Divide the decagon into 10 congruent isosceles triangles. Then find the area of a single such triangle.
About 3077.7 square units.
Practice makes perfect
We want to find the area of a decagon with a side length of 20 units. By drawing diagonals between opposite vertices, we can divide the decagon into 10 congruent isosceles triangles with a vertex angle that is 360^(∘) divided by 10.
360^(∘)/10=36^(∘)
Our first goal will be to find the area of a single such triangle. If we draw the height from the vertex angle of one triangle, it will bisect this angle and the base of the triangle, creating two right triangles. The leg opposite the bisected angle is 20 ÷ 2 = 10 units long, and the measure of the resulting angle is 36^(∘) ÷ 2 = 18^(∘).
We can find the length of the other leg of the right triangle by using the cotangent ratio.
The other leg, which is also the height falling onto one of the sides of the decagon, is about 30.777 units. Now, let's recall the formula for the area of a triangle.
A = 1/2 b h
Here, b is the triangle's base, and h is the height falling onto that base. In our case, b is 20 units, and the height h is about 30.777 units. Let's substitute these values into the formula to get the area of a single isosceles triangle.
The area of a single triangle is about 307.77 square units. Finally, to find the area of the decagon, we will multiply this value by 10.
10(307.77) = 3077.7
The decagon's area is about 3077.7 square units.
