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stiks02 [169]
1 year ago
14

Please look at the picture, I need help ASAP.

Mathematics
1 answer:
forsale [732]1 year ago
5 0

See below for the proof that the areas of the lune and the isosceles triangle are equal

<h3>How to prove the areas?</h3>

The area of the isosceles triangle is:

A_1 = \frac 12r^2\sin(\theta)

Where r represents the radius.

From the figure, we have:

\theta = 90

So, the equation becomes

A_1 = \frac 12r^2\sin(90)

Evaluate

A_1 = \frac 12r^2

Next, we calculate the length (L) of the chord as follows:

\sin(45) = \frac{\frac 12L}{r}

Multiply both sides by r

r\sin(45) = \frac 12L

Multiply by 2

L = 2r\sin(45)

This gives

L = 2r\times \frac{\sqrt 2}{2}

L = r\sqrt 2

The area of the semicircle is then calculated as:

A_2 = \frac 12 \pi (\frac{L}{2})^2

This gives

A_2 = \frac 12 \pi (\frac{r\sqrt 2}{2})^2

Evaluate the square

A_2 = \frac 12 \pi (\frac{2r^2}{4})

Divide

A_2 = \frac{\pi r^2}{4}

Next, calculate the area of the chord using

A_3 = \frac 12r^2(\theta - \sin(\theta))

Recall that:

\theta = 90

Convert to radians

\theta = \frac{\pi}{2}

So, we have:

A_3 = \frac 12r^2(\frac{\pi}{2} - \sin(\frac{\pi}{2}))

This gives

A_3 = \frac 12r^2(\frac{\pi}{2} - 1)

The area of the lune is then calculated as:

A = A_2 - A_3

This gives

A = \frac{\pi r^2}{4} -  \frac 12r^2(\frac{\pi}{2} - 1)

Expand

A = \frac{\pi r^2}{4} -  \frac{\pi r^2}{4} + \frac 12r^2

Evaluate the difference

A =  \frac 12r^2

Recall that the area of the isosceles triangle is

A_1 = \frac 12r^2

By comparison, we have:

A = A_1 = \frac 12r^2

This means that the areas of the lune and the isosceles triangle are equal

Read more about areas at:

brainly.com/question/27683633

#SPJ1

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