Given: Right Pyramid, ABCD is a rhombus
AB = 6, mBAD = 50, mSKO = 40
Segment SO is perpendicular to ABCD, Segment OK is perpendicular to segment DC
Find: Volume and Surface Area of the figure
To find the volume and surface area of the given figure (a right pyramid), we need to calculate the dimensions of the pyramid.
First, let's find the length of segment SK, which is the slant height of the pyramid. To do this, we can use the law of cosines in triangle SKO:
cos(∠SKO) = (SO² + OK² - SK²) / (2 * SO * OK)
Given that ∠SKO = 40°, and the lengths of SO and OK are unknown, we need to find them.
In rhombus ABCD, we have the following information:
AB = 6 (side length of the rhombus)
∠BAD = 50° (an interior angle of the rhombus)
Using the Law of Sines, we can find the length of side AD, which will also be the length of segment SO:
sin(∠BAD) / AB = sin(∠ABD) / AD
sin(50°) / 6 = sin(∠ABD) / AD
Simplifying, we get:
AD = (6 * sin(∠ABD)) / sin(50°)
To find the length of segment OK, we can use the Pythagorean theorem in triangle ODK:
OK² + DK² = 6²
Since we know that segment OK is perpendicular to segment DC, DK will be half the length of DC. Since DC is the diagonal of the rhombus, we can find its length using:
DC = 2 * AB * cos(50°)
Now that we have DK, we can calculate OK:
OK = sqrt(6² - (0.5 * DC)²)
Finally, we can substitute the values of SO and OK in the law of cosines to find SK:
cos(40°) = (SO² + OK² - SK²) / (2 * SO * OK)
Now that we have the lengths of SO, OK, and SK, we can calculate the volume and surface area of the pyramid.
The volume of a right pyramid is given by the formula:
V = (1/3) * base area * height
The base area can be calculated using the formula for the area of a rhombus:
base area = (1/2) * diagonal₁ * diagonal₂
The height of the pyramid can be found using the Pythagorean theorem in triangle SOK:
height = sqrt(SK² - OK²)
Once we have the base area and height, we can calculate the volume and surface area using the formulas.
Now let's go through these steps to find the volume and surface area of the given figure.
The area of the base is 36sin50 = 27.58
KS is the slant height of the pyramid.
OK is the height of the base, 6sin50 = 4.60
So, OK/KS=sin40 ==> KS=4.6/sin40 = 7.15
OS is the height of the pyramid, and OS^2 = KS^2-OK^2 = 29.96, so OS=5.47
The volume of the pyramid is 1/3 Bh = 1/3 * 27.58 * 5.47 = 50.33
The area is the base plus four sides.
Each side is an isosceles triangle with base=6 and height=KS=7.15
So, the surface area is 27.58+4*(6*7.15)/2 = 113.38
Check my math...