can someone please help me find the surface area of the pyramid with an equilateral triangle for a base
1 answer:
To obtain the total surface we have to calculate the surface of the 4 triangles and add up the areas (remember that the area of a triangle is (b*h)/2 , b is the base, h is the height ).
We will caculate first the area of the base triangle for that we considerer the fact that it is an equilateral triangle with sides of lenght 6 cm, now we calculate the height, I am going to draw please wait a moment
using the pythagorean theorem we have that
Then, the area of the triangle is 6*h/2 = 3h = 15.59 cm^2.
Now we calculate the area of the other 3 triangles, notice that those triangles have the same base and height so we will calculate for one of them and multiply by 3. From the image we know that the height is 15cm and the base is 6 cm so the area is 45cm^2, and 45*3 cm^2 = 135cm^2.
Finally we add up all the areas:
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So hmm notice the graph below
based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"
now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be
well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either
what's the "p" distance? well, we just found it, is just 1/8
so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8
and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic
based on where the focus point is at, and the directrix, then, the parabola is opening upwards, meaning the squared variable is the "x"
now, keep in mind, the vertex is at coordinates h,k
the vertex itself is half-way between the focus and directrix
the directrix is at y=15/8 and the focus is at y=17/8
so, half-way will then be
well, so is 1/4 between the focus point and the directrix, half of that is 1/8
so, if you move from the focus point 1/8 down, you'll get the y-coordinate for the vertex, or 1/8 up from the directrix, since the vertex is equidistant to either
what's the "p" distance? well, we just found it, is just 1/8
so, the x-coordinate is obviously -4, get the y-coordinate by 17/8 - 1/8 or 15/8 + 1/8
and plug your values (x-h)² = 4p(y-k) and then solve for "y", that's the equation of the quadratic