What is the volume of this cone? work shown plsss ❤️
1 answer:
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According to given conditions, m+n is equal to 409.
Consider the diagram below.
In the hexagon ABCDEF, let
AB = BC = CD = 3;
DE = EF = FA = 5;
Arc BAF is equal to one-third of the circle's circumference.
Hence, ∠BCF = ∠BEF = 60°;
Similarly, ∠CBE = ∠CFE = 60°;
Let the point of intersection of BE and CF be P, BE and AD be Q and CF and AD be R.
∴ Δ EFP and Δ BCP are equilateral, and so Δ PQR is also equilateral.
Also, ∠ BAD and ∠ BED subtend the same arc and so do ∠ ABE and ∠ ADE.
∴ Δ ABQ is similar to Δ EDQ
Also,
and
On solving these simultaneous equations, we get AD = 360/49
∴ m + n = 409.
To learn more about similarity of triangles, refer to this link:
#SPJ4
Answer:
Step-by-step explanation:
Hello There!
To solve for the area of this figure we need to split the figure into 3 different parts:
A rectangle with a length of 9 ft and a width of 7 ft
a rectangle with a width of 9ft + 7ft and a length of 25 ft
a rectangle with a width of 18 ft and a length of 22 ft
To find the area of a rectangle we use the formula
where w = width and l = length
for the first one we plug in the values
A = 9 * 7
9 * 7 = 63 so the area of the smallest rectangle is 63ft²
Now lets find the area of the larger rectangle
The dimensions are l = 25 ft and w = 9 + 7 (16 ft)
Now we can plug in the values into the area formula
A = 16 * 25
16*25=400 so the area of the larger rectangle is 400 ft²
Now lets find the area of the last rectangle
The dimensions are l = 22 ft and w = 18 ft
now lets plug in the values to the formula
A = 22 * 18 =396
so the area of the last rectangle is 396 ft²
Finally we want to add all of the areas together
396 + 400 + 63 = 859
So the area of the figure is 859 ft²