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class 10 introduction of chapter triangle​

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Answer:

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

All congruent figures are similar but all similar figures are not congruent.

SIMILAR POLYGONS

Two polygons are said to be similar to each other, if:

(i) their corresponding angles are equal, and

(ii) the lengths of their corresponding sides are proportional

Example:

Any two line segments are similar since length are proportional

Triangles Class 10 Notes Maths Chapter 6 Q1.1

Any two circles are similar since radii are proportional

Triangles Class 10 Notes Maths Chapter 6 Q2.1

Any two squares are similar since corresponding angles are equal and lengths are proportional.

Triangles Class 10 Notes Maths Chapter 6 Q3.1

Note:

Similar figures are congruent if there is one to one correspondence between the figures.

∴ From above we deduce:

Any two triangles are similar, if their

Triangles Class 10 Notes Maths Chapter 6 Q4.1

(i) Corresponding angles are equal

∠A = ∠P

∠B = ∠Q

∠C = ∠R

(ii) Corresponding sides are proportional

ABPQ=ACPR=BCQR

THALES THEOREM OR BASIC PROPORTIONALITY THEORY

Theorem 1:

State and prove Thales’ Theorem.

Statement:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Triangles Class 10 Notes Maths Chapter 6 Q5.1

Given: In ∆ABC, DE || BC.

To prove: ADDB=AEEC

Const.: Draw EM ⊥ AD and DN ⊥ AE. Join B to E and C to D.

Proof: In ∆ADE and ∆BDE,

ar(ΔADE)ar(ΔBDE)=12×AD×EM12×DB×EM=ADDB ……..(i) [Area of ∆ = 12 x base x corresponding altitude

In ∆ADE and ∆CDE,

ar(ΔADE)ar(ΔCDE)=12×AE×DN12×EC×DN=AEEC

∵ DE || BC …[Given

∴ ar(∆BDE) = ar(∆CDE)

…[∵ As on the same base and between the same parallel sides are equal in area

From (i), (ii) and (iii),

ADDB=AEEC

CRITERION FOR SIMILARITY OF TRIANGLES

Two triangles are similar if either of the following three criterion’s are satisfied:

AAA similarity Criterion. If two triangles are equiangular, then they are similar.

Corollary(AA similarity). If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

SSS Similarity Criterion. If the corresponding sides of two triangles are proportional, then they are similar.

SAS Similarity Criterion. If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.

Results in Similar Triangles based on Similarity Criterion:

Ratio of corresponding sides = Ratio of corresponding perimeters

Ratio of corresponding sides = Ratio of corresponding medians

Ratio of corresponding sides = Ratio of corresponding altitudes

Ratio of corresponding sides = Ratio of corresponding angle bisector segments.

AREA OF SIMILAR TRIANGLES

Theorem 2.

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

Given: ∆ABC ~ ∆DEF

To prove: ar(ΔABC)ar(ΔDEF)=AB2DE2=BC2EF2=AC2DF2

Const.: Draw AM ⊥ BC and DN ⊥ EF.

Proof: In ∆ABC and ∆DEF

Triangles Class 10 Notes Maths Chapter 6 Q6.1

ar(ΔABC)ar(ΔDEF)=12×BC×AM12×EF×DN=BCEF.AMDN …(i) ……[Area of ∆ = 12 x base x corresponding altitude

∵ ∆ABC ~ ∆DEF

∴ ABDE=BCEF …..(ii) …[Sides are proportional

∠B = ∠E ……..[∵ ∆ABC ~ ∆DEF

∠M = ∠N …..[each 90°

∴ ∆ABM ~ ∆DEN …………[AA similarity

∴ ABDE=AMDN …..(iii) …[Sides are proportional

From (ii) and (iii), we have: BCEF=AMDN …(iv)

From (i) and (iv), we have: ar(ΔABC)ar(ΔDEF)=BCEF.BCEF=BC2EF2

Similarly, we can prove that

ar(ΔABC)ar(ΔDEF)=AB2DE2=AC2DF2

∴ar(ΔABC)ar(ΔDEF)=AB2DE2=BC2EF2=AC2DF2

Results based on Area Theorem:

Ratio of areas of two similar triangles = Ratio of squares of corresponding altitudes

Ratio of areas of two similar triangles = Ratio of squares of corresponding medians

Ratio of areas of two similar triangles = Ratio of squares of corresponding angle bisector segments.

Note:

If the areas of two similar triangles are equal, the triangles are congruent.

PYTHAGORAS THEOREM

Theorem 3:

State and prove Pythagoras’ Theorem.

Statement:

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

Given: ∆ABC is a right triangle right-angled at B.

To prove: AB² + BC² = AC²

Const.: Draw BD ⊥ AC

Proof: In ∆s ABC and ADB,

Triangles Class 10 Notes Maths Chapter 6 Q7.1

∠A = ∠A …[common

∠ABC = ∠ADB …[each 90°

∴ ∆ABC ~ ∆ADB …[AA Similarity

∴ ABAD=ACAB ………[sides are proportional]

⇒ AB² = AC.AD

Now in ∆ABC and ∆BDC

∠C = ∠C …..[common]

∠ABC = ∠BDC ….[each 90°]

∴ ∆ABC ~ ∆BDC …..[AA similarity]

∴ BCDC=ACBC ……..[sides are proportional]

BC² = AC.DC …(ii)

On adding (i) and (ii), we get

AB² + BC² = ACAD + AC.DC

⇒ AB² + BC² = AC.(AD + DC)

AB² + BC² = AC.AC

∴AB² + BC² = AC²

CONVERSE OF PYTHAGORAS THEOREM

Theorem 4:

State and prove the converse of Pythagoras’ Theorem.

Statement:

Prove that, 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.

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