701/VB03. problem / Math - Solved Math Problems

MR-701 / 19. problem

An isosceles trapezoid of base 40 cm and 10 cm is described around a circle. Calculate the area of the triangle whose vertices are the points of contact of the legs and the smaller base.

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Source: Vene T. Bogoslavov: Collection of Solved Problems #3 - 19..)

Geometry → Two-dimensional geometric shapes → Quadrilaterals

Since a circle can be inscribed in a trapezoid, the trapezoid is a tangent quadrilateral. Therefore, the following relation can be written:

$a + b = c + c$

$40 + 10 = 2 c$

$2 c = 50$

$c = \frac{50}{2}$

$c = 25$

The Pythagorean theorem that can be written on the AED right triangle:

$h^{2} + {(\frac{a - b}{2})}^{2} = c^{2}$

$h^{2} + {(\frac{40 - 10}{2})}^{2} = 25^{2}$

$h^{2} + {(15)}^{2} = 25^{2}$

$h^{2} = 25^{2} - 15^{2}$

$h^{2} = 625 - 225$

$h^{2} = 400$

$h = \sqrt{400}$

$h = 20$

The Pythagorean theorem that can be written on the CFG right triangle:

$5^{2} = p^{2} + q^{2}$

$p^{2} + q^{2} = 25$

${(5 + q)}^{2} + {(10 - p)}^{2} = r^{2}$

${(5 + q)}^{2} + {(10 - p)}^{2} = 10^{2}$

$25 + 2 \cdot 5 \cdot q + q^{2} + 100 - 2 \cdot 10 \cdot p + p^{2} = 100$

$25 + 10 q + q^{2} - 20 p + p^{2} = 0$

$25 + 10 q - 20 p + 25 = 0$

$10 q - 20 p + 50 = 0$

$q - 2 p + 5 = 0$

$q = 2 p - 5$

$p^{2} + q^{2} = 25$

$p^{2} + {(2 p - 5)}^{2} = 25$

$p^{2} + 4 p^{2} - 2 \cdot 2 p \cdot 5 + 25 = 25$

$5 p^{2} - 20 p = 0$

$5 p (p - 4) = 0$

$p - 4 = 0$

$p = 4$

$q = 2 p - 5 = 2 \cdot 4 - 5 = 8 - 5$

$q = 3$

$A_{△} = \frac{a_{△} \cdot h_{△}}{2} = \frac{(b + 2 q) \cdot p}{2} = \frac{(10 + 2 \cdot 3) \cdot 4}{2}$

$A_{△} = \frac{16 \cdot 4}{2} = \frac{16 \cdot 2 \cdot 2}{2}$

$A_{△} = 32$

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