The answer is 7 R 4! hoped this helps!
We know that
[area of a regular hexagon]=6*[area of one <span>equilateral triangle]
</span>210.44=6*[area of one equilateral triangle]
[area of one equilateral triangle]=210.44/6-----> 35.07 cm²
[area of one equilateral triangle]=b*h/2
h=7.794 cm
b=2*area/h------> b=2*35.07/7.794------>b= 9 cm
the length side of a regular hexagon is 9 cm
<span>applying the Pythagorean theorem
</span>r²=h²+(b/2)²------>r²=7.794²+(4.5)²------> r²=81--------> r=9 cm
<span>this last step was not necessary because the radius is equal to the hexagon side------> (remember the equilateral triangles)
</span>
the answer is
the radius is 9 cm
The distance between two points on the plane is given by the formula below
![\begin{gathered} A=(x_1,y_1),B=(x_2,y_2) \\ \Rightarrow d(A,B)=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%28x_1%2Cy_1%29%2CB%3D%28x_2%2Cy_2%29%20%5C%5C%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B%28x_1-x_2%29%5E2%2B%28y_1-y_2%29%5E2%7D%20%5Cend%7Bgathered%7D)
Therefore, in our case,
Thus,
![\begin{gathered} \Rightarrow d(A,B)=\sqrt[]{(-1-5)^2+(-3-2)^2}=\sqrt[]{6^2+5^2}=\sqrt[]{36+25}=\sqrt[]{61} \\ \Rightarrow d(A,B)=\sqrt[]{61} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B%28-1-5%29%5E2%2B%28-3-2%29%5E2%7D%3D%5Csqrt%5B%5D%7B6%5E2%2B5%5E2%7D%3D%5Csqrt%5B%5D%7B36%2B25%7D%3D%5Csqrt%5B%5D%7B61%7D%20%5C%5C%20%5CRightarrow%20d%28A%2CB%29%3D%5Csqrt%5B%5D%7B61%7D%20%5Cend%7Bgathered%7D)
Therefore, the answer is sqrt(61)
In general,
Remember that
Therefore,
Answer:
127.253 m^3
Step-by-step explanation:
To find the volume, we start by getting the area of the square base
Mathematically, that will be 4.9^2 m^2
To complete the volume, we multiply the area of the base by the height
= 4.9^2 * 5.3 = 127.253 m^3
