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3 years ago

Hello i need help with these questions asap!!

Mathematics

1 answer:

makvit [3.9K]3 years ago

3 0

We use direct proportionality:

360⁰......Pi×R² (full circle)

120⁰.......x

x=120×pi×R²/360=pi×R²/3

x=pi×4²/3=16.76mi²

A=3×pi cm², for an angle of 2×pi/3=120⁰

In a similar way we compute the area of the full circle:

360.........x

120..........3×pi

x=3×pi×360/120=9×pi=3²×pi.

Thus the radius is 3cm.

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What are the endpoint coordinates for the midsegment of △PQR that is parallel to PQ¯¯¯¯¯?

andriy [413]

Answer:

M(x₄ ,y₄) = (-3.5 , 0.5) and

N (x₅ ,y₅) = ( -1 , -0.5 )

Step-by-step explanation:

Let the endpoint coordinates for the mid segment of △PQR that is parallel to PQ be

M(x₄ ,y₄) and N(x₅ ,y₅) such that MN || PQ

point P( x₁ , y₁) ≡ ( -3 ,3 )

point Q( x₂ , y₂) ≡ (2 , 1 )

point R( x₂ , y₂) ≡ (-4 , -2)

To Find:

M(x₄ ,y₄) = ? and

N (x₅ ,y₅) = ?

Solution:

We have Mid Point Formula as

$Mid\ point(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

As M is the mid point of PR and N is the mid point of RQ so we will have

$Mid\ pointM(x_{4} ,y_{4})=(\frac{x_{1}+x_{3} }{2}, \frac{y_{1}+y_{3} }{2})$

$Mid\ pointN(x_{5} ,y_{5})=(\frac{x_{2}+x_{3} }{2}, \frac{y_{2}+y_{3} }{2})$

Substituting the given value in above equation we get

$Mid\ pointM(x_{4} ,y_{4})=(\frac{-3+-4 }{2}, \frac{3+-2} }{2})$

∴ $Mid\ pointM(x_{4} ,y_{4})=(\frac{-7} }{2}, \frac{1}{2})$

∴ $Mid\ pointM(x_{4} ,y_{4})=(-3.5, 0.5)$

Similarly,

$Mid\ pointN(x_{5} ,y_{5})=(\frac{2+-4 }{2}, \frac{1+-2 }{2})$

∴ $Mid\ pointN(x_{5} ,y_{5})=(\frac{-2 }{2}, \frac{-1}{2})$

∴ $Mid\ pointN(x_{5} ,y_{5})=(-1, -0.5)$

∴ M(x₄ ,y₄) = (-3.5 , 0.5) and

N (x₅ ,y₅) = ( -1 , -0.5 )

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Answer:

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