Area Problem Solving

When you know the simple area formulas for shapes like rectangles, triangles, and circles, you can often use these in combination to find the area of more complicated shapes. You can add or subtract parts of shapes in order to build up to more complex shapes.

Subtracting to find the area of a shape

Example 1

What is the area of the shaded region of the following rectangle?

The area of the shaded region will be the difference between the entire outer rectangle and the area of the square that is cut out of the middle.

The dimensions of the rectangle are 6 by 8, making the area $L\times W=6\times 8$ , which is 48 square units.

Each side of the inner square measures 2 square units, making the area $S^2=2\times 2$ square units, or 4 square units.

So, the shaded part of the rectangle has an area of $48-4=44$ square units.

Adding on to find the area of a shape

Example 2

Find the total area of the shaded region. In the figure, all angles are right angles.

The given figure can be divided into 3 squares and one rectangle, as shown.

The total area will be the sum of the area of the rectangle and the squares.

The dimensions of the rectangle are 3 by 12, so the area is $3\times 12$ . That is 36 square units.

All three squares have sides that are 3 units long, so the area of each of them is $3\times 3$ , or 9 square units.

Therefore, the total area of the shape is $36+(3\times 9)=63$ square units.

There are other ways to split up the figure into known shapes, so if your first instinct was to slice it vertically into 3 rectangles and one square, that would work just fine too!

Example 3

If $∆ABC$ is a right triangle and BC is a semicircle, find the total area of the figure.

The total area is the sum of the areas of the triangle and the semicircle.

We know the length and altitude of the triangle, so we know the area of the triangle is

$\frac{1}{2}\times 4\times 3$ square units.

Since arc BC is a semicircle, BC is a diameter. But BC is also the hypotenuse of the right triangle $∆ABC$ .

Use the Pythagorean theorem to find the length BC.

$\mathrm{BC}=\sqrt{3^2+4^2}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$=\sqrt{5}$

So the diameter of the semicircle is 5 units, making the radius 2.5 units.

The area of the semicircle with a radius of 2.5 units is $\frac{1}{2}\times \pi \times {2.5}^2$ is approximately 9.82 square units.

Therefore, the total area of the shape is about $6+9.82$ , or $15.82$ square units.

Topics related to the Area Problem Solving

Perimeter, Area and Volume

Surface Area

Area

Flashcards covering the Area Problem Solving

5th Grade Math Flashcards

Common Core: 5th Grade Math Flashcards

Practice tests covering the Area Problem Solving

MAP 5th Grade Math Practice Tests

Common Core: 5th Grade Math Diagnostic Tests

Get help learning about area problem solving

Considering the different formulas your student needs to know and the creative applications they need to use to find the areas of oddly shaped objects, it can be kind of tricky to master all the skills alone. If your student needs help figuring out how to find the area of objects other than plain rectangles, circles, and triangles, the help of an expert tutor can be useful. To learn more about how tutoring can help your student understand area problem solving, contact the Educational Directors at Varsity Tutors today.