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In the trapezoid $ABCD$ where $AB||CD$, $S$ is the point of intersection of the diagonals. What is the area of the trapezoid if the area of the triangles ADS and ABS is respectively 5 and 7?

I know that the areas of the ADS and the CBS are equal, but I don't know how to get the area of the CDS, even though I know it is similar to the ABS. I suppose you could use the equation from finding the area of a trapezoid using only 2 of the 4 triangles that makes up its interior but I don't know the name of the theorem and would prefer to not use it. Instead, I would like to apply the law of cosines or of sines, but I have no idea how to do it here.

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triangles ADS and ABS have same height (drawn from A). so $DS:BS = 5:7$, similarly $CS:AS=5:7$, triangles ABS and CDS are similar (same angles) So $DC:AB = 5:7$. Note that the height of CDS and ABS (drawn from S) are also in the ratio of $5:7$, so area of CDS =$(\dfrac57)(\dfrac57)$ multiplied by area of ABS that is $(\dfrac57)(\dfrac57)7 = \dfrac{25}7$.

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  • $\begingroup$ it would be great if you used mathjax to format your answer . here is a quick tutorial on how to use it $\endgroup$ Jun 5 '18 at 18:57
  • $\begingroup$ @The Integrator, Thanks I am trying to use it. $\endgroup$
    – amitava
    Jun 5 '18 at 19:14
  • $\begingroup$ Note: you can type fractions using \frac{}{} for example \frac{a}{b} gives $\frac{a}{b}$ $\endgroup$ Jun 5 '18 at 19:19
  • $\begingroup$ @The Integrator, thanks! $\endgroup$
    – amitava
    Jun 6 '18 at 20:38
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enter image description here

This is what I got. I used similar triangles and ratio of lengths to get the area ratio.

Area $\Delta$ADS = Area$\Delta$BCS ($\Delta$ABD $-7 = \Delta$ABC$ – 7 = 5)$

$\frac{1}{2}AB(h-x) = 7$

$\frac{1}{2}AB(h) = 12$

$\frac{(h-x)}{h} = \frac{7}{12}$

$7h = 12h – 12x$

$5h = 12x$

$x = \frac{5h}{12}$

Therefore $h-x = \frac{7h}{12}$

Because $\Delta$CDS is similar to $\Delta$ABS (three angles the same), then the ratio of their lengths is $\frac{5}{7}$

The ratio of their areas is $\frac{25}{49}$. The area of $\Delta$CDS is therefore $7(\frac{25}{49}) = \frac{25}{7}$

The area of the trapezoid is therefore $5+5+7+\frac{25}{7} = 20\frac{4}{7}$

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  • $\begingroup$ it would be great if you used mathjax to format your answer . here is a quick tutorial on how to use it $\endgroup$ Jun 5 '18 at 20:15
  • $\begingroup$ @The Integrator. My answer was done in MS Word as it includes a graphic and is a picture file. I know how to use mathjax but that doesn't work in Word. I guess I should do the extra work and do the actual calculation in the answer space. Thanks for the reminder. $\endgroup$
    – Phil H
    Jun 5 '18 at 20:21
  • $\begingroup$ you could just include the graphic but type the calculations, would be good $\endgroup$ Jun 5 '18 at 20:25
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Here is a solution without using the law of sines and cosines. enter image description here

Denote by $PQ$ the distance between two points $P$ and $Q$ Notice that the are of the trapezoid is given by $$\frac{AB + CD }{2} \cdot r,$$

$r$ being the orthogonal distance between $AB$ and $CD$ as shown in the picture.

The Triangle $ABD$ has an area of $7+5$, thus we get $$\frac{AB}{2} \cdot r = 12.$$

Therefore $r = \frac{24}{AB}$. Plugging this into $\frac{CD}{2}\cdot r$, we get $$\frac{CD}{2}\cdot r = \frac{CD \cdot 12}{AB}$$

By the intercept theorem $$\frac{CD}{AB} = \frac{SD}{BS} = \frac{SD}{BS} \cdot \frac{h/2}{h/2} = \frac{5}{7},$$ where $h$ is the orthogonal distance between $A$ and $BD$.

Thus

$$\frac{AB + CD }{2} \cdot r = \frac{AB}{2} \cdot r + \frac{CD}{2}\cdot r = \frac{AB}{2} \cdot r + \frac{CD \cdot 12}{AB} = 12 + 12 \cdot \frac{5}{7} =\frac{144}{7}.$$

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