Area of the new circular park = sum of the areas of the 2 smaller parks
⇒ π (6/2)² + π (8/2)² = π(9+ 16) = 25π ⇒ 25 π = π R². ∴ R² = 25 ⇒ R = 5 m

Let x and y be the length and breadth of the rectangle respectively.
Then, x – 4 = y + 3 or x – y = 7 —-(i)
Area of the rectangle =xy; Area of the square = (x – 4) (y + 3)
(x – 4) (y + 3) =xy <=> 3x – 4y = 12 —-(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

Area of rectangle = 460 m²

Length = 115% of Breadth

Formula used:

Area of a rectangle = Length × breadth

Calculation:

Let the breadth be x m

Then the breadth will be 1.15x

The area of the rectangle will be

x × 1.5x =  460

⇒ 1.15x² = 460

⇒ x² = 460/1.15

⇒ x² = 400

⇒ x = 20 m

∴ The breadth of the rectangle is 20 m.

Area of circle = πd²/4

diameters increased by 100%, then new area = π(2d)² /4 => πd²
Then, New area = (πd² – πd²/4) => 3πd²/4

increase percent =[(3πd²/4) / ( πd²/4) ]*100 %
= [(3πd²/4) x (4/πd²)]*100 %
= 300%

d = a√2
12√2 = a√2
a = 12
l * b = a² = (12²) = 144
l – b = 7 ; l = b + 7
(b + 7)*(b) = 144
b² + 7b – 144 = 0
b = 9; l = 16
2(l + b) = 2(16 + 9) = 50m

Let the original length be l and the width be b
Therefore, the area = l*b
Now, as the length is increased by 25%, the new length is (1.25*l) and let the new width be x.
As the area is same, 1.25*l*x = l*b
x = b/1.25 = 0.8b
Therefore, the width is to be decreased by 20%.

15 * 15 = 225 sq m

Let a rectangular field ABCD
A/Q a man cross rectangular field diagonally and he had to walk 20m
So,Diagonal of Rectangle =20m
Length=16m (given)
We know that Each angle of a rectangle is 90°
ADC forms a right angled triangle
AC=16m(perpendicular)
AD=20m(Hypotenus)
CD=?(base)
According to Pythagorean Theorem
h2=b2+p2
(20)2=b2+(16)2
400=b2+256
b2=144
b=√144
b=12m

Parallel sides of a trapezium = 6 cm, and 10 cm

Area of trapezium =1/2 ( sum of the parallel sides ) × distance between the parallel sides

32=1/2 ( 6+10 ) × distance

32=8 × distance

Distance = 32/8 = 4 cm

So, the distance between the parallel lines of trapezium = 4 cm.

Height = 4 cm

Step-by-step explanation:

Parallel sides = 6 cm and 10 cm

Area of trapezium = 32 cm²

⇒  *(sum of || sides)* h = 32

⇒  * (6 + 10) * h = 32

⇒ 16 / 2 * h = 32

⇒ 8 * h = 32

⇒ h = 32 / 8

⇒ h = 4

When parallelogram is on same and located between the same parallels then its area will be equal. It is given rhombus and square are two parallelograms on same base and located between the same parallels.
∴ Area of rhombus = Area of square
∴ Ratio = 1:1

let 4x and 3x be sides of the rectangle
We know that area of rectangle = l×b
4x × 3x = 972
12×2 = 972
X2 = 81
X = 9
Therefore length = 4x = 4×9 = 36m
Breadth = 3x = 3×9 = 27m
Therefore perimeter = 2 (l+b) = 2(36+27) = 126m

per the question = 2(L+B)/B = 10/2
2(L+B)/B = 5/1
2L + 2B = 5B
2B – 5B = – 2L
– 3B = – 2L
B = 2/3 L

Given Area = 430 sq. cm.

Area = Length * Breadth
430 = L * 2/3 L
430 = 2/3 L²

Or L² = 1290 / 2
L² = 645
L = 25.4 cm

Number of cubes= (100x100x100) / (10x10x10)= 1000

(OR)

The volume of cubical box =100^3 = 1000000cubic cm
Volume of cubes =10^3 =1000cubic cm
Now, no. Of cubes that can be fit in the cubical box = [(The volume of cubical box)/(volume of 1 cube)]
= 1000000/1000
= 1000 cubes

(or)

Given: The volume of a wall is 12.8 m³.

Height is 5 times the breadth and length is 8 times the height.

Let breadth = x

∴ Height = 5x

⇒ Length = 8 × (5x) = 40x

The volume of a Cuboid is given by l × b × h (here l, b, h are, length, breadth, height respectively)

∴ l × b × h = 12.8

⇒ x × 5x × 40x = 12.8

⇒ 200x³ = 12.8

⇒ x³ = 12.8/200 m³ = 8/125

⇒ x= ∛ (8/125)m = 2/5 m = 0.4 m

∴ x = 0.4m = 0.4 × 100 = 40cm

That is Breadth is: 40cm

No. of bricks= volume of the wall/volume of 1 brick

= (800 x 600 x 22.5) / (25 x 11.25 x 6) = 6400

Area of the 4 walls of a room = 2 (Length + Breadth) * Height

Given that, Length = 7 meters, Width or Breadth = 8 meters and Height = 9 meters

So, the required area, A = 2 (7 + 8) * 9

= 2 * 15 * 9

= 270 sq. m.

The diameter of the largest circle that can be drawn inside a square = length of the side of the square

As per the question: The side of the square = 20 cm.

So, radius of the circle = 20/2 = 10 cm

Area of a circle = π × radius²

The area of the largest circle = π × 102 = 100 π cm²

The perimeter: 2 (L + B) = 210 or L+ B = 105 …… (i)

As per the question: (L – B) = 25 ……. (ii)

Solving the above two equations:

L + B = 105

B = 105 – L

Put the value of B in (ii).

L – (105 – L) = 25

L – 105 + L = 25

2L = 25 + 105

2L = 130

L = 65

So, – B = 105 – 65 = 40

Now, area of a rectangle: L *B= 65 * 40 = 2600 m²

When d₁ and d₂ are the diagonals of rhombus then,
The Side of rhombus = 

Squaring both sides

20² = 16² + d₂²

400 = 256 + d₂²

d₂² = 400 – 256

d₂ = √144

d₂ = 12

Area of rhombus =1/2( d₁ × d₂ )

=1/2 ( 16 × 12 )=96 cm²

Area of the entire grass field including the path = Length (85 + 5 +5) * Breadth (65 + 5+5)
= 95 * 75 = 7125 sq. m.

Area of the grass field excluding the path = 85 * 65 = 5525 sq. m.

So, area of the path = 7125 – 5525 = 1600 sq. m.

The cost of constructing is Rs. 5 per square meter.

So, the cost of constructing the entire path = 1600 * 5 = Rs. 8000

Area of a square:
(Side)² = × (diagonal)²

= x (16)²

= x 256

= 128 m²

Let the length of the rectangle be 5x, and breadth be 3x,
Perimeter of a rectangle = 2(l +b) = 480
2(5x + 3x) = 480
2 x 8x = 480
16x = 480
x = 480/16 = 30

∴ Length = 5 x 30 = 150 m
Breadth = 3 x 30 = 90 m

And, Area of rectangle = L x B
= 150 x 90 = 13500 m²