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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.

enter image description here

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where is M point located?random place? –  dato datuashvili Mar 13 '13 at 11:17
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Maybe $M$ as in "midpoint"? –  Gerry Myerson Mar 13 '13 at 12:10
    
Sorry for my late reply. I think it´s midpoint. But it doesnt say that in the task –  Sophia Mar 13 '13 at 12:50

3 Answers 3

up vote 3 down vote accepted

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 $[F]=28 \text{ cm}^2$

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by the way is 7cm length of left side? –  dato datuashvili Mar 13 '13 at 11:22
    
$7$ cm is the height of the trapezium –  user31280 Mar 13 '13 at 11:25
    
i doubt that given information is not enough to solve this problem,for triangles we dont have enough information about their lengths,which makes nonsens in solving this problem –  dato datuashvili Mar 13 '13 at 11:26
    
@dato you're right, the given information is not enough –  user31280 Mar 13 '13 at 11:32
    
Sometimes introducing the right variable is part of problem solving... Simplified area of F is $35/2+3\cdot a$, with same notation (and same method) as "merge me". –  Jean-Claude Arbaut Mar 13 '13 at 11:40

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 $s=t=1$.

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Thank you for your help!! –  Sophia Mar 14 '13 at 14:36

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:

$\hspace{2cm}$enter image description here

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 $=\frac12\cdot7\cdot\frac12(5+11)=28$.

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Thank you for your help!! –  Sophia Mar 14 '13 at 14:37

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