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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°). |