688/VB03. problem / Math - Solved Math Problems

MR-688 / 7. problem

The heights of the parallelogram are related as 2:3, its circumference is 40 cm, and the acute angle is 30°. Calculate the area of the parallelogram.

$h_{a} : h_{b} = 2 : 3$

$3 h_{a} = 2 h_{b}$

$h_{b} = \frac{3}{2} h_{a}$

$P = 2 (a + b)$

$\sin 30 ° = \frac{h_{a}}{b} = \frac{h_{b}}{a}$

$\frac{1}{2} = \frac{h_{a}}{b} = \frac{h_{b}}{a}$

$b = 2 h_{a}$

$a = 2 h_{b}$

$P = 2 (2 h_{b} + 2 h_{a})$

$P = 4 (h_{b} + h_{a})$

$P = 4 (h_{a} + \frac{3}{2} h_{a})$

$40 = 4 (h_{a} + \frac{3}{2} h_{a})$

$4 \cdot 10 = 4 \cdot (h_{a} + \frac{3}{2} h_{a})$

$10 = h_{a} + \frac{3}{2} h_{a}$

$10 = h_{a} + \frac{3}{2} h_{a} \cdot 2$

$10 \cdot 2 = h_{a} \cdot 2 + \frac{3}{2} h_{a} \cdot 2$

$20 = 2 h_{a} + 3 h_{a}$

$20 = 5 h_{a}$

$h_{a} = \frac{20}{5}$

$h_{a} = 4$

$h_{b} = \frac{3}{2} \cdot h_{a} = \frac{3}{2} \cdot 4 = \frac{3}{2} \cdot 2 \cdot 2$

$h_{b} = 6$

$a = 2 \cdot h_{b} = 2 \cdot 6 = 12$

$A = a \cdot h_{a} = 12 \cdot 4$

$A = 48$

MathReference - Problems