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vesna_86
2 months ago
6

Let $DEF$ be an equilateral triangle with side length $3.$ At random, a point $G$ is chosen inside the triangle. Compute the pro

bability that the length $DG$ is less than or equal to $1.$

Mathematics
2 answers:
AnnZ [12.3K]2 months ago
8 0

|\Omega|=(\text{the area of the triangle})=\dfrac{a^2\sqrt3}{4}=\dfrac{3^2\sqrt3}{4}=\dfrac{9\sqrt3}{4}\\|A|=(\text{the area of the sector})=\dfrac{\alpha\pi r^2}{360}=\dfrac{60\pi \cdot 1^2}{360}=\dfrac{\pi}{6}\\\\\\P(A)=\dfrac{\dfrac{\pi}{6}}{\dfrac{9\sqrt3}{4}}\\\\P(A)=\dfrac{\pi}{6}\cdot\dfrac{4}{9\sqrt3}\\\\P(A)=\dfrac{2\pi}{27\sqrt3}\\\\P(A)=\dfrac{2\pi\sqrt3}{27\cdot3}\\\\P(A)=\dfrac{2\pi\sqrt3}{81}\approx13.4\%

Zina [12.3K]2 months ago
3 0

Answer:

13.44%

Step-by-step explanation:

For the length of DG to be at most 1, point G needs to be located within a circular sector centered on D, having a radius of 1, and subtending an angle of 60°.

The area of the sector is

A_s = \dfrac{n}{360^\circ}\pi r^2

A_s = \dfrac{60^\circ}{360^\circ} \times 3.14159 \times 1^2

A_s = 0.5254

The area of the triangle is

A_t = \dfrac{1}{2}ef \sin D

A_t = \dfrac{1}{2}\times 3 \times 3 \sin 60^\circ

A_t = 3.8971

To find the probability, we take the area of the sector and divide it by the area of the triangle.

p = \dfrac{A_s}{A_t} = \dfrac{0.5254}{3.8971} = 0.1344

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2 months ago
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0.14 s

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x = \frac{-40\pm\sqrt{40^2 - 4\times (-2.7) \times (-5.5)}} {2(-2.7)}

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