Exercise 8.2 — Similarity Criteria

Criteria for similarity of triangles.

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Exercise 8.2 — Criteria for Similarity of Triangles (Class 10 Maths)

Exercise 8.2 is the heart of Chapter 8, Similar Triangles, in Class 10 Mathematics (CBSE, Telangana, and Andhra Pradesh boards). While Exercise 8.1 established the Basic Proportionality Theorem, this exercise introduces the three formal criteria for proving triangles similar — AAA (Angle-Angle-Angle), SSS (Side-Side-Side), and SAS (Side-Angle-Side). It then builds on these with 13 questions covering proofs, numerical problems, real-world applications, and geometric constructions.

This is the most frequently examined section of Chapter 8 in board exams. The AA similarity shortcut, the perimeter-ratio property, the shadow problem, the perpendicular-altitudes proof, and the similar-triangle construction are all classic exam questions. Mastering Exercise 8.2 gives students a significant advantage in the geometry section of the Class 10 board paper.

AA / AAA Similarity SSS Similarity SAS Similarity Perimeter Ratio Shadow & Lamp Post Problem

The Three Similarity Criteria — What They Mean and When to Use Each

These three criteria are the tools for every proof in Exercise 8.2. Each one provides a different minimal set of conditions that guarantee two triangles are similar (and therefore have proportional sides).

AAA / AA
All three corresponding angles are equal (or just two — the third follows automatically from the angle sum property)
→ Corresponding sides are proportional
SSS
All three pairs of corresponding sides are in the same ratio: AB/DE = BC/EF = CA/FD
→ Corresponding angles are equal
SAS
One angle is equal AND the sides including that angle are in proportion
→ The two triangles are similar
AAA Criterion (full statement)
If in two triangles all three pairs of corresponding angles are equal, then the sides opposite to the equal angles are in the same ratio and hence the triangles are similar.
Since angles in a triangle sum to 180°, if two angles match, the third automatically matches. This gives us the practical AA rule.
SSS Criterion
If the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are equal and hence the triangles are similar.
Converse: If corresponding angles are equal (AAA), then the corresponding sides are in the same ratio.
SAS Criterion
If one angle of a triangle equals one angle of another, and the sides including those angles are proportional, then the two triangles are similar.
Key ratio property: If △ABC ∼ △DEF, then
AB/DE = BC/EF = CA/FD = Perimeter(△ABC) / Perimeter(△DEF)
💡 The perimeter-ratio shortcut: When two triangles are similar, the ratio of their perimeters equals the ratio of any pair of corresponding sides. This is used directly in Exercise 8.2, Q2.

Try This 1 — Identify the Similarity Criterion for Each Pair

These warm-up problems train students to quickly recognise which criterion applies and to write the similarity relation in the correct symbolic form (vertex order matters!).

PartKey observationsCriterionRelationSimilar?
(i) ∠FHG = ∠IHK (VOA); FG ∥ IK → ∠F = ∠K, ∠G = ∠I (alt. int.) AAA △FHG ∼ △KHI ✔ Yes
(ii) PQ/LM = 6/3 = 2; QR/MN = 10/4 = 5/2 — ratios differ ✘ No
(iii) AX/AB = AY/AC = 2/5; ∠A = ∠A (common) SAS △AXY ∼ △ABC ✔ Yes
(iv) AP/AB = AJ/AC = 3/8; ∠A = ∠A (common) SAS △APJ ∼ △ABC ✔ Yes
(v) ∠A = ∠B = 90°; ∠AOQ = ∠BOP (VOA) AA △OAQ ∼ △OBP ✔ Yes
(vi) ∠A = ∠Q (40°); ∠B = ∠P (60°); ∠C = ∠R (80°) AAA △ABC ∼ △QPR ✔ Yes
(vii) AB/PQ = 2/5; BC/QR = 1/2 — ratios differ ✘ No
(viii) ∠A = ∠P = 70° but AB/PQ ≠ AC/PR (12/5 ≠ 2) ✘ No
📌 Vertex order in similarity notation is critical. △FHG ∼ △KHI means ∠F = ∠K, ∠H = ∠H, ∠G = ∠I. Writing △FGH ∼ △KIH would be wrong even if the triangles are similar — the correspondence must be stated precisely. Board examiners check this.

Try This 2 — Find the Value of x in Similar Triangle Problems

Once similarity is given, corresponding sides are proportional. Set up the ratio equation, cross-multiply, and solve. The challenge is identifying which sides correspond correctly — always match vertices in the order stated by the similarity relation.

PartGiven SimilarityRatio UsedWorkingx =
(i) △PRQ ∼ △LST QR/TS = PQ/LT 3/4.5 = 5/x → x = 5×4.5/3 7.5
(ii) △ABC ∼ △PQC AB/PQ = BC/QC 5/x = 6/3 → x = 5×3/6 2.5
(iii) △ABC ∼ △EDC AB/ED = BC/DC 24/14 = 22/x → x = 22×14/24 12.8
(iv) △RAB ∼ △RST RA/RS = AB/ST 6/8 = 9/x → x = 8×9/6 12
(v) △PMN ∼ △PQR MN/QR = PN/PR 5/15 = 4/(4+x) → 4+x = 12 8
(vi) △XAB ∼ △XZY XA/XZ = AB/ZY x/(x+7.5) = 12/18 → 3x = 2x+15 15
(vii) △ABC ∼ △EDC AB/ED = BC/DC 1.6/x = 1.5/15 → x = 1.6×10 16
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Question 1 — ∠ADE = ∠B: Prove △ABC ∼ △ADE and Find DE

Question 1 · Part (i)
Show that △ABC ∼ △ADE given ∠ADE = ∠B
In △ABC and △ADE: ∠A = ∠A (common angle) ∠ADE = ∠B (given) Two angles equal → by AA similarity: ∴ △ABC ∼ △ADE ✓
Question 1 · Part (ii)
AD = 3.8, AE = 3.6, BE = 2.1, BC = 4.2 cm. Find DE.
△ABC ∼ △ADE → AB/AD = BC/DE AB = AE + BE = 3.6 + 2.1 = 5.7 cm 5.7/3.8 = 4.2/DE DE = (4.2 × 3.8) / 5.7 = 15.96 / 5.7 DE = 2.8 cm ✓

Question 2 — Perimeters 30 cm and 20 cm; One Side 12 cm: Find the Corresponding Side

Question 2 · Solution
△ABC ∼ △DEF. Perimeters 30 cm and 20 cm. AB = 12 cm. Find DE.
Key property: AB/DE = Perimeter(△ABC) / Perimeter(△DEF) 12/DE = 30/20 DE = (12 × 20) / 30 = 240/30 DE = 8 cm ✓
💡 The perimeter ratio equals the sides ratio. This is a direct consequence of the fact that all three pairs of sides share the same scale factor. If the scale factor is k, every side of △DEF = k × corresponding side of △ABC, and the perimeters scale by the same factor k.

Question 3 — AB ∥ CD ∥ EF: Find x and y

AB = 7.5, DC = y, EF = 4.5, BC = x, CF = 3. Two pairs of similar triangles are created by the parallel lines — one sharing vertex C with vertically opposite angles, another sharing vertex B.

Question 3 · Solution
AB ∥ CD ∥ EF. Given AB=7.5, EF=4.5, CF=3. Find x (=BC) and y (=DC).
Finding x: In △ABC and △EFC: ∠ACB = ∠ECF (VOA); ∠BAC = ∠FEC (alt. int.); ∠ABC = ∠EFC (alt. int.) By AAA: △ABC ∼ △EFC AB/EF = BC/FC → 7.5/4.5 = x/3 x = (7.5 × 3)/4.5 = 22.5/4.5 x = 5 cm Finding y: In △BDC and △BEF: ∠B = ∠B (common); ∠D = ∠E (corr.); ∠C = ∠F (corr.) By AAA: △BDC ∼ △BEF BC/BF = DC/EF → 5/(5+3) = y/4.5 → 5/8 = y/4.5 y = (4.5 × 5)/8 = 22.5/8 y = 2.8 cm ✓

Question 4 — Girl Walking Away from Lamp Post: Find the Shadow Length

This is a classic real-world application of AA similarity. The lamp post and the girl act as the two parallel vertical sides of two nested similar triangles, with the ground as the base and the light rays forming the hypotenuse.

A (lamp) B C (girl) D E BD = 4.8 m DE = x 3.6 m 0.9 m
Lamp post and girl — nested similar triangles
△ABE ∼ △CDE by AA (right angles + common ∠E)
Question 4 · Solution
Lamp height = 3.6 m, girl height = 0.9 m, speed = 1.2 m/s. Find shadow length after 4 s.
Distance walked in 4 s: BD = 4 × 1.2 = 4.8 m Let shadow length DE = x m, so BE = BD + DE = 4.8 + x In △ABE and △CDE: ∠E = ∠E (common); ∠ABE = ∠CDE = 90° By AA: △ABE ∼ △CDE AB/CD = BE/DE → 3.6/0.9 = (4.8 + x)/x 4 = (4.8 + x)/x → 4x = 4.8 + x → 3x = 4.8 x = 1.6 m ∴ Shadow length = 1.6 m ✓

Question 5 — △ABC ∼ △PQR: CM and RN are Medians — Three Sub-Parts

Question 5 · Part (i)
Prove △AMC ∼ △PNR
CM and RN are medians → AM = ½AB, PN = ½PQ In △AMC and △PNR: ∠A = ∠P (△ABC ∼ △PQR) AC/PR = AB/PQ (△ABC ∼ △PQR) = (½AB)/(½PQ) = AM/PN So AC/PR = AM/PN and ∠A = ∠P By SAS similarity: ∴ △AMC ∼ △PNR ✓
Question 5 · Part (ii)
Prove CM/RN = AB/PQ
From part (i): △AMC ∼ △PNR CM/RN = AM/PN = (½AB)/(½PQ) = AB/PQ ∴ CM/RN = AB/PQ ✓
Question 5 · Part (iii)
Prove △CMB ∼ △RNQ
∠B = ∠Q (△ABC ∼ △PQR) BC/QR = AB/PQ = (½AB)/(½PQ) = BM/QN So BC/QR = BM/QN and ∠B = ∠Q By SAS similarity: ∴ △CMB ∼ △RNQ ✓

Question 6 — Trapezium Diagonals: Prove OA/OC = OB/OD Using Similarity

Question 6 · Solution
ABCD is a trapezium, AB ∥ DC. Diagonals meet at O. Show OA/OC = OB/OD.
In △OAB and △OCD: ∠AOB = ∠COD (vertically opposite angles) ∠OAB = ∠OCD (alternate interior angles, AB ∥ DC) ∠OBA = ∠ODC (alternate interior angles) By AAA similarity: △OAB ∼ △OCD ∴ OA/OC = OB/OD (ratios of corresponding sides) Hence proved. ✓
📌 This result — trapezium diagonals divide each other in the same ratio — is equivalent to what was proved using BPT in Exercise 8.1 Q8. The similarity approach is more elegant and is the preferred method for this exercise.

Question 7 — AB, CD, PQ ⊥ BD: Prove 1/x + 1/y = 1/z

This is one of the most beautiful proofs in the exercise. Three perpendiculars stand on a common base. By finding two pairs of similar triangles and adding their ratio equations, a reciprocal identity emerges.

Question 7 · Solution
AB ⊥ BD (= x), PQ ⊥ BD (= z), CD ⊥ BD (= y). Prove 1/x + 1/y = 1/z.
∠B = ∠Q = ∠D = 90° → AB ∥ PQ ∥ CD In △BQP and △BDC: ∠B = ∠B (common); ∠Q = ∠D = 90°; ∠P = ∠C (corresponding) By AAA: △BQP ∼ △BDC → BQ/BD = PQ/CD → BQ/BD = z/y ...(1) In △DQP and △DBA: ∠D = ∠D (common); ∠Q = ∠B = 90°; ∠P = ∠A (corresponding) By AAA: △DQP ∼ △DBA → QD/BD = PQ/AB → QD/BD = z/x ...(2) Adding (1) and (2): (BQ + QD)/BD = z/y + z/x BD/BD = z(1/y + 1/x) → 1 = z(1/y + 1/x) ∴ 1/x + 1/y = 1/z ✓

Question 8 — Flag Pole and Building Shadow: Find Building Height

Question 8 · Solution
Flag pole = 4 m, shadow = 6 m. Building shadow = 24 m. Find building height.
At the same time of day, sun angle is equal for both objects. In △ABC (flag) and △PQR (building): ∠A = ∠P (same sun angle) and ∠B = ∠Q = 90° By AA: △ABC ∼ △PQR AB/PQ = BC/QR → 4/h = 6/24 h = (4 × 24)/6 = 4 × 4 h = 16 m ✓
Real-world application: This is how surveyors historically estimated heights of unreachable objects — by measuring shadows at the same time of day and using similar triangle ratios. The same method applies to measuring tree heights, building heights, or even mountain heights indirectly.

Question 9 — Angle Bisectors CD and GH in Similar Triangles △ABC ∼ △FGE

Question 9 · Solution
△ABC ∼ △FGE, CD and GH are angle bisectors. Prove (i) CD/GH = AC/FG (ii) △DCB ∼ △HGE (iii) △DCA ∼ △HGF
△ABC ∼ △FGE → ∠ACB = ∠FGE CD bisects ∠ACB → ∠ACD = ½∠ACB GH bisects ∠FGE → ∠FGH = ½∠FGE ∴ ∠ACD = ∠FGH ...(1) and ∠DCB = ∠HGE ...(2) (i) In △ACD and △FGH: ∠A = ∠F (△ABC ∼ △FGE); ∠ACD = ∠FGH [from (1)] By AA: △ACD ∼ △FGH → CD/GH = AC/FG ✓ (ii) In △DCB and △HGE: ∠B = ∠E (△ABC ∼ △FGE); ∠DCB = ∠HGE [from (2)] By AA: △DCB ∼ △HGE ✓ (iii) In △DCA and △HGF: ∠A = ∠F; ∠ACD = ∠FGH [from (1)] By AA: △DCA ∼ △HGF ✓

Question 10 — Altitudes AX and DY of Similar Triangles: Prove AX:DY = AB:DE

Question 10 · Solution
△ABC ∼ △DEF, AX ⊥ BC, DY ⊥ EF. Prove AX:DY = AB:DE.
In △ABX and △DEY: ∠B = ∠E (△ABC ∼ △DEF) ∠X = ∠Y = 90° (altitudes) By AA similarity: △ABX ∼ △DEY AX/DY = AB/DE (ratios of corresponding sides) ∴ AX : DY = AB : DE ✓
💡 Generalisation: In similar triangles, all corresponding lengths — altitudes, medians, angle bisectors, perimeters — are in the same ratio as the corresponding sides. Only areas are in the ratio of the square of the corresponding sides (covered in Exercise 8.3).

Questions 11, 12, 13 — Constructing Similar Triangles

The last three questions are geometric constructions. The method is identical in all cases: draw a ray from a vertex, mark equal arcs, join the last point to the opposite vertex, and draw a parallel line at the required point to create the new triangle. The ratio used (m:n where the new triangle is m/n times the original) determines how many arcs to mark and which point to use.

Question 11
Construct a triangle similar to △ABC with sides = (5/3) of corresponding sides.
  1. Draw △ABC with given measurements.
  2. Draw ray BX making an acute angle with BC.
  3. Mark 5 equal arcs B₁, B₂, B₃, B₄, B₅ on BX (use compass, same radius).
  4. Join B₃ to C (since denominator = 3 → join 3rd point to C).
  5. Draw a parallel to B₃C through B₅ — this meets extended BC at C₁.
  6. Draw a parallel to AC through C₁ — this meets extended BA at A₁.
  7. △A₁BC₁ is the required similar triangle with sides = 5/3 of △ABC.
📌 Rule: When the ratio is m/n where m > n (enlargement): mark m arcs, join at n, draw parallel at m. When m < n (reduction): mark n arcs, join at n, draw parallel at m. Q12 uses m=2, n=3 (reduction); Q13 uses m=3, n=2 (enlargement, since 1½ = 3/2).
Question 12
Draw △ABC with AB=4, BC=5, CA=6 cm. Construct a similar triangle with sides = (2/3) of △ABC.
  1. Draw △ABC with AB = 4 cm, BC = 5 cm, CA = 6 cm.
  2. Draw ray AX at an acute angle to AB.
  3. Mark 3 equal arcs A₁, A₂, A₃ on AX.
  4. Join A₃ to B (denominator = 3).
  5. Draw parallel to A₃B through A₂ — meets AB at B₁.
  6. Draw parallel to BC through B₁ — meets AC at C₁.
  7. △AB₁C₁ is the required triangle with sides = 2/3 of △ABC.
Question 13
Draw isosceles △ABC with base AB=8 cm, altitude=4 cm. Construct a similar triangle with sides = 1½ = (3/2) times △ABC.
  1. Convert 1½ to 3/2 (mark 3 arcs, join at 2nd).
  2. Draw isosceles △ABC with AB = 8 cm and altitude CD = 4 cm.
  3. Draw ray AX at an acute angle to AB.
  4. Mark 3 equal arcs A₁, A₂, A₃ on AX.
  5. Join A₂ to B (denominator = 2).
  6. Draw parallel to A₂B through A₃ — meets extended AB at B₁.
  7. Draw parallel to BC through B₁ — meets extended AC at C₁.
  8. △AB₁C₁ has sides = 3/2 of △ABC.

All 13 Questions — Quick Reference at a Glance

Q#TopicCriterion / ToolKey Result
Try This 1Identify similarity criterion (8 parts)AA / SAS / SSS / Nonei,iii,iv,v,vi=Similar; ii,vii,viii=Not Similar
Try This 2Find x (7 parts)Ratio of corresponding sides7.5, 2.5, 12.8, 12, 8, 15, 16
Q1∠ADE = ∠B → △ABC ∼ △ADE; find DEAADE = 2.8 cm
Q2Perimeters 30 & 20, one side = 12Perimeter ratioCorresponding side = 8 cm
Q3AB ∥ CD ∥ EF, find x and yAAA (twice)x = 5, y = 2.8
Q4Shadow length — lamp post problemAA (right angles)Shadow = 1.6 m
Q5△ABC ∼ △PQR with medians CM, RNSAS (twice)CM/RN = AB/PQ
Q6Trapezium diagonals: OA/OC = OB/ODAAA△OAB ∼ △OCD
Q7Three perpendiculars: 1/x + 1/y = 1/zAAA (twice, add)Elegant reciprocal identity
Q8Flag pole shadow → building heightAABuilding = 16 m
Q9Angle bisectors in similar trianglesAA (three times)CD/GH = AC/FG
Q10Altitudes AX:DY = AB:DEAA△ABX ∼ △DEY
Q11Construct △ similar to △ABC, sides = 5/3ConstructionMark 5 arcs, join at 3, parallel at 5
Q12Construct with sides = 2/3ConstructionMark 3 arcs, join at 3, parallel at 2
Q13Isosceles, sides = 3/2 (= 1½)ConstructionMark 3 arcs, join at 2, parallel at 3
Q1DE = 2.8 cm
Q2Side = 8 cm
Q3x=5, y=2.8
Q4Shadow = 1.6 m
Q71/x+1/y = 1/z
Q8Building = 16 m

Common Mistakes to Avoid in Exercise 8.2

  • Wrong vertex order in similarity notation: △ABC ∼ △PQR means A↔P, B↔Q, C↔R. Writing the similarity in the wrong order makes all subsequent ratios incorrect. Always match angles in order when stating the similarity.
  • Using SAS with wrong sides: In SAS similarity, the sides that must be proportional are the ones that include the equal angle. Including an angle between two sides that are not the ones given is incorrect.
  • In Q3, trying to find y before x: You must find x = BC first (from △ABC ∼ △EFC), then use x to find y (from △BDC ∼ △BEF). The second similarity depends on BF = x + 3.
  • In construction questions (Q11–Q13), wrong arc count for enlargements: For m/n where m > n (scale up), mark m arcs and join at n. Students often join at m and draw parallel at n — the opposite — which gives a reduction instead of an enlargement.
  • Forgetting the "~ symbol" vertex correspondence: When writing △ABC ∼ △DEF, the statement itself carries the complete angle and side correspondence. Never use ≅ (congruent) when you mean ∼ (similar).
High-risk board exam trap (Q7): Students try to prove 1/x + 1/y = 1/z by working with a single pair of similar triangles. You need two separate similarity relationships (△BQP ∼ △BDC and △DQP ∼ △DBA) and then add the resulting equations. Stopping after just one pair will not give the result.

What Exercise 8.2 Prepares You For

The AA, SSS, and SAS similarity criteria proved here are the foundation for Exercise 8.3, which introduces the ratio of areas of similar triangles (= square of the side ratio). The perimeter-ratio relationship from Q2 extends naturally to the area ratio in 8.3. The shadow problem structure (Q4, Q8) reappears in trigonometry applications in Chapter 12.

The construction technique from Q11–Q13 is also the foundation for all geometric constructions involving scale — drawing maps, architectural plans, and scale models all use the same principles. In coordinate geometry, the section formula is the algebraic version of the same proportion idea.

For Telangana and Andhra Pradesh board exams, Q4 (shadow problem), Q7 (1/x + 1/y = 1/z), Q5 (medians of similar triangles), and Q2 (perimeter ratio) are among the most frequently tested questions. Q11–Q13 constructions appear as 5-mark problems every few years.

📐 Board Exam Tip (Telangana & AP): For any similarity proof, always follow this four-step format: (i) state which two triangles you're comparing, (ii) list the equal angles or proportional sides with reasons in parentheses, (iii) state the criterion (AA / SSS / SAS), (iv) write the similarity statement with correct vertex order. Examiners award marks for each step individually.
🎯 Revision checklist before your exam:
  • Can you state the AA, SSS, and SAS criteria from memory with correct conditions?
  • Do you know the perimeter-ratio property and can you apply it quickly?
  • Can you draw the shadow problem diagram and set up the ratio equation?
  • Do you remember the construction rule for m/n — which arc to join, which to draw parallel at?
  • Can you prove Q7 (1/x + 1/y = 1/z) end-to-end without notes?
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