Multiple choice

If in a $\Delta ABC$, $b=12$ units, $c=5$ units and $\Delta = 30$ sq. units, then the distance between vertex $A$ and incentre of the triangle is equal to?

  1. $2$ units
  2. $2\sqrt2 $ units
  3. $\sqrt2 $ units

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Incenter distance from vertex A is given by r / sin(A/2). Area = 30, s = (a+b+c)/2. Need a. Area = 0.5 * b * c * sin(A) => 30 = 0.5 * 12 * 5 * sin(A) => sin(A) = 1. So A = 90 degrees. a = sqrt(12^2 + 5^2) = 13. s = (13+12+5)/2 = 15. r = Area/s = 30/15 = 2. Distance = 2 / sin(45) = 2 / (1/sqrt(2)) = 2 * sqrt(2).

AI explanation

Given the area of triangle ABC is 30 square units and the sides b = 12 and c = 5, the formula Area = (1/2) * b * c * sin(A) gives 30 = (1/2) * 12 * 5 * sin(A), which means sin(A) = 1 and angle A = 90 degrees. The semi-perimeter s is found by first determining the hypotenuse a = sqrt(12^2 + 5^2) = 13, making s = (13 + 12 + 5) / 2 = 15. The distance from vertex A to the incentre is given by the formula AI = r / sin(A/2), where the inradius r = Area / s = 30 / 15 = 2. Since A is 90 degrees, sin(45 degrees) = 1/sqrt(2), so AI = 2 / (1/sqrt(2)) = 2 * sqrt(2) units.

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