Question : $\triangle$ABC is an equilateral triangle in which D, E, and F are the points on sides BC, AC, and AB, respectively, such that AD $\perp$ BC, BE $\perp$ AC and CF $\perp$ AB. Which of the following is true?
Option 1: 4AC$^2$ = 5BE$^2$
Option 2: 3AC$^2$ = 4BE$^2$
Option 3: 2AB$^2$ = 3AD$^2$
Option 4: 7AB$^2$ = 9AD$^2$
Correct Answer: 3AC$^2$ = 4BE$^2$
Solution : As we know, Height of the equilateral triangle = $(\frac{\sqrt{3}}{2})$ × (Sides) Height = AD = BE = CF Now, $(\frac{\sqrt{3}}{2}$) × AC = BE ⇒ $\sqrt{3}$ AC = 2BE Squaring on both sides ⇒ 3AC$^2$ = 4BE$^2$ Hence, the correct answer is 3AC$^2$ = 4BE$^2$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : ABC is an equilateral triangle points D, E and F are taken in sides AB, BC and CA respectively so that, AD = BE = CF. Then DE, EF, and FD enclose a triangle which is:
Question : In a triangle ABC, AD, BE, and CF are three medians. The perimeter of ABC is:
Question : The sides AB, BC, and AC of a $\triangle {ABC}$ are 12 cm, 8 cm, and 10 cm respectively. A circle is inscribed in the triangle touching AB, BC, and AC at D, E, and F respectively. The difference between the lengths of AD and CE is:
Question : If AB = 5 cm, AC = 12 cm, and AB$\perp$ AC, then the radius of the circumcircle of $\triangle ABC$ is:
Question : In a $\triangle ABC$, D and E are two points on AB and AC respectively such that DE || BC, DE bisects the $\triangle ABC$ in two equal areas. Then the ratio DB : AB is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile