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VladimirAG [237]
2 years ago
6

Helpp I'm struggling​

Mathematics
1 answer:
Alona [7]2 years ago
7 0

Answer:

The surface area is 198.75\ ft^2

Step-by-step explanation:

<u>Step 1:  Determine the area of the bottom</u>

A = s^2

A = (7.5\ ft)^2

A = 56.25\ ft^2

<u>Step 2:  Determine the area of the triangles</u>

A = \frac{h * b}{2}

A = \frac{9.5\ ft * \frac{7.5\ ft}{2}}{2}

A=\frac{9.5\ ft*3.75\ ft}{2}

A = 17.8125\ ft^2

There are two of these right triangles on every side which means that we have 8 of these triangles so we multiply the area that we got by 8.

A = 17.8125\ ft^2 * 8

A=142.5\ ft^2

<u>Step 3:  Combine the area to get the surface area</u>

A_{total}=56.25\ ft^2+142.5\ ft^2

A_{total}=198.75\ ft^2

Answer: The surface area is 198.75\ ft^2

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Can someone explain this to me please
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Answer:

c. 36·x

Step-by-step explanation:

Part A

The details of the circle are;

The area of the circle, A = 12·π cm²

The diameter of the circle, d = \overline {AB}

Given that \overline {AB} is the diameter of the circle, we have;

The length of the arc AB = Half the the length of the circumference of the circle

Therefore, we have;

A = 12·π = π·d²/4 = π·\overline {AB}²/4

Therefore;

12 = \overline {AB}²/4

4 × 12 = \overline {AB}²

\overline {AB}² = 48

\overline {AB} = √48 = 4·√3

\overline {AB} = 4·√3

The circumference of the circle, C = π·d = π·\overline {AB}

Arc AB = Half the the length of the circumference of the circle = C/2

Arc AB = C/2 = π·\overline {AB}/2

\overline {AB} = 4·√3

∴ C/2 = π·4·√3/2 = 2·√3·π

The length of arc AB = 2·√3·π cm

Part B

The given parameters are;

The length of \overline {OF} = The length of \overline {FB}

Angle D = angle B

The radius of the circle = 6·x

The measure of arc EF = 60°

The required information = The perimeter of triangle DOB

We have;

Given that the base angles of the triangles DOB are equal, we have that ΔDOB is an isosceles triangle, therefore;

The length of \overline {OD} = The length of \overline {OB}

The length of \overline {OB} = \overline {OF} + \overline {FB} = \overline {OF} + \overline {OF} = 2 × \overline {OF}

∴ The length of \overline {OD} = 2 × \overline {OF} = The length of \overline {OB}

Given that arc EF = 60°, and the point 'O' is the center of the circle, we have;

∠EOF = The measure of arc EF = 60° = ∠DOB

Therefore, in ΔDOB, we have;

∠D + ∠B = 180° - ∠DOB = 180° - 60° = 120°

∵ ∠D = ∠B, we have;

∠D + ∠B = ∠D + ∠D = 2 × ∠D = 120°

∠D = ∠B = 120°/2 = 60°

All three interior angles of ΔDOB = 60°

∴ ΔDOB is an equilateral triangle and all sides of ΔDOB are equal

Therefore;

The length of \overline {OD} = The length of \overline {OB} = The length of \overline {DB}  = 2 × \overline {OF}

The perimeter of ΔDOB = The length of \overline {OD} + The length of \overline {OB} + The length of \overline {DB} = 2 × \overline {OF} + 2 × \overline {OF} + 2 × \overline {OF} = 6 × \overline {OF}

∴ The perimeter of ΔDOB = 6 × \overline {OF}

The radius of the circle = \overline {OF} = 6·x

∴ The perimeter of ΔDOB = 6 × 6·x = 36·x

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