Log in Register

Login to your account

Username *
Password *
Remember Me

Create an account

Fields marked with an asterisk (*) are required.
Name *
Username *
Password *
Verify password *
Email *
Verify email *

Newsletter

Sign up for our newsletter!
Please wait

Areas of Complex Shapes

Worked out solutions to the worksheet: Areas of Complex Shapes Review


Areas_of_complex_shapes_vis_1 

1) Your neighbor’s pool is shaped like an oval with two straight sides and two circular edges.  The measurements are given in the diagram to the left.  Find the area of the pool and the distance around the pool.

 

 

Solution #1: 

Areas_of_complex_shapes_vis_2

 

The figure can be separated into a central rectangle and a left and right semicircle.  Note that the two semicircles are exactly the same as each other.  This fact will be used later in the problem to find area and perimeter.

 

 

 

Areas_of_complex_shapes_vis_3

 

Since the height of the figure is 14, the radius of each circle must be 7.  The width of the central rectangle can be found by subtracting the radii of the circles from the total of 25:  25 - 2·7 = 25 – 14 = 11.

 

 

 

The two semicircles have the same radius and can therefore be combined into one whole circle.  The total area of the pool is equal to the sum of the areas of the rectangle and the circle to the right.

Area of the rectangle: A = L·W = 14·11 = 154 ft2

Areas_of_complex_shapes_vis_4

Area of the circle: A = (pi)r2 = (pi) ·82 = 64(pi)  = 200.96ft2

Total area: A = 154 + 200.96 = 354.96 ft2



 

The distance around the pool can be found by adding the circumference of the circle to the two straight edges.  The circle has a circumference of 2 (pi)r = 2(pi) 8 = 16(pi) = 50.24 ft.  Each edge is 11 ft long, so the total distance around the pool is 50.24 + 11 + 11 = 72.24 ft.



Areas_of_complex_shapes_vis_5

 

2) One can find the area of a regular hexagon using interior equilateral triangles.  Find the area of the regular hexagon to the right.

 

 

 

Solution #2:

Areas_of_complex_shapes_vis_6 

A hexagon can be divided into six individual equilateral triangles.  One of these triangles is shown to the left.  Since the triangle is equilateral, all three sides of the triangle measure 8 cm.

 

 

 

Areas_of_complex_shapes_vis_7
A closer look at the equilateral triangle reveals that it can be split into two smaller right triangles.  Each of these triangles has a base of 4 and a hypotenuse of 8.  Using the Pythagorean Theorem, you know that c2 – a2 = b2 (a variation of a2 + b2 = c2), so 82 – 42 = b2.  

b2 = 64 – 16 = 48, so b is equal to the square root of 48.  b = 6.93.

 

The height of the equilateral triangle is 6.93 cm and its base is 8 cm.  The area of the triangle is ½ (6.93)(8) = 27.72 cm2.  Since the hexagon is made from 6 of these triangles, its area is (6)(27.72) = 166.32 cm2.



Areas_of_complex_shapes_vis_8 

 

 

3) Find the area and perimeter of the trapezoid to the right.  (Hint: use Pythagorean Theorem or distance formula to find the diagonal length)

 

 

 

 

 

 

Solution #3:

To find the perimeter of the trapezoid, find the lengths of each side.  The lengths of the horizontal and vertical sides:  RU = 10, UT = 8, and TS = 4.  The length of the diagonal side (RS) can be found using the Pythagorean theorem or distance formula.  62 + 82 = c2, so 36 + 64 = c2 and 100 = c2.  The square root of 100 is 10, so RS = 10.  The perimeter of the trapezoid can be found by adding 10, 8, 4, and 10.

The perimeter of RSTU = 32 units.

 

The area of RSTU can be found by first finding the average of the left and right lengths.  RU = 10 and ST = 4, so the average is 14/2 = 7.  The width of the figure is 8, so its area is equal to the width times the average length, and 8 × 7 = 56.

The area of RSTU = 56 units2.


 


 

Areas_of_complex_shapes_vis_9 

 

4) Find the area and perimeter of the rectangle to the left.  (Hint: use the Pythagorean Theorem or distance formula to find the diagonal length)

 

 

 

 

 

 

 

 

Solution #4:

First, find one of the lengths.  PM can be found using the Pythagorean Theorem or distance formula.  PM can be thought of as the hypotenuse of the a right triangle whose height is 3 and base is 4.  32 + 42 = 9 + 16 = 25, so the length of PM is the square root of 25, which equals 5.

All other sides can be found similarly and all are equal to 5.  The figure is actually a square and its perimeter is 5 + 5 + 5 + 5 = 20 units.

The area of the square can be found by taking the length times the width.  5 × 5 = 25, so the area is 25 units2.

Copyright © 2014. Free Math Resource.
All Rights Reserved.