Question

Can some one please help on 26 and 27 please thank you

Answer 1

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:


`A_(1)= (bh)/(2)`

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:


`H= (4)/(sin(45))=4√(2)`


`b=Hcos(45)=4√(2)cos(45)=4`

Therefore:


`A_(1)= (4* 4)/(2)=8cm^(2)`

(2) The area of a rectangle is given by:


`A_(2)=L1 * L2 = 8 * 4=32cm^(2)`

(3) This is also a triangle, so:


`A_(3)= (2* 4)/(2)=4cm^(2)`

The total are is:


`A=A_(1)+A_(2)+A_(3)= (8+32+4) \rightarrow \boxed{A=44cm^(2)}`

Part 2)

If you know the lengths of the two diagonals of a kite, the area is half the product of the diagonals. So:


`A= (d_(1)d_(2))/(2)= (10.2 * 8)/(2) \rightarrow \boxed{A=40.8ft^(2)}`