683/VB03. problem / Math - Solved Math Problems

MR-683 / 3. problem

In an acute triangle with a base of 10 cm and a height corresponding to the base of 8 cm, a rectangle is inscribed whose two vertices belong to the base of the triangle, and the other two on the other two sides. If the area of the rectangle is 15 cm², calculate its sides.

Source: Vene T. Bogoslavov: Collection of Solved Problems #3 - 3..)

Geometry → Two-dimensional geometric shapes → Triangle

Triangle ABC is similar to triangle A₁B₁C₁.

$∆ A B C ~ ∆ A_{1} B_{1} C$

$10 : y = 8 : (8 - x)$

$8 y = 10 \cdot (8 - x)$

$A_{□} = x \cdot y = 15$

$y = \frac{15}{x}$

$8 \cdot \frac{15}{x} = 80 - 10 x | \cdot x$

$120 = 80 x - 10 x^{2}$

$10 x^{2} - 80 x + 120 = 0 | \cdot \frac{1}{10}$

$x^{2} - 8 x + 12 = 0$

$x_{1, 2} = \frac{- (- 8) \pm \sqrt{{(- 8)}^{2} - 4 \cdot 1 \cdot 12}}{2}$

$= \frac{8 \pm \sqrt{64 - 48}}{2}$

$= \frac{8 \pm \sqrt{16}}{2}$

$= \frac{8 \pm 4}{2}$

$x_{1} = \frac{8 + 4}{2} = \frac{12}{2}$

$x_{1} = 6$

$x_{2} = \frac{8 - 4}{2} = \frac{4}{2}$

$x_{2} = 2$

$y_{1} = \frac{15}{x_{1}} = \frac{15}{6} = \frac{3 \cdot 5}{3 \cdot 2}$

$y_{1} = \frac{5}{2} = 2, 5$

$y_{2} = \frac{15}{x_{2}} = \frac{15}{2} = 7, 5$

$(x_{1} = 6; y_{1} = 2, 5) (x_{2} = 2; y_{2} = 7, 5)$