Question : The length of each side of an equilateral triangle is $14 \sqrt{3}$ cm. The area of the incircle (in cm2), is:
Option 1: 450
Option 2: 308
Option 3: 154
Option 4: 77
Correct Answer: 154
Solution : Given that the length of each side of an equilateral triangle is $14 \sqrt{3}\ cm$ BD = DC = $7\sqrt3$ cm By Pythagoras theorem, $AD=\sqrt{AB^2-BD^2}$ $= \sqrt{(14\sqrt3)^2-(7\sqrt3)^2}$ = 21 cm Since the radius of the incircle is one-third of the height of the equilateral triangle, OD = radius of incircle = $\frac{1}{3}\times$ 21 = 7 cm So, the area of incircle = $\pi r^2=\frac{22}{7}\times 7^2=154$ cm2 Hence, the correct answer is 154.
Application | Eligibility | Selection Process | Result | Cutoff | Admit Card | Preparation Tips
Question : The total surface area of a regular triangular pyramid with each edge of length 1 cm is:
Option 1: $4 \sqrt{3}$ cm2
Option 2: $\frac{4}{3} \sqrt{3}$ cm2
Option 3: $\sqrt{3}$ cm2
Option 4: $4$ cm2
Question : The area of an equilateral triangle is $4 \sqrt{3} \mathrm{~cm}^2$. Find the side (in cm) of the triangle.
Option 1: $2$
Option 2: $4$
Option 3: $\sqrt{3}$
Option 4: $2 \sqrt{3}$
Question : What is the area of a triangle having a perimeter of 32 cm, one side of 11 cm, and the difference between the other two sides is 5 cm?
Option 1: $8\sqrt{30}$ cm2
Option 2: $5\sqrt{35}$ cm2
Option 3: $6\sqrt{30}$ cm2
Option 4: $8\sqrt{2}$ cm2
Question : The area of an equilateral triangle is 173.2 cm2. Its side will be __________.
Option 1: 20 cm
Option 2: 22 cm
Option 3: 17.32 cm
Option 4: 21.32 cm
Question : If each side of an equilateral triangle is 12 cm, then its altitude is equal to:
Option 1: $6 \sqrt{3}\ \text{cm}$
Option 2: $3 \sqrt{6}\ \text{cm}$
Option 3: $6 \sqrt{2}\ \text{cm}$
Option 4: $3 \sqrt{2}\ \text{cm}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile