| --- |
| source: ncert |
| topic: geometry |
| class_level: class_10 |
| chapter: ch6 |
| difficulty: intermediate |
| --- |
| |
| Class 10 | Ch6: Triangles (Similarity) |
| SIMILAR TRIANGLES: same shape, different size. Corresponding angles equal, corresponding sides proportional. |
| Symbol: triangle ABC ~ triangle DEF. |
| BASIC PROPORTIONALITY THEOREM (BPT / Thales Theorem): |
| If a line is drawn parallel to one side of triangle, it divides the other two sides proportionally. |
| If DE || BC in triangle ABC: AD/DB = AE/EC. |
| CONVERSE: If a line divides two sides of triangle proportionally, it is parallel to third side. |
| CRITERIA FOR SIMILARITY: |
| AAA (AA): Two angles equal → triangles similar. |
| SSS: All three sides proportional → triangles similar. |
| SAS: Two sides proportional and included angle equal → triangles similar. |
| AREAS OF SIMILAR TRIANGLES: |
| Ratio of areas = Square of ratio of corresponding sides. |
| If triangle ABC ~ triangle DEF: Area(ABC)/Area(DEF) = (AB/DE)^2 = (BC/EF)^2 = (CA/FD)^2. |
| PYTHAGORAS THEOREM: |
| In right triangle, square of hypotenuse = sum of squares of other two sides. |
| c^2=a^2+b^2. Proof using similar triangles (altitude on hypotenuse). |
| CONVERSE: If c^2=a^2+b^2, angle opposite c is 90°. |
| SOLVED EXAMPLES: |
| Example 1: In triangle ABC, DE||BC. AD=2cm, DB=3cm, AE=4cm. Find EC. |
| By BPT: AD/DB=AE/EC → 2/3=4/EC → EC=6cm. |
| Example 2: Triangle ABC ~ triangle DEF. AB=4,BC=5,CA=6. DE=8. Find EF and FD. |
| Ratio AB/DE=4/8=1/2. So all sides doubled. |
| EF=2×BC=10. FD=2×CA=12. |
| Example 3: Areas of two similar triangles are 25cm^2 and 100cm^2. If base of smaller=3cm, find base of larger. |
| Area ratio=25/100=1/4=(side ratio)^2. Side ratio=1/2. Larger base=2×3=6cm. |
| COMMON MISTAKES: |
| BPT: line must be PARALLEL to one side (not just any line). |
| Area ratio=(side ratio)^2 NOT side ratio directly. |
| AA criterion: only need 2 angles equal (third is automatic since sum=180°). |
