In a right-angled triangle ABC, the median drawn from the vertex of the right angle C is 4
In a right-angled triangle ABC, the median drawn from the vertex of the right angle C is 4, and the median drawn to the larger leg is 2√7. Find the area of the triangle.
Since CH is the median drawn from the top of the right angle, then BH = AH = CH = 4 cm, then AB = 2 * AH 2 * 4 = 8 cm.
In a right-angled triangle ABC, according to the Pythagorean theorem, BC ^ 2 = AB ^ 2 – AC ^ 2 = 64 – AC^2.
BC = √ (64 – AC ^ 2). (1).
In a right-angled triangle ACM, according to the Pythagorean theorem, CM ^ 2 = AM ^ 2 – AC ^ 2 = 28 – AC ^ 2. (2).
CM = √ (28 – AC ^ 2). (2).
Multiply equation 2 by two, then 2 * CM = 2 * √ (28 – AC ^ 2).
Since AM is the median, then CM = BM = BC / 2, and BC = 2 * CM.
Then: √ (64 – AC ^ 2) = 2 * √ (28 – AC ^ 2).
64 – AC ^ 2 = 4 * (28 – AC ^ 2).
3 * AC ^ 2 = 48.
AC ^ 2 = 48/3 = 16.
AC = 4 cm.
Let’s define the BC leg. BC ^ 2 = AB ^ 2 – AC ^ 2 = 64 – 16 = 48.
BC = √48 = 4 * √3 cm.
Determine the area of the triangle ABC.
Savs = BC * AC / 2 = 4 * √3 * 4/2 = 8 * √3 cm2.
Answer: The area of the triangle is 8 * √3 cm2.