Prove in full that the square 9.9cm diameter fits the circle.
Prove here again if you can help id appreciate it

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Others unanswered 100 points in 3 questions.

see below

Step-by-step explanation:

We need to find the diameter of the square

We can find this using the Pythagorean theorem

a^2+b^2 = c^2  where the legs are 7 and 7 and the diameter is c

7^2 +7^2 = c^2

49+49 = c^2

98 = c^2

Taking the square root of each side

sqrt(98) = sqrt(c^2)

9.899494937 = c

Since 9.9 is less than 11 which is the diameter of the circle it will never touch the circle.

Since the longest part of the square is less than the diameter of the circle, the square will fit  inside the circle without touching

Diagonal < Diameter

Step-by-step explanation:

The square will fit inside without touching if the diagonal length is less than the diameter

Diagonal = sqrt(7² + 7²)

7sqrt(2)

Diameter = 11

7sqrt(2) < 11

Because

(7sqrt(2))² < 11²

98 < 121

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,

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step-by-step explanation: