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uranmaximum [27]
3 years ago
9

A home builder wants to decorate the post at the bottom of a stairway. One way to do this is to glue a square-based pyramid to t

he top of the post. The pyramid is made from a single, folded sheet of copper. The square is 10 inches on a side, and the triangular sides have a central height of 6 inches.
How many square inches of copper sheet are needed to make on pyramid?

Mathematics
1 answer:
AysviL [449]3 years ago
4 0

We have been given net of a pyramid. We are asked to find the total surface area of the pyramid.

The surface area of the pyramid will be area of base plus area of 4 triangular sides.

We have been given that base of pyramid is square, so area of base will be square of base side.

\text{Area of base}=(\text{10 in})^2=100\text{ in}^2

Let us find area of one triangular side.

\text{Area of triangle}=\frac{1}{2}\times \text{Base}\times\text{Height}

\text{Area of triangle}=\frac{1}{2}\times \text{10 in}\times\text{6 in}

\text{Area of triangle}=\text{5 in}\times\text{6 in}

\text{Area of triangle}=30\text{ in}^2

Now we will multiply area of one triangular face by 4 to find area of 4 triangular faces.

\text{Area of 4 triangular faces}=4\times 30\text{ in}^2

\text{Area of 4 triangular faces}=120\text{ in}^2

Total surface area of pyramid would be area of base plus area of 4 triangular faces.

\text{Total surface area of pyramid}=100\text{ in}^2+120\text{ in}^2

\text{Total surface area of pyramid}=220\text{ in}^2

Therefore, 220 square inches of copper sheet are needed to make one pyramid.

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let's check how much is it after 2 years firstly.


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Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}


\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}

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