# In rhombus ABCD from the vertex of obtuse angle B, the height BH is drawn to the side AD.

**In rhombus ABCD from the vertex of obtuse angle B, the height BH is drawn to the side AD. It intersects the diagonal AC at point M. The side of the rhombus is 15 and its area is 135. Find the area of triangle AMH.**

The area of the rhombus is equal to the product of the height and the base.

Savsd = AD * BH, then:

135 = 15 * BH.

BH = 135/15 = 9 cm.

From the right-angled triangle ABH, by the Pythagorean theorem, we determine the length of the leg AH.

AH ^ 2 = AB ^ 2 – BH ^ 2 = 15 ^ 2 – 9 ^ 2 = 225 – 81 = 144.

AH = 12 cm.

Let the length of the segment MH = X cm, then BM = (9 – X) cm.

Since the diagonals of the rhombus are the bisectors of the angles at the vertices of the rhombus, AM is the bisector of the triangle ABH. Then, by the property of the bisector of the triangle: AH / MH = AB / BM.

12 / X = 15 / (9 – X).

15 * X = 108 – 12 * X.

27 * X = 108.

X = MH = 108/27 = 4 cm.

Determine the area of the triangle AMH.

Samn = AH * MH / 2 = 12 * 4/2 = 24 cm2.

Answer: The area of the AMH triangle is 24 cm2.