**Geometry Formula**

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**Square Properties :**

P = **Perimeter**

A = **Area**

S = **Side**

d = **diameter**

P = 4 x s

A = S^{2}

d = a x √2

**Rectangle Properties :**

P = **Perimeter**

A = **Area**

d = diameter

P = 2 x ( a + b )

A = a x b

d = √a^{2} + b^{2}

**Triangle Properties :**

P = Perimeter

A = Area

P = a + b + c

A = b x h / 2

A = √s(s-a)(s-b)(s-c);

s = a + b + c / 2 = p / 2.

α + β + γ = 180^{o}

**Circle Properties :**

P = Perimeter

A = Area

P = 2πr

A = πr^{2}

π = 3.14

**Parallelogram Properties :**

P = (a + b) x 2

P = 2a + 2b

A = bh = ab sin α

**Circular Sector Properties :**

L = πr = θ / 180 ^{0}

A = πr^{2} θ/360 ^{0}

**Pythagorean Theorem :**

a^{2} + b^{2} = c^{2}

c = √a^{2} + √b^{2}

**Circular Ring Properties :**

A = π (R^{2} – r^{2})

**Sphere Properties :**

S = 4πr^{2}

V = ^{4}/_{3}πr^{3}

**Trapezoid Properties :**

P = a + b + c + d

A = h x a + b / 2

**Rectangular Box Properties :**

A = 2ab + 2ac + 2bc

V = abc

**Right Circular Cone :**

A = πr^{2} + πrs

S = √r^{2} +√h^{2}

V = 1 x πr^{2} h / 3

**Cube Properties :**

A = 6l^{2}

V = l^{3}

**Cylinder Properties :**

A = 2πr( r + h)

V = πr^{2} h

**Frustum of a Cone Properties :**

V = 1 x πh (r^{2} + rR + R^{2}) / 3

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y =2x ,y=-2x and y=6 se paribadh tribhugh ka area kya hoga please give me this question answer

Volume of sphere should be 4/3*pi*r^3..

Very Good Formula

am very poor please help me maths subject

Thanks for your good work it help me thanks again