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3 years ago

△ABC is a right isosceles triangle, the centers of two arcs are the midpoint of AB

Mathematics

1 answer:

mel-nik [20]3 years ago

4 0

Answer:

Step-by-step explanation:

Let's define 3 areas:

S = area of semicircle with radius 6 in (diameter AB)
T = area of quarter circle with radius 6√2 in (radius AC)
U = area of triangle ABC (side lengths 6√2)

The white space between the "moon" and the triangle has area ...

white = T - U

Then the area of the "moon" shape is ...

moon = S -white = S -(T -U) = S -T +U

The area we're asked to find is ...

moon - triangle = (S -T +U) -U = S -T

The formula for the area of a circle of radius r is ...

A = πr²

So, ...

S = (1/2)π(6 in)² = 18π in²

and

T = (1/4)π(6√2 in)² = 18π in²

The difference in areas is S -T = (18π in²) -(18π in²) = 0.

There is no difference between the areas of the "moon" and the triangle.

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Answer:

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Sergio works at a bakery and needs to cover

Elan Coil [88]

Answer:

The surface area that Sergio needs to frost on each cake is 3,238 square centimeters

Step-by-step explanation:

The picture of the question in the attached figure

we know that

The surface area of the cylindrical cake is equal to the lateral area plus the area of the top of cylinder (remember that the bottom of the cake does not need frosting)

$SA=\pi r^{2}+2\pi r h$

we have

$r=40.2/2=20.1\ cm$ ---> the radius is half the diameter

$h=15.6\ cm$

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substitute the given values

$SA=(3.14)(20.1)^{2}+2(3.14)(20.1)(15.6)$

$SA=1,268.59+1,969.16\\SA=3,237.75\ cm^2$

Round up to the nearest whole number

$SA=3,238\ cm^2$

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3 years ago

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Murljashka [212]

1) 1/2 * 1/2 = 1/4

2) 43/100 of mine were heads, and (.43)^2 = .1849

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