

$ar(\Delta DEC) =\frac {1}{2} \times EC \times DM$

$\frac {ar( \Delta ADE) }{ar( \Delta DEC)} =\frac { 1/ 2 \times AE \times DM)}{ (1/ 2 \times EC \times DM)} = \frac {AE}{ EC}$ ..........(2)

Consider $\Delta BDE \; and \; \Delta CED$

$ar(\Delta BDE) = ar(\Delta CED)$ .........(3)

[Since $\Delta BDE \; and \; \Delta CED$ are on the same base, DE, and between the same parallels, BC and DE.]

Therefore

$\frac {AD}{DB}= \frac {AE}{EC}$ [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 proportion (or are in the same ratio).

Corresponding angles are equal

$\angle A=\angle D,\angle B= \angle E,\angle C=\angle F$

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

$ \frac {AB}{DE} = \frac {BC}{EF} = \frac {AC}{DF}$

Symbolically, we write the similarity of these two triangles as

$\Delta ABC \sim \Delta DEF$

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 criterion 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 criterion 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 $\Delta POQ \sim \Delta SOR$

So, $\angle P = \angle S$ (As per Alternate angles)

and $\angle Q = \angle R$

Also, $\angle POQ = \angle SOR$ (As they are Vertically opposite angles)

Therefore, $\Delta POQ \sim \Delta SOR$ (AAA similarity criterion)

2) The side lengths of $\Delta PQR$ are 16, 8, and 18, and the side lengths of $\Delta XYZ$ 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 $\Delta ABC,\; \angle A = 22^0 \; and \; \angle B = 68^0$. In $\Delta DEF, \angle D = 22^0 \; and \; \angle F = 90^0$

For $\Delta ABC$,three angles are

$\angle A = 22^0$ and $\angle B = 68^0$

$\angle C= 180 - ( \angle A+ \angle B) = 90$

For $\Delta DEF$,three angles are

$ \angle D = 22^0 \; and \; \angle F = 90^0$

$\angle E= 180 - ( \angle D+ \angle F) = 68$

So from AAA similarity criterion, triangles are similar

$\Delta ABC \sim \Delta DEF$

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

.

Given $\Delta ABC \sim \Delta DEF$

$\frac {ar (ABC)}{ar (DEF)} =\frac {AB^2}{DE^2} = \frac {BC^2}{EF^2} = \frac {AC^2}{DF^2}$

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 $\Delta ABC ~\sim \Delta DEF$ and their areas be, respectively, 64 cm

We know that

$ \frac {ar( ABC)}{ar( DEF)}= \frac {BC^2}{EF^2}$

$\frac {64}{121}=\frac {BC^2}{15.4^2}$

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.

$AC^2 = AB^2 +BC^2$

- Take a card board of size say 20 cm x 20 cm.
- Cut any right angled triangle and paste it on the cardboard. Suppose that its sides are a, b and c.
- Cut a square of side a cm and place it along the side of length a cm of the right angled triangle.
- Similarly cut squares of sides b cm and c cm and place them along the respective sides of the right angles triangle.
- Color the diagram as shown in below figure. You can choose any other color if you want
- Join BH and AI. These are two diagonals of the square ABIH. The two diagonals intersect each other at the point O.
- Through O, draw $RS || BC$.
- Draw PQ, the perpendicular bisector of RS, passing through O.
- Now the square ABIH is divided in four quadrilaterals. Colour them as shown in figure.
- From the square ABIH cut the four quadrilaterals.
- Now arrange them as given above and You will see the sum of square of a and b matches square of c in second figure

Verify if these are sides of Right angle triangle

- 3,4,5
- 5,12,13
- 7,24,25
- 20,21,29
- 12,16,20
- 15,36,39
- 10,24,26
- 4,5,8

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 triangle 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,

$\angle B = 90$ and BD = AC, then

(i) $\Delta ADB \sim \Delta ABC$

(ii) $\Delta BDC \sim \Delta ABC$

(iii) $\Delta ADB \sim \Delta BDC$

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.

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.