Formulas for geometry, related to area and perimeter of triangles, rectangles, cercles, sectors, and volume of sphere, cone, cylinder are presented. Here are Geometry Formulas for perimeter, circumference, area, surface area, and volume. Scroll down the page for more examples and solutions using the geometry formulas.

Right Triangle and Pythagora’s theorem

Pythagora’s theorem: The two sides a and b of a right triangle and the hypotenuse c are related by

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

Area and Perimeter of Triangle

Perimeter = a + b + c

There are several formulas for the area.

If the base b and the corresponding height h are known, we use the formula

Area = (1 / 2) * b * h.

If two sides and the angle between them are known, we use one of the formulas, depending on which side and which angle are known

Area = (1 / 2)* b * c sin A

Area = (1 / 2)* a * c sin B

Area = (1 / 2)* a * b sin C .

If all three sides are known, we may use Heron’s formula for the area.

Area = sqrt [ s(s – a)(s – b)(s – c) ] , where s = (a + b + c)/2.

Area and Perimeter of Rectangle

Perimeter = 2L + 2W Area = L * W

Area of Parallelogram

Area = b * h

Area of Trapezoid

Area = (1 / 2)(a + b) * h

Circumference of a Circle and Area of a Circular Region

Circumference = 2*Pi*r Area = Pi*r^{ 2}

Arclength and Area of a Circular Sector

Arclength: s = r*t

Area = (1/2) *r^{ 2} * t

where t is the central angle in RADIANS.

Volume and Surface Area of a Rectangular Solid

Volume = L*W*H Surface Area = 2(L*W + H*W + H*L)

Volume and Surface Area of a Sphere

Volume = (4/3)* Pi * r^{ 3} Surface Area = 4 * Pi * r^{ 2}

Volume and Surface Area of a Right Circular Cylinder

Volume = Pi * r^{ 2} * h Surface Area = 2 * Pi * r * h

Volume and Surface Area of a Right Circular Cone

Volume = (1/3)* Pi * r^{ 2} * h Surface Area = Pi * r * sqrt (r^{ 2} + h^{ 2})

Questions and Answers For Geometry Formulas

1. A rectangle has a perimeter of 320 meters and its length L is 3 times its width W. Find the dimensions W and L, and the area of the rectangle.

Solution and answer:

Use the formula of the perimeter to write.
2 L + 2 W = 320

We now rewrite the statement “its length L is 3 times its width W” into a mathematical equation as follows:
L = 3 W

We substitute L in the equation 2 L + 2 W = 320 by 3 W.
2(3 W) + 2 W = 320

Expand and group like terms.
8 W = 320

Solve for W.
W = 40 meters

Use the equation L = 3 W to find L.
L = 3 W = 120 meters

Use the formula of the area.
Area = L W = 120 * 40 = 4800 meters^{ 2}

2. The perimeter of a rectangle is 50 feet and its area is 150 feet^{ 2}. Find the length L and the width W of the rectangle, such that L > W.

Solution and answer:

Use the formula of the perimeter to write
2 L + 2 W = 50

and the formula of the area to write
L W = 150

Divide all terms in the equation 2 L + 2 W = 50 by 2 to obtain
L + W = 25

Solve the above for W
W = 25 – L

Substitute W by 25 – L in the equation L W = 150
L(25 – L) = 150

Expand the above equation and rewrite with right term equal to zero.
-L^{ 2} + 25 L – 150 = 0