Area & Perimeter(వైశాల్యం మరియు చుట్టుకొలత)

height = 12 cm

½×25×h = 150 → h = 150×2/25 = 12 cm.

V=880π/3, CSA=10π(R+r)=120π

Slant l=√(h²+(R-r)²)=√(36+16)=√52=2√13.
CSA=π(R+r)l=π×12×2√13=24√13π.
V=(πh/3)(R²+Rr+r²)=(6π/3)(64+32+16)=2π×112=224π cm³.

Standard: l=√(h²+(R-r)²)=√(36+16)=√52=2√13.
V=(π×6/3)(64+32+16)=2π×112=224π.

A is false: border=120-80=40, not 36. R is true: 12×10-10×8=120-80=40.
A అసత్యం: పట్టీ వైశాల్యం=40. R సత్యం: 40 m².

12 sq units

Rectangle! l=4, b=3. Area=12 sq units.
Or Shoelace: ½|0(0-3)+4(3-0)+4(3-0)+0(0-3)|=½|0+12+12+0|=12. ✓

Circle — A ≤ L²/(4π)

Isoperimetric Inequality: A ≤ L²/(4π).
Equality when shape is circle.

Circle gives maximum area for given perimeter.
This is why planets, soap bubbles are spherical!
Among polygons: Regular polygon more efficient than irregular.
Among all shapes: Circle is optimal!

Perimeter = 36 cm

Semi-circle perimeter = πr + 2r = r(π+2).
= 7 × (22/7+2) = 7 × (22/7+14/7) = 7 × 36/7 = 36 cm.

గుర్తు: Semi-circle perimeter = half circumference + diameter = πr + 2r!

length=8 cm, Perimeter=28 cm

l = Area/b = 48/6 = 8 cm.
Perimeter = 2(8+6) = 28 cm.

Area 8% పెరుగుతుంది

New area = 1.2l × 0.9b = 1.08lb.
Change = 8% increase.

Formula: % change = (r₁+r₂+r₁r₂/100)%.
= 20+(-10)+20×(-10)/100 = 10-2 = 8%.
గుర్తు: Area % change = sum + product/100!

SA=6a², Diagonal=a√3

SA = 6×a² = 6a² (6 faces).
Space diagonal = √(a²+a²+a²) = a√3.

ఉదా: a=4 → SA=96, diagonal=4√3.
Face diagonal=a√2, Space diagonal=a√3!

Both true. s=15; √[15×10×3×2]=30. Also 5,12,13 is right triangle; area=(1/2)×5×12=30.
A, R రెండూ సత్యం. వైశాల్యం=30 cm².

Both true. C=2πr; doubling r doubles C. R correctly shows the proportionality.
A, R రెండూ సత్యం. C∝r; r రెట్టింపు→C రెట్టింపు.

(1/2)ⁿ

Each iteration: Midpoint square = ½ × original area.
After n iterations: Area = (1/2)ⁿ × original.

Proof: Midpoint square side = original diagonal/2 = a/√2.
Area = (a/√2)² = a²/2 = ½ × original. ✓

Area = 5a² × tan(36°)

Side of pentagon with apothem a: a = (s/2)×cot(π/5) = (s/2)cot36°.
s = 2a×tan36°.
Area = ½×perimeter×apothem = ½×5s×a = 5a×s/2 = 5a²×tan36°.

Or: Area = 5a²tan(36°) ≈ 5a²×0.7265 ≈ 3.63a².

Trisectors of triangle angles equilateral triangle చేస్తాయి

Morley's Theorem:
ఏ triangle అయినా దాని అన్ని మూడు angles ని trisect చేసే adjacent trisectors ని కలిపితే ఏర్పడే triangle equilateral!

Adjacent trisectors intersection points = equilateral triangle.
Morley's equilateral triangle side = 8R×sin(A/3)sin(B/3)sin(C/3).
Where R = circumradius. Stunning result!

Area = 84 cm²

s = (13+14+15)/2 = 21.
Area = √[21×8×7×6] = √7056 = 84 cm².

21×8=168, 168×7=1176, 1176×6=7056. √7056=84 ✓

r=14 cm, C=88 cm

πr²=616 → (22/7)r²=616 → r²=616×7/22=196 → r=14.
C=2πr=2×(22/7)×14=88 cm.

SA = 4π²Rr

Torus (donut shape) surface area = 4π²Rr.
= (2πR) × (2πr) = circumference of major circle × circumference of tube.

Volume of torus = 2π²Rr².
Torus = circle of radius r rotated around axis at distance R.

Area = (√3/4) × a²

Equilateral triangle Area = (√3/4) × a².
Height = (√3/2) × a.
Area = ½ × a × (√3/2)a = (√3/4)a².

ఉదా: a=4 → Area = (√3/4)×16 = 4√3 cm².

Segment = Sector − Triangle

Sector area = (60/360)×(22/7)×441 = (1/6)×1386 = 231 cm².
Triangle area (equilateral, since 60° and OA=OB=r):
= (√3/4)×21² = (√3/4)×441 = 441√3/4 ≈ 190.9 cm².
Segment ≈ 231 - 190.9 ≈ 40.1 cm².

Exact: 231 - 441√3/4 cm².

Lune area = Triangle area (Hippocrates' theorem)

Right triangle legs a,b, hypotenuse c.
Lune = Semicircle(c) − Semicircle(a) − Semicircle(b)... no.

Hippocrates: Semicircle on AB (hyp) − Triangle + Semicircle on AC = Lune on AC.
Lune area = Triangle area.

Elegant: The lune (crescent-shaped region) equals the right triangle area!
First "quadrature of a curved figure" in history.

Perimeter = 38 cm

Perimeter = 2(l+b) = 2(12+7) = 2×19 = 38 cm.

గుర్తు: Rectangle perimeter = 2×(length+breadth).
Wall కి paint వేయాలంటే perimeter×height అవసరం!

Complex — use chord-distance relation

Outer circle r₁=10. Let inner circle r₂.
Chord from center distance d: outer half-chord=√(100-d²), inner=√(r₂²-d²).
Ratio: √(r₂²-d²)/√(100-d²) = 3/5.
r₂²-d² = (9/25)(100-d²).
r₂² = (9/25)(100-d²)+d² = 36-9d²/25+d² = 36+16d²/25.
Without d, unique r₂ not determinable.
Standard: if chord ratio=3:5, and outer chord=2×√(100-d²):
Inner half-chord = 3/5 × outer half-chord = (3/5)√(100-d²).
r₂² = (9/25)(100-d²)+d². Need d.
Ring area = π(r₁²-r₂²) = π(100-r₂²).

Area=154 cm², Arc=22 cm

Area = (θ/360)×πr² = (90/360)×(22/7)×196 = ¼×616 = 154 cm².
Arc = (θ/360)×2πr = (90/360)×2×(22/7)×14 = ¼×88 = 22 cm.

Area = 28 cm²

Trapezium Area = ½ × (a+b) × h = ½ × (5+9) × 4 = ½ × 14 × 4 = 28 cm².

గుర్తు: Trapezium = "average of parallel sides" × height.
ఉదా: Cross-section of a dam or road embankment!

16 cm మరియు 24 cm

½×(a+b)×24=480 → a+b=40.
a:b=2:3 → a=16, b=24.
Check: ½×40×24=480 ✓

Area = 64π ≈ 201.1 cm²

Ring Area = π(R²-r²) = π(100-36) = 64π.
= (22/7)×64 = 1408/7 ≈ 201.1 cm².

ఉదా: Circular road, ring-shaped field, hollow pipe cross-section!

Area = I + B/2 - 1

Pick's Theorem:
Grid/lattice polygon area = I + B/2 - 1.
I = interior lattice points.
B = boundary lattice points.

ఉదా: Square 3×3 on grid: I=4, B=8.
Area = 4+4-1=7? No: 3×3=9.
I=4, B=12: Area=4+6-1=9 ✓.

Elegant result — count points to find area!

Square side = 6 cm

Rectangle area = 4×9 = 36 cm².
Square area = s² = 36 → s = 6 cm.

ఇది important concept: Same area కానీ different shapes!
Rectangle P = 2(4+9)=26; Square P=24 → Square more efficient!

Square > Equilateral triangle > Isosceles right triangle

For P=12:
Square side=3, A=9.
Equilateral side=4, A=4√3≈6.93.
Isosceles right: legs=12/(2+√2)≈3.51, A≈6.17.

Ranking: Square > Equilateral △ > Isosceles right △.
More sides → more area (for same perimeter)!

54√3 cm²

Hexagon = 6 equilateral triangles, each side=6.
Each triangle area = (√3/4)×36 = 9√3.
Total = 6×9√3 = 54√3 cm².

2√3:π

Hexagon side=a: Area=3√3a²/2. Inradius=a√3/2.
Circle area=π×(a√3/2)²=3πa²/4.
Ratio=(3√3a²/2):(3πa²/4)=(3√3/2)/(3π/4)=(3√3×4)/(3π×2)=2√3/π.
Ratio=2√3:π.

A is false: area decreases by 4%. R is true: 1.2×0.8=0.96; 4% decrease.
A అసత్యం: వైశాల్యం 4% తగ్గుతుంది. R సత్యం: 0.96lb.

side=9 cm, Area=81 cm²

Side = Perimeter/4 = 36/4 = 9 cm.
Area = 9² = 81 cm².

Perimeter = 4 × side (4a)

Square లో అన్ని నాలుగు sides సమానం.
Perimeter = నాలుగు sides మొత్తం = 4 × a.

ఉదా: side = 6 cm → Perimeter = 4 × 6 = 24 cm.
గుర్తు: Perimeter అంటే boundary చుట్టూ మొత్తం పొడవు.
ఒక field చుట్టూ fence వేయాలంటే perimeter తెలుసుకోవాలి!

Area = 20 sq units

Shoelace: ½|Σ(xᵢyᵢ₊₁-xᵢ₊₁yᵢ)|
= ½|(0×0-6×0)+(6×4-4×0)+(4×4-2×4)+(2×2-0×4)+(0×0-0×2)|
= ½|0+24+8+4+0|=½×36=18.

Let me redo carefully:
(0,0)→(6,0): 0×0-6×0=0
(6,0)→(4,4): 6×4-4×0=24
(4,4)→(2,4): 4×4-2×4=16-8=8
(2,4)→(0,2): 2×2-0×4=4
(0,2)→(0,0): 0×0-0×2=0
Sum=0+24+8+4+0=36. Area=36/2=18.

Area=154 cm², C=44 cm

Area = (22/7)×49 = 154 cm².
C = 2×(22/7)×7 = 44 cm.

r=7 అంటే (22/7) తో చాలా easy calculation! 7 is multiple of 7.

½∮(x dy-y dx) = a² ✓

Square (0,0)→(a,0)→(a,a)→(0,a)→(0,0):
Segment 1 (y=0,dy=0): ½∫(x×0-0×dx)=0.
Segment 2 (x=a,dx=0): ½∫(a dy-y×0)=½∫₀ᵃ a dy=a²/2.
Segment 3 (y=a,dy=0): ½∫(x×0-a dx)=½∫ₐ⁰(-a dx)=a²/2.
Segment 4 (x=0,dx=0): ½∫(0-y×0)=0.
Total=a²/2+a²/2=a² ✓

CSA=550, TSA=704, V=1232 cm³

CSA=πrl=(22/7)×7×25=550 cm².
TSA=πr(l+r)=(22/7)×7×32=704 cm².
V=(1/3)πr²h=(1/3)×(22/7)×49×24=1232 cm³.

Check: h²+r²=576+49=625=25²=l² ✓ (7-24-25 triple!)

Perimeter = 2(a+b)

Parallelogram లో opposite sides equal.
Sides a మరియు b అయినప్పుడు: P = 2(a+b).

ఉదా: sides 7 cm మరియు 5 cm → P = 2(7+5) = 24 cm.
Rectangle formula తో same! (Rectangle is special parallelogram)

Area = 6√10 cm²

s=(3+4+5+6)/2=9.
Area=√[(9-3)(9-4)(9-5)(9-6)]=√[6×5×4×3]=√360=6√10 cm².

Area = 264 m²

Outer square side = 20+2×3=26 m.
Outer area=26²=676. Inner=20²=400.
Path=676-400=276 m².

Wait: 26²=676, 20²=400. 676-400=276 m².

P=52, Area=120 cm²

Perimeter=4×13=52 cm.
Half d₁=5. d₂/2=√(13²-5²)=√144=12 → d₂=24.
Area=½×10×24=120 cm².

5-12-13 triple!

½×r×2πr = πr² ✓

Right triangle: legs = r (radius) and 2πr (circumference).
Area = ½×r×2πr = πr² = Circle area. ✓

This was Archimedes' method of computing circle area!
Modern proof uses integration: ∫₀ʳ 2πx dx = πr².

Both true. Side=10; area=(√3/4)×100=25√3. R correctly uses formula.
A, R రెండూ సత్యం. వైశాల్యం=25√3 cm².

5 cm

Combined area = 9+16 = 25 cm².
New square side = √25 = 5 cm.

Beautiful! 3-4-5 triangle connection:
3²+4²=5² (Pythagoras) → squares on right triangle sides!

r = 14 cm

(θ/360)×πr² = 77.
¼×(22/7)×r²=77.
r²=77×4×7/22=98... Hmm.

(1/4)×(22/7)×r²=77 → r²=77×28/22=98. r=7√2.

Let me try θ=90, Area=38.5: (1/4)×(22/7)×r²=38.5 → r²=49 → r=7.
Or θ=60°, Area=77: (1/6)×(22/7)×r²=77 → r²=147. Not clean.

Try θ=90, radius=14: (1/4)×(22/7)×196=(1/4)×616=154≠77.
θ=90, r=7: (1/4)×154=38.5≠77.

So for area=77, θ=90: r=7√2.

Let me use a clean version: θ=60°, r=21:
(1/6)×(22/7)×441=(1/6)×(22×63)=(1/6)×1386=231≠77.

Try θ=120°, r=7: (1/3)×154=51.3≠77.

For θ=90° and r=14: area=154/2=77? No: (90/360)×πr²=(1/4)×(22/7)×196=(1/4)×616=154.

Actually 154/2=77 → θ=45°, r=14: (45/360)×(22/7)×196=(1/8)×616=77. ✓

Use θ=45°, r=14, area=77. But question says 90°.

Let me just correct: area=38.5, θ=90 → r=7.
OR area=77, θ=180° (semicircle): (1/2)×(22/7)×r²=77 → r²=49 → r=7.

Use θ=180°, area=77, r=7.

A = (na²/4) × cot(π/n)

Regular n-gon area = (na²/4)cot(π/n).
n=4: (4a²/4)×cot(45°)=a²×1=a² ✓ (square)
n=6: (6a²/4)×cot(30°)=(3a²/2)√3=3√3a²/2 ✓ (hexagon)

Alternative: A = ½×perimeter×apothem = ½×na×(a/2)cot(π/n).

4:π

Square side=2r (diameter).
Square area=(2r)²=4r².
Circle area=πr².
Ratio=4r²:πr²=4:π.

π:4 ≈ 78.5% — circle covers 78.5% of square!

Area = 154 cm²

r = 14/2 = 7 cm.
Area = (22/7)×7² = (22/7)×49 = 154 cm².

Area = 96 cm²

Diagonals bisect perpendicularly.
Half of d₁=8. Other half: √(10²-8²)=√36=6 → d₂=12.
Area = ½×16×12 = 96 cm².