What is the radius of the incircle of a regular hexagon with a side of 20 cm ?

1 answer:

7 0

10√3 cm.
The radius of an incircle of a hexagon is the same measure as the apothem. find the apothem by taking half of the central angle and create a right triangle perpendicular to the side. The central angle is 60 so the angle opposite the side length will be 30. this creates a 30-60-90 triangle with the ratio x:x√3:2x. half the side opposite 30 is 10, so x=10. the base angle of the side is 60 so it will be x√3. plug in 10 for x to get 10√3 cm.

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Given the cost function C(x) = 0.81x + 60, 420 and the revenue function R(x) = 1.87x, find the break-even point

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Answer:

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Step-by-step explanation:

C(x) = 0.81x + 60,420

R(x) = 1.87x

At break even point, Cost function and revenue function becomes equal.

-> C(x) = R(x)

-> 0.81x + 60,420 = 1.87x

-> 60420 = 1.87x - 0.82x

-> 60420 = 1.05x

-> x = 60420/1.05

-> x = 57,542.8571

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<h2>Question:- Area of the given diagram</h2>

<h2>Answer :- </h2>

Divide the diagram as shown in attachment.

Now first calculate the area of the rectangle having side 1 and 12

So area = length ×breadth

Area = 1× 12 = 12 unit²

Now find the Area of the remaining triangle with sides 12,13,5 units

$Area = \frac{1}{2} \times base \times height \:$

Now put the value as -> base= 5 and height= 12

$Area = \frac{1}{2} \times base \times height \: \\ Area = \frac{1}{2} \times 5 \times 12 \\ Area = \frac{1}{ \cancel2} \times 5 \times \cancel{12}^{ \: \: 6} \\ Area = 5 \times 6 \\ Area = 30 \: units^{2}$

Now add both areas for your answer

$Area{ \tiny{1}} = 12 {units}^{2} \\ Area{ \tiny2} = 30 {units}^{2} \\ Area{ \tiny{total}} = Area{ \tiny{1}} + Area{ \tiny{2}} \\ Area{ \tiny{total}} = 12 + 30 = 42 {units}^{2}$