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3 years ago

Find the Lateral area for the regular pyramid. (Keep it as a Square Root)

Mathematics

1 answer:

Shtirlitz [24]3 years ago

3 0

1. You must apply the following formula:

LA=1/2 ×(Perimeter of the base)(Slant length)

2. The slant length is (By the Pythagorean Theorem):

a^2=b^2+c^2

a= b^2+c^2

a=√(1^2+3^2)

a=√10

3. The perimeter of the base is:

P=2+2+2+2

P=8

4. Then, you have:

LA=1/2 ×(Perimeter of the base)(Slant Length)

LA=1/2 ×(8)(√10)

LA=(8√10)/2

LA=4√10

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Answer:

Step-by-step explanation:

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Even I am surprised by what I wrote. Hopefully, it is right and helps you ^-^ Please comment if you have any questions.

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3 years ago

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Artyom0805 [142]

Take

$\begin{cases}u=x-5y\\v=8x-y\end{cases}$

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$\begin{cases}\mathbf x(u,v)=\dfrac{-u+5v}{39}\\\\\mathbf y(u,v)=\dfrac{-8u+v}{39}\end{cases}$

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$|\det J|=\left|\begin{vmatrix}\mathbf x_u&\mathbf x_v\\\mathbf y_u&\mathbf y_v\end{vmatrix}\right|=\dfrac1{32}$

So upon transforming the coordinates to the u-v plane, you have (and I'm guessing on what the integrand actually is)

$\displaystyle\iint_R(10x-5y)(8x-y)\,\mathrm dA=\frac1{32}\int_{u=0}^{u=2}\int_{v=5}^{v=10}\frac5{13}(2u+3v)v\,\mathrm dv\,\mathrm du$
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4 years ago

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3 years ago