Class 6 Maths Chapter 6 NCERT Solutions Hinglish mein: Perimeter and Area

Ganita Prakash (Reprint 2026-27) ke Chapter 6, Perimeter and Area, ke har “Figure it Out” sawal aur puzzle ke jawab. Page numbers book se match karte hain, taaki har sawal jaldi mil jaaye. Short notes: Chapter 6 notes.

Figure it Out: perimeter (page 132)

Question 1: missing terms nikaalo

• (a) Rectangle ka perimeter = 14 cm, breadth = 2 cm. Length + breadth = 14 ÷ 2 = 7 cm, to length = 7 − 2 = 5 cm.
• (b) Square ka perimeter = 20 cm. Side = 20 ÷ 4 = 5 cm.
• (c) Rectangle ka perimeter = 12 m, length = 3 m. Length + breadth = 12 ÷ 2 = 6 m, to breadth = 6 − 3 = 3 m.

Question 2: taar ka rectangle, phir square

Taar ki lambai rectangle ka perimeter hai: 2 × (5 + 3) = 16 cm. Usi taar se square bana, to uski side = 16 ÷ 4 = 4 cm.

Question 3: triangle ki teesri side

Teesri side = 55 − (20 + 14) = 55 − 34 = 21 cm.

Question 4: park ki fencing ka kharcha

Perimeter = 2 × (150 + 120) = 2 × 270 = 540 m. Kharcha = 540 × ₹40 = ₹21,600.

Question 5: 36 cm ki dori

• (a) Square: 36 ÷ 4 = 9 cm har side.
• (b) Barabar sides wala triangle: 36 ÷ 3 = 12 cm har side.
• (c) Barabar sides wala hexagon: 36 ÷ 6 = 6 cm har side.

Question 6: rassi ke teen chakkar

Khet ka perimeter = 2 × (230 + 160) = 2 × 390 = 780 m. Teen chakkar ke liye 3 × 780 = 2,340 m rassi chahiye.

Matha Pachchi: do running tracks (page 133)

Akshi ka bahar wala track 70 m × 40 m hai, to ek chakkar = 2 × (70 + 40) = 220 m. Toshi ka andar wala track 60 m × 30 m hai, to ek chakkar = 2 × (60 + 30) = 180 m.

Question 1

Akshi ke 5 chakkar = 5 × 220 = 1,100 m.

Question 2

Toshi ke 7 chakkar = 7 × 180 = 1,260 m. Toshi zyada daudi, 160 m zyada.

Question 3: positions mark karo

Dono apne-apne track ke neeche-daayein (bottom-right) kone se shuru karti hain aur pehle neeche wali side pe daudti hain, jaise arrows dikhate hain. Har track pe dots 10 m door hain, to points mark karne ke liye dots gin sakte ho.

• (a) A, Akshi 250 m ke baad: 250 = 220 + 30. Ek poora chakkar, phir 30 m aur: A neeche wali side pe, start se 30 m pe hai.
• (b) B, Akshi 500 m ke baad: 500 = 2 × 220 + 60. B neeche wali side pe, start se 60 m pe (neeche-baayein kone se 10 m pehle).
• (c) C, Akshi 1,000 m ke baad: 1,000 = 4 × 220 + 120. Usne 4 poore chakkar lagaye. Extra 120 m = 70 m (neeche) + 40 m (baayein side) + 10 m, to C upar wali side pe, upar-baayein kone se 10 m pe hai.
• (d) X, Toshi 250 m ke baad: 250 = 180 + 70. Extra 70 m = 60 m (neeche) + 10 m, to X baayein side pe, neeche-baayein kone se 10 m upar hai.
• (e) Y, Toshi 500 m ke baad: 500 = 2 × 180 + 140. Extra 140 m = 60 + 30 + 50, to Y upar wali side pe, upar-baayein kone se 50 m pe hai.
• (f) Z, Toshi 1,000 m ke baad: 1,000 = 5 × 180 + 100. Usne 5 poore chakkar lagaye. Extra 100 m = 60 + 30 + 10, to Z upar wali side pe, upar-baayein kone se 10 m pe hai.

Deep Dive: ek hi finishing line (page 134)

Jhande dono square tracks ki neeche wali side ke beech mein hain, aur daudne wale neeche wali side pe jhandon ki taraf aate hain, jaise arrows dikhate hain. Har ek ko theek 350 m daudna hai.

• A (andar wala track, 100 m sides): andar wale track ke neeche-daayein kone se shuru karo. Daayein side upar (100 m), upar wali side (100 m), baayein side neeche (100 m) aur neeche wali side ka aadha (50 m): 100 + 100 + 100 + 50 = 350 m.
• B (bahar wala track, 150 m sides): upar wali side pe, upar-daayein kone se 25 m pe shuru karo. 125 m upar-baayein kone tak, 150 m baayein side neeche, aur 75 m (neeche wali side ka aadha) jhande tak: 125 + 150 + 75 = 350 m.

Kya triangle ka perimeter 9 units hai? (page 134)

Toshi sahi hai: ye 9 units se zyada hai. Do sides seedhi hain, 3-3 units, lekin tirchi side 3 squares ko diagonal mein paar karti hai. Unit square ka diagonal uski side se lamba hota hai, to tirchi side 3 units se zyada hai. Book ke short form mein perimeter 6s + 3d hai.

Straight aur diagonal units (page 135)

Har figure ke chaaron taraf straight units (s) aur diagonal units (d) gin kar, jis order mein dikhaye gaye hain:

1. 8s + 2d
2. 4s + 6d
3. 12s + 6d
4. 18s + 6d

Regular polygon ka perimeter (page 135–136)

Square aur equilateral triangle mein samaanta: dono regular polygons hain, saari sides barabar aur saare angles barabar. Isliye dono ka perimeter = sides ki sankhya × ek side: square ke liye 4 × side aur triangle ke liye 3 × side.

General rule: regular polygon ka perimeter = sides ki sankhya × ek side ki lambai. 4 cm side wale regular pentagon ka perimeter 5 × 4 = 20 cm hai.

Split and rejoin (page 136)

Har tukda 6 cm × 2 cm hai, jiska perimeter 2 × (6 + 2) = 16 cm hai. Jab do tukde kisi lambai pe chhoote hain, wo lambai dono tukdon ke bahar wale kinaare se gayab ho jaati hai. Isliye:

Arrangement ka perimeter = 16 + 16 − 2 × (jitni lambai pe tukde chhoo rahe hain)

• (a) 2 cm pe chhoo rahe: 32 − 4 = 28 cm (jaisa book mein diya hai).
• (b) L-shape, 2 cm pe chhoo rahe: 32 − 4 = 28 cm.
• (c) T-shape, 2 cm pe chhoo rahe: 32 − 4 = 28 cm.
• (d) Agal-bagal, 3 cm pe chhoo rahe: 32 − 6 = 26 cm.

22 cm perimeter wali shape: 32 − 2 × (chhoone ki lambai) = 22 chahiye, to tukdon ko 5 cm pe chhoona hoga. Dono ko unki lambi 6 cm wali sides se agal-bagal rakho aur ek tukde ko doosre ke saath 1 cm khiska do.

Figure it Out: area (page 138)

Question 1

Width = area ÷ length = 300 ÷ 25 = 12 m.

Question 2

Area = 500 × 200 = 1,00,000 sq m = 100 sq m ke 1,000 hisse. Kharcha = 1,000 × ₹8 = ₹8,000.

Question 3

Bagaan ka area = 100 × 50 = 5,000 sq m. Pedon ki sankhya = 5,000 ÷ 25 = 200 ped.

Question 4: rectangles mein todna

• (a) Shape ko chaar rectangles mein todo: 3 m × 4 m (12 sq m), 4 m × 1 m (4 sq m), 3 m × 2 m (6 sq m) aur 2 m × 3 m (6 sq m). Area = 12 + 4 + 6 + 6 = 28 sq m.
• (b) Poora 5 m × 3 m rectangle lo (15 sq m) aur beech ka 3 m × 2 m khaali hissa ghatao (6 sq m). Area = 15 − 6 = 9 sq m.

Tangram (page 139)

Tukdon ko ek-doosre pe rakhne se pata chalta hai: A aur B (bade triangles) barabar hain; C aur E (chhote triangles) barabar hain; aur D, F aur G har ek do chhote triangles ke barabar hai. C ke area ko 1 hissa maano: C = E = 1, D = F = G = 2, A = B = 4.

1. Same area wale tukde: A aur B; C aur E; D, F aur G.
2. D, C ka 2 guna hai. C aur E milkar D ko poora dhak lete hain, to D = C + E.
3. D aur F ka area same hai, 2-2 hisse.
4. F aur G ka area same hai, 2-2 hisse, bhale hi unki shape alag hai.
5. A, G se do guna bada hai: 4 hisse aur 2 hisse.
6. Bada square = C ke area ka 16 guna (4 + 4 + 1 + 1 + 2 + 2 + 2 = 16).
7. Rectangle ka area bhi C ka 16 guna hai, kyunki wo unhi saat tukdon se bina gap aur overlap ke bana hai.
8. Perimeters alag hain. Rectangle square se lamba aur patla hai, isliye uski boundary lambi hai, bhale hi dono ka area same hai.

Squared paper pe area ka andaaza (page 140)

Poore squares ko 1 aur diagonal se kate aadhe squares ko ½ gino. Chaaron figures, order mein, ka area 4, 9, 10 aur 11 sq units hai.

Let’s Explore: 24 sq units area wale rectangles (page 141–142)

Poore numbers wali sides ke rectangles hain 1 × 24, 2 × 12, 3 × 8 aur 4 × 6 (aur yahi ghuma kar).

• (a) Sabse zyada perimeter: 1 × 24, yaani 2 × (1 + 24) = 50 units.
• (b) Sabse kam perimeter: 4 × 6, yaani 2 × (4 + 6) = 20 units.
• (c) 32 sq cm area: rectangles hain 1 × 32, 2 × 16 aur 4 × 8. Sabse zyada perimeter 1 cm × 32 cm (66 cm) aur sabse kam 4 cm × 8 cm (24 cm).

Andaaza: kisi bhi area ke liye sabse lamba-patla rectangle (1 × area) sabse zyada perimeter rakhta hai, aur square ke sabse kareeb wala sabse kam.

Grid paper pe triangle ka area (page 143)

Rectangle ABCD 5 squares lamba aur 4 squares ooncha hai, to uska area = 20 sq units.

1. Triangle BAD rectangle ka aadha hai: 10 sq units.
2. Triangle ABE: rectangle AFED ka aadha (3 × 4 = 12, aadha = 6) + rectangle BFEC ka aadha (2 × 4 = 8, aadha = 4) = 10 sq units.

Nishkarsh: jis triangle ka base rectangle ki ek side ho aur teesra kona saamne wali side pe ho, uska area rectangle ka aadha hota hai.

Figure it Out: grid paper pe area (page 144)

Har figure ko rectangles aur triangles mein todo (har triangle ek rectangle ka aadha) aur jodo:

• (a) 4 × 5 rectangle (20) + do triangles, har ek 4 × 1 ka aadha (2 + 2) = 24 sq units.
• (b) 4 × 5 rectangle (20) + 4 × 3 ka aadha triangle (6) + 4 × 2 ka aadha triangle (4) = 30 sq units.
• (c) Upar ka triangle 6 × 2 ka aadha (6) + 3 × 8 rectangle (24) + chhota triangle 3 × 2 ka aadha (3) + bada triangle 3 × 10 ka aadha (15) = 48 sq units.
• (d) 4 × 3 rectangle (12) + 1 × 2 ka aadha triangle (1) + 3 × 2 ka aadha triangle (3) = 16 sq units.
• (e) Side wale kono se guzarti seedhi line se todo: 4 × 2 ka aadha triangle (4) + 4 × 4 ka aadha triangle (8) = 12 sq units.

Making it ‘More’ or ‘Less’ (page 145–146)

Har unit square ki 4 sides hain. Do squares ke beech ki har shared side boundary se do baar chhup jaati hai, isliye 9 squares ke liye perimeter = 36 − 2 × (shared sides ki sankhya).

1. Sabse kam perimeter: 12 units, 3 × 3 square se.
2. Sabse zyada perimeter: 20 units, jaise 9 squares ki seedhi patti.
3. 18 units perimeter: 4 squares ka 2 × 2 block, aur baaki 5 squares usse judi ek row mein.
4. Doosri shapes: 18 aur 20 units ke liye kai alag shapes ban sakti hain. 12 units ke liye sirf 3 × 3 square, kyunki 9 squares ki koi aur arrangement itni sides share nahi karti.

24-unit figure mein naya square jodna: agar wo figure ko 1 side pe chhoota hai, perimeter 2 badhta hai; 2 sides pe, perimeter same rehta hai; 3 sides pe, perimeter 2 ghatta hai.

House plans (page 146–147)

Charan ka ghar (30 ft gehra)

Chaudai = 15 ft (master bedroom) + 5 ft (toilet) + 15 ft (kitchen) = 35 ft.

• Small bedroom: 180 ÷ 15 = 12, to 15 ft × 12 ft.
• Utility: kitchen ke upar; 15 ft − 12 ft = 3 ft oonchi, to 15 ft × 3 ft = 45 sq ft.
• Garden: 30 − 15 − 12 = 3 ft ooncha aur 15 + 5 = 20 ft chauda, to 20 ft × 3 ft = 60 sq ft.
• Parking: 15 ft × 3 ft = 45 sq ft.
• Hall: kitchen ke neeche wala hissa 20 ft × 12 ft = 240 sq ft. Plan mein toilet aur hall ke beech 5 ft × 5 ft ka raasta bhi hai, jo hall ke saath gino to 25 sq ft aur judta hai.
• Ghar ka area = 35 ft × 30 ft = 1,050 sq ft.

Sharan ka ghar (42 ft chauda)

Gehrai = 15 ft (master bedroom) + 10 ft (small bedroom) = 25 ft. Hall 23 ft chauda aur kitchen 18 ft, to toilet 23 − 18 = 5 ft chauda hai, aur utility wala column 42 − 12 − 23 = 7 ft chauda.

• Toilet: 5 ft × 10 ft = 50 sq ft.
• Utility: 70 ÷ 7 = 10, to 7 ft × 10 ft.
• Hall: 25 − 10 = 15 ft gehra, to 23 ft × 15 ft = 345 sq ft.
• Entrance: 7 ft × 15 ft = 105 sq ft.
• Small bedroom: 12 ft × 10 ft = 120 sq ft.
• Ghar ka area = 42 ft × 25 ft = 1,050 sq ft.

Dono ki tulna: dono gharon ka area same hai, 1,050 sq ft. Charan ke ghar ka perimeter 2 × (35 + 30) = 130 ft aur Sharan ka 2 × (42 + 25) = 134 ft hai, to Sharan ke ghar ka perimeter zyada hai.

Area Maze Puzzles (page 148)

• (a) 30 sq cm. Upar ke dono rectangles ki oonchai same hai, aur 26, 13 ka do guna hai, to daayan column baayein se do guna chauda hai. Neeche ke rectangles ki oonchai bhi same hai, to missing area 2 × 15 = 30 sq cm.
• (b) 9 sq cm. Neeche wala rectangle 2 cm ooncha aur 10 sq cm area ka hai, to 5 cm chauda hai. Lamba rectangle 5 − 3 = 2 cm chauda hai, aur 10 sq cm area ke saath 5 cm ooncha. Shaded rectangle 3 cm chauda aur 5 − 2 = 3 cm ooncha hai: 3 × 3 = 9 sq cm.
• (c) 16 sq cm. Beech wala rectangle: 42 ÷ 6 = 7 cm chauda. Neeche wala 7 + 5 = 12 cm chauda hai, to 60 ÷ 12 = 5 cm ooncha. Shaded hissa 15 − 6 − 5 = 4 cm ooncha aur 7 − 3 = 4 cm chauda: 4 × 4 = 16 sq cm.
• (d) 5 cm. Chhota rectangle 18 ÷ 5 = 3.6 cm ooncha hai, to bada 4 + 3.6 = 7.6 cm ooncha hai aur 38 ÷ 7.6 = 5 cm chauda.

Figure it Out: mixed sawal (page 149)

Question 1

Kul area = 5 × 10 + 2 × 7 = 50 + 14 = 64 sq m. Ek jawab 8 m × 8 m hai. Aur bhi ho sakte hain: 16 m × 4 m aur 32 m × 2 m.

Question 2

Width = 1,000 ÷ 50 = 20 m.

Question 3

Farsh = 5 × 4 = 20 sq m, carpet = 3 × 3 = 9 sq m. Bina carpet ka hissa = 20 − 9 = 11 sq m.

Question 4

Garden = 15 × 12 = 180 sq m. Phoolon ki kyaariyan = 4 × (2 × 1) = 8 sq m. Lawn = 180 − 8 = 172 sq m.

Question 5

Jaise, Shape A 1 × 18 ki patti ho sakti hai (area 18 sq units, perimeter 38 units) aur Shape B 4 × 5 rectangle (area 20 sq units, perimeter 18 units). A ka area kam hai lekin perimeter zyada.

Question 6

Jawab tumhare page ke size pe depend karta hai. Agar page W cm chauda aur H cm ooncha hai, to border (W − 3) cm chauda aur (H − 2) cm ooncha hoga, to uska perimeter = 2 × ((W − 3) + (H − 2)). Jaise 20 cm × 27 cm ke page pe border 17 cm × 25 cm hoga aur perimeter 2 × 42 = 84 cm.

Question 7

Bahar wale rectangle ka area 12 × 8 = 96 sq units hai, to andar wale ko 48 sq units chahiye. Beech mein 8 × 6 ka rectangle banao: ye baayein-daayein 2-2 units aur upar-neeche 1-1 unit ki jagah chhodta hai, to bahar wale rectangle ko nahi chhoota.

Question 8

(c) hamesha sahi hai. 10 cm side ka square lo: uska perimeter 40 cm hai. Aadha kaatne pe do 10 cm × 5 cm ke rectangles milte hain, har ek ka perimeter 30 cm. Dono milkar 60 cm, jo 40 cm ka 1½ guna hai. Baaki statements galat hain: har rectangle square ka aadha hai (bada nahi), dono ke perimeter milkar square se zyada hain, aur square ka area dono rectangles ke barabar hai (teen guna nahi).