Find the size of the area (triangle in a trapezoid)

Question

Can anyone please help me to solve this task? The area "F" has to be found.

I thought I could find the size of the area of the trapezoid and then subtract the size of the two triangles. But I cannot find the size of the triangles. The trapezoid is 5 cm* 11cm, divided thorugh 2 and then all together * 7 makes then 192.5 square centimeters.

Answer 1

Area of trapezium is $\cfrac12(5+11)\cdot 7$

Area of the upper and lower triangles are $\cfrac 12 \cdot 5\cdot a$ and $\cfrac12 \cdot 11 \cdot b$ with heights $a$ and $b$ respectively such that $a+b=7$ $$[F]=\cfrac12(5+11)\cdot 7-\cfrac 12 \cdot 5\cdot a-\cfrac12 \cdot 11 \cdot b$$ Unless $M$ is a special (known or fixed) point, there are several solutions for the area of $F$.

If $M$ is a midpoint then $a=b=3.5 \text{ cm}$ and $[F]=28 \text{ cm}^2$

Answer 2

Call the vertices of the trapezoid $A,B,C,D$. The labelling goes counterclockwise, and $A$ is at the lower left corner.

Draw line $MN$ parallel to $AB$, and meeting $BC$ at $N$.

Our region is made up of two triangles $MBN$ and $MNC$. If we view them as having base $MN$, then the sum of their heights is $7$. Thus if $MN=z$, then the area of the shaded triangle is $\dfrac{7z}{2}$.

We cannot find an explicit numerical answer without making some assumptions. Suppose that $M$ divides the line segment $AD$ in the ratio $s:t$. Then $$z=MN=\frac{11t+5s}{s+t}.$$ In particular, if $M$ is the midpoint of $AD$, as the symbol $M$ perhaps suggests, then we can take $s=t=1$.

Answer 3

If $M$ is the midpoint of the side it's on, then the horizontal width at that altitude is $\frac{5+11}{2}=8$. The areas of both the top and the bottom halves of $F$ are $\frac12\cdot8\cdot\frac72$, so the whole area is $8\cdot\frac72=28\text{ cm}^2$.

Justification:

$\hspace{2cm}$

Assume $\overline{AB}\,\|\,\overline{DC}\,\|\,\overline{MN}$ and $M$ is the midpoint of $\overline{AD}$. By similar triangles, $|\overline{MP}|=\frac12|\overline{AB}|$ and $|\overline{PN}|=\frac12|\overline{DC}|$. Thus, $|\overline{MN}|=\frac12\left(|\overline{AB}|+|\overline{DC}|\right)$.

The sum of the altitudes of $\triangle BMN$ and $\triangle CMN$ on base $\overline{MN}$ is $7$. Thus, $|F|$, the sum of the areas of the triangles, is half the sum of the altitudes $\times$ the base $=\frac12\cdot7\cdot\frac12(5+11)=28$.