# 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.