Question

В трапеции ABCD основания AD и BC равны 4 и 3 соответственно, а её площадь равна 84. Найдите площадь трапеции BCNM , где MN – средняя линия трапеции ABCD .

Answer 1

Sabcd=1/2 (AD+BC)*BH=MN*BH
MN=(AD+BC)/2=7/2=3.5
т.L - т.пересечения BH с MN
BL=LH=BH/2
Sbcnm=1/2 * MN * BL=1/2 MN*1/2 BH= 1/2(MN*BH)=1/2 * Sabcd=42

Answer 2

I can help you with your math problem. Here is a detailed solution:

The area of a trapezoid is given by the formula $$S = \frac{a + b}{2}h$$, where $$a$$ and $$b$$ are the bases and $$h$$ is the height. In this case, we have $$a = 4$$, $$b = 3$$ and $$S = 84$$. Solving for $$h$$, we get $$h = \frac{2S}{a + b} = \frac{2 \cdot 84}{4 + 3} = 24$$.

The midline of a trapezoid is the line segment that connects the midpoints of the non-parallel sides. Its length is equal to the average of the bases, that is $$m = \frac{a + b}{2}$$. In this case, we have $$m = \frac{4 + 3}{2} = 3.5$$.

The trapezoid $$BCNM$$ is similar to the trapezoid $$ABCD$$, since they have the same angles and proportional sides. Therefore, the ratio of their areas is equal to the square of the ratio of their corresponding sides, that is $$\frac{S_1}{S_2} = \left(\frac{b}{m}\right)^2$$, where $$S_1$$ is the area of $$BCNM$$ and $$S_2$$ is the area of $$ABCD$$. Solving for $$S_1$$, we get $$S_1 = S_2 \left(\frac{b}{m}\right)^2 = 84 \left(\frac{3}{3.5}\right)^2 = 64.8$$.

Therefore, the area of the trapezoid $$BCNM$$ is $$64.8$$.

You can find more information about trapezoids and their properties on these websites: [1](https://ru.onlinemschool.com/math/formula/trapezium_right/), [2](https://oge.sdamgia.ru/problem?id=333309&print=true), [3](https://mat.1sept.ru/view_article.php?ID=201000511). I hope this helps.