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

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!

• Draw a diagram. Diagonals will make $45^0$ angle with parallel sides. $a+b = 2m$ Jun 1 '21 at 20:00
• @MathLover, I do have a diagram. Why $a+b=2m$? Thank you!
– Medi
Jun 1 '21 at 20:02
• 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 $\angle AOB = 90^0$. Jun 1 '21 at 20:04
• 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 $2m$. Jun 1 '21 at 20:13
• @MathLover, thank you, I got it.
– Medi
Jun 1 '21 at 20:19

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.
The trapezoid is isoceles, hence $$\,\overline{AC} = \overline{BD}\,$$ in:
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: $$\left(1,-\frac{a/2+b/2}{m}\right)\cdot\left(1,+\frac{a/2+b/2}{m}\right) = 1-\frac{(a/2+b/2)^2}{m^2} = 0 \\ \Longrightarrow \quad m = \frac{a+b}{2}$$ $$\mbox{Area} = m^2 = \left(\frac{a+b}{2}\right)^2$$