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AURORKA [14]
4 years ago
7

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

Mathematics
1 answer:
Marta_Voda [28]4 years ago
4 0
Part 1)<span>

<span>Let's divide this problem into three shape and next we will compute the area for each shape:</span>

(1) First triangle
(2) Rectangle
(3) Second Triangle

(1) The best known and simplest formula is:

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

<span>where b</span> 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:

<span>A_{1}= \frac{4\times 4}{2}=8cm^{2}</span>

(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:

<span>A_{3}= \frac{2\times 4}{2}=4cm^{2}</span>

The total are is:

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

Part 2)<span>
</span>
If you know the lengths of the two diagonals of a kite, the area is half the product of the diagonals. So:

</span>A= \frac{d_{1}d_{2}}{2}= \frac{10.2 \times 8}{2} \rightarrow \boxed{A=40.8ft^{2}}<span>
</span>
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