**Geometry Formulas**

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**

**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.

^{ 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

Sources: PinterPandai, Wikipedia, Pioneer Mathematics

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