We want to determine the given polygon's and the . Let's start by finding the perimeter.
Perimeter
The polygon has one unknown side. To determine the perimeter, we need to know this last side as well. For this purpose, we will add a segment to the diagram which we label b.
We can find the length of the segment by using the .
a^2+b^2=c^2
6^2+b^2= 10^2
b=± 8
b=8
As we can see, b is 8 meters long. Let's add this to the diagram.
Now we have enough information to find the length of the final side. Again, we will use the Pythagorean Theorem.
a^2+b^2=c^2
5^2+b^2= 8^2
b=± sqrt(39)
b=sqrt(39)
The last side is sqrt(39). With this information, we can determine the perimeter of the shape by adding up all the sides.
P = 5+sqrt(39)+4+10+10 ⇔ P = sqrt(39) + 29
We can also find the approximate perimeter using the calculator.
P = sqrt(39) + 29 ≈ 35m
Area
Next, we want to find the area of the given figure. Notice that we can divide it into three simple shapes - two right triangles and one rectangle.
The area of the whole figure is the sum of the areas of the smaller figures.
A_(figure) = A_(1^(st)△)+A_(rectangle)+ A_(2^(nd)△)
Let's find the areas of the smaller figures.
1^(st) Triangle's Area
Let's take a closer look at the first triangle.
As we can see, this is a right triangle with lengths measuring 5 m and sqrt(39) m. Let's recall the formula for the .
A = 1/2bh
In the formula above, b is a side of a triangle and h is the falling on this side. In the case of a right triangle, because the legs are perpendicular to each other b can be one leg and h can be the other. Let's substitute 5 for b and sqrt(39) for h to find the area.
A_(1^(st) △)= 1/2bh
A_(1^(st) △)= 1/2( 5 sqrt(39))
A_(1^(st) △)= 5sqrt(39)/2
Rectangle's Area
The rectangle's sides are 4 m and 8 m.
Let's recall the formula for the .
A = l w
In the formula above l and w are the two side lengths of the rectangle. Therefore, if we substitute 4 for l and 8 for w we will get the rectangle's area in meters.
A_(rectangle) = l w
A_(rectangle) = 4( 8)
A_(rectangle) = 32
2^(nd) Triangle's Area
This time, the legs of the triangle measure 6 and 8 meters.
Let's calculate the area of this triangle like we did for the first one.
A_(2^(nd)△)=1/2bh
A_(2^(nd)△)=1/2( 6)( 8)
A_(2^(nd)△) = 24
Area of The Whole Figure
Now that we know the area's of all of the smaller figures, let's calculate the area of the whole area by adding them.
A_(figure) = A_(1^(st)△)+A_(rectangle)+ A_(2^(nd)△)
A_(figure) = 5sqrt(39)/2 + 32 + 24
A_(figure) = 5sqrt(39)/2 + 56
A_(figure) = 56+ 5sqrt(39)/2
