# The height drawn from the top of the obtuse corner of the rhombus is 24cm and divides the side in the ratio 7:18

**The height drawn from the top of the obtuse corner of the rhombus is 24cm and divides the side in the ratio 7:18, counting from the top of the acute angle. Find the area of the parts into which this height divides the rhombus.**

Let the side of the rhombus be equal to X cm. AB = BC + CD = AD = X cm.

By condition, AH / BH = 7/18.

7 * VN = 18 * AN.

BH = 18 * AH / 7.

AH = X – BH = X – 18 * AH / 7.

25 * AN / 7 = X.

AH = 7 * X / 25.

BH = X – 7 * X / 25 = 18 * X / 25.

From the right-angled triangle АНD, according to the Pythagorean theorem, we express the hypotenuse АD.

AD ^ 2 = AH ^ 2 + DH ^ 2.

X ^ 2 = (7 * X / 25) ^ 2 + 242.

X ^ 2 – 49 * X ^ 2/625 = 576.

576 * X ^ 2 = 567 * 625.

X = AB = BC = CD = AD = 25 cm.

Determine the area of the rhombus.

Savsd = AB * DN = 25 * 24 = 600 cm2.

The area of the triangle ADH is equal to:

Sadn = AN * DH / 2 = (7 * 25/25) * 24/2 = 84 cm2.

Let us determine the area of the quadrilateral DНВС.

Sdnvs = Savsd – Sadn = 600 – 84 = 516 cm2.

Answer: The areas of the parts of the rhombus are 84 cm2 and 516 cm2.