Perimeter and Area

(i) Length (l) of rectangular land = 500 m
Breadth (b) of rectangular land = 300 m
Area of a rectangle = length × breadth = l × b
Area = 500 × 300 = 1,50,000 m²

(ii) Cost of 1 m² of the land = Rs 10,000
Total cost of the land = Total Area × Rate per m²
Total Cost = 1,50,000 × 10,000 = Rs 1,500,000,000 (Rs 1500 crores).

Perimeter of a square = 320 m
Formula for perimeter of a square = 4 × side
4 × side = 320
side = 320 / 4 = 80 m.
The length of the side of the square park is 80 m.

Area of a square = side × side
Area = 80 × 80 = 6400 m².

Area of rectangular plot = 440 m²
Length (l) = 22 m
Area = length × breadth
440 = 22 × breadth
Breadth = 440 / 22 = 20 m.

Perimeter of a rectangle = 2(length + breadth)
Perimeter = 2(22 + 20) = 2(42) = 84 m.

Perimeter of rectangular sheet = 100 cm
Length (l) = 35 cm
Perimeter = 2(length + breadth)
100 = 2(35 + breadth)
50 = 35 + breadth
Breadth = 50 - 35 = 15 cm.

Area of rectangle = length × breadth
Area = 35 × 15 = 525 cm².

Side of the square park = 60 m
Area of the square park = side × side = 60 × 60 = 3600 m².

Given: Area of rectangular park = Area of square park
Area of rectangular park = 3600 m².
Length of rectangular park = 90 m
Area = length × breadth
3600 = 90 × breadth
Breadth = 3600 / 90 = 40 m.

Length of rectangular wire = 40 cm
Breadth of rectangular wire = 22 cm
Length of wire = Perimeter of rectangle = 2(l + b)
= 2(40 + 22) = 2(62) = 124 cm.

Now, the wire is rebent into a square.
Perimeter of square = Length of wire = 124 cm
4 × side = 124
side = 124 / 4 = 31 cm.

Area of rectangle = l × b = 40 × 22 = 880 cm².
Area of square = side × side = 31 × 31 = 961 cm².

Since 961 cm² > 880 cm², the square shape encloses more area.

Perimeter of rectangle = 130 cm
Breadth (b) = 30 cm
Perimeter = 2(length + breadth)
130 = 2(length + 30)
65 = length + 30
Length = 65 - 30 = 35 cm.

Area of rectangle = length × breadth
Area = 35 × 30 = 1050 cm².

Length of door = 2 m, Breadth of door = 1 m
Area of door = 2 × 1 = 2 m².

Length of wall = 4.5 m, Breadth of wall = 3.6 m
Total Area of wall = 4.5 × 3.6 = 16.2 m².

Area to be white washed = Area of wall - Area of door
Area to be white washed = 16.2 - 2 = 14.2 m².

Rate of white washing = Rs 20 per m².
Total cost = Area × Rate = 14.2 × 20 = Rs 284.

Formula: Area of a parallelogram = base × height

(a) Area = 7 × 4 = 28 cm²

(b) Area = 5 × 3 = 15 cm²

(c) Area = 2.5 × 3.5 = 8.75 cm²

(d) Area = 5 × 4.8 = 24 cm²

(e) Area = 2 × 4.4 = 8.8 cm²

Formula: Area of a triangle = ½ × base × height

(a) Area = ½ × 4 × 3 = 6 cm²

(b) Area = ½ × 5 × 3.2 = 8 cm²

(c) Area = ½ × 3 × 4 = 6 cm²

(d) Area = ½ × 3 × 2 = 3 cm²

Formula: Area = base × height

(a) height = Area / base = 246 / 20 = 12.3 cm

(b) base = Area / height = 154.5 / 15 = 10.3 cm

(c) base = Area / height = 48.72 / 8.4 = 5.8 cm

(d) height = Area / base = 16.38 / 15.6 = 1.05 cm

Formula: Area = ½ × base × height ⇒ base × height = 2 × Area

(a) height = (2 × 87) / 15 = 174 / 15 = 11.6 cm

(b) base = (2 × 1256) / 31.4 = 2512 / 31.4 = 80 mm

(c) height = (2 × 170.5) / 22 = 341 / 22 = 15.5 cm

(a) For parallelogram PQRS, taking base SR = 12 cm. Then height QM = 7.6 cm.
Area = base × height = 12 × 7.6 = 91.2 cm².

(b) We know Area = 91.2 cm².
Now taking base PS = 8 cm. Then corresponding height is QN.
Area = PS × QN
91.2 = 8 × QN
QN = 91.2 / 8 = 11.4 cm.

Area of parallelogram = 1470 cm².

Taking base AB = 35 cm, the corresponding height is DL.
Area = AB × DL
1470 = 35 × DL
DL = 1470 / 35 = 42 cm.

Taking base AD = 49 cm, the corresponding height is BM.
Area = AD × BM
1470 = 49 × BM
BM = 1470 / 49 = 30 cm.

Since the triangle is right angled at A, sides AB and AC form the base and height.
Taking base AB = 5 cm and height AC = 12 cm:
Area of ΔABC = ½ × base × height
Area = ½ × 5 × 12 = 30 cm².

Now, let's take base BC = 13 cm and height AD.
Area of ΔABC = ½ × BC × AD
30 = ½ × 13 × AD
60 = 13 × AD
AD = 60 / 13 = 4.61 cm (approx).

Base BC = 9 cm, Height AD = 6 cm.
Area of ΔABC = ½ × BC × AD
Area = ½ × 9 × 6 = 27 cm².

Now taking base AB = 7.5 cm, let height from C to AB be CE.
Area = ½ × AB × CE
27 = ½ × 7.5 × CE
54 = 7.5 × CE
CE = 54 / 7.5 = 7.2 cm.

Circumference of a circle = 2πr

(a) r = 14 cm
C = 2 × (22/7) × 14 = 2 × 22 × 2 = 88 cm.

(b) r = 28 mm
C = 2 × (22/7) × 28 = 2 × 22 × 4 = 176 mm.

(c) r = 21 cm
C = 2 × (22/7) × 21 = 2 × 22 × 3 = 132 cm.

Area of a circle = πr²

(a) r = 14 mm
Area = (22/7) × 14 × 14 = 22 × 2 × 14 = 616 mm².

(b) d = 49 m ⇒ r = 49/2 m = 24.5 m
Area = (22/7) × 24.5 × 24.5 = (22/7) × (49/2) × (49/2) = 11 × 7 × 24.5 = 1886.5 m².

(c) r = 5 cm
Area = (22/7) × 5 × 5 = (22/7) × 25 = 550 / 7 = 78.57 cm² (approx).

Circumference = 154 m
2πr = 154
2 × (22/7) × r = 154
r = (154 × 7) / (2 × 22) = (154 × 7) / 44 = 7 × 7 / 2 (since 154 / 22 = 7)
r = 49 / 2 = 24.5 m.

Area = πr²
Area = (22/7) × 24.5 × 24.5 = 11 × 7 × 24.5 = 1886.5 m².

Diameter of the circular garden (d) = 21 m
Radius (r) = 21 / 2 = 10.5 m
Circumference of the garden = πd = (22/7) × 21 = 22 × 3 = 66 m.

The gardener makes 2 rounds of fence.
Length of rope required = 2 × Circumference = 2 × 66 = 132 m.

Cost of rope per meter = Rs 4
Total Cost = 132 × 4 = Rs 528.

Radius of outer circular sheet (R) = 4 cm
Radius of inner removed circle (r) = 3 cm

Area of remaining sheet = Outer Area - Inner Area
= πR² - πr² = π(R² - r²)
= 3.14 × (4² - 3²)
= 3.14 × (16 - 9)
= 3.14 × 7 = 21.98 cm².

Diameter of the circular cover (d) = 1.5 m
Length of lace required = Circumference = πd
= 3.14 × 1.5 = 4.71 m.

Cost of 1 meter lace = Rs 15
Total Cost = 4.71 × 15 = Rs 70.65.

Diameter = 10 cm ⇒ Radius (r) = 5 cm.
Perimeter of the figure = Length of the semicircular arc + Length of diameter
Perimeter = πr + d
= (22/7) × 5 + 10
= 110/7 + 10 = 15.71 + 10 = 25.71 cm.

Diameter = 1.6 m ⇒ Radius (r) = 0.8 m.
Area of the circular table-top = πr²
= 3.14 × (0.8)² = 3.14 × 0.64 = 2.0096 m².

Rate of polishing = Rs 15 per m².
Cost of polishing = Area × Rate
Cost = 2.0096 × 15 = Rs 30.14 (approx).

For Circle:
Length of wire = Circumference of circle = 44 cm
2πr = 44
2 × (22/7) × r = 44
(44/7) × r = 44
r = (44 × 7) / 44 = 7 cm.
Area of circle = πr² = (22/7) × 7 × 7 = 154 cm².

For Square:
Length of wire = Perimeter of square = 44 cm
4 × side = 44
side = 44 / 4 = 11 cm.
Area of square = side × side = 11 × 11 = 121 cm².

Comparing areas: 154 cm² (Circle) > 121 cm² (Square).
Thus, the circle encloses more area.

Radius of large circular sheet (R) = 14 cm
Area = πR² = (22/7) × 14 × 14 = 22 × 2 × 14 = 616 cm².

Radius of one small circle (r) = 3.5 cm = 7/2 cm
Area of two small circles = 2 × [πr²] = 2 × [(22/7) × (7/2) × (7/2)] = 2 × [77/2] = 77 cm².

Area of rectangle = length × breadth = 3 × 1 = 3 cm².

Total area removed = Area of 2 circles + Area of rectangle = 77 + 3 = 80 cm².

Area of the remaining sheet = Total area - Area removed
= 616 - 80 = 536 cm².

Area of square aluminium sheet = side × side = 6 × 6 = 36 cm².

Area of circle cut out = πr² = 3.14 × 2 × 2 = 3.14 × 4 = 12.56 cm².

Leftover area = Area of square - Area of circle
= 36 - 12.56 = 23.44 cm².

Circumference = 31.4 cm
2πr = 31.4
2 × 3.14 × r = 31.4
6.28 × r = 31.4
r = 31.4 / 6.28 = 5 cm.

Area = πr² = 3.14 × (5)² = 3.14 × 25 = 78.5 cm².

Length of garden (l) = 90 m, Breadth of garden (b) = 75 m
Area of garden = l × b = 90 × 75 = 6750 m².

Since 1 hectare = 10000 m²
Area of garden in hectares = 6750 / 10000 = 0.675 hectare.

Path width = 5 m
Outer length = 90 + 5 + 5 = 100 m
Outer breadth = 75 + 5 + 5 = 85 m
Outer area = 100 × 85 = 8500 m².

Area of path = Outer Area - Inner Area of garden
= 8500 - 6750 = 1750 m².

Inner length = 125 m, Inner breadth = 65 m
Inner area = 125 × 65 = 8125 m².

Path width = 3 m
Outer length = 125 + 3 + 3 = 131 m
Outer breadth = 65 + 3 + 3 = 71 m
Outer area = 131 × 71 = 9301 m².

Area of path = Outer Area - Inner Area
= 9301 - 8125 = 1176 m².

Outer length (cardboard) = 8 cm
Outer breadth = 5 cm
Outer Area = 8 × 5 = 40 cm².

Margin width = 1.5 cm
Inner length (picture) = 8 - (1.5 × 2) = 8 - 3 = 5 cm
Inner breadth = 5 - (1.5 × 2) = 5 - 3 = 2 cm
Inner Area = 5 × 2 = 10 cm².

Total area of margin = Outer Area - Inner Area
= 40 - 10 = 30 cm².

Inner room length = 5.5 m, Inner breadth = 4 m
Inner Area = 5.5 × 4 = 22 m².

Width of verandah = 2.25 m
Outer length = 5.5 + (2.25 × 2) = 5.5 + 4.5 = 10 m
Outer breadth = 4 + (2.25 × 2) = 4 + 4.5 = 8.5 m
Outer Area = 10 × 8.5 = 85 m².

(i) Area of the verandah = Outer Area - Inner Area
= 85 - 22 = 63 m².

(ii) Rate of cementing = Rs 200 per m².
Cost of cementing = 63 × 200 = Rs 12,600.

Outer side of square garden = 30 m
Outer Area = 30 × 30 = 900 m².

Path width = 1 m. Path is inside.
Inner side = 30 - (1 × 2) = 30 - 2 = 28 m
Inner Area = 28 × 28 = 784 m².

(i) Area of the path = Outer Area - Inner Area
= 900 - 784 = 116 m².

(ii) Area of remaining portion (inner area) to plant grass = 784 m².
Rate = Rs 40 per m².
Cost = 784 × 40 = Rs 31,360.

Length (l) = 700 m, Breadth (b) = 300 m.
Area of the park = 700 × 300 = 2,10,000 m².

Width of cross roads = 10 m.
Area of road parallel to length = 700 × 10 = 7000 m².
Area of road parallel to breadth = 300 × 10 = 3000 m².
Area of common central square intersection = 10 × 10 = 100 m².

Area of the roads = Area of road 1 + Area of road 2 - Common area
= 7000 + 3000 - 100 = 10000 - 100 = 9900 m².

Area of roads in hectares = 9900 / 10000 = 0.99 hectares.

Area of park excluding cross roads = Area of park - Area of roads
= 2,10,000 - 9900 = 2,00,100 m².

Excluding area in hectares = 200100 / 10000 = 20.01 hectares.

Length = 90 m, Breadth = 60 m, Road width = 3 m.

(i) Area of road along length = 90 × 3 = 270 m².
Area of road along breadth = 60 × 3 = 180 m².
Common intersection area = 3 × 3 = 9 m².

Total area covered by the roads = 270 + 180 - 9
= 450 - 9 = 441 m².

(ii) Cost of constructing roads = Area × Rate
= 441 × 110 = Rs 48,510.

Length of cord required around circular pipe = Circumference = 2πr
= 2 × 3.14 × 4 = 25.12 cm.

Length of cord required around the square box = Perimeter = 4 × side
= 4 × 4 = 16 cm.

Since 25.12 cm > 16 cm, yes, she had cord left over.

Cord left = 25.12 - 16 = 9.12 cm.

Let dimensions of rectangular lawn: Length = 10 m, Breadth = 5 m.
Let radius of circular flower bed = 2 m.

(i) Area of whole land (rectangle) = l × b
= 10 × 5 = 50 m².

(ii) Area of flower bed (circle) = πr²
= 3.14 × 2 × 2 = 12.56 m².

(iii) Area of lawn excluding flower bed = Total Area - Flowerbed Area
= 50 - 12.56 = 37.44 m².

(iv) Circumference of flower bed = 2πr
= 2 × 3.14 × 2 = 12.56 m.

Let length = 5x and breadth = 4x.
Perimeter = 2(l + b) = 2(5x + 4x) = 2(9x) = 18x.
18x = 108 ⇒ x = 108 / 18 = 6.
Length = 5(6) = 30 m.
Breadth = 4(6) = 24 m.

Area = l × b
3584 = 64 × b ⇒ b = 3584 / 64 = 56 m.
Perimeter of the field = 2(l + b) = 2(64 + 56) = 2(120) = 240 m.
Distance covered in 5 rounds = 5 × 240 = 1200 m = 1.2 km.
Speed = 6 km/h.
Time = Distance / Speed = 1.2 / 6 = 0.2 hours.
0.2 hours = 0.2 × 60 minutes = 12 minutes.

Let breadth = x. Then length = 2x.
Area = l × b = (2x)(x) = 2x².
2x² = 288 ⇒ x² = 144 ⇒ x = 12 cm.
Breadth = 12 cm, Length = 24 cm.
Perimeter = 2(24 + 12) = 2(36) = 72 cm.

Perimeter of field = Total Cost / Rate = 3200 / 16 = 200 m.
Side of square = Perimeter / 4 = 200 / 4 = 50 m.
Area of the field = 50 × 50 = 2500 m².
Cost of reaping per 100 m² = Rs 35.
Total Cost = (2500 / 100) × 35 = 25 × 35 = Rs 875.

Let base = x. Then altitude = 2x.
Area = base × altitude = x × 2x = 2x².
2x² = 338 ⇒ x² = 169 ⇒ x = 13 cm.
Base = 13 cm.
Altitude = 2(13) = 26 cm.

Area = ½ × base × height
120 = ½ × 15 × h
240 = 15h
h = 240 / 15 = 16 cm.

Let base = 4x and height = 5x.
Area = ½ × 4x × 5x = 10x².
10x² = 90 ⇒ x² = 9 ⇒ x = 3 cm.
Base = 4(3) = 12 cm.
Height = 5(3) = 15 cm.

This forms two cross paths parallel to length and breadth.
Area of path parallel to length = 30 × 2 = 60 m².
Area of path parallel to breadth = 24 × 2 = 48 m².
Area of common square = 2 × 2 = 4 m².
Total Area = 60 + 48 - 4 = 108 - 4 = 104 m².

Total area of park = 60 × 40 = 2400 m².
Let width of road be w.
Area of paths = Total Area - Lawn Area = 2400 - 2109 = 291 m².
Area of paths = (60w) + (40w) - (w²) = 100w - w².
w² - 100w + 291 = 0.
w² - 97w - 3w + 291 = 0 ⇒ w(w - 97) - 3(w - 97) = 0.
(w - 3)(w - 97) = 0. Since w cannot be 97 (greater than width), w = 3 m.

2πr = 44
2 × (22/7) × r = 44
r = (44 × 7) / 44 = 7 cm.
Area = πr² = (22/7) × 7 × 7 = 154 cm².

C1 = 2π(20) = 40π
C2 = 2π(13) = 26π
Sum of circumferences = 40π + 26π = 66π.
Let radius of new circle be R. Then 2πR = 66π.
2R = 66 ⇒ R = 33 cm.

Inner circumference = 220 m
2 × (22/7) × r = 220 ⇒ r = (220 × 7) / 44 = 35 m.
Outer radius (R) = inner radius + width = 35 + 7 = 42 m.
Outer circumference = 2 × (22/7) × 42 = 2 × 22 × 6 = 264 m.
Cost of fencing = Outer circumference × Rate = 264 × 20 = Rs 5280.

Length of wire = Circumference of circle = 2πr
= 2 × (22/7) × 28 = 2 × 22 × 4 = 176 cm.
Perimeter of square = Length of wire = 176 cm.
4 × side = 176 ⇒ side = 176 / 4 = 44 cm.

C1 / C2 = 3 / 5 ⇒ (2πr1) / (2πr2) = 3 / 5 ⇒ r1 / r2 = 3 / 5.
Ratio of areas = (πr1²) / (πr2²) = (r1 / r2)²
= (3 / 5)² = 9 / 25.
Ratio is 9 : 25.

In 12 hours, the hour hand completes exactly 1 full rotation.
Radius (r) = 4.5 cm.
Distance covered = 1 circumference = 2πr
= 2 × (22/7) × 4.5 = (44 × 4.5) / 7 = 198 / 7 ≈ 28.28 cm.

Radius (r) = 28 cm = 0.28 m.
Distance per revolution = Circumference = 2 × (22/7) × 0.28 = 2 × 22 × 0.04 = 1.76 m.
Number of revolutions = Total Distance / Distance per rev
= 704 / 1.76 = 400 revolutions.

Length of wire = Perimeter of rectangle = 2(40 + 26) = 2(66) = 132 cm.
Circumference of formed circle = 132 cm
2πr = 132 ⇒ 2 × (22/7) × r = 132
(44/7) × r = 132 ⇒ r = (132 × 7) / 44 = 3 × 7 = 21 cm.

Let outer radius be R and inner radius r = 14 m.
Area of track = πR² - πr² = π(R² - r²)
1232 = (22/7)(R² - 14²)
R² - 196 = (1232 × 7) / 22 = 56 × 7 = 392.
R² = 392 + 196 = 588.
R = √588 ≈ 24.25 m.
Width of track = R - r = 24.25 - 14 = 10.25 m.

Perimeter of a semicircle = πr + 2r (arc length + diameter).
P = (22/7) × 14 + 2(14)
P = (22 × 2) + 28 = 44 + 28 = 72 cm.

Area of equilateral triangle = (√3 / 4) × side².
16√3 = (√3 / 4) × side²
16 = side² / 4
side² = 64
side = 8 cm.