Question : The side of an equilateral triangle is $12 \text{ cm}$. What is the area (in $\text{cm}^2$, rounded off to 2 decimal places) of the triangle? (Given $\sqrt{3} = 1.732$)
Option 1: 68.07
Option 2: 63.89
Option 3: 62.35
Option 4: 65.23
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Correct Answer: 62.35
Solution : Given: The side of an equilateral triangle is $12 \text{ cm}$. Area of the equilateral triangle: $=\frac{\sqrt{3}}{4} × (\text{side})^{2}$ $=\frac{\sqrt{3}}{4} × 12^{2}$ $=\frac{\sqrt{3}}{4} × 144$ $=\sqrt{3} × 36$ $= 1.732 × 36 $ $= 62.35 \text{ cm}^2$ Hence, the correct answer is 62.35.
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Question : The side of an equilateral triangle is 12 cm. What is the radius of the circle circumscribing this equilateral triangle?
Option 1: $6 \sqrt{3}\ \text{cm}$
Option 2: $4\sqrt{3}\ \text{cm}$
Option 3: $9 \sqrt{3}\ \text{cm}$
Option 4: $5\sqrt{3}\ \text{cm}$
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: $\frac{15\sqrt 3}{2}\text{ cm}$
Question : The side of an equilateral triangle is 9 cm. What is the radius of the circle circumscribing this equilateral triangle?
Option 1: $2\sqrt{3}$ cm
Option 2: $5\sqrt{3}$ cm
Option 3: $4\sqrt{3}$ cm
Option 4: $3\sqrt{3}$ cm
Question : The centroid of an equilateral triangle PQR is L. If PQ = 6 cm, the length of PL is:
Option 1: $2 \sqrt{3}\ \text{cm}$
Option 2: $4 \sqrt{3}\ \text{cm}$
Option 3: $5 \sqrt{3}\ \text{cm}$
Option 4: $3 \sqrt{3}\ \text{cm}$
Question : The height of an equilateral triangle is $9 \sqrt{3} \mathrm{~cm}$. What is the area of this equilateral triangle?
Option 1: $92 \sqrt{3} \mathrm{~cm}^2$
Option 2: $67 \sqrt{3} \mathrm{~cm}^2$
Option 3: $49 \sqrt{3} \mathrm{~cm}^2$
Option 4: $81 \sqrt{3} \mathrm{~cm}^2$
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