Mathematics, 06.06.2020 23:02 mari8493

45 POINTS FOR SOLVING THIS!!

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Answer from: jade5801

157.98 in^2

Step-by-step explanation:

Given: volume of pyramid is equal to 100 cubic inches and height of the pyramid is 5 inches

To find: surface area of the pyramid

Solution:

Let h denotes the height of the pyramid and 'a' denotes side of the square base of the pyramid.

Volume of the pyramid = $\frac{1}{3}a^2h$

Also,

Volume of the pyramid = 100 cubic inches

$\frac{1}{3}a^2h=100\\\frac{1}{3}a^2(5)=100\\a^2=\frac{100\times 3}{5}=60\\a=\sqrt{60}$

Surface area of the pyramid $(S) =a\sqrt{4h^2+a^2}+a^2$

So,

$S=\sqrt{60}\sqrt{100+60}+60\\=\sqrt{60}\sqrt{160}+60\\=\sqrt{9600}+60\\=97.9796+60\\=157.9796\\\approx 157.98$

Surface area = 157.98 in^2

Answer from: FailingstudentXD

Surface area of the pyramid = 157.99 in²

Step-by-step explanation:

Volume of a pyramid is given by the formula,

Volume = $\frac{1}{3}(\text{Area of the base})\times \text{height}$

100 = $\frac{1}{3}\text{(Side)}^{2} \times \text{height}$

300 = s² × 5

s² = 60

s = √60

s = 2√15 in ≈ 7.746 in

Now surface area of the pyramid = Area of base + 4×(Area of one lateral side)

Area of square base = (Side)² = 60 in

Area of one lateral side = $\frac{1}{2}(\text{Base})(\text{Lateral height})$

Since Lateral height = $\sqrt{(h)^{2}+(\frac{S}{2})^{2}}$ [By applying Pythagoras theorem in the given triangle]

= $\sqrt{(5)^{2}+(3.873)^{2}}$

= $\sqrt{25+15}$

= $\sqrt{40}$

= 6.325 in.

Now area of lateral side = $\frac{1}{2}(7.746)(6.325)$

= 24.497 in²

Surface area of the pyramid = 60 + (4×24.497)

= 60 + 97.987

= 157.987 in²

≈ 157.99 in²

Answer from: Quest

30005

yes ok

Answer from: Quest

answer: 7. if the functions had a negative leading coefficient and is of even degree, which statement about the graph is true?

step-by-step explanation:

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