MHB Areas of Plane Figures: 63.0 & 96.0 | Need Help w/ 3 & 4

[Etrujillo]

I think the answer to number 1 is 63.0 or 60 if rounded to the nearest tenth, and i think number 2 is. 96.0 or 100 if rounded to the nearest tenth. Need help with 3 and 4. I think for 3 i would add 6 and 8 as the vertical diagonal and the horizontal diagonal would be 5 . 6+8=14 so the area would =35(rounded to nearest tenth would be 40) am i correct? Number 4 is where i get completely lost.

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[MarkFL]

1.) The area for a trapezoid is given by:

$$A=\frac{h}{2}(B+b)$$

Plugging in the given values, we find:

$$A=\frac{6\text{ m}}{2}(11+10)\text{ m}=63\text{ m}^2\approx60\text{ m}^2\quad\checkmark$$

2.) The area for a parallelogram is given by:

$$A=bh$$

Plugging in the given values, we find:

$$A=(12\text{ in})(8\text{ in})=96\text{ in}^2\approx100\text{ in}^2\quad\checkmark$$

3.) The area for a kite can be found from the product of its diagonals. We can see one is 10 cm in length, and we can use Pythagoras to get the other:

$$\sqrt{6^2-5^2}+\sqrt{8^2-5^2}=\sqrt{6}+\sqrt{39}$$

And so the area is:

$$A=(10\text{ cm})((\sqrt{6}+\sqrt{39})\text{ cm})=10(\sqrt{6}+\sqrt{39})\text{ cm}^2\approx90\text{ cm}^2$$

4.) We have a trapezoid, and we know the big base \(B\) and the little base \(b\), but we don't know the height \(h\). But, we can find it using Pythagoras. Consider the right triangle making up the left part of the diagram. We are given the hypotenuse, and the smaller leg must be half the difference between the big base and the little base. Can you continue to find the height?

 

[Etrujillo]

If i use a2 + b2 = c2
I get 8^+4^=80
80 would be the other leg?
Would this be considered a right triangle?
So one leg would =4 (10-16) and the other 80?

 

[MarkFL]

I just noticed you are instructed to round to the nearest tenth, not ten. :)

To find the height you need to use:

$$3^2+h^2=8^2\implies h=\sqrt{8^2-3^2}=\sqrt{55}$$

Now, just plug in the numbers into the formula I gave in 1.) (Smile)