A tetrahedron is a triangular pyramid. It has four triangular faces and four vertices.
In the question, we are told that the length of the edges are all equal = 20m and the length from one corner to the center of the base = 11.5m
We sketch a diagram of a tetrahedron with the given measurement as shown below
To find the height, we will use the Pythagoras theorem
20² = h² + 11.5²
h² = 20² - 11.5²
h² = 267.75
h = √267.75
h = 16.36 (rounded to two decimal places)
Step-by-step explanation:
we have
length =8in
breadth =6in
since . The length and width of the actual flower bed will be 24 times larger than the length and width in the drawing.
new
length=8×24=192in
breadth =6×24=144in
(b) What is the perimeter of the actual flower bed ? Show your work.
<em>answer:the perimeter of the</em><em> </em><em>actual flower bed</em><em> =2(l+b)</em>
<em> =2(l+b)=2(192+144)=672</em><em>in</em>
(a) What is the perimeter of the drawing? Show your work
<em>answer</em><em> </em><em>:</em><em> </em><em>the perimeter of the </em><em>drawing</em>
<em>=</em><em>2</em><em>(</em><em>8</em><em>+</em><em>6</em><em>)</em><em>=</em><em>9</em><em>6</em><em>in</em>
<em>(c) What is the effect on the perimeter of the flower bed with the dimensions are multiplied by 24? Show your work</em>
<em>p</em><em>e</em><em>r</em><em>i</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em> </em><em>o</em><em>f</em><em> </em><em>flower </em><em>bed</em><em> </em><em>/</em><em>p</em><em>e</em><em>r</em><em>i</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em> </em><em>o</em><em>f</em><em> </em><em>d</em><em>r</em><em>a</em><em>w</em><em>i</em><em>n</em><em>g</em>
<em>=</em><em>9</em><em>6</em><em>/</em><em>6</em><em>7</em><em>2</em><em>=</em><em>1</em><em>/</em><em>7</em>
<h3>
<em>perimeter</em><em> </em><em>of</em><em> </em><em>drawing</em><em> </em><em>is</em><em> </em><em>increased</em><em> </em><em>by</em><em> </em><em>7</em><em>t</em><em>i</em><em>m</em><em>e</em><em>s</em><em> </em><em>of</em><em> </em><em>perimeter</em><em> </em><em>of</em><em> </em><em>flower</em><em> </em><em>bed</em></h3>
Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
here's the solution,
we know :

So,

so.. let the perpendicular be 12x and hypotenuse be 13x
now,
by applying pythagoras theorem,
where,
- b = base
- h = hypotenuse
- p = perpendicular
So,
so,
hope it helps !!