Triangles
Triangles : Formulas, Theorems & Examples (Class 10 Maths)
Triangles and Similarity
Triangles are fundamental geometric shapes that form the building blocks for many other polygons. In this chapter, we will delve into the fascinating world of similar triangles. Similarity is a concept that describes shapes that are the same in proportion but may differ in size. Understanding similar triangles is crucial in geometry as it has wide applications in various fields like architecture, engineering, map-making, and even art!
In this chapter, you will learn about:
• Similar Figures and Similar Polygons
• Basic Proportionality Theorem (Thales’ Theorem)
• Criteria for Similarity of Triangles (AAA, AA, SSS, SAS)
• Areas of Similar Triangles
• Pythagoras Theorem and its Converse
1. SIMILAR FIGURES
• DEFINITION: Similar figures are figures that have the same shape but may have different sizes. This means one is an enlargement or reduction of the other.
• Examples:
o Two circles of different radii.
o Two squares of different side lengths.
o Photographs of different sizes of the same object.
o Maps and scale models of buildings.
• Congruent vs. Similar:
o Important Note: All congruent figures are always similar (they have the same shape and the same size!).
o However, the converse is not true: similar figures are not necessarily congruent (they have the same shape but may have different sizes).
2. SIMILAR POLYGONS
• DEFINITION: Two polygons are said to be similar to each other if:
o (i) Corresponding Angles are Equal: This means that if you match up the vertices of the two polygons in the ‘same order’, the angles at those corresponding vertices must be equal.
o (ii) Lengths of Corresponding Sides are Proportional: This means the ratio of the lengths of any two corresponding sides is the same for all pairs of corresponding sides. If polygon A and polygon B are similar, then:
(Side 1 of A) / (Corresponding Side 1 of B) = (Side 2 of A) / (Corresponding Side 2 of B) = … = a constant ratio (called the scale factor).
• Examples:
o Line Segments: Any two line segments are similar because their lengths are always proportional.
o Circles: Any two circles are similar because the ratio of their circumferences to their diameters is always π, and the ratio of their radii acts as the scale factor.
o Squares: Any two squares are similar because all angles are 90 degrees and the ratio of their side lengths is constant.
o Example (Non-Similar Rectangles): Consider a rectangle with sides 2cm and 4cm, and another with sides 3cm and 5cm. While all angles are 90 degrees, the ratio of sides is not constant (2/3 ≠ 4/5). Therefore, these rectangles are NOT similar.
Key Takeaways: Similar Polygons
* Same shape, sides are proportional, corresponding angles are equal.
* Conditions for similarity: (i) Equal corresponding angles, (ii) Proportional corresponding sides.
* Not all polygons with equal angles or proportional sides are necessarily similar (need both conditions to be met).
3. SIMILAR TRIANGLES
• Deduction from Similar Polygons: Two triangles are similar if:
o (i) Corresponding Angles are Equal:
If ΔABC and ΔPQR are similar, then:
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
o (ii) Corresponding Sides are Proportional:
If ΔABC and ΔPQR are similar, then:
AB/PQ = AC/PR = BC/QR
Key Takeaways: Similar Triangles (Definition)
* Two triangles are similar if: * (i) All corresponding angles are equal (∠A=∠P, ∠B=∠Q, ∠C=∠R) * (ii) All corresponding sides are proportional (AB/PQ = AC/PR = BC/QR)
* Both conditions MUST be satisfied for triangles to be similar (by definition).
4. THEOREM 1: BASIC PROPORTIONALITY THEOREM (THALES’ THEOREM)
• Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then it divides the other two sides in the same ratio.
• Diagram:
• Given: In ΔABC, DE || BC, where D is a point on AB and E is a point on AC.
• To prove: AD/DB = AE/EC
• Construction: Draw EM ⊥ AD and DN ⊥ AE. Join B to E and C to D. (To use the area of triangles to establish ratios, we draw perpendiculars (altitudes) EM and DN to sides AD and AE respectively. Joining BE and CD helps us form triangles ABDE and ACDE for comparison.)
• Proof:
o Step 1: Area Ratios in ΔADE and ΔBDE:
In ΔADE and ΔBDE,
ar(ΔADE) / ar(ΔBDE) = (1/2 × AD × EM) / (1/2 × DB × EM) [Area of a triangle = 1/2 × base × height]
= AD / DB …(i)
[Explanation: Here, we consider AD and DB as bases of triangles ADE and BDE respectively. Notice that both triangles share the same altitude EM from vertex E to the line containing AB.]
o Step 2: Area Ratios in ΔADE and ΔCDE:
In ΔADE and ΔCDE,
ar(ΔADE) / ar(ΔCDE) = (1/2 × AE × DN) / (1/2 × EC × DN) [Area of a triangle = 1/2 × base × height]
= AE / EC …(ii)
[Explanation: Similarly, here AE and EC are bases of triangles ADE and CDE, and they share the same altitude DN from vertex D to the line containing AC.]
o Step 3: Equal Areas:
ar(ΔBDE) = ar(ΔCDE) …(iii)
[Reason: ΔBDE and ΔCDE are on the same base DE and between the same parallel lines DE and BC. Triangles on the same base and between the same parallels are equal in area.]
o Step 4: Equating Ratios:
From (i), (ii) and (iii),
Since ar(ΔBDE) = ar(ΔCDE), we have:
AD/DB = AE/EC
Hence Proved.
THEOREM 1: BASIC PROPORTIONALITY THEOREM (BPT) / THALES’ THEOREM
• Example:
In triangle ABC, DE || BC. If AD = 2cm, DB = 3cm, and AE = 3cm, find EC.
AD=2, DB=3, AE=3, EC=?
Solution:
Since DE || BC, by Thales’ Theorem (BPT), we have:
AD/DB = AE/EC
Substituting the given values:
2/3 = 3/EC
Cross-multiplying:
2 * EC = 3 * 3
2 * EC = 9
EC = 9/2 = 4.5 cm
Therefore, EC = 4.5 cm.
Key Takeaways: Thales’ Theorem (BPT)
* If a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally.
* Formula: AD/DB = AE/EC (when DE || BC in ΔABC, and D is on AB, E is on AC)
* Useful for finding unknown side lengths in triangles when parallel lines are involved.
5. CRITERION FOR SIMILARITY OF TRIANGLES
• Introduction: While the definition of similar triangles requires both corresponding angles to be equal AND corresponding sides to be proportional, we have shortcuts or criteria that allow us to prove similarity by checking fewer conditions. These are:
(a) AAA (Angle-Angle-Angle) Similarity Criterion
Explanation: If all three pairs of corresponding angles are equal, it guarantees that the triangles have the same shape. The sides will automatically be in proportion.
Theorem Statement: If in two triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F, then ΔABC ~ ΔDEF.
Diagram (Conceptual): Imagine two triangles where all angles match, they are just scaled versions of each other.
(b) AA (Angle-Angle) Similarity Criterion (Corollary to AAA)
Explanation: If two pairs of angles are equal, the third pair of angles must also be equal (because the sum of angles in a triangle is always 180°). Therefore, AA similarity is sufficient.
Theorem Statement: If in two triangles ABC and DEF, ∠A = ∠D, and ∠B = ∠E, then ΔABC ~ ΔDEF.
Diagram (Conceptual): Same as AAA.
(c) SSS (Side-Side-Side) Similarity Criterion
Explanation: If all three pairs of corresponding sides are in the same ratio, it forces the triangles to have the same shape. The corresponding angles will automatically be equal.
Theorem Statement: If in two triangles ABC and DEF, AB/DE = AC/DF = BC/EF, then ΔABC ~ ΔDEF.
Diagram (Conceptual): Imagine sides of one triangle being scaled versions of the other, the angles must also match to close the shape.
(d) SAS (Side-Angle-Side) Similarity Criterion
Explanation: If we have two sides in proportion and the angle between those sides is equal, it fixes the shape enough for similarity.
Theorem Statement: If in two triangles ABC and DEF, AB/DE = AC/DF and ∠A = ∠D, then ΔABC ~ ΔDEF.
Diagram (Conceptual): Imagine two sides scaled and the angle between them fixed, the triangle shape is defined.
• Examples of Applying Similarity Criteria:
o (a) AA Similarity:
In ΔPQR and ΔXYZ, ∠P = ∠X = 60° and ∠Q = ∠Y = 80°. Therefore, by AA similarity criterion, ΔPQR ~ ΔXYZ.
o (b) SSS Similarity:
In ΔABC and ΔDEF, AB/DE = 2/4 = 1/2, BC/EF = 3/6 = 1/2, AC/DF = 4/8 = 1/2. Since AB/DE = BC/EF = AC/DF, by SSS similarity criterion, ΔABC ~ ΔDEF.
o (c) SAS Similarity:
In ΔLMN and ΔUVW, LM/UV = 3/6 = 1/2, LN/UW = 5/10 = 1/2, and ∠L = ∠U = 50°. Since LM/UV = LN/UW and ∠L = ∠U (included angle), by SAS similarity criterion, ΔLMN ~ ΔUVW.
Key Takeaways: Similarity Criteria
* AA Similarity: Two pairs of equal angles.
* SSS Similarity: Three pairs of proportional sides.
* SAS Similarity: Two pairs of proportional sides and the included angle equal.
* These criteria provide shortcuts to prove triangles are similar without checking ALL conditions from the definition.
6. RESULTS IN SIMILAR TRIANGLES
• If two triangles are similar (e.g., ΔABC ~ ΔDEF), then:
o 1. Ratio of corresponding sides = Ratio of corresponding perimeters:
(Perimeter of ΔABC) / (Perimeter of ΔDEF) = AB/DE = BC/EF = AC/DF
[Explanation: Perimeter is the sum of sides, so if sides are scaled, the perimeter is scaled by the same factor.]
o 2. Ratio of corresponding sides = Ratio of corresponding medians:
(Length of median of ΔABC) / (Length of corresponding median of ΔDEF) = AB/DE = …
[Explanation: Medians are line segments within the triangle, they scale proportionally with the sides.]
o 3. Ratio of corresponding sides = Ratio of corresponding altitudes:
(Length of altitude of ΔABC) / (Length of corresponding altitude of ΔDEF) = AB/DE = …
[Explanation: Altitudes are heights, they also scale proportionally.]
o 4. Ratio of corresponding sides = Ratio of corresponding angle bisector segments:
(Length of angle bisector of ΔABC) / (Length of corresponding angle bisector of ΔDEF) = AB/DE = …
[Explanation: Angle bisector segments also scale proportionally.]
Key Takeaways: Ratios in Similar Triangles
If ΔABC ~ ΔDEF, then:
* (Perimeter of ΔABC) / (Perimeter of ΔDEF) = AB/DE = BC/EF = AC/DF
* (Median of ΔABC) / (Corresponding Median of ΔDEF) = AB/DE = …
* (Altitude of ΔABC) / (Corresponding Altitude of ΔDEF) = AB/DE = …
* (Angle Bisector of ΔABC) / (Corresponding Angle Bisector of ΔDEF) = AB/DE = …
* All these ratios are equal to the ratio of corresponding sides (scale factor).
7. THEOREM 2: AREA OF SIMILAR TRIANGLES THEOREM
• Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Formula: Area of Similar Triangles Theorem
If ΔABC ~ ΔDEF, then:
ar(ΔABC) / ar(ΔDEF) = (AB/DE)² = (BC/EF)² = (AC/DF)²
• Explanation: Area is a two-dimensional measure. When linear dimensions (sides) scale by a factor ‘k’, the area scales by k².
• Results based on Area Theorem:
o 1. Ratio of areas of two similar triangles = Ratio of squares of corresponding altitudes.
o 2. Ratio of areas of two similar triangles = Ratio of squares of corresponding medians.
o 3. Ratio of areas of two similar triangles = Ratio of squares of corresponding angle bisector segments.
• Example:
If ΔABC ~ ΔPQR and ar(ΔABC) = 64 cm², ar(ΔPQR) = 121 cm². If PQ = 15.4 cm, find AB.
Solution:
Since ΔABC ~ ΔPQR,
ar(ΔABC) / ar(ΔPQR) = (AB/PQ)²
64 / 121 = (AB/15.4)²
√(64/121) = AB/15.4
8/11 = AB/15.4
AB = (8/11) * 15.4 = 11.2 cm
Key Takeaways: Area of Similar Triangles Theorem
* Ratio of areas of similar triangles = square of the ratio of corresponding sides.
* ar(ΔABC) / ar(ΔDEF) = (AB/DE)² = (BC/EF)² = (AC/DF)²
* Also, ratio of areas = square of ratio of corresponding altitudes, medians, angle bisectors.
* Useful for finding areas or side lengths when dealing with similar triangles and their areas.
8. THEOREM 3: PYTHAGORAS THEOREM
• Statement: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
• Diagram:
• Given: ΔABC is a right triangle right-angled at B.
• To prove: AB² + BC² = AC²
• Construction: Draw BD ⊥ AC (altitude from B to hypotenuse AC). (To use similarity of triangles to establish side ratios, we draw altitude BD from vertex B to the hypotenuse AC.)
• Proof:
o Step 1: Similarity of ΔABC and ΔADB:
In ΔABC and ΔADB,
∠A = ∠A [Common angle]
∠ABC = ∠ADB = 90° [Given ∠ABC=90° and Construction BD⊥AC]
∴ ΔABC ~ ΔADB [By AA similarity criterion]
o Step 2: Side Ratios from Similarity (ΔABC ~ ΔADB):
∴ AB/AD = AC/AB [Corresponding sides of similar triangles are proportional]
⇒ AB² = AC × AD …(i)
o Step 3: Similarity of ΔABC and ΔBDC:
Now in ΔABC and ΔBDC,
∠C = ∠C [Common angle]
∠ABC = ∠BDC = 90° [Given ∠ABC=90° and Construction BD⊥AC]
∴ ΔABC ~ ΔBDC [By AA similarity criterion]
o Step 4: Side Ratios from Similarity (ΔABC ~ ΔBDC):
∴ BC/DC = AC/BC [Corresponding sides of similar triangles are proportional]
⇒ BC² = AC × DC …(ii)
o Step 5: Adding Equations (i) and (ii):
On adding (i) and (ii), we get
AB² + BC² = AC × AD + AC × DC
= AC × (AD + DC) [Taking AC common]
= AC × AC [From diagram, AD + DC = AC]
= AC²
∴ AB² + BC² = AC²
Hence Proved.
THEOREM 3: PYTHAGORAS THEOREM
• Example:
In a right-angled triangle PQR, right-angled at Q, if PQ = 5cm and QR = 12cm, find PR.
PQ=5, QR=12, PR=?
Solution:
Using Pythagoras Theorem in right-angled ΔPQR:
PR² = PQ² + QR²
PR² = 5² + 12²
PR² = 25 + 144
PR² = 169
PR = √169 = 13 cm
Therefore, PR = 13 cm.
Key Takeaways: Pythagoras Theorem
* Applicable only to right-angled triangles.
* (Hypotenuse)² = (Side 1)² + (Side 2)²
* Formula: AC² = AB² + BC² (if ∠B = 90° in ΔABC)
* Used to find side lengths in right triangles.
9. THEOREM 4: CONVERSE OF PYTHAGORAS THEOREM
• Statement: In a triangle, if the 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.
• Explanation: The converse theorem reverses the statement of the original theorem. If a² + b² = c², then it implies a right triangle with the right angle opposite to side ‘c’.
Formula (for checking if a triangle is right-angled): Converse of Pythagoras Theorem
If in ΔABC, AC² = AB² + BC², then ∠B = 90°
• Example:
Check if a triangle with sides 7cm, 24cm, and 25cm is a right-angled triangle.
Solution:
Let a=7cm, b=24cm, c=25cm (c is the longest side).
Check if a² + b² = c²
a² + b² = 7² + 24² = 49 + 576 = 625
c² = 25² = 25² = 625
Since a² + b² = c², by the Converse of Pythagoras Theorem, the triangle is a right-angled triangle, and the angle opposite the side of length 25cm is the right angle.
Key Takeaways: Converse of Pythagoras Theorem
* Used to check if a triangle is right-angled given its side lengths.
* If (Longest Side)² = (Side 1)² + (Side 2)², then the triangle is right-angled.
* Formula: If AC² = AB² + BC² in ΔABC, then ∠B = 90°.
10. RESULTS BASED ON PYTHAGORAS THEOREM
• (i) Result on Obtuse Triangles:
If in ΔABC, ∠B is obtuse (greater than 90°), and AD ⊥ BC (with D on BC extended), then:
AC² = AB² + BC² + 2 BC.BD
E (Altitude from A to BC extended)
• (ii) Result on Acute Triangles:
If in ΔABC, ∠B is acute (less than 90°), and AD ⊥ BC (with D on BC), then:
AC² = AB² + BC² – 2 BD.BC
Key Takeaways: Pythagoras Theorem Extensions
* Obtuse Triangle (∠B > 90°): AC² = AB² + BC² + 2 BC.BD (where AD ⊥ BC extended)
* Acute Triangle (∠B < 90°): AC² = AB² + BC² – 2 BD.BC (where AD ⊥ BC)
* These are extensions of Pythagoras theorem to non-right triangles, relating sides and projections.
11. SUMMARY OF KEY TAKEAWAYS:
• Similar Figures and Polygons: Shapes with the same form but possibly different sizes. Similar polygons have equal corresponding angles and proportional corresponding sides.
• Basic Proportionality Theorem (Thales’ Theorem): A line parallel to one side of a triangle divides the other two sides proportionally. (AD/DB = AE/EC if DE || BC).
• Similarity Criteria for Triangles:
o AA Similarity: Two pairs of equal angles.
o SSS Similarity: Three pairs of proportional sides.
o SAS Similarity: Two pairs of proportional sides and the included angle equal.
• Ratios in Similar Triangles: Ratios of corresponding sides are equal to the ratio of perimeters, medians, altitudes, and angle bisectors.
• Area of Similar Triangles Theorem: The ratio of areas of similar triangles is the square of the ratio of their corresponding sides.
• Pythagoras Theorem: In a right triangle, (Hypotenuse)² = (Side 1)² + (Side 2)². (AC² = AB² + BC² in right ΔABC, ∠B=90°).
• Converse of Pythagoras Theorem: If in a triangle, (Longest Side)² = (Sum of squares of other two sides), then the angle opposite the longest side is 90°.
• Pythagoras Extensions (Obtuse/Acute Triangles): Formulas to relate sides in obtuse and acute triangles using projections.