Inside a square with side length 10, two congruent equilateral triangles are drawn such that they share one side and each has on
e vertex on a vertex of the square. What is the side length of the largest square that can be inscribed in the space inside the square and outside of the triangles?
1 answer:
Answer:
5 length
Step-by-step explanation:
The diagram attached shows two equilateral triangles ABC & CDE. Since both squares share one side of the square BDFH of length 10, then their lengths will be 5 each. To obtain the largest square inscribed inside the original square BDFH, it makes sense to draw two other equilateral triangles AGH & EFG at the upper part of BDFH with length equal to 5.
So, the largest square that can be inscribe in the space outside the two equilateral triangles ABC & CDE and within BDFH is the square ACEG.
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Answer:
Option C. Yes; y=2x
Step-by-step explanation:
we know that
A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form or
In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin
so
In this problem
For x=0, y=0
That means ----> the line passes through the origin
<em>Find the value of k</em>
For x=2, y=4
---->
For x=4, y=8
---->
The values of k are equal
therefore
The table represent a direct variation
The equation is equal to
Given:
Radius of the cone = 5 ft
Slant height of the cone = 21 ft
To find:
The surface area of the cone
Solution:
Surface area of cone formula:
where r is the radius and l is the slant height.
The surface area of the cone is 130π square feet.