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3 years ago

A rectangular pyramid has a volume of 160 cubic feet. Find two possible sets of measurements for the base area and height of the

pyramid.

Mathematics

1 answer:

muminat3 years ago

4 0

$\bf \textit{volume of a pyramid}\\\\ V=\cfrac{1}{3}Bh~~ \begin{cases} B=area~of\\ \qquad its~base\\ h=height\\ \cline{1-1} V=160 \end{cases}\implies 160=\cfrac{Bh}{3}\implies 480=Bh \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} 480=&2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 5\\ &8\cdot 60\\ &32\cdot 15 \end{cases}~\hspace{5em}\stackrel{B}{8}\times\stackrel{h}{60}\qquad \qquad \stackrel{B}{32}\times\stackrel{h}{15}$

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andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5) and

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Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)

To Find:

M(x₄ ,y₄) = ? and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

$Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

As M is the mid point of PR and N is the mid point of RQ so we will have

$Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})$

$Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})$

Substituting the given value in above equation we get

$Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})$

∴ $Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})$

∴ $Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)$

Similarly,

$Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})$

∴ $Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})$

∴ $Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)$

∴ M(x₄ ,y₄) = (-3.5 , 0.5) and

N (x₅ ,y₅) = ( -1 , -0.5 )

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Step-by-step explanation:

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