The diagonal of the square is equal to the diameter of the circle, which is 20 cm. Side of the square = diagonal / √2 = 20 cm / √2 = 10√2 cm. Area = (10√2 cm)² = 200 cm².

Perimeter of a square = 4 × side, so side = perimeter / 4 = 40 cm / 4 = 10 cm. Area = side × side = 10 cm × 10 cm = 100 cm².

Area = side² = 8² = 64 cm²

Area = 1/2 × (base1 + base2) × height = 1/2 × (10 cm + 6 cm) × 4 cm = 32 cm².

Area = 1/2 * (base1 + base2) * height = 1/2 * (5 + 7) * 4 = 24 cm²

Area = 1/2 × (base1 + base2) × height = 1/2 × (10 cm + 6 cm) × 4 cm = 32 cm².

The area of a triangle is given by (1/2) * base * height. Here, base = 4 and height = 3, so area = (1/2) * 4 * 3 = 6.

The area of a triangle is given by the formula (1/2) * base * height. Thus, Area = (1/2) * 10 * 5 = 25 cm².

Area = 1/2 * base * height = 1/2 * 10 * 5 = 25 square units.

Area = 1/2 * base * height = 1/2 * 12 * 5 = 30 cm².

Area = 1/2 × base × height = 1/2 × 6 cm × 4 cm = 12 cm².

The area of a triangle is given by A = 1/2 * base * height. Here, A = 1/2 * 8 * 5 = 20 cm².

Area = (1/2) * base * height = (1/2) * 10 * 5 = 25.

Area = 1/2 * base * height = 1/2 * 10 * 5 = 25 cm².

The area is calculated as (1/2) * base * height = (1/2) * 4 * 3 = 6.

Area = 0.5 * base * height = 0.5 * 1 * 1 = 0.5 square units.

The area of an ellipse is given by A = πab. Here, A = π * 7 * 4 = 28π.

Area = (√3/4) * side² = (√3/4) * 6² = (√3/4) * 36 = 9√3 cm².

The area of an equilateral triangle is given by the formula (√3/4)a².

Area of base = πr² = π(3²) = 9π square units.

Area of base = πr² = π(4)² = 16π square units.

Area = 1/2 * base * height = 1/2 * 4 * 3 = 6.

The area is given by the integral from 1 to 2 of 1/x dx. This evaluates to [ln(x)] from 1 to 2 = ln(2) - ln(1) = ln(2).

The area under the curve is given by ∫(from 1 to 4) (1/x) dx = [ln(x)] from 1 to 4 = ln(4) - ln(1) = ln(4).

The area under the curve is given by ∫(from 0 to 2) (2x^2 + 3) dx = [(2/3)x^3 + 3x] from 0 to 2 = (16/3 + 6) = 10.

The area is given by the integral from 0 to π/2 of cos(x) dx. This evaluates to [sin(x)] from 0 to π/2 = 1 - 0 = 1.

The area is given by the integral from 0 to π of sin(x) dx. This evaluates to [-cos(x)] from 0 to π = [1 - (-1)] = 2.

The area is given by the integral ∫_0^3 x^2 dx = [x^3/3]_0^3 = 27/3 = 9.

The area is ∫(1 to 3) x^2 dx = [1/3 * x^3] from 1 to 3 = (27/3 - 1/3) = 26/3.

The area under the curve y = x^3 from x = 1 to x = 2 is given by ∫(from 1 to 2) x^3 dx = [x^4/4] from 1 to 2 = (16/4) - (1/4) = 4 - 0.25 = 3.75.