These are the answers for Math unit 2 lesson 6
Using Pythagorean theorem with 3-D figures
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First, find the height of the pyramid by using the Pythagorean Theorem
Imagine a right triangle going through the middle of the pyramid and perpendicular to the base. The slant of the pyramid is the hypotenuse.
The formula for the height of the pyramid is h²=l²-(s/2)², the slant height takes the place of c, the leg of the right triangle takes the place of b, and the height of the triangle takes the place of a
square root both sides: h=√20²-(22/2)²
simplify exponents: h=√400-121
square root: h≈16.7
Then, solve for the volume like normal 1/3·b·h
Given in the question,
side of square = 10
To find the slant height pyramid we will use pythagorus theorem
13² = 5² + height²
height = √13² - 5²
Now find the height of the pyramid
12² = 5² + height²
height = 10.91
Base area of the pyramid
area = l²
Volume = 1/3(height)(basearea)
we know that
The volume of the square pyramid is equal to
----> the length of the base of the square pyramid
---> the height of the pyramid
substitute the values in the formula
solve the equation below for the
volume of square volume of triangle
v=l*w*h v= (1/2 bh)h
v= ? v= ?
The volume of the square pyramid is given by;
Where l=12cm is the length of the square base and h=10cm is the height of the pyramid.
We substitute the values into the formula to get;
This simplifies to,
Third option is correct.
I have the Same one But Different lengths would you help me?
the answer would have to be 7.8
the probability is 0