Quadrilateral MNPQ is shown on the grid below.

Mathematics

1 answer:

AfilCa [17]2 years ago

5 0

Answer:

<em>42 square units</em>

Step-by-step explanation:

Find the diagram attached

Area of the diagram = Area of the triangle + Area of rectangle

Area of rectangle = Length * Width

Area of rectangle = 3 * 6

Area of rectangle = 18 square units

Area of triangle = 1/2 * base * height

Get the base of the tringle using the pythagoras theorem;

b^2 = 10^2-6^2

b^2 = 100 - 36

b^2 = 64

b = 8units

height = 6 units

Area of the triangle = 1/2 * 8 * 6

Area of the triangle = 48/2

Area of the triangle = 24 square units

Area of the figure = 18+24

<em>Area of the figure = 42 square units</em>

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The owner of a cattle ranch would like to fence in a rectangular area of 3000000 m^2 and then divide it in half with a fence dow

Reika [66]

If you notice the picture below, the amount of fencing, or perimeter, that will be used will be 3w + 2l

now $\bf \begin{cases}
A=l\cdot w\\\\\
A=3000000\\
----------\\
3000000=l\cdot w\\\\
\frac{3000000}{w}=l
\end{cases}\qquad thus
\\\\\\
P=3w+2l\implies P=3w+2\left( \cfrac{3000000}{w} \right)\implies P=3w+\cfrac{6000000}{w}
\\\\\\
P=3w+6000000w^{-1}\\\\
-----------------------------\\\\
now\qquad \cfrac{dP}{dw}=3-\cfrac{6000000}{w^2}\implies \cfrac{dP}{dw}=\cfrac{3w^2-6000000}{w^2}
\\\\\\
\textit{so the critical points are at }
\begin{cases}
0=w^2\\\\
0=\frac{3w^2-6000000}{w^2}
\end{cases}$

solve for "w", to see what critical points you get, and then run a first-derivative test on them, for the minimum

notice the $0=w^2\implies 0=w$ so. you can pretty much skip that one, though is a valid critical point, the width can't clearly be 0

so.. check the critical points on the other