# Basic Geometry – Area of Complex Shapes

## Area and Perimeter of Complex Shapes

Complex figures can be divided into several smaller shapes where the perimeter or area formula is known, then added.

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## Basic Math Practice

Fractions, Percent

and Decimals

Area and Perimeter of Complex Shapes

**Math Test Prep**

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**Example – Composite 2-D shapes **

To determine the area of any composite figure, simply ** ADD** the areas of each component basic figure. Be sure to write your final answer with square units.

**Determine the area of the given shape. **

The original shape can be redrawn as a rectangle and a triangle. Rectangles have opposite sides that are congruent (exactly the same).

Area _{Composite} = Area _{Triangle} + Area _{Rectangle}

Area _{Triangle} = (1/2)(Base)(Height) = (1/2)(3m)(1.5m) = 2.25 m^{2}

Area _{Rectangle }= (Base)(Height) = (3m)(1.5m) = 4.5 m^{2}

Area _{Composite} = (2.25m^{2}) + (4.5m^{2}) = 6.75 m^{2}

** Determine the Surface Area of a Composite 3-D Solid**

To determine the surface area of any composite solid, simply **add** the surface areas of each component basic solid. You must also subtract the area of any internal face. Be sure to write your final answer with square units.

**Ex. Determine the surface area of the given shape. Leave the final answer in terms of pi.**

The original shape can be redrawn as a cylinder and a cone. We will have to subtract the area of the circle where the figures meet from each surface area equation because they are “inside” the solid.

SurfaceArea _{Composite} = S.Area _{Cone} + S.Area _{Cylinder}

S.Area _{Cone} = (Base Area)+(1/2)(Perimeter)(Height) = (1/2)(dπ)(h) = (1/2)(6π)(2) = 6π ft^{2}

S.Area _{Cylinder }= 2(Base Area)+(Perimeter)(Height) = (πr^{2})+(dπ)(h) = (π3^{2})+(6π)(5) = 39π ft^{2}

S.Area _{Composite} = (6π ft^{2}) + (39π ft^{2}) = 45π ft^{2}

**Written by:** Brian Stocker MA, Complete Test Preparation Inc.

**Modified: ** December 13th, 2018

**Published: ** October 9th, 2017