Dosto, "Algebra" naam sunte hi darr lagta hai — par CET Group D mein isse bahut basic level ke sawal aate hain. Kuch identities (jaise (a+b)²) aur simple equations, bas!
Geometry mein bhi poora theorem-proof nahi chahiye — sirf angle ke rules: triangle ke teeno angle 180°, polygon ke interior angles, aur circle ke do-teen basic facts.
CET mein isse 1–2 questions aate hain. Ye chapter chhota hai — 6 identities aur 8-10 geometry rules ratt lo, kaam ho gaya. Chalo shuru! 💪
1. Algebra ki 6 Zaroori Identities ⭐
| # | Identity | Kab kaam aati hai |
|---|---|---|
| 1 | (a + b)² = a² + 2ab + b² | Square kholne mein |
| 2 | (a − b)² = a² − 2ab + b² | Minus wale square mein |
| 3 | a² − b² = (a + b)(a − b) | ⭐ Sabse zyada use — bade numbers ka antar |
| 4 | (a + b)³ = a³ + b³ + 3ab(a + b) | Cube kholne mein |
| 5 | a³ + b³ = (a + b)(a² − ab + b²) | Cube factorise karne mein |
| 6 | a³ − b³ = (a − b)(a² + ab + b²) | Wahi, minus ke saath |
✏️ Solved Example — 51² − 49² = ?
Step 1: a² − b² = (a+b)(a−b) lagao
Step 2: = (51 + 49)(51 − 49) = 100 × 2 = 200 ✓
Bina identity ke 2601 − 2401 karna padta — identity se 5 second!
Step 2: = (51 + 49)(51 − 49) = 100 × 2 = 200 ✓
Bina identity ke 2601 − 2401 karna padta — identity se 5 second!
2. Do Kaam ke Derived Formulas ⭐
⭐ a + b aur ab diye hon to:
a² + b² = (a + b)² − 2ab · (a − b)² = (a + b)² − 4ab
Solved Example: a + b = 7 aur ab = 12 → a² + b² = ?
Step 1: 7² − 2(12) = 49 − 24 = 25 ✓ (check: a,b = 3 aur 4 → 9+16 = 25 ✓)
Solved Example: a + b = 7 aur ab = 12 → a² + b² = ?
Step 1: 7² − 2(12) = 49 − 24 = 25 ✓ (check: a,b = 3 aur 4 → 9+16 = 25 ✓)
⭐ "x + 1/x" wale famous sawal
Agar x + 1/x = k ho to:
x² + 1/x² = k² − 2 · x³ + 1/x³ = k³ − 3k
Solved Example: x + 1/x = 5 → x² + 1/x² = 25 − 2 = 23 ✓ (aur x³+1/x³ = 125 − 15 = 110)
x² + 1/x² = k² − 2 · x³ + 1/x³ = k³ − 3k
Solved Example: x + 1/x = 5 → x² + 1/x² = 25 − 2 = 23 ✓ (aur x³+1/x³ = 125 − 15 = 110)
3. Simple Equations (samikaran)
- Ek variable: jo number x ke saath jud raha hai use dusri taraf le jaao (sign badal ke), phir divide karo.
- Do variable: ek ki value di ho to seedha rakh do; warna ek equation se x nikaal ke dusri mein daalo (substitution).
✏️ Solved Examples
3x + 5 = 20 → 3x = 20 − 5 = 15 → x = 5 ✓
2x + 3y = 12, x = 3 → 6 + 3y = 12 → 3y = 6 → y = 2 ✓
2x + 3y = 12, x = 3 → 6 + 3y = 12 → 3y = 6 → y = 2 ✓
4. GEOMETRY — Angles ke Rules
| Naam | Rule |
|---|---|
| Complementary (purak) | Do angle ka jod = 90° |
| Supplementary (sampurak) | Do angle ka jod = 180° |
| Vertically opposite | Aamne-saamne ke angle barabar |
| Straight line par | Sab angle ka jod = 180° |
| Ek point ke charon taraf | Jod = 360° |
5. Triangle ke Rules ⭐
- Teeno angle ka jod = 180° (sabse zyada poocha jata hai).
- Exterior angle = do opposite interior angles ka jod.
- Triangle banne ke liye: do chhoti bhujaon ka jod teesri se bada hona chahiye.
- Types: Equilateral (teeno 60°) · Isosceles (do barabar) · Scalene (sab alag) · Right-angled (ek 90°).
✏️ Solved Examples
Do angle 50° aur 60° → teesra = 180 − 110 = 70° ✓
Exterior angle 110°, ek opposite interior 45° → dusra = 110 − 45 = 65° ✓
Exterior angle 110°, ek opposite interior 45° → dusra = 110 − 45 = 65° ✓
6. Polygon (bahubhuj) ke Formulas ⭐
| Kya | Formula | Example |
|---|---|---|
| Interior angles ka jod | (n − 2) × 180° | Hexagon (6) → 720° |
| Exterior angles ka jod | Hamesha 360° | Kisi bhi polygon ka |
| Regular polygon ka ek exterior angle | 360° / n | Pentagon → 72° |
| Regular polygon ka ek interior angle | (n−2)×180 / n = 180 − exterior | Hexagon → 120° |
| Quadrilateral ka jod | 360° | (n=4 wala hi case) |
✏️ Solved Example — Quadrilateral ke teen angle 80°, 90°, 100°. Chautha?
Step 1: Jod = 360° → Step 2: 360 − (80+90+100) = 360 − 270 = 90° ✓
7. Circle ke 4 Basic Facts
- Semicircle mein bana angle = 90° (Thales) — ye sabse zyada poocha jata hai!
- Ek hi segment ke angles barabar hote hain.
- Tangent (sparsh rekha) radius par 90° banati hai.
- Centre par bana angle = us arc par circle ke kisi point par bane angle ka dugna.
8. Parallel Lines aur Transversal
- Corresponding angles (ek jaisi jagah) = barabar.
- Alternate angles (aar-paar tirchhe) = barabar.
- Co-interior (ek hi taraf andar) = jod 180°.
✏️ Type 1 — a² − b² wala shortcut
"51² − 49²" ya "(103)² − (97)²" → (a+b)(a−b) lagao. Bade numbers ka square nikalne ki zaroorat hi nahi — ye CET ka favourite hai.
✏️ Type 2 — "a+b aur ab diya hai"
a² + b² poocha → (a+b)² − 2ab. (a−b)² poocha → (a+b)² − 4ab.
Aur "x + 1/x = k" wale mein x² + 1/x² = k² − 2. Ye teen formula hi kaafi hain.
Aur "x + 1/x = k" wale mein x² + 1/x² = k² − 2. Ye teen formula hi kaafi hain.
✏️ Type 3 — Triangle ka teesra angle / exterior angle
Do angle diye → teesra = 180 − dono ka jod.
Exterior angle diya → wo do opposite interior ka jod hota hai (110 = 45 + 65).
Exterior angle diya → wo do opposite interior ka jod hota hai (110 = 45 + 65).
✏️ Type 4 — Regular polygon ka angle
Ek exterior angle = 360/n (pentagon 72°, hexagon 60°, octagon 45°).
Ek interior angle = 180 − exterior (pentagon 108°, hexagon 120°, octagon 135°).
Ulta sawal: "ek exterior 40° hai to bhujaayein kitni?" → n = 360/40 = 9.
Ek interior angle = 180 − exterior (pentagon 108°, hexagon 120°, octagon 135°).
Ulta sawal: "ek exterior 40° hai to bhujaayein kitni?" → n = 360/40 = 9.
✏️ Type 5 — Circle ka angle
"Semicircle mein bana angle?" → hamesha 90°.
"Centre par 100° hai to circle ke point par?" → aadha = 50°. Ye do facts se hi zyadatar circle sawal ban jaate hain.
"Centre par 100° hai to circle ke point par?" → aadha = 50°. Ye do facts se hi zyadatar circle sawal ban jaate hain.
Trick 1: a²−b² sabse kaam ka
Bade square ka antar = (jod) × (antar)
51²−49² = 100 × 2 = 200. 103²−97² = 200 × 6 = 1200. Calculation ka jhanjhat khatam!
Bade square ka antar = (jod) × (antar)
51²−49² = 100 × 2 = 200. 103²−97² = 200 × 6 = 1200. Calculation ka jhanjhat khatam!
Trick 2: "x + 1/x" wale teen formula
k²−2 aur k³−3k
x+1/x = k diya ho to square wala k²−2, cube wala k³−3k. Bas dono ratt lo, poora topic khatam.
k²−2 aur k³−3k
x+1/x = k diya ho to square wala k²−2, cube wala k³−3k. Bas dono ratt lo, poora topic khatam.
Trick 3: Polygon mein exterior pehle nikalo
Exterior = 360/n · Interior = 180 − exterior
Interior ka lamba formula (n−2)×180/n yaad karne ki zaroorat nahi — exterior nikalo, 180 mein se ghatao. Bas!
Exterior = 360/n · Interior = 180 − exterior
Interior ka lamba formula (n−2)×180/n yaad karne ki zaroorat nahi — exterior nikalo, 180 mein se ghatao. Bas!
Trick 4: Regular polygon ke ready angles
Pentagon 108° · Hexagon 120° · Octagon 135°
Interior angles. Aur exterior: 72°, 60°, 45°. Ye 6 numbers exam mein baar-baar aate hain.
Pentagon 108° · Hexagon 120° · Octagon 135°
Interior angles. Aur exterior: 72°, 60°, 45°. Ye 6 numbers exam mein baar-baar aate hain.
Trick 5: Equation mein "ulta karke" le jao
Plus → minus, guna → bhaag
3x + 5 = 20 → 5 udhar jaake minus (15), 3 udhar jaake bhaag (x = 5). Har simple equation isi se hal.
Plus → minus, guna → bhaag
3x + 5 = 20 → 5 udhar jaake minus (15), 3 udhar jaake bhaag (x = 5). Har simple equation isi se hal.
❌ Galti 1: (a+b)² = a² + b² likh dena
2ab bhi aata hai! (2+3)² = 25 hai, 4+9 = 13 nahi. Ye Maths ki sabse common galti hai.
❌ Galti 2: a²+b² ke liye seedha (a+b)² lagana
a² + b² = (a+b)² − 2ab. 2ab ghatana bhoolna answer galat kar dega (49 vs 25).
❌ Galti 3: Polygon mein interior aur exterior mila dena
Exterior ka jod hamesha 360° (kitni bhi bhujaayein hon), interior ka jod (n−2)×180°. Hexagon: interior jod 720°, ek interior 120°, ek exterior 60°.
❌ Galti 4: Exterior angle ko galat samajhna
Triangle mein exterior angle = do opposite interior ka jod (uske saath wale ka nahi). 110 = 45 + 65 — ye rule seedha answer de deta hai.
❌ Galti 5: Centre wala angle aadha-dugna palat dena
Centre par bana angle dugna hota hai (circle ke point par bane angle se). Centre 100° → point par 50°, na ki 200°.
(a + b)² = ?tap 👆
a² + 2ab + b²
(a − b)² = ?tap 👆
a² − 2ab + b²
a² − b² = ?tap 👆
(a + b)(a − b)
a³ + b³ = ?tap 👆
(a + b)(a² − ab + b²)
(a + b)³ = ?tap 👆
a³ + b³ + 3ab(a + b)
a² + b² (a+b aur ab se)?tap 👆
(a + b)² − 2ab
x + 1/x = k → x² + 1/x²?tap 👆
k² − 2
x + 1/x = k → x³ + 1/x³?tap 👆
k³ − 3k
Triangle ke angles ka jod?tap 👆
180°
Exterior angle = ?tap 👆
Do opposite interior angles ka jod
Quadrilateral ke angles ka jod?tap 👆
360°
Polygon ke interior angles ka jod?tap 👆
(n − 2) × 180°
Exterior angles ka jod?tap 👆
Hamesha 360°
Regular polygon ka ek exterior angle?tap 👆
360/n
Regular hexagon ka interior angle?tap 👆
120° (pentagon 108°, octagon 135°)
Semicircle mein bana angle?tap 👆
90° (Thales)
Tangent aur radius ka angle?tap 👆
90°
Complementary & supplementary?tap 👆
Jod 90° aur 180°
| Cheez | Formula / Value |
|---|---|
| (a±b)² | a² ± 2ab + b² |
| a² − b² | (a+b)(a−b) |
| a³ ± b³ | (a±b)(a² ∓ ab + b²) |
| a² + b² (a+b, ab se) | (a+b)² − 2ab |
| x + 1/x = k | x²+1/x² = k²−2 · x³+1/x³ = k³−3k |
| Triangle angles | 180° · exterior = do opposite ka jod |
| Quadrilateral | 360° |
| Polygon interior sum | (n−2) × 180° |
| Polygon exterior sum | 360° (hamesha) |
| Regular: ek exterior · interior | 360/n · 180 − exterior |
| Pentagon · Hexagon · Octagon (interior) | 108° · 120° · 135° |
| Semicircle ka angle · tangent-radius | 90° dono |
Practice (exam-style)
Q1. If a + b = 7 and ab = 12, then a² + b² is:
💡 (a+b)² − 2ab = 49 − 24 = 25 (a,b = 3 aur 4).
Practice (exam-style)
Q2. The value of 51² − 49² is:
💡 (51+49)(51−49) = 100 × 2 = 200.
Practice (exam-style)
Q3. If x + 1/x = 5, then x² + 1/x² is:
💡 k² − 2 = 25 − 2 = 23.
Practice (exam-style)
Q4. Solve: 3x + 5 = 20.
💡 3x = 15 → x = 5.
Practice (exam-style)
Q5. If 2x + 3y = 12 and x = 3, then y is:
💡 6 + 3y = 12 → 3y = 6 → y = 2.
Practice (exam-style)
Q6. (x + 3)(x − 3) is equal to:
💡 (a+b)(a−b) = a² − b² = x² − 9.
Practice (exam-style)
Q7. Two angles of a triangle are 50° and 60°. The third angle is:
💡 180 − (50+60) = 70°.
Practice (exam-style)
Q8. An exterior angle of a triangle is 110° and one of the opposite interior angles is 45°. The other opposite interior angle is:
💡 Exterior = dono opposite interior ka jod → 110 − 45 = 65°.
Practice (exam-style)
Q9. The sum of the interior angles of a hexagon is:
💡 (n−2) × 180 = 4 × 180 = 720°.
Practice (exam-style)
Q10. Each exterior angle of a regular pentagon is:
💡 360/n = 360/5 = 72° (interior 108° hota hai).
Practice (exam-style)
Q11. Each interior angle of a regular hexagon is:
💡 Exterior = 360/6 = 60 → interior = 180 − 60 = 120°.
Practice (exam-style)
Q12. The angle in a semicircle is:
💡 Thales ka rule — semicircle mein bana angle hamesha 90°.
Practice (exam-style)
Q13. The supplement of 110° is:
💡 Supplement = 180 − 110 = 70° (complement hota 90 − angle).
Practice (exam-style)
Q14. Three angles of a quadrilateral are 80°, 90° and 100°. The fourth angle is:
💡 360 − (80+90+100) = 90°.
Algebra mein 6 identities kaafi hain — sabse kaam ki a² − b² = (a+b)(a−b) (51²−49² = 200 seedha!). Do derived formula ratt lo: a²+b² = (a+b)² − 2ab aur x+1/x = k → x²+1/x² = k²−2. Sabse badi galti: (a+b)² mein 2ab bhoolna. Geometry mein triangle 180°, quadrilateral 360°, polygon interior (n−2)×180° par exterior ka jod hamesha 360° — isliye regular polygon mein pehle exterior = 360/n nikalo, phir interior = 180 − exterior (hexagon 120°, pentagon 108°). Aur circle ke do facts: semicircle ka angle 90° aur centre ka angle dugna. 🚀