Find the area of a rectangle whose sides are 3: 4, and the perpendicular drawn from the top

Find the area of a rectangle whose sides are 3: 4, and the perpendicular drawn from the top of the rectangle to the diagonal is 12 cm.

Let ABCD be a given rectangle. BH – perpendicular to the AC diagonal, BH = 12 cm. AB: BC = 3: 4, let AB = 3x, BC = 4x.

Triangle ABC is rectangular, according to the Pythagorean theorem: AC² = AB² + BC² = (3x) ² + (4x) ² = 9x² + 16x² = 25x². Hence, AC = 5x.

The ABC triangle is equal to the ACD triangle, which means that their areas are equal. That is, the area of ​​the ABCD rectangle is twice the area of ​​the ABC triangle.

S (ABC) = 1/2 * BH * AC = 1/2 * 12 * 5x = 30x.

Hence, S (АBСD) = 30x * 2 = 60x.

Also, the area of ​​a rectangle ABCD can be expressed as the product of length and width:

S (ABСD) = AB * BC = 3x * 4x = 12x².

It turns out that 12x² = 60x;

12x² – 60x = 0;

12x (x – 5) = 0;

x = 0 (cannot be),

x = 5.

Hence, AB = 3x = 3 * 5 = 15 cm.

BC = 4x = 4 * 5 = 20 cm.

Therefore, S (АBСD) = 15 * 20 = 300 cm².

Answer: the area of ​​the ABCD rectangle is 300 cm².