Question : What is the altitude of an equilateral triangle whose side is 15 cm?
Option 1: $15\sqrt3\text{ cm}$
Option 2: $10\sqrt 3\text{ cm}$
Option 3: $\frac{9\sqrt 3 }{2}\text{ cm}$
Option 4: $\frac{15\sqrt 3}{2}\text{ cm}$
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Correct Answer: $\frac{15\sqrt 3}{2}\text{ cm}$
Solution : Given: The side of the triangle is 15 cm. Let the altitude be $x$ cm. The length of the altitude of an equilateral triangle = $\frac{\sqrt3}{2}$ × side of the triangle. ⇒ $x = \frac{\sqrt3}{2}\times15$ ⇒ $x = \frac{15\sqrt3}{2}$ Hence, the correct answer is $\frac{15\sqrt3}{2}\text{ cm}$.
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Question : G is the centroid of the equilateral triangle ABC. If AB = 10 cm, then the length of AG (in cm) is:
Question : $G$ is the centroid of the equilateral triangle $ABC$. If $AB$ is $9\text{ cm}$, then $AG$ is equal to:
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