Exercise 8.2 — Similarity Criteria
Criteria for similarity of triangles.
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.
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).
Since angles in a triangle sum to 180°, if two angles match, the third automatically matches. This gives us the practical AA rule.
Converse: If corresponding angles are equal (AAA), then the corresponding sides are in the same ratio.
Key ratio property: If △ABC ∼ △DEF, thenAB/DE = BC/EF = CA/FD = Perimeter(△ABC) / Perimeter(△DEF)
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!).
| Part | Key observations | Criterion | Relation | Similar? |
|---|---|---|---|---|
| (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 |
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.
| Part | Given Similarity | Ratio Used | Working | x = |
|---|---|---|---|---|
| (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 |
Question 1 — ∠ADE = ∠B: Prove △ABC ∼ △ADE and Find DE
Question 2 — Perimeters 30 cm and 20 cm; One Side 12 cm: Find the Corresponding Side
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 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.
Question 5 — △ABC ∼ △PQR: CM and RN are Medians — Three Sub-Parts
Question 6 — Trapezium Diagonals: Prove OA/OC = OB/OD Using Similarity
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 8 — Flag Pole and Building Shadow: Find Building Height
Question 9 — Angle Bisectors CD and GH in Similar Triangles △ABC ∼ △FGE
Question 10 — Altitudes AX and DY of Similar Triangles: Prove AX:DY = AB:DE
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.
- Draw △ABC with given measurements.
- Draw ray BX making an acute angle with BC.
- Mark 5 equal arcs B₁, B₂, B₃, B₄, B₅ on BX (use compass, same radius).
- Join B₃ to C (since denominator = 3 → join 3rd point to C).
- Draw a parallel to B₃C through B₅ — this meets extended BC at C₁.
- Draw a parallel to AC through C₁ — this meets extended BA at A₁.
- △A₁BC₁ is the required similar triangle with sides = 5/3 of △ABC.
- Draw △ABC with AB = 4 cm, BC = 5 cm, CA = 6 cm.
- Draw ray AX at an acute angle to AB.
- Mark 3 equal arcs A₁, A₂, A₃ on AX.
- Join A₃ to B (denominator = 3).
- Draw parallel to A₃B through A₂ — meets AB at B₁.
- Draw parallel to BC through B₁ — meets AC at C₁.
- △AB₁C₁ is the required triangle with sides = 2/3 of △ABC.
- Convert 1½ to 3/2 (mark 3 arcs, join at 2nd).
- Draw isosceles △ABC with AB = 8 cm and altitude CD = 4 cm.
- Draw ray AX at an acute angle to AB.
- Mark 3 equal arcs A₁, A₂, A₃ on AX.
- Join A₂ to B (denominator = 2).
- Draw parallel to A₂B through A₃ — meets extended AB at B₁.
- Draw parallel to BC through B₁ — meets extended AC at C₁.
- △AB₁C₁ has sides = 3/2 of △ABC.
All 13 Questions — Quick Reference at a Glance
| Q# | Topic | Criterion / Tool | Key Result |
|---|---|---|---|
| Try This 1 | Identify similarity criterion (8 parts) | AA / SAS / SSS / None | i,iii,iv,v,vi=Similar; ii,vii,viii=Not Similar |
| Try This 2 | Find x (7 parts) | Ratio of corresponding sides | 7.5, 2.5, 12.8, 12, 8, 15, 16 |
| Q1 | ∠ADE = ∠B → △ABC ∼ △ADE; find DE | AA | DE = 2.8 cm |
| Q2 | Perimeters 30 & 20, one side = 12 | Perimeter ratio | Corresponding side = 8 cm |
| Q3 | AB ∥ CD ∥ EF, find x and y | AAA (twice) | x = 5, y = 2.8 |
| Q4 | Shadow length — lamp post problem | AA (right angles) | Shadow = 1.6 m |
| Q5 | △ABC ∼ △PQR with medians CM, RN | SAS (twice) | CM/RN = AB/PQ |
| Q6 | Trapezium diagonals: OA/OC = OB/OD | AAA | △OAB ∼ △OCD |
| Q7 | Three perpendiculars: 1/x + 1/y = 1/z | AAA (twice, add) | Elegant reciprocal identity |
| Q8 | Flag pole shadow → building height | AA | Building = 16 m |
| Q9 | Angle bisectors in similar triangles | AA (three times) | CD/GH = AC/FG |
| Q10 | Altitudes AX:DY = AB:DE | AA | △ABX ∼ △DEY |
| Q11 | Construct △ similar to △ABC, sides = 5/3 | Construction | Mark 5 arcs, join at 3, parallel at 5 |
| Q12 | Construct with sides = 2/3 | Construction | Mark 3 arcs, join at 3, parallel at 2 |
| Q13 | Isosceles, sides = 3/2 (= 1½) | Construction | Mark 3 arcs, join at 2, parallel at 3 |
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).
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.
- 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?