Curriculum
Mathematics Grade 10
Grade 10 Mathematics
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Chapter 1. Real Numbers
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Chapter 2. Polynomials
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Chapter 3: PAIR OF LINEAR EQUATIONS IN TWO VARIABLES
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Chapter 4: QUADRATIC EQUATIONS
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Chapter 5: ARITHMETIC PROGRESSIONS
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Chapter 6: TRIANGLES
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Chapter 7: COORDINATE GEOMETRY
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Chapter 8: INTRODUCTION TO TRIGONOMETRY
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Chapter 9: SOME APPLICATIONS OF TRIGONOMETRY
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Chapter 10: CIRCLES
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Chapter 11: AREAS RELATED TO CIRCLES
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Chapter 12: SURFACE AREAS AND VOLUMES
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Chapter 13: STATISTICS
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Chapter 14: PROBABILITY
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Interactive Worksheets
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Chapter Wise Worksheets
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Solved Problems and solutions
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Chapter 6: TRIANGLES
Grade 10 · Mathematics · Chapter 6
Triangles
From similar figures to the Thales Theorem, similarity criteria and the Pythagoras Theorem — master every concept with proofs, worked examples and practice.
| Similar Figures | Thales Theorem | AAA · SSS · SAS | Pythagoras |
What You Will Learn
| ✦ Similar vs congruent figures | ✦ Basic Proportionality (Thales) Theorem & its converse |
| ✦ AAA, AA, SSS and SAS similarity criteria | ✦ Worked Examples 1–10 (fully solved) |
| ✦ Areas of similar triangles theorem | ✦ Pythagoras Theorem with proof |
| ✦ Converse of Pythagoras Theorem | ✦ Practice Sets A–D with collapsible answers |
| ✦ Applications — heights, shadows, map scales | ✦ Chapter summary and 8-point exam check |
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Section 1 Similar Figures — Introduction |
Two figures are congruent if they have the same shape and the same size. Two figures are similar if they have the same shape but not necessarily the same size. This chapter focuses on similarity — a more general relationship.
Key Observations from Everyday Life
| ▶ | All circles are similar — any circle can be enlarged/reduced to match any other. |
| ▶ | All squares are similar — all have equal angles (90°) and equal sides in ratio. |
| ▶ | All equilateral triangles are similar — all angles are 60° and all sides are equal. |
| ▶ | Photographs of the same monument in different sizes are similar — same shape, different size. |
| ▶ | A square and a rectangle are NOT similar — equal angles but sides not proportional. |
| ▶ | A square and a rhombus are NOT similar — sides proportional but angles not equal. |
Definition — Similar Polygons
Two polygons of the same number of sides are similar if:
(i) All corresponding angles are equal, AND
(ii) All corresponding sides are in the same ratio (proportion).
Both conditions must hold — neither alone is sufficient.
| Pair of Figures | Angles Equal? | Sides Proportional? | Similar? |
|---|---|---|---|
| Any two circles | Yes (360°) | Yes (ratio of radii) | Yes ✓ |
| Any two equilateral triangles | Yes (all 60°) | Yes | Yes ✓ |
| Square (3cm) & Rectangle (3×3.5cm) | Yes (all 90°) | No | No ✗ |
| Square (2.1cm) & Rhombus (4.2cm) | No | Yes (ratio 1:2) | No ✗ |
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Section 2 Basic Proportionality Theorem (Thales Theorem) |
Thales (640–546 BC), a Greek mathematician, gave an important result about equiangular triangles: the ratio of any two corresponding sides in two equiangular triangles is always the same. This leads to the Basic Proportionality Theorem.
Theorem 6.1 — Basic Proportionality Theorem (BPT / Thales Theorem)
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.
In △ABC, if DE ∥ BC then AD/DB = AE/EC
Proof (Area Method)
| 1. | Given: In △ABC, DE ∥ BC, with D on AB and E on AC. |
| 2. | Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB. |
| 3. | ar(ADE)/ar(BDE) = (½ × AD × EN) / (½ × DB × EN) = AD/DB …(1) |
| 4. | ar(ADE)/ar(DEC) = (½ × AE × DM) / (½ × EC × DM) = AE/EC …(2) |
| 5. | △BDE and △DEC share the same base DE and lie between the same parallels BC and DE, so ar(BDE) = ar(DEC) …(3) |
| 6. | From (1), (2) and (3): AD/DB = AE/EC ✓ |
Theorem 6.2 — Converse of BPT
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
AD/DB = AE/EC ⇒ DE ∥ BC
Visual Diagram — BPT in Action
| A (apex) | |
| D ———— E | (DE ∥ BC) |
| B —————— C | AD/DB = AE/EC |
WORKED EXAMPLE 1 — Using BPT
If DE ∥ BC in △ABC, prove that AD/AB = AE/AC.
⇒ DB/AD = EC/AE (inverting)
⇒ DB/AD + 1 = EC/AE + 1
⇒ (DB+AD)/AD = (EC+AE)/AE
⇒ AB/AD = AC/AE ⇒ AD/AB = AE/AC ✓
WORKED EXAMPLE 2 — Trapezium with a parallel line
ABCD is a trapezium with AB ∥ DC. E and F are on sides AD and BC such that EF ∥ AB. Prove that AE/ED = BF/FC.
Join AC, meeting EF at G.
In △ADC: EG ∥ DC ⇒ AE/ED = AG/GC …(1) (BPT)
In △CAB: GF ∥ AB ⇒ CG/GA = CF/BF
⇒ AG/GC = BF/FC …(2)
From (1) and (2): AE/ED = BF/FC ✓
WORKED EXAMPLE 3 — Isosceles triangle proof
In △PQR, PS/SQ = PT/TR and ∠PST = ∠PRQ. Prove PQR is isosceles.
⇒ ∠PST = ∠PQR (corresponding angles) …(1)
But ∠PST = ∠PRQ (given) …(2)
From (1) and (2): ∠PQR = ∠PRQ
⇒ PQ = PR (sides opposite equal angles) ⇒ PQR is isosceles. ✓
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Section 3 Criteria for Similarity of Triangles |
Two triangles are similar (~) when corresponding angles are equal AND corresponding sides are proportional. We write △ABC ~ △DEF and always use correct vertex order.
| Criterion | What to Check | Theorem |
|---|---|---|
| AAA | All 3 corresponding angles equal | Theorem 6.3 |
| AA | Only 2 angles equal (3rd follows automatically) | Corollary of Thm 6.3 |
| SSS | All 3 pairs of corresponding sides proportional | Theorem 6.4 |
| SAS | 2 sides proportional AND the included angle equal | Theorem 6.5 |
Theorem 6.3 — AAA Similarity
If in two triangles, all three corresponding angles are equal, then their corresponding sides are proportional and the triangles are similar.
Theorem 6.4 — SSS Similarity
If the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are equal and the triangles are similar.
Theorem 6.5 — SAS Similarity
If one angle of a triangle equals one angle of another, and the sides including these angles are in the same ratio, then the triangles are similar.
Note — RHS Similarity
In two right triangles, if hypotenuse and one side of one are proportional to hypotenuse and one side of the other, then the triangles are similar.
WORKED EXAMPLE 4 — AAA Similarity
In a figure, PQ ∥ RS. Prove that △POQ ~ △SOR.
⇒ ∠P = ∠S (alternate angles) and ∠Q = ∠R (alternate angles)
∠POQ = ∠SOR (vertically opposite angles)
∴ △POQ ~ △SOR (AAA similarity criterion) ✓
WORKED EXAMPLE 5 — SSS Similarity, find ∠P
△ABC: AB=3.8, BC=6, CA=3√3. △PQR: RQ=7.6, QP=12, PR=6√3. Find ∠P.
All three ratios equal 1/2 ⇒ △ABC ~ △RQP (SSS similarity)
∴ ∠C = ∠P (corresponding angles)
∠C = 180° − 80° − 60° = 40° ⇒ ∠P = 40° ✓
WORKED EXAMPLE 6 — SAS Similarity
OA · OB = OC · OD. Show ∠A = ∠C and ∠B = ∠D.
∠AOD = ∠COB (vertically opposite angles) …(2)
From (1) and (2): △AOD ~ △COB (SAS similarity)
∴ ∠A = ∠C and ∠D = ∠B (corresponding angles of similar triangles) ✓
WORKED EXAMPLE 7 — Shadow and Lamp-Post Problem
A girl (height 90 cm) walks 4.8 m from a 3.6 m lamp. Find the length of her shadow.
In △ABE and △CDE: ∠B = ∠D = 90°, ∠E = ∠E (common) ⇒ AA similarity
BE/DE = AB/CD ⇒ (4.8 + x)/x = 3.6/0.9 = 4
4.8 + x = 4x ⇒ 3x = 4.8 ⇒ x = 1.6
Shadow length = 1.6 m ✓
WORKED EXAMPLE 8 — Medians of Similar Triangles
△ABC ~ △PQR with medians CM and RN. Prove: (i) △AMC ~ △PNR (ii) CM/RN = AB/PQ (iii) △CMB ~ △RNQ
AB = 2AM, PQ = 2PN (medians bisect) ⇒ AM/PN = CA/RP and ∠A=∠P
∴ △AMC ~ △PNR (SAS similarity) ✓
(ii) From (i): CM/RN = CA/RP = AB/PQ ⇒ CM/RN = AB/PQ ✓
(iii) CM/RN = AB/PQ = BC/QR = 2BM/2QN ⇒ CM/RN = BC/QR = BM/QN
∴ △CMB ~ △RNQ (SSS similarity) ✓
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Section 4 Areas of Similar Triangles |
Theorem 6.6 — Areas of Similar Triangles
If two triangles are similar, then the ratio of their areas equals the square of the ratio of their corresponding sides.
ar(△ABC) / ar(△PQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)²
Proof Outline
Draw altitudes AM and PN. ar(ABC)/ar(PQR) = (BC × AM)/(QR × PN)
In △ABM and △PQN: ∠B=∠Q (similar triangles), ∠M=∠N=90° ⇒ △ABM ~ △PQN (AA)
⇒ AM/PN = AB/PQ
∴ ar(ABC)/ar(PQR) = (BC/QR) × (AM/PN) = (BC/QR) × (AB/PQ)
Since AB/PQ = BC/QR: ar(ABC)/ar(PQR) = (BC/QR)² = (AB/PQ)² ✓
WORKED EXAMPLE 9 — Area Ratio Problem
D is a point on BC such that ∠ADC = ∠BAC. Show that CA² = CB · CD.
∠BAC = ∠ADC (given)
∠C = ∠C (common)
∴ △ABC ~ △DAC (AA similarity)
⇒ CA/CD = CB/CA ⇒ CA² = CB · CD ✓
WORKED EXAMPLE 10 — Vertical Pole and Tower Shadow
A vertical pole 6 m casts a shadow 4 m. At the same time a tower casts a shadow 28 m. Find the tower’s height.
△ (pole + shadow) ~ △ (tower + shadow) (AA: right angle + same sun angle)
Height of tower / Height of pole = Shadow of tower / Shadow of pole
Height / 6 = 28 / 4 = 7
Height of tower = 42 m ✓
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Section 5 Pythagoras Theorem and Its Converse |
Theorem 6.7 — Pythagoras Theorem
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
AC² = AB² + BC²
(where ∠B = 90° and AC is the hypotenuse)
Proof Using Similarity
| 1. | Given: △ABC with ∠B = 90°. Draw BD ⊥ AC (altitude from right angle to hypotenuse). |
| 2. | In △ADB and △ABC: ∠A = ∠A (common), ∠ADB = ∠ABC = 90° ⇒ △ADB ~ △ABC (AA) ⇒ AD/AB = AB/AC ⇒ AB² = AD × AC …(1) |
| 3. | In △BDC and △ABC: ∠C = ∠C (common), ∠BDC = ∠ABC = 90° ⇒ △BDC ~ △ABC (AA) ⇒ DC/BC = BC/AC ⇒ BC² = DC × AC …(2) |
| 4. | Add (1) and (2): AB² + BC² = AC × (AD + DC) = AC × AC = AC² ✓ |
Theorem 6.8 — Converse of Pythagoras Theorem
In a triangle, if the square of one side equals the sum of squares of the other two sides, then the angle opposite to the first side is a right angle.
If AC² = AB² + BC² ⇒ ∠B = 90°
Proof of Converse
Given: AC² = AB² + BC² in △ABC. Construct △PQR with ∠Q = 90°, PQ = AB, QR = BC.
By Pythagoras: PR² = PQ² + QR² = AB² + BC² = AC² ⇒ PR = AC
So △ABC ≅ △PQR (SSS congruence) ⇒ ∠B = ∠Q = 90° ✓
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Section 6 Practice Exercises with Answers |
Set A — Similar Figures and BPT
Basic identification and application of Thales Theorem.
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(1) In △ABC, DE ∥ BC, AD = 1.5 cm, DB = 3 cm, AE = 1 cm. Find EC.
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(2) In △ABC, DE ∥ BC, AD = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm. Find AB.
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(3) Is EF ∥ QR? PE=3.9 cm, EQ=3 cm, PF=3.6 cm, FR=2.4 cm.
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(4) Is EF ∥ QR? PE=4 cm, QE=4.5 cm, PF=8 cm, RF=9 cm.
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(5) ABCD trapezium with AB∥DC and diagonals meeting at O. Show AO/OC = BO/OD.
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(6) Fill blanks: All _____ are similar. All _____ triangles are similar. (circles / equilateral)
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Reveal Answers ▼
(3) PE/EQ = 3.9/3 = 1.3; PF/FR = 3.6/2.4 = 1.5 — ratios not equal. EF is NOT ∥ QR | (4) PE/QE = 4/4.5; PF/RF = 8/9 = 4/4.5 — equal. EF ∥ QR ✓
(5) Use AA: △AOB ~ △COD (AB∥DC) ⇒ AO/OC = BO/OD | (6) circles; equilateral
Set B — Similarity Criteria
State the criterion used and write triangles in symbolic form.
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(1) △ABC: ∠A=60°, ∠B=80°, ∠C=40°. △PQR: ∠P=60°, ∠Q=80°, ∠R=40°. Similar?
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(2) △ABC: AB=2, BC=2.5, CA=3. △PQR: PQ=4, QR=5, RP=6. Similar?
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(3) △ABC: AB=2.5, ∠A=80°, AC=3. △DEF: DE=5, ∠D=80°, DF=6. Similar?
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(4) In △ODC ~ △OBA, ∠BOC=125° and ∠CDO=70°. Find ∠DOC, ∠DCO and ∠OAB.
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(5) △ABC: AB=3, BC=6, CA=8. △DEF: DE=4.5, EF=9, FD=12. Find ratio of areas.
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(6) E is on side AD produced of parallelogram ABCD and BE intersects CD at F. Prove △ABE ~ △CFB.
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Reveal Answers ▼
(3) Sides 2.5/5=1/2, 3/6=1/2 with included ∠A=∠D=80° — Yes, SAS. △ABC ~ △DEF
(4) ∠DOC = 180°−125° = 55°; ∠DCO = 180°−70°−55° = 55°; ∠OAB = ∠ODC = 70°
(5) Ratio of sides = 2:3; ratio of areas = 4:9 | (6) AB∥CD ⇒ ∠ABE=∠FCB and ∠AEB=∠FBC ⇒ AA similarity
Set C — Pythagoras Theorem
Apply Pythagoras Theorem and its converse.
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(1) ∠B = 90° in △ABC. AB = 6 cm, BC = 8 cm. Find AC.
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(2) In △ABC, AB = 5, BC = 12, CA = 13. Is ∠B = 90°?
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(3) An equilateral triangle has side 2a. Find its height using Pythagoras.
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(4) In △ABC, ∠C = 90°. D is midpoint of AB. Show that AB² = 4CD² − 4AC² + … (Hint: use Pythagoras twice)
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(5) A ladder 10 m long reaches a wall 8 m high. How far is the foot of the ladder from the wall?
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(6) In right △ABC (∠C=90°) with altitude CD to hypotenuse, prove AC² = AD × AB.
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Reveal Answers ▼
(3) Height = √(4a²−a²) = a√3 | (4) CD² = AC² + AD² (Pythagoras in △ACD); AB = 2AD; manipulate to get result
(5) Distance = √(100−64) = √36 = 6 m | (6) △ACD ~ △ABC (AA: ∠A common, ∠C=∠D=90°) ⇒ AC/AB = AD/AC ⇒ AC² = AD×AB ✓
Set D — Mixed Problems and Proofs
Combine multiple concepts. Show all reasoning.
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(1) A girl 1.2 m tall stands 4 m from a 6 m lamp. Find the length of her shadow and the distance of the tip of the shadow from the lamp base.
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(2) In △ABC, altitudes AD and CE intersect at P. Prove: △AEP ~ △CDP and △ABD ~ △CBE.
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(3) If AD and PM are medians of △ABC and △PQR, and △ABC ~ △PQR, prove AB/PQ = AD/PM.
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(4) The areas of two similar triangles are 64 cm² and 100 cm². If one side of the first is 12 cm, find the corresponding side of the second.
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(5) In △ABC, ∠B = 90°. D is a point on BC such that BD = 3 cm, DC = 4 cm, ∠ADB = 90°. Find AC and the area of △ABC.
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(6) A map uses scale 1:50000. If two towns are 8 cm apart on the map, what is the actual distance in km?
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Reveal Answers ▼
(2) ∠AEP = ∠CDP = 90° and ∠APE = ∠CPD (vert. opp.) ⇒ AA; same logic for second pair
(3) △ABC ~ △PQR ⇒ AB/PQ = BC/QR; BC = 2BD, QR = 2QM ⇒ BD/QM = AB/PQ; ∠B = ∠Q ⇒ △ABD ~ △PQM ⇒ AB/PQ = AD/PM ✓
(4) (side²/area): ratio of areas = 64/100 ⇒ ratio of sides = 8/10; corresponding side = 12×(10/8) = 15 cm
(5) AD² = BD×DC = 12 ⇒ AD = 2√3; AB² = BD×BC = 3×7 = 21 ⇒ AB=√21; AC = √(21+49) = √70; Area = ½×√21×7 = 7√21/2 cm²
(6) Actual = 8 × 50000 = 400000 cm = 4 km
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Section 7 Chapter Summary |
All Key Results at a Glance
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Similar Figures Same shape, not necessarily same size. Congruent ⇒ Similar (not vice versa). All circles, all squares, all equilateral triangles are similar. |
BPT (Thales Theorem) DE ∥ BC in △ABC ⇔ AD/DB = AE/EC. Converse equally important — know both directions. |
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Similarity Criteria AAA (or AA): 3 (or 2) equal angles. SSS: all sides proportional. SAS: 2 sides proportional + included angle equal. In triangles, one condition implies the other. |
Areas of Similar Triangles ar(△1)/ar(△2) = (side ratio)². If sides are in ratio k:1, areas are in ratio k²:1. |
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Pythagoras Theorem ∠B=90° ⇒ AC²=AB²+BC². Proved using similarity of triangles formed by the altitude to the hypotenuse. |
Converse of Pythagoras AC²=AB²+BC² ⇒ ∠B=90°. Use this to verify if a triangle is right-angled from its side lengths. |
8-Point Quick-Check Before Your Exam
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1For BPT, always write the ratio as part/part (AD/DB), not part/whole (AD/AB), unless the question specifically uses that form.
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2When writing △ABC ~ △DEF, make sure the vertex order is correct: A↔D, B↔E, C↔F. Wrong order = wrong answer.
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3For triangles, you do NOT need to check both angle and side conditions — proving one set is enough (unlike polygons in general).
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4Area ratio = square of side ratio. If sides are in ratio 3:4, areas are in ratio 9:16. Don’t forget to square!
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5Pythagoras is only valid when you know one angle is 90°. Use the Converse when you want to prove an angle is 90° from given side lengths.
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6The altitude from the right angle to the hypotenuse creates three similar triangles: the whole triangle and the two smaller ones formed.
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7In proof questions, always state the theorem or criterion used: “by AA similarity criterion,” “by Thales Theorem,” etc. Naming earns marks.
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8For application problems (shadows, heights, maps): draw a diagram, label the similar triangles, write the proportion, then solve.
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Grade 10 Maths Chapter 6 — Triangles
Topics covered: similar figures, definition of similar polygons, scale factor, Basic Proportionality Theorem (Thales Theorem), converse of BPT, similarity of triangles, AAA similarity criterion, AA similarity criterion, SSS similarity criterion, SAS similarity criterion, RHS similarity criterion, areas of similar triangles theorem, Pythagoras Theorem proof using similarity, converse of Pythagoras Theorem, applications — shadow problems, heights of towers, map scales. CBSE Class 10 Maths Chapter 6 board exam revision.