Part
1)
Let's divide this problem into three shape and next we will
compute the area for each shape:
(1) First triangle
(2) Rectangle
(3) Second Triangle
(1) The best known and simplest formula is:
[tex]A_{1}= \frac{bh}{2}[/tex]
where b is the length of the base of the triangle, and h is the height or altitude of the triangle. Using trigonometry the
base is given by:
[tex]H= \frac{4}{sin(45)}=4\sqrt{2}[/tex]
[tex]b=Hcos(45)=4\sqrt{2}cos(45)=4[/tex]
Therefore:
[tex]A_{1}= \frac{4\times 4}{2}=8cm^{2}[/tex]
(2) The area of a rectangle is given by:
[tex]A_{2}=L1 \times L2 = 8 \times 4=32cm^{2}[/tex]
(3) This is also a triangle, so:
[tex]A_{3}= \frac{2\times 4}{2}=4cm^{2}[/tex]
The total are is:
[tex]A=A_{1}+A_{2}+A_{3}= (8+32+4) \rightarrow \boxed{A=44cm^{2}}[/tex]
Part 2)
If you know the lengths of the two diagonals of a kite, the area is half
the product of the diagonals. So:
[tex]A= \frac{d_{1}d_{2}}{2}= \frac{10.2 \times 8}{2} \rightarrow \boxed{A=40.8ft^{2}} [/tex]
