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4 years ago

The side of an Equileteral triangle is 12cm. What is its Area?

Mathematics

1 answer:

elena55 [62]4 years ago

3 0

Answer:

A = 62.35 cm²

Step-by-step explanation:

Use the area formula A = $\frac{\sqrt{3}a^2}{4}$ , where a is the side length.

Plug in the values:

A = $\frac{\sqrt{3}(12^2)}{4}$

A = $\frac{\sqrt{3}(144)}{4}$

A = 62.35 cm²

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At 3pm, the length of the shadow of a thin vertical pole standing on level ground is the same as the height of the pole. A while

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Answer: The height of the pole is 175.97 ( approx )

Step-by-step explanation:

Let the height of the pole is x cm,

Also, let the angle of elevation of the sun to the pole at 3 pm is $\theta$

Thus, by the question,

$tan\theta = \frac{\text{ The height of the pole}}{\text{ The height of the shadow}}$

$\implies tan\theta = \frac{x}{x}$ ( at 3pm the height of shadow = height of the pole)

$\implies tan\theta = 1$

$\implies \theta = 45^{\circ}$

Again, according to the question,

When the angle of elevation is $(\theta - 12^{\circ})$ ,

The height of shadow = x + 95,

$\implies tan(45-12)^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}=\frac{x}{x+95}$

$\implies tan 33^{\circ}x + 95\times tan33^{\circ}=x$

$\implies tan 33^{\circ}x - x = - 95\times tan33^{\circ}$

$\implies -0.3505924068 x = - 61.6937213538$

$x=175.969930202\approx 175.97\text{ cm}$

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