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lbvjy [14]
1 year ago
8

1. The figure shows the regular triangular pyramid SABC. The base of the pyramid has an edge AB = 6 cm and the side wall has an

apothem SM = √15 cm. Calculate the pyramid: 1) the base elevation AM; 2) the elevation SO; 3) the area of the base; 4) the area of the side surface; 5) the total surface area; 6) volume.

Mathematics
1 answer:
Musya8 [376]1 year ago
7 0

Given:

• AB = 6 cm

,

• SM = √15 cm

Let's solve for the following:

• 1) the base elevation AM.

Given that we have a regular triangular pyramid, the length of the three bases are equal.

AB = BC = AC

BM = BC/2 = 6/2 = 3 cm

To solve for AM, which is the height of the base, apply Pythagorean Theorem:

\begin{gathered} AM=\sqrt{AB^2-BM^2} \\  \\ AM=\sqrt{6^2-3^2} \\  \\ AM=\sqrt{36-9} \\  \\ AM=\sqrt{27} \\  \\ AM=5.2\text{ cm} \end{gathered}

The base elevation of the pyramid is 5.2 cm.

• (2)., The elevation SO.

To find the elevation of the pyramid, apply Pythagorean Theorem:

SO=\sqrt{SM^2-MO^2}

Where:

SM = √15 cm

MO = AM/2 = 5.2/2 = 2.6 cm

Thus, we have:

\begin{gathered} SO=\sqrt{(\sqrt{15})^2-2.6^2} \\  \\ SO=\sqrt{15-6.76} \\  \\ SO=2.9\text{ cm} \end{gathered}

Length of SO = 2.9 cm

• (3). Area of the base:

To find the area of the triangular base, apply the formula:

A=\frac{1}{2}*BC*AM

Thus, we have:

\begin{gathered} A=\frac{1}{2}*6^*5.2 \\  \\ A=15.6\text{ cm}^2 \end{gathered}

The area of the base is 15.6 square cm.

• (4). Area of the side surface.

Apply the formula:

SA=\frac{1}{2}*p*h

Where:

p is the perimeter

h is the slant height, SM = √15 cm

Thus, we have:

\begin{gathered} A=\frac{1}{2}*(6*3)*\sqrt{15} \\  \\ A=34.86\text{ cm}^2 \end{gathered}

• (5). Total surface area:

To find the total surface area, apply the formula:

TSA=base\text{ area + area of side surface}

Where:

Area of base = 15.6 cm²

Area of side surface = 34.86 cm²

TSA = 15.6 + 34.86 = 50.46 cm²

The total surface area is 50.46 cm²

• (6). Volume:

To find the volume, apply the formula:

V=\frac{1}{3}*area\text{ of base *height}

Where:

Area of base = 15.6 cm²

Height, SO = 2.9 cm

Thus, we have:

\begin{gathered} V=\frac{1}{3}*15.6*2.9 \\  \\ V=15.08\text{ cm}^3 \end{gathered}

The volume is 15.08 cm³.

ANSWER:

• 1.) 5.2 cm

,

• 2.) 2.9 cm

,

• 3.) 15.6 cm²

,

• 4.) 34.86 cm²

,

• (5). 50.46 cm²

,

• 6). 15.08 cm³.

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Step-by-step explanation:

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For square one, we note that the centre square is shared by two diagonals, so we saved one square from the two diagonals.

The side length is 3 for square 1, 4 for square 2, and so on.

Let

n= square number (1, 2,3...)

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G1(n) = function that gives the number of green squares for square n, n=odd

G2(n) = function that gives the number of green squares for square n, n=even

side length, L=n+2   ................(1)

G1(n) = twice the side length less one, as discussed above

G1(n) = 2L-1       now substitute L=n+2

G1(n) = 2(n+2) -1    simplify

G1(n) = 2n + 3

Check:

for n=1, square 1 has 2*1+3 = 5 green squares ... checks

for n=3, square 3 has 2*3+3 = 9... checks

for n=5, square 5 has 2*5+3 = 13 ....checks

For even squares, it is even easier, because

G2(n) = 2L = 2(n+2)

check:

for n=2, square 2 has 2(2+2) = 8 green squares........checks

for n=4, square 4 has 2(4+2) = 12 green squares........checks.

Fincally, we apply our formula to n=20 and n=21

square 20 : G2(20) = 2(n+2) = 2(20+2) = 44 green quars

square 21 : G1(21) = 2n+3 = 2(21)+3 = 45 green squares

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