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Find the area of an isosceles trapezoid $ABCD$ $(AB\parallel CD)$ with height $m$ and perpendicular diagonals $(AC\perp BD)$.

The area of $ABCD$ is given by $$S_{ABCD}=\dfrac{a+b}{2}.h=\dfrac{a+b}{2}.m$$ where $a$ and $b$ are the two bases $(a>b)$. So we have to find $a+b$ in terms of $m$ or maybe even $\dfrac{a+b}{2}$. Any help would be appreciated. Thank you in advance!

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  • $\begingroup$ Draw a diagram. Diagonals will make $45^0$ angle with parallel sides. $a+b = 2m$ $\endgroup$
    – Math Lover
    Jun 1 '21 at 20:00
  • $\begingroup$ @MathLover, I do have a diagram. Why $a+b=2m$? Thank you! $\endgroup$
    – Medi
    Jun 1 '21 at 20:02
  • $\begingroup$ Say diagonals meet at $O$ and one of the bases is $AB$. What are the angles where diagonals meet? All of them are $90^0$. As it is isosceles trapezoid, $OA = OB$ and $\angle AOB = 90^0$. $\endgroup$
    – Math Lover
    Jun 1 '21 at 20:04
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    $\begingroup$ Drop a perp from $O$ to $AB$. Say it meets at $H$. Then $\triangle OHB$ is a right angled triangle with one of the angles being $45^0$ so the other is $45^0$ too. $OH = HB = AB/2$ and $OH = x$ So $AB = 2x$. Similarly $CD = 2(m-x)$. Add them to get $2m$. $\endgroup$
    – Math Lover
    Jun 1 '21 at 20:13
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    $\begingroup$ @MathLover, thank you, I got it. $\endgroup$
    – Medi
    Jun 1 '21 at 20:19
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COMMENT.-The problem seems indeterminate because in the area between two parallels whose distance is equal to $m$, if only the fact that they are perpendicular diagonals is given as additional data and the problem were determined then the answer would be equal to $m^2$ because a square side $m$ satisfies the conditions of the problem. But it does not appear to be the case for any other trapezoid as shown in the accompanying figure.

enter image description here

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The trapezoid is isoceles, hence $\,\overline{AC} = \overline{BD}\,$ in:

enter image description here

Given the above coordinate system, we have for the diagonals, with a bit of analytic geometry: $$ \overline{AD} \; : \; \begin{cases} x = -a/2 + (b/2+a/2)t \\ y = mt \quad ; \quad 0 \le t \le 1 \end{cases} \quad \Longrightarrow \quad x-(a/2+b/2)y/m=-a/2 \\ \overline{BC} \; : \; \begin{cases} x = +a/2 - (b/2+a/2)t \\ y = mt \quad ; \quad 0 \le t \le 1 \end{cases} \quad \Longrightarrow \quad x+(a/2+b/2)y/m=+a/2 $$ The diagonals are perpendicular to each other if the dot product of the normals of the line segments $\overline{AD}$ and $\overline{BC}$ is zero: $$ \left(1,-\frac{a/2+b/2}{m}\right)\cdot\left(1,+\frac{a/2+b/2}{m}\right) = 1-\frac{(a/2+b/2)^2}{m^2} = 0 \\ \Longrightarrow \quad m = \frac{a+b}{2} $$ $$ \mbox{Area} = m^2 = \left(\frac{a+b}{2}\right)^2 $$

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