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:

$A_{1}= \frac{bh}{2}$

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:

$H= \frac{4}{sin(45)}=4\sqrt{2}$

$b=Hcos(45)=4\sqrt{2}cos(45)=4$

Therefore:

$A_{1}= \frac{4\times 4}{2}=8cm^{2}$

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

$A_{2}=L1 \times L2 = 8 \times 4=32cm^{2}$

(3) This is also a triangle, so:

$A_{3}= \frac{2\times 4}{2}=4cm^{2}$

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:

$A= \frac{d_{1}d_{2}}{2}= \frac{10.2 \times 8}{2} \rightarrow \boxed{A=40.8ft^{2}}$