ABC26GN4266 · Area

Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:

Find the length of the altitude of an equilateral triangle of side $3 \sqrt{3}$ cm.
Answer
SELF-PRACTICE — the source book printed no answer.

Nothing is invented here, so this question has no answer on record.

Explanation
Area of the triangle $=\frac{\sqrt{3}}{4} \times (3\sqrt{3})^{2}=\frac{27\sqrt{3}}{4}$. Let the height be $h$. Then, $\frac{1}{2} \times 3\sqrt{3} \times h=\frac{27\sqrt{3}}{4} \Leftrightarrow h=\frac{27\sqrt{3}}{4} \times \frac{2}{3\sqrt{3}}=\frac{9}{2}=4.5\ \text{cm}$.

Explanation as extracted from the printed page; notation may be imperfect.

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