Find the area of the following figures.
The area A of a circle with radius r equals:
A = \pi r^2
The radius is the distance from the center \left(2{,}1\right) to any point on the circle. Here, we can choose \left(0{,}1\right).
We have r=2.
Therefore, the area equals:
\pi \times 2^2 = 4\pi
The area of the circle is 4\pi.
The area of a rectangle is the length times the width. That is:
BC \times AB
Use the pythagoras theorem to find the length of BC and the width of AB:
\left(BC\right)^2 = 3^2 + 3^2 =18
BC =\sqrt{18}
\left(AB\right)^2 = 2^2 + 2^2 =8
AB = \sqrt8
Therefore, the area equals:
\sqrt{18} \times\sqrt8 = \sqrt{144} = 12
The area of the rectangle is 12.
The area of a trapezoid is the average of the bases times the altitude:
\dfrac{AB+BC}{2}\times AH
where H=\left(-1{,}1\right).
Here we have:
- AB=5
- BC=9
- AH=3
Therefore, the area equals:
\dfrac{5+9}{2}\times 3 = 21
The area of the trapezoid is 21.
The area of a triangle is equal to half of the base times the height:
\dfrac{1}{2}\times BC \times AH
where H=\left(4{,}1\right)
Therefore the area equals:
\dfrac{1}{2} \times 4 \times 3 = 6
The area of the triangle is 6.
The area of the pentagon is equal to sum of the areas of two trapezoids (EFCD and FGBC) and a triangle (AGE). We have:
- F=\left(2{,}2\right)
- G=\left(2{,}4\right)
The area of FECD equals:
\dfrac{FC+ED}{2}\times FE
\dfrac{5+4}{2} \times 1 = \dfrac{9}{2}
The area of FGBC equals:
\dfrac{FC+GB}{2}\times FG
\dfrac{5+3}{2}\times 2= 8
The area of AGE equals:
\dfrac{1}{2}\times AG \times EG
\dfrac{1}{2}\times 1 \times 3 = \dfrac{3}{2}
Therefore, the area of the pentagon equals:
\dfrac{9}{2}+8+\dfrac{3}{2}=14
The area of the pentagon equals 14.
The area of a parallelogram equals the base time the height.
Here, we can see that the segment AC is perpendicular to the side BC. Therefore, it can serve as the height of the parallelogram to the base BC.
Therefore, the area equals:
AC \times BC = 3 \times 3 = 9
The area of the parallelogram equals 9.
The area of a triangle is equal to half of the base times the height.
The segment AH, where H=\left(6{,}1\right), is perpendicular to the continuation of the side BC. Therefore, it can serve as the height of the triangle to the base BC.
Therefore the area equals:
\dfrac{1}{2}\times BC \times AH
\dfrac{1}{2}\times 4 \times 3 = 6
The area of the triangle equals 6.