RS Aggarwal Solutions: Perimeter and Area of Plane Figures- 2 | RS Aggarwal Solutions for Class 10 Mathematics PDF Download

Given: Length of hall (l) = 5 + breadth(b) = 5 + b
Area of hall = 750 m²

We know that,
Area of rectangle = Length × Breadth
⇒ 750 = (5 + b) × b
⇒ 750 = b² + 5b
⇒ b² + 5b – 750 = 0
⇒ b² + 30b – 25b – 750 = 0
⇒ b(b + 30) – 25(b + 30) = 0
⇒ (b + 30) (b – 25) = 0
This gives us two equations,
i. b + 30 = 0
⇒ b = -30
ii. b – 25 = 0
⇒ b = 25
Since, the length of the rectangle cannot be negative
Therefore, b = 25 m
⇒ l = (b + 5) m
⇒ l = (25 + 5) m
⇒ l = 30 m

Given: Length of field (l) = 23 + breadth(b) = 23 + b
Perimeter of field = 206 m

We know that,
Perimeter = 2(l + b)
⇒ 206 = 2(23 + b + b)
⇒ 206 = 2(23 + 2b)
⇒ 206 = 46 + 4b
⇒ 4b = 206 – 46
⇒ 4b = 160
⇒ b = 40 m
Therefore,
Length of field = 23 + b
= 23 + 40
= 63 m
Now,
Area of rectangle = Length × Breadth
= 63 × 40
= 2520 m²

Given:
Length = 12 m
Length of diagonal = 15 m

We know that,
Base² + Perpendicular² = Hypotenuse²
⇒ 12² + Perpendicular² = 15²
⇒ Perpendicular² = 15² – 12²
⇒ Perpendicular² = 225– 144
⇒ Perpendicular² = 81
⇒ Perpendicular² = 9
That is,
Breadth = 9 m
Now,
Area = Length × Breadth
= 12 m × 9 m
= 108 m²

Given: Length of rectangle (l) = 3 × breadth(b) = 3b
Diagonal of rectangle = 8√10 m

We know that,
Base² + Perpendicular² = Hypotenuse²
⇒ b² + (3b)² = (8√10)²
⇒ b² + 9b² = 640
⇒ 10b² = 640

⇒ b² = 64
⇒ b = 8 cm
Therefore,
l = 3b = 24 cm
Hence,
Perimeter of a rectangle = 2(length + breadth)
= 2(24 + 8)
= 64 cm

Let the length be l
And, breadth be b
Now,
Area = l × b = lb
Increase in length = 20% of length + length



Decrease in breadth = breadth - 20% of breadth




Since,

Therefore,
The area is decreased



= 4%
Hence change in area = 4% decrease

Given:
Length = 80 m
Breadth = 50 m
Width of the path = 1m
Area of path = 1911 m²

Length of field with path = 80 + (1 + 1)
= 82 m
Breadth of field with path = 50 + (1 + 1)
= 52 m
Area of field with path = Length of field with path × Breadth of field with path
= 82 m × 52 m
= 4264 m²
Area of field without path = Length without path × Breadth without path
= 80 m × 50 m
= 4000 m²
Now,
Area of path = Area of field - Area of field without path
= 4264 m² – 4000 m²
= 264 m²

Given:
Length of diagonal = 10√2 cm
Let the side of square = x cm

We know that,
Hypotenuse² = Base² + Perpendicular²
⇒ (10√2)² = x² + x²
⇒ 200 = 2x²

⇒ x² = 100
⇒ x = 10 cm
Now,
Area of a square = side²
= (10 cm)²
= 100 cm²

Given:
Area of square field = 6050 m²
Let the side of square = x m

We know that,
Area of a square = side²
⇒ 6050 = x²
⇒ x = 55√2
Now,
Hypotenuse² = Base² + Perpendicular²
= (55√2)² + (55√2)²
= 6050 + 6050
= 12100 m²
Therefore,
Diagonal = √12100
= 110 m

Given:
Area of square field = 0.5 hectare = 5000 m²
Let the side of square = x m

We know that,
Area of a square = side²
⇒ 5000 = x²
⇒ x = 50√2
Now,
Hypotenuse² = Base² + Perpendicular²
= (50√2)² + (50√2)²
= 5000 + 5000
= 10000 m²
Therefore,
Diagonal = √10000
= 100 m

Given:
Area of equilateral triangle = 4√3 cm²

We know that,
Area of equilateral triangle = 


⇒ side² = 4 × 4
⇒ side² = 16
⇒ side = 4 cm
Now,
Perimeter of triangle = 3 × side
= 3 × 4 cm
= 12 cm

Given:
Side of equilateral triangle = 8 cm

We know that,
Area of equilateral triangle = 


= 16√3 cm²

Given: Side of an equilateral triangle = 6√3 cm

Height of equilateral triangle = 

= 9 cm²

Given:
Height of equilateral triangle = 3√3 cm

We know that,
Height of equilateral triangle = 


⇒ Side = 6 cm
Now,
Area of equilateral triangle = 


= 9√3 cm²

Given:
Base: Height = 3: 4
Area = 216 cm²
Let,
Base = 3x
Height = 4x

We know that,
Area of a triangle = 1/2 × base × height
⇒ 216 = 1/2 × 3x × 4x
⇒ 216 × 2 = 12x²
⇒ 12 x² = 432

⇒ x² = 36
⇒ x = 6 cm
Therefore,
Height = 4x
= 24 cm

Given:
Rate = Rs 9 per m²
Side a = 20 m
Side b = 21 m
Side c = 29 m

Area of a scalene triangle = √(s(s-a)(s-b)(s-c))



⇒ s = 35 m
Now,
Area of triangular field = √(35 m × (35-20)m × (35-21)m × (35-29)m)
= √(35 m × 15 m × 14 m × 6 m)
= √44100 m²
= 210 m²
Total Cost = 210 m² × Rs 9 per m²
= Rs 1890

Let the side be x
Now,
Area of equilateral triangle = 

And,
Area of square = side²
= x²



Therefore, the ratio is 4:√3

Let the side = radius = x
Now,
Area of circle = πr²


⇒ x² = 49
⇒ x = 7 cm
Area of equilateral triangle = 

Given:
Area of rhombus = 480 cm²
Length of diagonal 1 (d₁) = 20 cm

Let, Length of diagonal 2 be d₂
Area of rhombus = 1/2 × d₁ × d₂
⇒ 480 = 1/2 × 20 × d₂

⇒ d₂ = 48 cm
Now,
Side of rhombus = 1/2 × √(48² + 20²)
= 1/2 × √(2304 + 400)
= 1/2 × √2704
= 1/2 × 52
= 26 cm

Given:
Side = 24 cm
Length of diagonal 1 (d₁) = 20 cm

Let, Length of diagonal 2 be d₂
We know that,
Side of rhombus = 1/2 × √(d₁² + d₂²)
⇒ 20 = 1/2 × √(24² + d²²)
⇒ 20 × 2 = √(576 + d₂²)
⇒ 40 = √(576 + d₂²)
Squaring both sides,
⇒ 1600 = 576 + d₂²
⇒ d₂² = 1024
⇒ d₂ = 32 cm
Now,
Area of rhombus = 1/2 × d₁ × d₂
= 1/2 × 24 × 32
= 384 cm²

Given: Length of rectangle (l) = 3 × breadth(b) = 3b
Diagonal of rectangle = 8√10 m

We know that,
Base² + Perpendicular² = Hypotenuse²
⇒ b² + (3b)² = (8√10)²
⇒ b² + 9b² = 640
⇒ 10b² = 640

⇒ b² = 64
⇒ b = 8 cm
Therefore,
l = 3b = 24 cm
Hence,
Perimeter of a rectangle = 2(length + breadth)
= 2(24 + 8)
= 64 cm